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Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition

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Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition

Samacheer Kalvi 11th Bio Botany Mineral Nutrition Text Book Back Questions and Answers

Question 1.
Identify correct match.

1. Die back disease of citrus (i) Mo
2. Whip tail disease (ii) Zn
3. Brown heart of turnip (iii) Cu
4. Little leaf (iv) B

(a) 1. (iii), 2. (ii), 3. (iv), 4. (i).
(b) 1. (iii), 2. (i), 3. (iv), 4. (ii).
(c) 1. (i), 2. (iii), 3. (ii), 4. (iv).
(d) 1. (iii), 2. (iv), 3. (ii), 4. (i).
Answer:
(b) 1. (iii), 2. (i), 3. (iv), 4. (ii).

Question 2.
If a plant is provided with all mineral nutrients but, Mn concentration is increased, what will be the deficiency?
(a) Mn prevent the uptake of Fe, Mg but not Ca
(b) Mn increase the uptake of Fe, Mg and Ca
(c) Only increase the uptake of Ca
(d) Prevent the uptake Fe, Mg, and Ca
Answer:
(a) Mn prevent the uptake of Fe, Mg but not Ca

Samacheer Kalvi 11th Bio Botany Solutions 12 Mineral Nutrition

Question 3.
The element which is not remobilized?
(a) Phosphorus
(b) Potassium
(c) Calcium
(d) Nitrogen
Answer:
(c) Calcium

Question 4.
Match the correct combination.

Minerals

Role

(a) Molybdenum 1. Chlorophyll
(b) Zinc 2. Methionine
(c) Magnesium 3. Auxin
(d) Sulphur 4. Nitrogenase

(a) A – 1, B – 3, C – 4, D – 2
(b) A – 2, B – 1, C – 3, D – 4
(c) A – 4, B – 3, C – 1, D – 2
(d) A – 4, B – 2, C – 1, D – 3
Answer:
(c) A – 4, B – 3, C – 1, D – 2

Question 5.
Identify the correct statement:
(i) Sulphur is essential for amino acids Cystine and Methionine
(ii) Low level of N, K, S and Mo affect the cell division
(iii) Non – leguminous plant Alnus which contain bacterium Frankia
(iv) Denitrification carried out by nitrosomonas and nitrobacter.

(a) (i), (ii) are correct
(b) (i), (ii), (iii) are correct
(c) I only correct
(d) all are correct
Answer:
(b) (i), (ii), (iii) are correct

Question 6.
The nitrogen is present in the atmosphere in huge amount but higher plants fail to utilize it. Why?
Answer:
The higher plants do not have the association of bacteria or fungi, which are able to fix atmospheric nitrogen.

Question 7.
Why is that in certain plants deficiency symptoms appear first in younger parts of the plants while in others, they do so in mature organs?
Answer:
In certain plants, the deficiency symptom appears first in the younger part of the plant, due to the immobile nature of certain minerals like calcium, sulphur, iron, boron and copper.

Question 8.
Plant A in a nutrient medium shows whiptail disease plant B in a nutrient medium shows a little leaf disease. Identify mineral deficiency of plant A and B?
Answer:
Mineral deficiency of plant A and B:

  1. Plant A is deficient of the mineral molybdenum (Mo).
  2. Plant B is deficient of the mineral zinc (Zn).

Question 9.
Write the role of nitrogenase enzyme in nitrogen fixation?
Answer:
The role of nitrogenase enzyme in nitrogen fixation:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition 2

Question 10.
Explain the insectivorous mode of nutrition in angiosperms?
Answer:
Plants which are growing in nitrogen deficient areas develop insectivorous habit to resolve nitrogen deficiency.
(i) Nepenthes (Pitcher plant): Pitcher is a modified leaf and contains digestive enzymes. Rim of the pitcher is provided with nectar glands and acts as an attractive lid. When insect is trapped, proteolytic enzymes will digest the insect.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition 3

(ii) Drosera (Sundew): It consists of long club shaped tentacles which secrete sticky digestive fluid which looks like a sundew.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition 4

(iii) Utricularia (Bladder wort): Submerged plant in which leaf is modified into a bladder to collect insect in water.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition 5

(iv) Dionaea (Venus fly trap): Leaf of this plant modified into a colourful trap. Two folds of lamina consist of sensitive trigger hairs and when insects touch the hairs it will close.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition 6

Samacheer Kalvi 11th Bio Botany Mineral Nutrition Additional Questions & Answers

I. Choose the correct answer (1 Mark)
Question 1.
Plants naturally obtain nutrients from:
(a) atmosphere
(b) water
(c) soil
(d) all of these
Answer:
(d) all of these

Question 2.
Which of the following are included under micro nutrients:
(a) sodium, carbon and hydrogen
(b) magnesium, nitrogen and silicon
(c) sodium, cobalt and selenium
(d) calcium, sulphur and potassium
Answer:
(c) sodium, cobalt and selenium

Question 3.
Who coined the term ‘Hydroponics’:
(a) Julius Von Sachs
(b) William Frederick Goerick
(c) Liebig
(d) Wood word
Answer:
(b) William Frederick Goerick

Question 4.
Selenium is essential for plants:
(a) to prevent water lodging
(b) to enhance growth
(c) to resist drought
(d) to prevent transpiration
Answer:
(a) to prevent water lodging

Question 5.
Actively mobile minerals are:
(a) nitrogen and phosphorus
(b) iron and manganese
(c) sodium and cobalt
(d) silicon and selenium
Answer:
(a) nitrogen and phosphorus

Question 6.
Copper shows deficiency symptoms first that appear in young leaves due to:
(a) less active movement of minerals to younger leaves
(b) active movement of minerals
(c) the immobile nature of mineral
(d) none of the above
Answer:
(c) the immobile nature of mineral

Question 7.
Molybdenum is essential for the reaction of:
(a) hydrolase enzyme
(b) nitrogenase enzyme
(c) carboxylase enzyme
(d) dehydrogenase enzyme
Answer:
(b) nitrogenase enzyme

Question 8.
Match the following:

A. Magnesium (i) dehydrogenase
B. Nickel (ii) ion exchange
C. Zinc (iii) chlorophyll
D. Potassium (iv) urease

(a) A – (ii); B – (i); C – (iv); D – (iii)
(b) A – (iii); B – (ii); C – (i); D – (iv)
(c) A – (ii); B – (iv); C – (i); D – (iii)
(d) A – (iii); B – (iv); C – (i); D – (ii)
Answer:
(d) A – (iii); B – (iv); C – (i); D – (ii)

Question 9.
Nitrogen is the essential component of:
(a) carbohydrate
(b) fatty acids
(c) protein
(d) none of these
Answer:
(c) protein

Question 10.
Which of the element is involved in the synthesis of DNA and RNA:
(a) calcium
(b) magnesium
(c) sulphuric
(d) potassium
Answer:
(b) magnesium

Question 11.
The deficiency of magnesium is the plant, causes:
(a) necrosis
(b) interveinal chlorosis
(c) sand drown of tobacco
(d) all the above
Answer:
(d) all the above

Question 12.
Sulphur is an essential components of amino acids like:
(a) histidine, leucine and aspartic acid
(b) valene, alkaline and glycine
(c) cystine, cysteine and methionine
(d) none of the above
Answer:
(c) cystine, cysteine and methionine

Question 13.
Indicate the correct statements:
(i) Iron is the essential element for the synthesis of chlorophyll and carotenoid
(ii) Iron is the activator of carboxylene enzyme
(iii) Iton is the component of cytochrome
(iv) lvon is the component of plastocyanin

(a) (i) and (ii)
(b) (ii) and (iv)
(c) (ii) and (iii)
(d) (i) and (iii)
Answer:
(d) (i) and (iii)

Question 14.
Khaira disease of rice is caused by:
(a) deficiency of boron
(b) deficiency of zinc
(c) deficiency of iron
(d) deficiency of all the three
Answer:
(b) deficiency of zinc

Question 15.
Match the following:

A. Marginal chlorosis (i) nitrogen
B. Anthocyanin formation (ii) zinc
C. Hooked leaf tip (iii) potassium
D. Little leaf (iv) calcium

(a) A – (ii); B – (iii); C – (i); D – (iv)
(b) A – (iii), B – (ii); C – (iv); D – (i)
(c) A – (iii); B – (i); C – (iv); D – (ii)
(d) A – (iv); B – (iii); C – (i); D – (ii)
Answer:
(c) A – (iii); B – (i); C – (iv); D – (ii)

Question 16.
Increased concentration of manganese in plants will prevent the uptake of:
(a) calcium and potassium
(b) sodium and potassium
(c) boron and silicon
(d) iron and magnesium
Answer:
(d) iron and magnesium

Question 17.
Which of the statement is not correct?
(a) Aluminium toxicity causes the appearance of brown spots in the leaves.
(b) Aluminium toxicity causes the precipitation of nucleic acid.
(c) Aluminium toxicity inhibits ATPase activity
(d) Aluminium toxicity inhibits cell division.
Answer:
(a) Aluminium toxicity causes the appearance of brown spots in the leaves.

Question 18.
The techniques of Aeroponics was developed by:
(a) Goerick
(b) Amon and Hoagland
(c) Soifer Hillel and David Durger
(d) Von Sachs
Answer:
(c) Soifer Hillel and David Durger

Question 19.
Nitrogen occurs in atmosphere in the form of N2, two nitrogen atoms joined together by strong:
(a) di – covalent bond
(b) triple covalent bond
(c) non – valent bond
(d) none of these
Answer:
(b) triple covalent bond

Question 20.
The process of converting atmospheric nitrogen (N2) into ammonia is termed as:
(a) nitrogen cycle
(b) nitrification
(c) nitrogen fixation
(d) ammonification
Answer:
(c) nitrogen fixation

Question 21.
Find out the odd organism:
(a) Rhizobium
(b) Cyanobacteria
(c) Azolla
(d) Pistia
Answer:
(d) Pistia

Question 22.
The legume plants secretes phenolics to attract:
(a) Azolla
(b) Rhizobium
(c) Nitrosomonas
(d) Streptococcus
Answer:
(b) Rhizobium

Question 23.
Which are the organisms help in nitrogen fixation of lichens:
(a) Anabaena and Nostoc
(b) Anabaena alone
(c) Nostoc alone
(d) Anabaena azollae
Answer:
(a) Anabaena and Nostoc

Question 24.
Nitrogenase enzyme is active:
(a) only in aerobic condition
(b) only in anaerobic condition
(c) both in aerobic and anaerobic condition
(d) only in toxic condition
Answer:
(b) only in anaerobic condition

Question 25.
Ammonia (NH3+) is converted into nitrite (NO2) by a bacterium called:
(a) Nitrobacter bacterium
(b) Rhizobium
(c) Anabaena azollae
(d) Nitrosomonas
Answer:
(d) Nitrosomonas

Question 26.
Decomposition of organic nitrogen (proteins and amino acids) from dead plants and animals into ammonia is called:
(a) nitrification
(b) ammonification
(c) nitrogen fixation
(d) denitrification
Answer:
(b) ammonification

Question 27.
The bacteria involved in the denitrification process are:
(a) E.coli and Anabaena
(b) Streptococcus and Bacillus vulgaris
(c) Pseudomonas and Thiobacillus
(d) none of the above
Answer:
(c) Pseudomonas and Thiobacillus

Question 28.
In the process of ammonium assimilation:
(a) Ammonia is converted into nitrites
(b) Ammonia is converted into atmospheric nitrogen
(c) Ammonia is converted into ammonium ions
(d) Ammonia is converted into amino acids
Answer:
(d) Ammonia is converted into amino acids

Question 29.
The transfer of amino group (NH2) from glutamic acid to keto group of keto acid is termed as:
(a) Transamination
(b) Hydrogenation
(c) Nitrification
(d) Denitrification
Answer:
(a) Transamination

Question 30.
Monotrapa (Indian pipe) absorbs nutrients through:
(a) Rhizobium association
(b) mycorrhizal association
(c) microbial association
(d) animal association
Answer:
(b) mycorrhizal association

Question 31.
Cuscuta is a:
(a) partial parasite
(b) total root parasite
(c) obligate stem parasite
(d) partial stem parasite
Answer:
(c) obligate stem parasite

Question 32.
Indicate the correct statement:
(a) Loranthus grows on banana and coconut
(b) Loranthus grows on fig and mango trees
(c) Balanophora is a stem parasite
(d) Viscum is a root parasite
Answer:
(b) Loranthus grows on fig and mango trees

Question 33.
The association of mycorrhizae with higher plants is termed as:
(a) Parasitism
(b) Mutualism
(c) Symbiosis
(d) Saprophytic
Answer:
(c) Symbiosis

Question 34.
In Utricularia, the bladder is a modified form of:
(a) leaf
(b) stem
(c) tentacle
(d) lamina
Answer:
(a) leaf

Question 35.
Lichens are the indicators of:
(a) carbon monoxide
(b) nitrogen oxide
(c) sulphur di oxide
(d) hydrogen sulphide
Answer:
(c) sulphur di oxide

II. Answer the following (2 Marks)

Question 1.
Define micro nutrients of plants.
Answer:
Essential minerals which are required in less concentration called are as Micro nutrients.

Question 2.
Mention any two actively mobile minerals.
Answer:
Nitrogen and Phosphorus.

Question 3.
What is the role of molybdenum in the conversion of nitrogen into ammonia?
Answer:
Molybdenum (Mo) is essential for nitrogenase enzyme during reduction of atmospheric nitrogen into ammonia.

Question 4.
What is the role of potassium on osmotic potential of the cell?
Answer:
Potassium (K) plays a key role in maintaining osmotic potential of the cell. The absorption of water, movement of stomata and turgidity are due to osmotic potential.

Question 5.
What are the deficiency symptoms of nitrogen?
Answer:
Chlorosis, stunted growth, anthocyanin formation.

Question 6.
Explain the role of sulphur in plant biochemistry.
Answer:
Essential component of amino acids like cystine, cysteine and methionine, constituent of coenzyme A, Vitamins like biotin and thiamine, constituent of proteins and ferredoxin plants utilise sulphur as sulphate (SO4) ions.

Question 7.
Define the term Siderophores.
Answer:
Siderophores (iron carriers) are iron chelating agents produced by bacteria. They are used to chelate ferric iron (Fe3+) from environment and host.

Question 8.
List out any two iron deficiency symptoms in plants.
Answer:
Interveinal chlorosis, formation of short and slender stalk and inhibition of chlorophyll formation.

Question 9.
What is the role of Boron in plant physiology.
Answer:
Translocation of carbohydrates, uptake and utilisation of Ca++, pollen germination, nitrogen metabolism, fat metabolism, cell elongation and differentiation. It is absorbed as borate BO3-  ions.

Question 10.
Write down the deficiency symptoms of molybdenum in plants.
Answer:
Chlorosis, necrosis, delayed flowering, retarded growth and whip tail disease of cauliflower.

Question 11.
Explain briefly about aluminium toxicity on plants.
Answer:
Aluminium toxicity causes precipitation of nucleic acid, inhibition of ATPase, inhibition of cell division and binding of plasma membrane with Calmodulin.

Question 12.
Define Aeroponics.
Answer:
It is a system where roots are suspended in air and nutrients are sprayed over the roots by a motor driven rotor.

Question 13.
Define nitrogen fixation.
Answer:
The process of converting atmospheric nitrogen (N2) into ammonia is termed as nitrogen fixation. Nitrogen fixation can occur by two methods:

  1. Biological
  2. Non – Biological.

Question 14.
Mention any two ways of non – biological nitrogen fixation.
Answer:
Two ways of non – biological nitrogen fixation:

  1. Nitrogen fixation by chemical process in industry.
  2. Natural electrical discharge during lightening fixes atmospheric nitrogen.

Question 15.
Match the following.

A. Lichens (i) Anabaena Azolla
B. Anthoceros (ii) Frankia
C. Azolla (iii) Anabaena and Nostoc
D. Casuarina (iv) Nostoc

Answer:
A – (iii), B – (iv), C – (i), D – (ii).

Question 16.
Define the term Nitrate assimilation.
Answer:
The process by which nitrate is reduced to – ammonia is called nitrate assimilation and occurs during nitrogen cycle.

Question 17.
Explain.the term Transamination.
Answer:
Transfer of amino group (NH3+) from glutamic acid glutamate to keto group of keto acid. Glutamic acid is the main amino acid from which other amino acids are synthesised by transamination.

Question 18.
Explain briefly about total stem parasite.
Answer:
The leafless stem twine around the host and produce haustoria. eg: Cuscuta (Dodder), a rootless plant growing on Zizyphus, Citrus and so on.

Question 19.
Give two examples of symbiotic mode of nutrition.
Answer:
Two examples of symbiotic mode of nutrition:

  1. Lichens: It is a mutual association of Algae and Fungi. Algae prepares food and fungi absorbs water and provides thallus structure.
  2. Mycorrhizae: Fungi associated with roots of higher plants including Gymriosperms. eg: Pinus.

Question 20.
Explain briefly about insectivorous mode of nutrition.
Answer:
Plants which are growing in nitrogen deficient areas develop insectivorous habit to resolve nitrogen deficiency.

III. Answer the following (3 Marks)

Question 1.
What are the criteria required for essential minerals in plants?
Answer:
The criteria required for essential minerals in plants:

  1. Elements necessary for growth and development.
  2. They should have direct role in the metabolism of the plant.
  3. It cannot be replaced by other elements.
  4. Deficiency makes the plants impossible to complete their vegetative and reproductive phase.

Question 2.
Explain the unclassified minerals required for plants.
Answer:
Minerals like Sodium,Silicon, Cobalt and Selenium are not included in the list of essential nutrients but are required by some plants, these minerals are placed in the list of unclassified minerals. These minerals play specific roles for example, Silicon is essential for pest resistance, prevent water lodging and aids cell wall formation in Equisetaceae (Equisetum), Cyperaceae and Gramineae.

Question 3.
Distinguish between macro and micro nutrients?
Answer:
Macro nutrients:

  • Excess than 10 mmole Kg-1 in tissue concentration or 0.1 to 10 mg per gram of dry weight.
  • eg: C, H, O, N, P, K, Ca, Mg and S.

Micro nutrients:

  • Less than 10 mmole Kg-1 in tissue concentration or equal or less than 0.1 mg per gram of dry weight.
  •  eg: Fe, Mn, Cu, Mo, Zn, B, Cl and Ni.

Question 4.
Explain briefly the functions and deficiency symptoms of potassium.
Answer:
Functions: Maintains turgidity and osmotic potential of the cell, opening and closure of stomata, phloem translocation, stimulate activity of enzymes, anion and cation balance by ion – exchange. It is absorbed as K+ ions. Deficiency symptoms: Marginal chlorosis, necrosis, low cambial activity, loss of apical dominance, lodging in cereals and curled leaf margin.

Question 5.
What is meant by Chelating agents? Explain the role of EDTA as chemical chelating agent.
Answer:
Plants which are growing in alkaline soil when supplied with all nutrients including iron will show iron deficiency. To rectify this, we have to make iron into a soluble complex by adding a chelating agent like EDTA (Ethylene Diamine Tetra Acetic acid) to form Fe – EDTA.

Question 6.
Explain the term critical concentration of minerals.
Answer:
To increase the productivity and also to avoid mineral toxicity knowledge of critical concentration is essential. Mineral nutrients lesser than . critical concentration cause deficiency symptoms. Increase of mineral nutrients more than the normal concentration causes toxicity. A concentration, at which 10% of the dry weight of tissue is reduced, is considered as toxic critical concentration.

Question 7.
Describe the competitive behaviour of iron and manganese.
Answer:
Iron and Manganese exhibit competitive behaviour. Deficiency of Fe and Mn shows similar symptoms. Iron toxicity will affect absorption of manganese. The possible reason for iron toxicity is excess usage of chelated iron in addition with increased acidity of soil (pH less than 5.8) Iron and manganese toxicity will be solved by using fertilizer with balanced ratio of Fe and Mn.

Question 8.
Who are people responsible for developing hydroponics?
Answer:
Hydroponics or Soil less culture: Von Sachs developed a method of growing plants in nutrient solution. The commonly used nutrient solutions are Knop solution (1865) and Amon and Hoagland Solution (1940). Later the term Hydroponics was coined by Goerick (1940) and he also introduced commercial techniques for hydroponics. In hydroponics roots are immersed in the solution containing nutrients and air is supplied with help of tube.

Question 9.
List out the free living bacteria and fungi responsible for non-symbiotic nitrogen fixation.
Answer:
Free living bacteria and fungi also fix atmospheric nitrogen.

Aerobic Azotobacter, Beijerneckia and Derxia
Anaerobic Clostridium
Photosynthetic Chlorobium and Rhodospirillum
Chemosynthetic Disulfovibrio
Free living fungi Yeast and Pullularia
Cyanobacteria Nostoc, Anabaena and Oscillatoria.

Question 10.
Define the term Ammonification.
Answer:
Decomposition of organic nitrogen (proteins and amino acids) from dead plants and animals into ammonia is called ammonification. Organisim involved in this process are Bacillus ramosus and Bacillus vulgaris.

Question 11.
Explain briefly Catalytic amination.
Answer:
Glutamate amino acid combines with ammonia to form the amide glutamine.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition 8
Glutamine reacts with a ketoglutaric acid to form two molecules of glutamate.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition 9
(GOGAT – Glutamine – 2 – Oxoglutarate aminotransferase)

Question 12.
Compare the partial stem parasite and partial root parasite.
Answer:
The partial stem parasite and partial root parasite:

  1. Partial Stem Parasite: eg: Loranthus and Viscum (Mistletoe) Loranthus grows on fig and mango trees and absorb water and minerals from xylem.
  2. Partial root parasite: eg: Santalum album (Sandal wood tree) in its juvenile stage produces haustoria which grows on roots of many plants.

Question 13.
Explain the mode of nutrition in pitcher plant.
Answer:
Pitcher is a modified leaf and contains digestive enzymes. Rim of the pitcher is provided with nectar glands and acts as an attractive lid. When insect is trapped proteolytic enzymes will digest the insect.

Question 14.
What is meant by saprophytic mode of nutrition?
Answer:
Saprophytes derive nutrients from dead and decaying matter. Bacteria and fungus are main saprophytic organisms. Some angiosperms also follow saprophytic mode of nutrition. eg: Neottia. Roots of Neottia (Bird’s Nest Orchid) associate with mycorrhizae and absorb nutrients as a saprophyte. Monotropa (Indian Pipe) grow on humus rich soil found in thick forests. It absorbs nutrient through mycorrhizal association.

Question 15.
Describe briefly the method of nitrogen fixation in leguminous plants.
Answer:
Rhizobium bacterium is found in leguminous plants and fix atmospheric nitrogen. This kind of symbiotic association is beneficial for both the bacterium and plant. Root nodules are formed due to bacterial infection. Rhizobium enters into the-host cell and proliferates, it remains separated from the host cytoplasm by a membrane.

IV. Answer the following (5 Marks)

Question 1.
Write an essay on the functions and deficiency symptoms of macro nutrients.
Answer:
Macronutrients, their functions, their mode of absorption, deficiency symptoms and deficiency diseases are discussed here:
(i) Nitrogen (N): It is required by the plants in greatest amount. It is an essential component of proteins, nucleic acids, amino acids, vitamins, hormones, alkaloids, chlorophyll and cytochrome. It is absorbed by the plants as nitrates (NO3).

Deficiency symptoms: Chlorosis, stunted growth, anthocyanin formation.

(ii) Phosphorus (P): Constituent of cell membrane, proteins, nucleic acids, ATP, NADP, phytin and sugar phosphate. It is absorbed as H2PO4+ and HPO4 ions.

Deficiency symptoms: Stunted growth, anthocyanin formation, necrosis, inhibition of cambial activity, affect root growth and fruit ripening.

(iii) Potassium (K): Maintains turgidity and osmotic potential of the cell, opening and closure of stomata, phloem translocation, stimulate activity of enzymes, anion and cation balance by ion – exchange. It is absorbed as K+ ions.

Deficiency symptoms: Marginal chlorosis, necrosis, low cambial activity, loss of apical dominance, lodging in cereals and curled leaf margin.

(iv) Calcium (Ca): It is involved in synthesis of calcium pectate in middle lamella, mitotic spindle formation, mitotic cell division, permeability of cell membrane, lipid metabolism, activation of phospholipase, ATPase, amylase and activator of adenyl kinase. It is absorbed as Ca2+ exchangeable ions.

Deficiency symptoms: Chlorosis, necrosis, stunted growth, premature fall of leaves and flowers, inhibit seed formation, Black heart of Celery, Hooked leaf tip in Sugar beet, Musa and Tomato.

(v) Magnesium (Mg): It is a constituent of chlorophyll, activator of enzymes of carbohydrate metabolism (RUBP Carboxylase and PEP Carboxylase) and involved in the synthesis of DNA and RNA. It is essential for binding of ribosomal sub units. It is absorbed as Mg2+ ions.

Deficiency symptoms: litter veinal chlorosis, necrosis, anthocyanin (purple) formation and Sand drown of tobacco.

(vi) Sulphur (S): Essential component of amino acids like cystine, cysteine and methionine, constituent of coenzyme A, Vitamins like biotin and thiamine, constituent of proteins and ferredoxin. plants utilise sulphur as sulphate (SO4) ions.

Deficiency symptoms: Chlorosis, anthocyanin formation, stunted growth, rolling of leaf tip and reduced nodulation in legumes.

Question 2.
Describe the role of micro nutrients on plant health and function.
Answer:
Micronutrients even though required in trace amounts are essential for the metabolism of plants. They play key roles in many plants. eg: Boron is essential for translocation of sugars, molybdenum is involved in nitrogen metabolism and zinc is needed for biosynthesis of auxin. Here, we will study about the role of micro nutrients, their functions, their mode of absorption, deficiency symptoms and deficiency diseases.

(i) Iron (Fe): Iron is required lesser than macronutrient and larger than micronutrients, hence, it can be placed in any one of the groups. Iron is an essential element for the synthesis of chlorophyll and carotenoids. It is the component of cytochrome, ferredoxin, flavoprotein, formation of chlorophyll, porphyrin, activation of catalase, peroxidase enzymes. It is absorbed as ferrous (Fe2+) and ferric (Fe3+) ions. Mostly fruit trees are sensitive to iron.

Deficiency: Interveinal Chlorosis, formation of short and slender stalk and inhibition of chlorophyll formation.

(ii) Manganese (Mn): Activator of Carboxylases, oxidases, dehydrogenases and kinases, involved in splitting of water to liberate oxygen (photolysis). It is absorbed as manganous (Mn2+) ions.

Deficiency: Interveinal chlorosis, grey spot on oats leaves and poor root system.

(iii) Copper (Cu): Constituent of plastocyanin, component of phenolases, tyrosinase, enzymes involved in redox reactions, synthesis of ascorbic acid, maintains carbohydrate and nitrogen balance, part of oxidase and cytochrome oxidase. It is absorbed as cupric (Cu2+) ions,

Deficiency: Die back of citrus, Reclamation disease of cereals and legumes, chlorosis, necrosis and Exanthema in Citrus.

(iv) Zinc (Zn): Essential for the synthesis of Indole acetic acid (Auxin) activator of carboxylases, alcohol dehydrogenase, lactic dehydrogenase, glutamic acid dehydrogenase, carboxy peptidases and tryptophan synthetase. It is absorbed as Zn2+ ions.

Deficiency: Little leaf and mottle leaf due to deficiency of auxin, Inter veinal chlorosis, stunted growth, necrosis and Khaira disease of rice.

(v) Boron (B): Translocation of carbohydrates, uptake and utilisation of Ca++, pollen germination, nitrogen metabolism, fat metabolism, cell elongation and differentiation. It is absorbed as borate BO3- ions.

Deficiency: Death of root and shoot tips, premature fall of flowers and fruits, brown heart of beet root, internal cork of apple and fruit cracks.

(vi) Molybdenum (Mo): Component of nitrogenase, nitrate reductase, involved in nitrogen metabolism, and nitrogen fixation. It is absorbed as molybdate (Mo2+) ions.

Deficiency: Chlorosis, necrosis, delayed flowering, retarded growth and whip tail disease of cauliflower.

(vii) Chlorine (Cl): It is involved in Anion – Cation balance, cell division, photolysis of water. It is absorbed as Cl ions.
Deficiency: Wilting of leaf tips.

(viii) Nickel (Ni): Cofactor for enzyme urease and hydrogenase.

Deficiency: Necrosis of leaf tips.

Question 3.
Give the details of minerals and their deficiency symptoms.
Answer:
Name of the deficiency disease and symptoms:

  1. Chlorosis (Overall)
    • (a) Interveinal chlorosis
    • (b) Marginal chlorosis
  2. Necrosis (Death of the tissue)
  3. Stunted growth
  4. Anthocyanin formation
  5. Delayed flowering
  6. Die back of shoot, Reclamation disease, Exanthema in citrus (gums on bark)
  7. Hooked leaf tip
  8. Little Leaf
  9. Brown heart of turnip and Internal cork of apple
  10. Whiptail of cauliflower and cabbage
  11. Curled leaf margin

Deficiency minerals:

  1. Nitrogen, Potassium, Magnesium, Sulphur, Iron, Manganese, Zinc and Molybdenum. Magnesium, Iron, Manganese and Zinc Potassium
  2. Magnesium, Potassium, Calcium, Zinc, Molybdenum and Copper.
  3. Nitrogen, Phosphorus, Calcium, Potassium and Sulphur.
  4. Nitrogen, Phosphorus, Magnesium and Sulphur
  5. Nitrogen, Sulphur and Molybdenum
  6. Copper
  7. Calcium
  8. Zinc
  9. Boron
  10. Molybdenum
  11. Potassium

Question 4.
Give the schematic diagram of nitrogen cycle.
Answer:
The schematic diagram of nitrogen cycle:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition 7

Question 5.
Describe the modes of biological nitrogen fixation.
Answer:
Symbiotic bacterium like Rhizobium fixes atmospheric nitrogen. Cyanobacteria found in Lichens, Anthoceros, Azolla and coralloid roots of Cycas also fix nitrogen. Non – symbiotic (free living bacteria) like Clostridium also fix nitrogen. Symbiotic nitrogen fixation:
1. Nitrogen fixation with nodulation: Rhizobium bacterium is found in leguminous plants and fix atmospheric nitrogen. This kind of symbiotic association is beneficial for both the bacterium and plant. Root nodules are formed due to bacterial infection. Rhizobium enters into the host cell and proliferates, it remains separated from the host cytoplasm by a membrane.

2. Stages of Root nodule formation:

  • Legume plants secretes phenolics which attracts Rhizobium.
  • Rhizobium reaches the rhizosphere and enters into the root hair, infects the root hair and leads to curling of root hairs.
  • Infection thread grows inwards and separates the infected tissue from normal tissue.
  • A membrane bound bacterium is formed inside the nodule and is called bacteroid.
  • Cytokinin from bacteria and auxin from host plant promotes cell division and leads to nodule formation

3. Non – Legume: Alnus and Casuarina contain the bacterium Frankia Psychotria contains the bacterium Klebsiella.
Nitrogen fixation without nodulation. The following plants and prokaryotes are involved in nitrogen fixation:

  • Lichens – Anabaena and Nostoc
  • Anthoceros – Nostoc
  • Azolla – Anabaena azollae
  • Cycas – Anabaena and Nostoc.

Solution To Activity
Textbook Page No: 95

Question 1.
Collect leaves showing mineral deficiency. Tabulate the symptoms like Marginal Chlorosis, Interveinal Chlorosis, Necrotic leaves, Anthocyanin formation in leaf, Little leaf and Hooked leaf. (Discuss with your teacher about the deficiency of minerals)
Answer:
Symptoms:

  1. Marginal Chlorosis
  2. interveinal Chlorosis
  3. Necrotic leaves
  4. Anthocyanin formation in leaves
  5. Little leaf
  6. Hooked leaf

Minerals:

  1. Potassium (K)
  2. Magnesium (Mg)
  3. Nickel (Ni)
  4. Phosphorus (P)
  5. Zinc (Zn)
  6. Calcium (Ca)

Textbook Page No: 98

Question 1.
Preparation of Solution Culture to find out Mineral Deficiency
1. Take a glass jar or polythene bottle and cover with black paper (to prevent algal growth and roots reacting with light).
2. Add nutrient solution.
3. Fix a plant with the help of split cork.
4. Fix a tube for aeration.
5. Observe the growth by adding specific minerals.
Answer:
The deficiency of minerals like nitrogen, phosphorus, calcium, potassium and sulphur cause stunted growth in plants.

Textbook Page No: 99

Question 1.
Collect roots of legumes with root nodules.
• Take cross section of the root nodule.
• Observe under microscope. Discuss your observations with your teacher.
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition 1

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Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth

Students can Download Bio Botany Chapter 10 Secondary Growth Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth

Samacheer Kalvi 11th Bio Botany Secondary Growth Text Book Back Questions and Answers

Question 1.
Consider the following statements In spring season vascular cambium:
(i) is less active
(ii) produces a large number of xylary elements
(iii) forms vessels with wide cavities of these

(a) (i) is correct but (ii) and (iii) are not correct
(b) (i) is not correct but (ii) and (iii) are correct
(c) (i) and (ii) are correct but (iii) is not correct
(d) (i) and (ii) are not correct but (iii) is correct
Answer:
(b) (i) is not correct but (ii) and (iii) are correct

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 2.
Usually, the monocotyledons do not increase their girth, because:
(a) They possess actively dividing cambium
(b) They do not possess actively dividing cambium
(c) Ceases activity of cambium
(d) All are correct
Answer:
(b) They do not possess actively dividing cambium

Question 3.
In the diagram of lenticel identify the parts marked as A,B,C,D.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth 1
(a) A. Phellem, B. Complementary tissue, C. Phelloderm, D. Phellogen.
(b) A. Complementary tissue, B. Phellem, C. Phellogen, D. Phelloderm.
(c) A. Phellogen, B. Phellem, C. Pheiloderm, D. complementary tissue
(d) A. Phelloderm, B. Phellem, C. Complementary tissue, D. Phellogen
Answer:
(a) A. Phellem, B. Complementary tissue, C. Phelloderm, D. Phellogen.

Question 4.
The common bottle cork is a product of:
(a) Dermatogen
(b) Phellogen
(c) Xylem
(d) Vascular cambium
Answer:
(b) Phellogen

Question 5.
What is the fate of primary xylem in a dicot root showing extensive secondary growth?
(a) It is retained in the center of the axis
(b) It gets crushed
(c) May or may not get crushed
(d) It gets surrounded by primary phloem
Answer:
(b) It gets crushed

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 6.
In a forest, if the bark of a tree is damaged by the horn of a deer, How will the plant overcome the damage?
Answer:
When the bark is damaged, the phellogem forms a complete cylinder around the stem and it gives rise to ring barks.

Question 7.
In which season the vessels of angiosperms are larger in size, why?
Answer:
In spring season the vessels are larger in size, because the cambium cells are very active during spring season.

Question 8.
Continuous state of dividing tissue is called meristem. In connection to this, what is the role of lateral meristem?
Answer:
Apical meristems produce the primary plant body. In some plants, the lateral meristem increase the girth of a plant. This type of growth is secondary because the lateral meristem are not directly produced by apical meristems. Woody plants have two types of lateral meristems: a vascular cambium that produces xylem, phloem tissues and cork cambium that produces the bark of a tree.

Question 9.
A timber merchant bought 2 logs of wood from a forest & named them A & B, The log A was 50 year old & B was 20 years old. Which log of wood will last longer for the merchant? Why?
Answer:
The wood of 50 years old will last longer than 20 years old wood, because timber from hard wood is more durable and more resistant to the attack of micro organisms and insect than the timber from sap wood.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 10.
A transverse section of the trunk of a tree shows concentric rings which are known as growth rings. How are these rings formed? What are the significance of these rings?
Answer:
The annual ring denotes the combination of early wood and late wood and the ring becomes evident to our eye due to the high density of late wood. Sometimes annual rings are called growth rings but it should be remembered all the growth rings are not annual. In some trees more than one growth ring is formed with in a year due to climatic changes. Additional growth rings are developed within a year due to adverse natural calamities like drought, frost, defoliation, flood, mechanical injury and biotic factors during the middle of a growing season, which results in the formation of more than one annual ring.

Such rings are called pseudo – or false – annual rings. Each annual ring corresponds to one year’s growth and on the basis of these rings, the age of a particular plant can easily be calculated. The determination of the age of a tree by counting the annual rings is called dendrochronology.

Samacheer Kalvi 11th Bio Botany Secondary Growth Other Important Questions & Answers

I. Choose the correct answer. (I Marks)
Question 1.
The roots and stems grow in length with the help of:
(a) cambium
(b) secondary growth
(c) apical meristem
(d) vascular parenchyma
Answer:
(c) apical meristem

Question 2.
The increase in the girth of plant is called:
(a) primary growth
(b) tertiary growth
(c) longitudinal growth
(d) secondary growth
Answer:
(d) secondary growth

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 3.
The secondary vascular tissues include:
(a) secondary xylem and secondary phloem
(b) secondary xylem, cambium strip and secondary phloem
(c) secondary phloem and fascicular cambium
(d) secondary xylem and primary phloem
Answer:
(a) secondary xylem and secondary phloem

Question 4.
Choose the correct statements.
(i) A strip of vascular cambium is present between xylem and phloem of the vascular bundle.
(ii) Vascular cambium is believed originate from fusiform initials.
(iii) The vascular cambium is originated from procambium of vascular bundle
(iv) Vascular cambium is present between fusiform initials and ray initials

(a) (i) and (iv)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer:
(b) (i) and (iii)

Question 5.
Match the following:

A. Xylem (i) Treachery elements
B. Secondary xylem (ii) Water transport
C. Phloem (iii) Sieve elements
D. Secondary phloem (iv) Food transport

(a) B – (i); A – (ii); C – (iii); D – (iv)
(b) B – (ii); A – (iii); C – (i); D – (iv)
(c) A – (ii); B – (i); C – (iv); D – (iii)
(d) A – (i); B – (ii); C – (iii); D – (iv)
Answer:
(c) A – (ii); B – (i); C – (iv); D – (iii)

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 6.
The axial system of the secondary xylem includes:
(a) treachery elements, sieve elements, fibers and axial parenchyma
(b) treachery elements, fibers and axial parenchyma
(c) treachery elements and fibers
(d) sieve elements and axial parenchyma
Answer:
(b) treachery elements, fibers and axial parenchyma

Question 7.
The study of wood by preparing sections for microscopic observation is termed as:
(a) histology
(b) xylotomy
(c) phoemtomy
(d) anatomy
Answer:
(b) xylotomy

Question 8.
Ray cells are present between:
(a) primary xylem and phloem
(b) primary xylem and secondary xylem
(c) secondary xylem and phloem
(d) secondary phloem and cambium
Answer:
(c) secondary xylem and phloem

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 9.
The axial system Consists of vertical files of:
(a) treachery elements and sieve elements
(b) treachery elements and apical parenchyma
(c) sieve elements are fibers
(d) treachery elements, fibers and wood parenchyma
Answer:
(d) treachery elements, fibers and wood parenchyma

Question 10.
Morus rubra has:
(a) porous wood
(b) soft wood
(c) spring wood
(d) sap wood
Answer:
(a) porous wood

Question 11.
Which of the statement is not correct?
(a) In temperate regions, the cambium is very active in winter season.
(b) In temperate regions, the cambium is very active in spring season.
(c) In temperate regions, cambium is less active in winter season.
(d) In temperate regions early wood is formed in spring season.
Answer:
(a) In temperate regions, the cambium is very active in winter season.

Question 12.
Usually more distinct annual rings are formed:
(a) in tropical plants
(b) in seashore plants
(c) in temperate plants
(d) in desert plants
Answer:
(c) in temperate plants

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 13.
False annual rings are formed due to:
(a) rain
(b) adverse natural calamities
(c) severe cold
(d) none of the above
Answer:
(b) adverse natural calamities

Question 14.
determination of the age of a tree by counting the annual rings is called:
(a) chronology
(b) dendrochronology
(c) palaeology
(d) histology
Answer:
(c) palaeology

Question 15.
The age of American sequoiadendron tree is about:
(a) 350 years
(b) 3,000 years
(c) 3400 years
(d) 3500 years
Answer:
(d) 3500 years

Question 16.
The wood of Acer plant has:
(a) ring porous
(b) diffuse porous
(c) central porous
(d) none of the above
Answer:
(b) diffuse porous

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 17.
In fully developed tyloses:
(a) only starchy crystals are present
(b) resin and gums only are present
(c) oil and tannins are present
(d) starchy crystals, resins, gums, oils, tannins or colored substances are present
Answer:
(d) starchy crystals, resins, gums, oils, tannins or colored substances are present

Question 18.
In bombax:
(a) the sieve tubes are blocked by tylose like outgrowths
(b) the resin ducts are blocked by tylose like outgrowths
(c) the phloem tube is blocked by tylose like out growths
(d) none of the above
Answer:
(a) the sieve tubes are blocked by tylose like outgrowths

Question 19.
Which of the statement is not correct?
(a) Sap wood and heart wood can be distinguished in the secondary xylem
(b) Sap wood is paler in colour
(c) Heart wood is darker in colour
(d) The sap wood conducts minerals, while the heart wood conduct water
Answer:
(d) The sap wood conducts minerals, while the heart wood conduct water

Question 20.
Timber from heart wood is:
(a) more fragile and resistant to the attack of insects
(b) more durable and more resistant to the attack of micro organism and insects
(c) more hard and less resistant to the attack of micro organism
(d) less durable and more resistant to the attack of micro organism and insects
Answer:
(b) more durable and more resistant to the attack of micro organism and insects

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 21.
The dye, haematoxylin is obtained from:
(a) the heart wood of haematoxylum campechianum
(b) the sap wood of haematoxylum campechianum
(c) cambium cells of haematoxylum campechianum
(d) the seeds of haematoxylum campechianum
Answer:
(a) the heart wood of haematoxylum campechianum

Question 22.
Canada balsam is produced from:
(a) Pisum sativum
(b) resin of Arjuna plant
(c) Abies balsamea
(d) the root of Vinca rosea
Answer:
(c) Abies balsamea

Question 23.
Some commercially important phloem or bast fibres are obtained from:
(a) banana
(b) bamboo
(c) vinca rosea
(d) cannabis sativa
Answer:
(d) cannabis sativa

Question 24.
Phellogen comprises:
(a) homogeneous sclerenchyma cells
(b)homogeneous meristamatic cells
(c) homogeneous collenchyma cells
(d) none of the above cells
Answer:
(b)homogeneous meristamatic cells

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 25.
Phelloderm is otherwise called as:
(a) primary cortex
(b) cork wood
(c) secondary cortex
(d) rhytidome
Answer:
(c) secondary cortex

Question 26.
Lenticel is helpful in:
(a) transportation of food
(b) photosynthesis
(c) exchanges of gases and transpiration
(d) transportation of water
Answer:
(c) exchanges of gases and transpiration

Question 27.
The antimalarial compound quinine is, extracted from:
(a) seeds of cinchona
(b) bark of cinchona
(c) leaves of cinchona
(d) flowers of cinchona
Answer:
(b) bark of cinchona

Question 28.
Gum Arabic is obtained from:
(a) Hevea brasiliensis
(b) Acacia Senegal
(c) Pinus
(d) Dilonix regia
Answer:
(b) Acacia Senegal

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 29.
Turpentine used as thinner of paints is obtained from:
(a) Acacia Senegal
(b) Vinca rosea
(c) Hevea brasiliensis
(d) Pinus
Answer:
(d) Pinus

Question 30.
Rubber is obtained from:
(a) Bombax mori
(b) Hevea brasiliensis
(c) Quercus suber
(d) Morus rubra
Answer:
(b) Hevea brasiliensis

II. Answer the following. (2 Marks)

Question 1.
Define primary growth?
Answer:
The roots and stems grow in length with the help of apical meristems. This is called primary growth or longitudinal growth.

Question 2.
Mention the two lateral meristem responsible for secondary growth.
Answer:
The secondary growth in dicots and gymnosperms is brought about by two lateral meristems.

  1. Vascular Cambium and
  2. Cork Cambium

Question 3.
What is meant by vascular cambium?
Answer:
The vascular cambium is the lateral meristem that produces the secondary vascular tissues. i.e., secondary xylem and secondary phloem.

Question 4.
Define intrafascicular or fascicular cambium?
Answer:
A strip of vascular cambium that is believed to originate from the procambium is present between xylem and phloem of the vascular bundle. This cambial strip is known as intrafascicular or fascicular cambium.

Question 5.
Define interfascicular cambium?
Answer:
In between the vascular bundles, a few parenchymatous cells of the medullary rays that are in line with the fascicular cambium become meristematic and form strips of vascular cambium. It is called interfascicular cambium.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 6.
What is vascular cambial ring?
Answer:
This interfascicular cambium joins with the intrafascicular cambium on both sides to form a continuous ring. It is called a vascular cambial ring.

Question 7.
What is meant by stratified cambium?
Answer:
If the fusiform initials are arranged in horizontal tiers, with the end of the cells of one tier appearing at approximately the same level, as seen in Tangential Longitudinal Section (TLS), it is called storied (stratified) cambium.

Question 8.
Explain non – stratified cambium.
Answer:
In plants with long fusiform initials, they strongly overlap at the ends, and this type of cambium is called non – storied (non – startified) cambium.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 9.
Give a brief note on ray initials.
Answer:
These are horizontally elongated cells. They give rise to the ray cells and form the elements of the radial system of secondary xylem and phloem.

Question 10.
How does secondary xylem or wood form?
Answer:
The secondary xylem, also called wood, is formed by a relatively complex meristem, the vascular cambium, consisting of vertically (axial) elongated fusiform initials and horizontally (radially) elongated ray initials.

Question 11.
What is meant by spring wood?
Answer:
In the spring season, cambium is very active and produces a large number of xylary elements having vessels / tracheids with wide lumen. The wood formed during this season is called spring wood or early wood.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 12.
How does the autumn wood form?
Answer:
In winter, the cambium is less active and forms fewer xylary elements that have narrow vessels /  tracheids and this wood is called autumn wood or late wood.

Question 13.
Define growth rings?
Answer:
The annual ring denotes the combination of early wood and late wood and the ring becomes evident to our eye due to the high density of late wood. Sometimes annual rings are called growth rings.

Question 14.
Define dendroclimatology?
Answer:
It is a branch of dendrochronology concerned with constructing records of past climates and climatic events by analysis of tree growth characteristics, especially growth rings.

Question 15.
Explain diffuse porous woods with an example.
Answer:
Diffuse porous woods are woods in which the vessels or pores are rather uniform in size and distribution throughout an annual ring. eg: Acer

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 16.
What is meant by ring porous woods?
Answer:
The pores of the early wood are distinctly larger than those of the late wood. Thus rings of wide and narrow vessels occur.

Question 17.
Define tyloses?
Answer:
In many dicot plants, the lumen of the xylem vessels is blocked by many balloon like ingrowths from the neighbouring parenchymatous cells. These balloons like structure are called tyloses.

Question 18.
Mention two plants from which bast fibres are obtained.
Answer:
Two plants from which bast fibres are obtained:

  1. Flax – Linum ustitaissimum
  2. Hemp – Cannabis sativa

Question 19.
Define Rhytidome?
Answer:
Rhytidome is a technical term used for the outer dead bark which consists of periderm and isolated cortical or phloem tissues ? formed during successive secondary growth, eg: Quercus.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 20.
What is polyderm? Explain briefly.
Answer:
Polyderm is found in the roots and underground stems. eg: Rosaceae. It refers to a special type of protective tissues consisting of uniseriate suberized layer alternating with multiseriate nonsuberized cells in periderm.

Question 21.
Define’bark’?
Answer:
The term ‘bark’ is commonly applied to all the tissues outside the vascular cambium of stem (i.e., periderm, cortex, primary phloem and secondary phloem).

Question 22.
What are the functions of lenticel?
Answer:
Lenticel is helpful in exchange of gases and transpiration called lenticular transpiration.

Question 23.
Explain briefly phelloderm.
Answer:
It is a tissue resembling cortical living parenchyma produced centripetally (inward) from the phellogen as a part of the periderm of stems and roots in seed plants.

Question 24.
What is the function of secondary phloem?
Answer:
Secondary phloem is a living tissue that transports soluble organic compounds made during photosynthesis to various parts of plant.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 25.
what is periderm?
Answer:
Whenever stems and roots increase in thickness by secondary growth, the periderm, a protective tissue of secondary origin replaces the epidermis and Often primary cortex. The periderm consists of phellem, phellogen, and phelloderm.

III. Answer the following. (3 Marks)

Question 1.
Distinguish between primary and secondary growth.
Answer:
1. Primary growth: The plant organs originating from the apical meristems pass through a period of expansion in length and width. The roots and stems grow in length with the help of apical meristems. This is tailed primary growth or longitudinal growth.

2. Secondary growth: The gymnosperms and most angiosperms, including some monocots, show an increase in thickness of stems and roots by means of secondary growth or latitudinal growth.

Question 2.
Explain fusiform initials.
Answer:
These are vertically elongated cells. They give rise to the longitudinal or axial system of the secondary xylem (treachery elements, fibers, and axial parenchyma) and phloem (sieve elements, fibers, and axial parenchyma).

Question 3.
Explain briefly about false annual rings.
Answer:
Additional growth rings are developed within a year due to adverse natural calamities like drought, frost, defoliation, flood, mechanical injury and biotic factors during the middle of a growing season, which results in the formation of more than one annual ring. Such rings are called pseudo – or false – annual rings.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 4.
Write down the differences between spring wood and autumn wood.
Answer:
The differences between spring wood and autumn wood:

Spring wood or Early wood

Autumn wood or Late wood

1. The activity of cambium is faster. 1. Activity of cambium is slower.
2. Produces large number of xylem elements. 2. Produces fewer xylem elements.
3. Xylem vessels /  trachieds have wider lumen. 3. Xylem vessels / trachieds have narrow lumen.
4. Wood is lighter in colour and has lower density. 4. Wood is darker in colour and has a higher density.

Question 5.
How do you distinguish between sap wood and heart wood?
Answer:

Sap wood (Alburnum)

Heart wood (Duramen)

1. Living part of the wood. 1. Dead part of the wood.
2. It is situated on the outer side of wood. 2.It is situated in the certre part of wood.
3. It is less in coloured. 3. It is dark in coloured.
4. Very soft in nature. 4. Hard in nature.
Tyloses are absent.  Tyloses are present.
5. It is not durable and not resistant to microorganisms. 5. It is more durable and resists microorganisms.

Question 6.
What are fossil resins? Explain with an example.
Answer:
Plants secrete resins for their protective benefits. Amber is a fossilized tree resinespecially from the wood, which has been appreciated for its colour and natural beauty since neolithic times. Much valued from antiquity to the present as a gemstone, amber is made into a variety of decorative objects. Amber is used in jewellery. It has also been used as a healing agent in folk medicine.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 7.
Write briefly about Cork cambium.
Answer:
It is a secondary lateral meristem. It comprises homogenous meristematic cells unlike vascular cambium. It arises from epidermis, cortex, phloem or pericycle (extrastelar in origin). Its cells divide periclinally and produce radially arranged files of cells. The cells towards the outer side differentiate into phellem (cork) and those towards the inside as phelloderm (secondary cortex).

Question 8.
Explain the term lenticel.
Answer:
Lenticel is raised opening or pore on the epidermis or bark of stems and roots. It is formed during secondary growth in stems. When phellogen is more active in the region of lenticels, a mass of loosely arranged thin – walled parenchyma cells are formed. It is called complementary tissue or filling tissue. Lenticel is helpful in exchange of gases and transpiration called lenticular transpiration.

Question 9.
Mention the benefits of bark in a tree.
Answer:
Bark protects the plant from parasitic fungi and insects, prevents water loss by evaporation and guards against variations of external temperature. It is an insect repellent, decay proof, fireproof and is used in obtaining drugs or spices. The phloem cells of the bark are involved in conduction of food while secondary cortical cells involved in storage.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 10.
Distinguish between Intrafascicular Interfascicular cambium.
Answer:
Between Intrafascicular Interfascicular cambium:

Intrafascicular cambium

Interfascicular cambium

1. Present inside the vascular bundles 1. Present in between the vascular bundles.
2. Originates from the procambium. 2. Originates from the medullary rays.
3. Initially it forms a part of the primary meristem. 3. From the beginning it forms a part of the secondary meristem.

IV. Answer In detail
Question 1.
Describe the activity of vascular with the help of diagram.
Answer:
Activity of Vascular Cambium:
The vascular cambial ring, when active, cuts off new cells both towards the inner and outer side. The cells which are produced outward form secondary phloem and inward secondary xylem. At places, cambium forms some narrow horizontal bands of parenchyma which passes through secondary phloem and xylem. These are the rays. Due to the continued formation of secondary xylem and phloem through vascular cambial activity, both the primary xylem and phloem get gradually crushed.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth 6

Question 2.
Describe the formation of sap wood and heart wood with suitabie diagram.
Answer:
Sap wood and heart wood can be distinguished in the secondary xylem. In any tree the outer part of the wood, which is paler in colour, is called sap wood are alburnum. The centre part of the wood, which is darker in colour is called heart wood or duramen. The sap wood conducts water while the heart wood stops conducting water. As vessels of the heart wood are blocked by tyloses, water is not conducted through them.

Due to the presence of tyloses and their contents the heart wood becomes coloured, dead and the hardest part of the wood. From the economic point of view, generally the heartwood is more useful than the sapwood. The timber form the heartwood is more durable and more resistant to the attack of microorganisms and insects than the timber from sapwood.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth 4

Question 3.
Draw and label the transverse section of dicot stem showing the secondary growth.

Answer:
The transverse section of dicot stem showing the secondary growth:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth 5

Question 4.
Distinguish between Phellem and Phelloderm.
Answer:
Phellem (Cork):

  1. It is formed on the outer side of phellogen.
  2. Cells are compactly arranged in regular tires and rows without intercellular spaces.
  3. Protective in function.
  4. Consists of nonliving cells with suberized walls.
  5. Lenticels are present.

Phelloderm (Secondary cortex):

  1. It is formed on the inner side of phellogen.
  2. Cells are loosely arranged with intercellular spaces.
  3. As it contains chloroplast, it synthesises and stores food.
  4. Consists of living cells, parenchymatous in nature and does not have suberin.
  5. Lenticels are absent.

Question 5.
Write down the economic importance of tree bark.
Answer:
The economic importance of tree bark:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth 3

Question 5.
Draw the different stages of secondary growth in a dicot root and label the parts.
Answer:
Stages of secondary growth in a dicot root and label the parts:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth 2

Solution To Activity
Textbook Page No: 38
Question 1.
Generally monocots do not have secondary growth, but palms and bamboos have woody stems. Find the reason.
Answer:
Some of the monocots like palm and bamboos show an increase in thickness of stems by means of secondary growth or latitudinal growth.

Textbook Page No: 48
Question 2.
Be friendly with your environment (Eco friendly) Why should not we use the natural products which are made by plant fibres like rope, fancy bags, mobile pouch, mat and gunny bags etc., instead of using plastics or nylon?
Answer:
We should not use the natural products, which are made by plants fibres, because, if we use more of plant products the greedy people will exploit the plant resources for making plant products and thereby depleting the tree cover, which in turn causes reduction in rain fall.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

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Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Additional Questions

Students can Download Maths Chapter 3 Geometry Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Additional Questions

The midpoint calculator will take two coordinates in the Cartesian coordinate system and find the point directly in-between both of them.

Question 1.
In ΔPQR, PS is a median. If QS = 7 cm find the length of QR?
Solution:
PS is the median ⇒ S is the midpoint of QR
Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 add 1
Given
QS = 7 cm
∴ SR = 7 cm
QR = QS + SR
= 7 + 7 = 14cm

Question 2.
In ΔABC, G is the centroid. If AD = 6 cm, BC = 4 cm and BE = 9 cm find the perimeter of ΔBDG.
Solution:
In ΔABC, G is the centroid.
Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 add 2
If AD = 6 cm
⇒ GD = \(\frac{1}{3}\) of AD = \(\frac{1}{3}\) (6)
BE = 9 cm
⇒ BG = \(\frac{2}{3}\) of BE = \(\frac{2}{3}\) (9) = 6cm
Also D is the midpoint of BC ⇒ BD
= \(\frac{1}{1}\) of BC = \(\frac{1}{2}\) (4) = 2cm
∴ Perimeter of ΔBDG = BD + GD + BG = 2 + 2 + 6 = 10 cm

Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Additional Questions

Question 3.
Construct a rhombus FISH with FS = 8 cm and ∠F = 80°
Solution:
Given FS = 8 cm and ∠F = 80°
Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 add 3
Steps :
(i) Drawn a line segment FS = 7 cm.
(ii) At F, made ∠SFX = ∠SFY = 40° on either side of FS.
(iii) At S, made ∠FSP = ∠FSQ = 40° on either side of FS
(iv) Let FX and SP cut at H and FY and SQ cut at I.
(v) FISH is the required rhombus

Calculation of Area:
Area of the rhombus FISH = \(\frac{1}{2}\) × d1 × d2 sq.units = \(\frac{1}{2}\) × 5.9 × 7 cm² = 20.65 cm²

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Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Students can Download Physics Chapter 2 Kinematics Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Physics Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Samacheer Kalvi 11th Physics Kinematics Textual Questions Solved

Samacheer Kalvi 11th Physics Kinematics Multiple Choice Questions

Question 1.
Which one of the following Cartesian coordinate systems is not followed in physics?
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 2.
Identify the unit vector in the following:
(a) \(\hat{i}+\hat{j}\)
(b) \(\frac{\hat{i}}{\sqrt{2}}\)
(c) \(\hat{k}-\frac{\hat{j}}{\sqrt{2}}\)
(d) \(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\)
Answer:
(d) \(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\)

online acceleration calculator tool makes the calculation faster, and it displays the acceleration of the object in a fraction of seconds.

Question 3.
Which one of the following physical quantities cannot be represented by a scalar?
(a) Mass
(b) length
(c) momentum
(d) magnitude of acceleration
Answer:
(c) momentum

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Question 4.
Two objects of masses m1 and m2, fall from the heights h1 and h2 respectively. The ratio of the magnitude of their momenta when they hit the ground is [AIPMT 20121]
(a) \(\sqrt{\frac{h_{1}}{h_{2}}}\)
(b) \(\sqrt{\frac{m_{1} h_{1}}{m_{2} h_{2}}}\)
(c) \(\frac{m_{1}}{m_{2}} \sqrt{\frac{h_{1}}{h_{2}}}\)
(d) \(\frac{m_{1}}{m_{2}}\)
Answer:
(c) \(\frac{m_{1}}{m_{2}} \sqrt{\frac{h_{1}}{h_{2}}}\)

Question 5.
If a particle has negative velocity and negative acceleration, its speed
(a) increases
(b) decreases
(c) remains same
(d) zero
Answer:
(a) increases

Question 6.
If the velocity is\(\overrightarrow{\mathrm{v}}\) – 2\(\hat{i}\) +t2\(\hat{j}\) – 9\(\overrightarrow{\mathrm{k}}\) , then the magnitude of acceleration at t = 0.5 s is
(a) 1 m s-2
(b) 1 m
(c) zero
(d) -1 m s s-2
Answer:
(a) 1 m s-2

Question 7.
If an object is dropped from the top of a building and it reaches the ground at t = 4 s, then the height of the building is (ignoring air resistance) (g = 9.8 m s-2).
(a) 77.3 m
(b) 78.4 m
(c) 80.5 m
(d) 79.2 m
Answer:
(b) 78.4 m

Question 8.
A ball is projected vertically upwards with a velocity v. It comes back to ground in time t. Which v -1 graph shows the motion correctly?[NSEP 00 – 01]
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 9.
If one object is dropped vertically downward and another object is thrown horizontally from the same height, then the ratio of vertical distance covered by both objects at any instant is
(a) 1
(b) 2
(c) 4
(d) 0.5
Answer:
(a) 1

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Question 10.
A ball is dropped from some height towards the ground. Which one of the following represents the correct motion of the ball?
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 11.
If a particle executes uniform circular motion in the xy plane in clockwise direction, then the angular velocity is in
(a) +y direction
(b) +z direction
(c) -z direction
(d) -x direction
Answer:
(c) -z direction

Question 12.
If a particle executes uniform circular motion, choose the correct statement [NEET 2016]
(a) The velocity and speed are constant.
(b) The acceleration and speed are constant.
(c) The velocity and acceleration are constant.
(d) The speed and magnitude of acceleration are constant.
Answer:
(d) The speed and magnitude of acceleration are constant.

Question 13.
If an object is thrown vertically up with the initial speed u from the ground, then the time taken by the object to return back to ground is
(a) \(\frac{u^{2}}{2 g}\)
(b) \(\frac{u^{2}}{g}\)
(c) \(\frac{u}{2 g}\)
(d) \(\frac{2 u}{g}\)
Answer:
(d) \(\frac{2 u}{g}\)

Question 14.
Two objects are projected at angles 30° and 60° respectively with respect to the horizontal direction. The range of two objects are denoted as R30° and R60°– Choose the correct relation from the following:
(a) R30° = R60°
(b) R30° = 4R60°
(c) \(\mathrm{R}_{30^{\circ}}=\frac{\mathrm{R}_{60^{\circ}}}{2}\)
(d) R30° = 2R60°
Answer:
(a) R30° = R60°

Question 15.
An object is dropped in an unknown planet from height 50 m, it reaches the ground in 2 s. The acceleration due to gravity in this unknown planet is
(a) g = 20 m s-2
(b) g = 25 m s-2
(c) g = 15 m s-2
(d) g = 30 m s -2
Answer:
(a) g = 25 m s-2

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Samacheer Kalvi 11th Physics Kinematics Short Answer Questions

Question 1.
Explain what is meant by Cartesian coordinate system?
Answer:
At any given instant of time, the frame of reference with respect to which the position of the object is described in terms of position coordinates (x, y, z) is called Cartesian coordinate system.

Reference Angle Calculator is a free online tool that displays the reference angle for the given angle and its position.

Question 2.
Define a vector. Give examples.
Answer:
Vector is a quantity which is described by the both magnitude and direction. Geometrically a vector is directed line segment.
Example – force, velocity, displacement.

Question 3.
Define a scalar. Give examples.
Answer:
Scalar is a property which can be described only by magnitude.
Example – mass, distance, speed.

Question 4.
Write a short note on the scalar product between two vectors.
Answer:
The scalar product (or dot product) of two vectors is defined as the product of the magnitudes of both the vectors and the cosine of the angle between them. Thus if there are two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) having an angle 0 between them, then their scalar product is defined as \(\overrightarrow{\mathrm{A}}\) • \(\overrightarrow{\mathrm{B}}\) = AB cos 0. Here, AB and are magnitudes of \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\).

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Question 5.
Write a short note on vector product between two vectors.
Answer:
The vector product or cross product of two vectors is defined as another vector having a magnitude equal to the product of the magnitudes of two vectors and the sine of the angle between them. The direction of the product vector is perpendicular to the plane containing the two vectors, in accordance with the right hand screw rule or right hand thumb rule. Thus, if\(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are two vectors, then their vector product is written as \(\overrightarrow{\mathrm{A}}\) × \(\overrightarrow{\mathrm{B}}\) which is a vector C defined by \(\overrightarrow{\mathrm{c}}\) = \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) = (AB sin 0) \(\hat{n}\)
The direction \(\hat{n}\) of \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) , i.e., \(\overrightarrow{\mathrm{c}}\) is perpendicular to the plane containing the vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\).

Question 6.
How do you deduce that two vectors are perpendicular?
Answer:
If two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are perpendicular to each other than their scalar product \(\overrightarrow{\mathrm{A}}\) \(\overrightarrow{\mathrm{B}}\) = 0 because cos 90° = 0. Then he vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are said to be mutually orthogonal.

Question 7.
Define displacement and distance.
Answer:
Distance is the actual path length traveled by an object in the given interval of time during the motion. It is a positive scalar quantity. Displacement is the difference between the final and initial positions of the object in a given interval of time. It can also be defined as the shortest distance between these two positions of the object. It is a vector quantity.

SamacheerKalvi.Guru

Question 8.
Define velocity and speed.
Answer:
Speed is defined as the ratio of total distance covered to the total time taken, it is a scalar quantity and always it is positive. Velocity is defined as the ratio of the displacement vector to the corresponding time interval. It is a vector quantity or it can also be defined as rate of change of displacement.

Question 9.
Define acceleration.
Answer:
Acceleration of a particle is defined as the rate of change of velocity or it can also be defined as the ratio of change in velocity to the given interval of time.

The Average Velocity calculator computes the velocity (V) based on the change in position (Δx) and the change in time (Δt).

Question 10.
What is the difference between velocity and average velocity.
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 11.
Define a radian?
One radian is the angle subtended at the center of a circle by an arc that is equal in length to the radius of the circle.
1 rad = 57.295°

Question 12.
Define angular displacement and angular velocity.
Answer:
1. Angular displacement:
The angle described by the particle about the axis of rotation in a given time is called angular displacement.

2. Angular velocity:
The rate of change of angular displacement is called angular velocity.

SamacheerKalvi.Guru

Question 13.
What is non uniform circular motion?
If the speed of the object in circular motion is not constant, then we have non-uniform circular motion. For example, when the bob attached to a string moves in vertical circle, the speed of the bob is not the same at all time Whenever the speed is not same in circular motion, the particle will have both centripetal and tangential acceleration.

Question 14.
Write down the kinematic equations for angular motion.
Answer:
Kinematic equations for circular motion are –

  1. \(\omega=\omega_{0}+\alpha t\)
  2. \(\theta=\omega_{0} t+\frac{1}{2} \alpha t^{2}\)
  3. \(\omega^{2}=\omega_{o}^{2}+2 \alpha \theta\)
  4. \(\theta=\frac{\left(\omega_{0}+\omega\right)}{2} t\)

Here,
ω0  = initial angular velocity
ω = final angular velocity
θ = angular displacement
α = angular acceleration
t = time.

Question 15.
Write down the expression for angle made by resultant acceleration and radius vector in the non uniform circular motion.
Answer:
The angle made by resultant acceleration and radius vector in the non uniform circular motion is –
\(\tan \theta=\frac{a_{t}}{\left(\frac{V^{2}}{r}\right)}\) or \(\theta=\tan ^{-1}\left(\frac{a_{t}}{\left(\frac{V^{2}}{r}\right)}\right)\)

SamacheerKalvi.Guru

Samacheer Kalvi 11th Physics Kinematics Long Answer Questions

Question 1.
Explain in detail the triangle law of addition.
Answer:
Let us consider two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) as shown in figure. To find the resultant of the two vectors we apply the triangular.

Law of addition as follows:
present the vectors A and by the two adjacent sides of a triangle taken in the same order. Then the resultant is given by the third side of the triangle as shown in figure.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
To explain further, the head of the first vector \(\overrightarrow{\mathrm{A}}\) is connected to the tail of the second vect \(\overrightarrow{\mathrm{B}}\) Let O he the angle between\(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\). Then \(\overrightarrow{\mathrm{R}}\) is the resultant vector connecting the tail of the first vector \(\overrightarrow{\mathrm{A}}\) to the head of the second vector \(\overrightarrow{\mathrm{B}}\) The magnitude of \(\overrightarrow{\mathrm{R}}\). (resultant) given geometrically by the length of (OQ) and the direction of the resultant vector is the angle between \(\overrightarrow{\mathrm{R}}\). and \(\overrightarrow{\mathrm{A}}\). Thus we write
\(\overrightarrow{\mathrm{R}}\) = \(\overrightarrow{\mathrm{A}}\) +\(\overrightarrow{\mathrm{B}}\) \(\overrightarrow{\mathrm{OQ}}\) = \(\overrightarrow{\mathrm{OP}}\) + \(\overrightarrow{\mathrm{PQ}}\)

1. Magnitude of resultant vector:
The magnitude and angle of the resultant vector ar determined by using triangle law of vectors as follows.From figure, consider the triangle ABN, which is obtained by extending the side OA to ON. ABN is a right angled triangle.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
From figure, let R is the magnitude of the resultant of \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\).
cos θ = \(\frac { AN}{ B }\) ∴ AN = B cos θ and sinθ = \(\frac { BN}{ B }\) ∴BN = B sinθ
For ∆ OBN, we have OB2 = ON2 + BN2
⇒ R2 = (A + B cos θ)2 + (B sinθ)2
⇒ R2 = A2 + B2 cos2θ + 2ABcosθ B2 sin2θ
⇒ R2 = A2 + B2(cos2θ + sin2θ) + 2AB cos θ
⇒ R2 = \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)

2. Direction of resultant vectors:
If 0 is the angle between \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) then,
\(|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta}\)
If R makes an angle α with \(\overrightarrow{\mathrm{A}}\) , then in AOBN,
tan α = \(\frac { BN}{ ON }\) = \(\frac { BN}{ OA + AN }\)
tan α = \(\frac { B sinθ }{ A + B cosθ}\) ⇒ α = \(\tan ^{-1}\left(\frac{B \sin \theta}{A+B \cos \theta}\right)\)

Question 2.
Discuss the properties of scalar and vector products.
Answer:
Properties of scalar product of two vectors are:
(1) The product quantity \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{B}}\) is always a scalar. It is positive if the angle between the vectors is acute (i.e., < 90°) and negative if the angle between them is obtuse (i.e. 90°<0< 180°).

(2) The scalar product is commutative, i.e. \(\overrightarrow{\mathrm{A}}\) \(\overrightarrow{\mathrm{B}}\) ≠ \(\overrightarrow{\mathrm{B}}\). \(\overrightarrow{\mathrm{A}}\)

(3) The vectors obey distributive law i.e. \(\overrightarrow{\mathrm{A}}\)(\(\overrightarrow{\mathrm{B}}\) + \(\overrightarrow{\mathrm{C}}\)) = \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{B}}\) + \(\overrightarrow{\mathrm{A}}\) .\(\overrightarrow{\mathrm{C}}\)
(4) The angle between the vectors θ = \(\cos ^{-1}\left[\frac{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}}{\mathrm{AB}}\right]\)

(5) The scalar product of two vectors will be maximum when cos θ = 1, i.e. θ = 0°, i.e., when the vectors are parallel;
\((\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}})_{\max }=\mathrm{AB}\)

(6) The scalar product of two vectors will be minimum, when cos θ = -1, i.e. θ = 180°.
\((\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}})_{\min }=-\mathrm{AB}\) when the vectors are anti-parallel.

(7) If two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are perpendicular to each other than their scalar product \(\overrightarrow{\mathrm{A}}\) .\(\overrightarrow{\mathrm{B}}\) = 0, because cos 90° 0. Then the vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are said to be mutually orthogonal.

(8) The scalar product of a vector with itself is termed as self-dot product and is given by (\(\overrightarrow{\mathrm{A}}\))2 = \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{A}}\) = AA cos 0 = A2. Here angle 0 = 0°.
The magnitude or norm of the vector \(\overrightarrow{\mathrm{A}}\) is |\(\overrightarrow{\mathrm{A}}\)| = A = \(\sqrt{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{A}}}\).

(9) In case of a unit vector \(\hat{n}\)
\(\hat{n}\) . \(\hat{n}\) = 1 x 1 x cos 0 = 1. For example, \(\hat{i}\) – \(\hat{i}\) = \(\hat{j}\) . \(\hat{j}\) = \(\hat{k}\) . \(\hat{k}\) = 1.

(10) In the case of orthogonal unit vectors, \(\hat{i}\),\(\hat{j}\) and \(\hat{k}\),
\(\hat{i}\) . \(\hat{j}\) = \(\hat{j}\).\(\hat{k}\) = \(\hat{k}\) . \(\hat{i}\)= 1.1 cos 90° = 0

(11) In terms of components the scalar product of \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) can be written as
\(\overrightarrow{\mathrm{A}}\).\(\overrightarrow{\mathrm{B}}\) = (Ax\(\hat{i}\) + Ay\(\hat{j}\) + Az\(\hat{k}\)).(Bx\(\hat{i}\) + By\(\hat{j}\) + Bz\(\hat{k}\))
= A xBx + AyBy+ AzBz, with all other terms zero.
The magnitude of vector | \(\overrightarrow{\mathrm{A}}\) | is given by
| \(\overrightarrow{\mathrm{A}}\) | = A = \(\sqrt{\mathrm{A}_{x}^{2}+\mathrm{A}_{y}^{2}+\mathrm{A}_{z}^{2}}\)

Properties of vector product of two vectors are:
(1) The vector product of any two vectors is always another vector whose direction is perpendicular to the plane containing these two vectors, i.e., orthogonal to both the vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\), even though the vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) may or may not be mutually orthogonal.

(2) The vector product of two vectors is not commutative, i.e., \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) ≠ \(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\). But,
\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)=-\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\).
Here it is worthwhile to note that |\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)| =
|\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)| = AB sin 0 i.e., in the case of the product vectors \(\overrightarrow{\mathrm{B}}\)=-\(\overrightarrow{\mathrm{B}}\) and \(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\), the magnitudes are equal but directions are opposite to each other.

(3) The vector product of two vectors will have maximum magnitude when sin 0 = 1, i.e., 0 = 90° i.e., when the vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are orthogonal to each other.
\((\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}})_{\mathrm{max}}=\mathrm{AB} \hat{n}\) = AB \(\hat{n}\)

(4) The vector product of two non-zero vectors will be minimum when sin θ = 0, i.e θ = 0° or 180°
\((\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}})_{\min }=0\)
i. e., the vector product of two non – zero vectors vanishes, if the vectors are either parallel or anti parallel.

(5) The self – cross product, i.e., product of a vector with itself is the null vector
\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{A}}\) = AA sin 0° \(\hat{n}\) = \(\overrightarrow{\mathrm{0}}\) In physics the null vector 0 is simply denoted as zero.

(6) The self – vector products of unit vectors are thus zero.
\(\hat{i}\) x \(\hat{i}\) = \(\hat{ j}\) x \(\hat{j}\) = \(\hat{k}\) x \(\hat{k}\) = 0

(7) In the case of orthogonal unit vectors, \(\hat{i}\), \(\hat{j}\). \(\hat{k}\) , in accordance with the right hand screw rule:
\(\hat{i}\) x \(\hat{j}\) = \(\hat{k}\), \(\hat{j}\) x \(\hat{k}\) = \(\hat{i}\) and \(\hat{k}\) x \(\hat{i}\) = \(\hat{j}\)
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Also, since the cross product is not commutative,
\(\hat{j}\) x \(\hat{i}\) = –\(\hat{k}\), \(\hat{k}\) x \(\hat{j}\) = –\(\hat{i}\) and \(\hat{i}\) x \(\hat{k}\) = \(\hat{j}\)

(8) In terms of components, the vector product of two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) is –
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Note that in the \(\hat{j}^{\mathrm{th}}\) component the order of multiplication is different than \(\hat{i}^{\mathrm{th}}\) and \(\hat{k}^{\mathrm{th}}\) components.

(9) If two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) form adjacent sides in a parallelogram, then the magnitude of |\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)| will give the area of the parallelogram as represented graphically in figure.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

(10) Since we can divide a parallelogram into two equal triangles as shown in the figure, the area of a triangle with \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) as sides is \(\frac { 1 }{ 2 }\) |\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)| . This is shown in the Figure. A number of quantities used in Physics are defined through vector products. Particularly physical quantities representing rotational effects like torque, angular momentum, are defined through vector products.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 3.
Derive the kinematic equations of motion for constant acceleration.
Answer:
Consider an object moving in a straight line with uniform or constant acceleration ‘a’. Let u be the velocity of the object at time t = 0, and v be velocity of the body at a later time t.

Velocity – time relation:
(1) The acceleration of the body at any instant is given by the first derivative of the velocity with respect to time,
a = \(\frac {dv}{dt}\) or dv = a dt
Integrating both sides with the condition that as time changes from 0 to t, the velocity changes from u to v. For the constant acceleration,
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

(2) The velocity of the body is given by the first derivative of the displacement with respect to time.
v = \(\frac {ds}{dt}\) or ds = vdt
and since v = u + at,
we get ds = (u+ at ) dt
Assume that initially at time t = 0, the particle started from the origin. At a later time t, the particle displacement is s. Further assuming that acceleration is time-independent, we have
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Velocity – displacement relation:

(3) The acceleration is given by the first derivative of velocity with respect to time.
a = \(\frac {dv}{dt}\) = \(\frac {dv}{ds}\) = \(\frac {ds}{dt}\) = \(\frac {dv}{ds}\) v [since ds/dt = v] where s is displacement traverse
This is rewritten as a = \(\frac{1}{2} \frac{d v^{2}}{d s}\) or ds = \(\frac{1}{2 a} d\left(v^{2}\right)\) Integrating the above equation, using the fact when the velocity changes from u2 to v2, displacement changes from 0 to s, we get
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
We can also derive the displacement 5 in terms of initial velocity u and final velocity v. From equation we can
write,
at = v – u
Substitute this in equation, we get
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

SamacheerKalvi.Guru

Question 4.
Derive the equations of motion for a particle (a) falling vertically (b) projected vertically.
Answer:
’Equations of motion for a particle falling vertically downward from certain height. Consider an object of mass m falling from a height h. Assume there is no air resistance. For convenience, let us choose the downward direction as positive y – axis as shown in the figure. The object experiences acceleration ‘g’ due to gravity which is constant near the surface of the Earth. We can use kinematic equations to explain its motion. We have The acceleration \(\overrightarrow{\mathrm{a}}\) = g \(\hat{i}\)
By comparing the components, we get,
Equations of motion for a particle thrown vertically upwards,
ax = 0, ax = 0, ay = g Let us take for simplicity, ay = a = g

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
If the particle is thrown with initial velocity ‘u’ downward which is in negative y – axis, then velocity and position at of the particle any time t is given by
v = u + gt
v = ut + \(\frac {1}{2}\) – gt2
The square of the speed of the particle when it is at a distance y from the hill – top, is v2 = u2 + 2 gy
Suppose the particle starts from rest.
Then u = 0
Then the velocity v, the position of the particle and v2 at any time t are given by (for a point y from the hill – top)
v = gt …………(i)
y = \(\frac {1}{2}\) – gt2 …………(ii)
v2 = 2gy …………(iii)
The time (t = T) taken by the particle to reach the ground (for which y = h), is given by using equation (ii),
h = \(\frac {1}{2}\) – gT2 …………(iv)
T = \(\sqrt{\frac{2 h}{g}}\) …………(v)
The equation (iv) implies that greater the height (h), particle takes more time (T) to reach the ground. For lesser height (h), it takes lesser time to reach the ground. The speed of the particle when it reaches the ground (y = h) can be found using equation (iii), we get,
\(v_{\text {ground }}=\sqrt{2 g h}\) …………(vi)
The above equation implies that the body falling from greater height (h) will have higher velocity when it reaches the ground. The motion of a body falling towards the Earth from a small altitude (h<<R), purely under the force of gravity is called free fall. (Here R is radius of the Earth).

case (ii):
A body thrown vertically upwards:
Consider an object of mass m thrown vertically upwards with an initial velocity u. Let us neglect the air friction. In this case we choose the vertical direction as positive y axis as shown in the figure, then the acceleration a = -g (neglect air friction) and g points towards the negative y axis. The kinematic equations for this motion are,
The velocity and position of the object at any time t are,
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
v = u – gt ……………(vii)
s = ut – \(\frac {1}{2}\) – gt2 …………..(viii)
The velocity of the object at any position y (from the point where the object is thrown) is
v2 = u2 – 2gy …………..(ix)

Question 5.
Derive the equation of motion, range and maximum height reached by the particle thrown at an oblique angle 9 with respect to the horizontal direction.
Answer:
This projectile motion takes place when the initial velocity is not horizontal, but at some angle with the vertical, as shown in Figure.
(Oblique projectile)
Examples:

  • Water ejected out of a hose pipe held obliquely.
  • Cannon fired in a battle ground.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Consider an object thrown with initial velocity at an angle θ with the horizontal.
Then,
\(\overrightarrow{\mathrm{u}}\) = ux î + uy\(\hat{j}\) .
where ux = u cos θ is the horizontal component and uy = u sin θ the vertical component of velocity. Since the acceleration due to gravity is in the direction opposite to the direction of vertical component uy , this component will gradually reduce to zero at the maximum height of the projectile. At this maximum height, the same gravitational force will push the projectile to move downward and fall to the ground. There is no acceleration along the x direction throughout the motion. So, the horizontal component of the velocity (ux = u cos θ) remains the same till the object reaches the ground. Hence after the time t, the velocity along horizontal motion vx = ux + axt = ux = u cos θ. The horizontal distance travelled by projectile m time t is sx = \(u_{x} t+\frac{1}{2} a_{x} t^{2}\)
Here, sx = x, ux = u cos θ, ax = 0
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Thus, x = u cos θ or t = \(\frac {x}{u cos θ}\) ……..(i)
Next, for the vertical motion vy= uy + ayt
Here uy = u sin θ, ay = -g (acceleration due to gravity acts opposite to the motion).
Thus, vy= u sin θ – gt
The vertical distance traveled by the projectile in the same time t is
Here, sy = y, uy = u sin θ, ax = -g. Then
y = u sinθ t – \(\frac {1}{2}\) – gt2 ………..(ii)
Substitute the value of t from equation (i) in equation (ii), we have .
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Thus the path followed by the projectile is an inverted parabola Maximum height (hmax): The maximum vertical distance travelled by the projectile during the journey is called maximum height. This is determined as follows:
For the vertical part of the motion.
\(v_{y}^{2}=u_{y}^{2}+2 a_{y} s\)
Here, uy = u sin θ, a = -g, s = hmax, and at the maximum height vy = 0
Hence, (0)2 = u2 sin2 θ = 2 ghmax or \(h_{\max }=\frac{u^{2} \sin ^{2} \theta}{2 g}\)

Time of flight (Tf):
The total time taken by the projectile from the point of projection till it hits the horizontal plane is called time of flight. This time of flight is the time taken by the projectile to go from point O to B via point A as shown
we know that sy = y = 0 (net displacement in y-direction is zero),
uy = u sin θ, ay = -g , t = Tf Then
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Horizontal range (R):
The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range (R). This is found easily since the horizontal component of initial velocity remains the same. We can write Range R = Horizontal component of velocity x time of flight = u cos θ x \(\mathrm{T}_{f}=\frac{u^{2} \sin 2 \theta}{g}\) The horizontal range directly depends on the initial speed (u) and the sine of angle of projection (θ). It inversely depends on acceleration due to gravity ‘g’.
For a given initial speed u, the maximum possible range is reached when sin 2θ is maximum, sin 2θ = 1. This implies 2θ = π/2 or θ = \(\frac {π}{4}\) This means that if the particle is projected at 45 degrees with respect to horizontal, it attains maximum range, given by.
\(\mathrm{R}_{\max }=\frac{u^{2}}{g}\) ………..(vi)

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Question 6.
Derive the expression for centripetal acceleration.
Answer:
In uniform circular motion the velocity vector turns continuously without changing its magnitude (speed), as shown in figure.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Note that the length of the velocity vector is not changed during the motion, implying that the speed remains constant. Even though the velocity is tangential at every point in the circle, the acceleration is acting towards the center of the circle. This is called centripetal acceleration. It always points towards the center of the circle. This is shown in the figure.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
The centripetal acceleration is derived from a simple geometrical relationship between position and velocity vectors.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Let the directions of position and velocity vectors shift through the same angle θ in a small interval of time ∆t, as shown in figure. For uniform circular motion, r = \(\left|\vec{r}_{1}\right|\) = \(\left|\vec{r}_{2}\right|\) and v = \(\left|\vec{v}_{1}\right|\) = \(\left|\vec{v}_{2}\right|\). If the particle moves from position vector \(\vec{r}_{1}\) to \(\vec{r}_{2}\), the displacement is given by ∆\(\overrightarrow{\mathrm{r}}\) = \(\vec{r}_{2}\) – \(\vec{r}_{1}\) and the change in velocity from \(\vec{v}_{1}\) to\(\vec{v}_{2}\) is given by ∆\(\overrightarrow{\mathrm{v}}\) = \(\vec{v}_{2}\) – \(\vec{v}_{1}\),. The magnitudes of the displacement ∆r and of ∆v satisfy the following relation. \(\frac {∆r}{r}\) = \(\frac {-∆v}{v}\) = θ Here the negative sign implies that ∆v points radially inward, towards the center of the circle.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
For uniform circular motion v = cor, where co is the angular velocity of the particle about center. Then the centripetal acceleration can be written as.
a = -ω2r

Question 7.
Derive the expression for total acceleration in the non uniform circular motion.
Answer:
If the speed of the object in circular motion is not constant, then we have non-uniform circular motion. For example, when the bob attached to a string moves in vertical circle, the speed of the bob is not the same at all time. Whenever the speed is not same in circular motion, the particle will have both centripetal and tangential acceleration as shown in the figure.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
The resultant acceleration is obtained by vector sum of centripetal and tangential acceleration Since centripetal acceleration is \(\frac{v^{2}}{r}\), the magnitude of this resultant acceleration is given by –\(\dot{a}_{\mathrm{R}}=\sqrt{a_{t}^{2}+\left(\frac{v^{2}}{r}\right)^{2}}\)
This resultant acceleration makes an angle 0 with the radius vector as shown in figure.
This angle is given by tan θ = \(\frac{a_{t}}{\left(v^{2} / r\right)}\)

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Samacheer Kalvi 11th Physics Kinematics Numerical Questions

Question 1.
The position vector of the particle has length 1 m and makes 30° with the x-axis. What are the lengths of the x and y – components of the position vector?
Answer:
Given,
Length of position vector = 1 m
Angle made with x axis = 30
Solution:
Length of X component (OB) = OA cos θ
= 1 x cos 30°
= \(\frac{\sqrt{3}}{2}\) (or) 0.87 m
Length of Y component (AB) = OA sin θ = 1 x sin 30° = \(\frac { 1 }{ 2 }\) = 0.5 m.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 2.
A particle has its position moved from \(\vec{r}_{1}\) = 3\(\hat{i}\) + 4\(\hat{j}\) to r2 = \(\hat{i}\) + 2\(\hat{i}\). Calculate the displacement vector (∆\(\overrightarrow{\mathrm{r}}\) ) and draw the \(\vec{r}_{1}\), \(\vec{r}_{2}\) and ∆\(\overrightarrow{\mathrm{r}}\) vector in a two dimensional Cartesian coordinate system.
Answer:
Given,
Position vectors \(\vec{r}_{1}\) = 3\(\hat{i}\) + 4\(\hat{j}\)
\(\vec{r}_{1}\) = \(\hat{i}\) + 2\(\hat{j}\)
Solution:
Displacement vector:
∆r= \(\vec{r}_{2}\) – \(\vec{r}_{1}\) = (1 – 3)\(\hat{i}\) + (2 – 4) \(\hat{j}\)
∆r = -2\(\hat{i}\) -2\(\hat{j}\) = -2(\(\hat{i}\) + \(\hat{j}\))
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

The Average Velocity calculator computes the velocity (V) based on the change in position (Δx) and the change in time (Δt).

Question 3.
Calculate the average velocity of the particle whose position vector changes from \(\vec{r}_{1}\) = 5\(\hat{i}\) + 6\(\hat{j}\) to \(\vec{r}_{2}\) = 2\(\hat{i}\) + 3 \(\hat{j}\) in a time 5 seconds.
Answer:
Given,
Position vectors of a particle
\(\vec{r}_{1}\) = 5\(\hat{i}\) + 6\(\hat{j}\),
\(\vec{r}_{2}\) = 2\(\hat{i}\) + 35\(\hat{j}\)
time(t) = 5s
Solution:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 4.
Convert the vector \(\overrightarrow{\mathrm{r}}\) = 3\(\hat{i}\) + 2\(\hat{j}\) into a unit vector.
Answer:
Given:
Position vector\(\hat{r}\) = 3\(\hat{i}\) + 2\(\hat{j}\)
Solution:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 5.
What are the resultants of the vector product of two given vectors given by \(\overrightarrow{\mathrm{A}}\) = 4\(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\) and \(\overrightarrow{\mathrm{B}}\) = 5\(\hat{i}\) + 3\(\hat{j}\) – 4\(\hat{k}\) ?
Answer:
Given,
Vectors \(\overrightarrow{\mathrm{A}}\) = 4\(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\)
\(\overrightarrow{\mathrm{B}}\) = 5\(\hat{i}\) + 3\(\hat{j}\) – 4\(\hat{k}\)
Solution:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 6.
object at an angle such that the horizontal range is 4 times of the maximum height. What is the angle of projection of the object?
Answer:
Give,
Horizontal range = 4Hmax
Solution:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 7.
The following graphs represent velocity – time graph. Identity what kind of motion a particle undergoes in each graph.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
(a) At all the points, slope of the graph is constant.
∴ \(\overrightarrow{\mathrm{a}}\) = constant

(b) No change in magnitude of velocity with respect to time
∴ \(\overrightarrow{\mathrm{v}}\) = constant

(c) Slope of this graph is greater than graph (a) but constant
∴ \(\overrightarrow{\mathrm{a}}\) = constant but greater than the graph (a)

(d) At each point slope of the curve increases.
∴ \(\overrightarrow{\mathrm{a}}\) is a variable and object is in accelerated motion.

SamacheerKalvi.Guru

Question 8.
The following velocity – time graph represents a particle moving in the positive x-direction. Analyse its motion from 0 to 7 s. Calculate the displacement covered and distance travelled by the particle from 0 to 2 s.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
As per graph,
(a) From 0 to 1.5 s the particle moving in a opposite direction.

  • From 1.5 s to 2 s the particle is moving with increasing velocity.
  • From 2 s to 5 s velocity of the particle is constant of magnitude 1 ms -1
  • From 5 s to 6 s velocity of the particle is decreasing.
  • From 6 s to 7 s the particle is at rest.

(b) Distance covered by the particle – Area covered under (v -t) graph
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Displacement of the particle
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 9.
A particle is projected at an angle of θ with respect to the horizontal direction. Match the following for the above motion.
(a) vx – decreases and increases
(b) vy  – remains constant
(c) Acceleration – varies
(d) Position vector – remains downward
Answer:
(a) vx = remains constant
(b) vy = decreases and increases
(c) a = remains downward
(d) r = varies

Question 10.
A water fountain on the ground sprinkles water all around it. If the speed of the water coming out of the fountain is v, calculate the total area around the fountain that gets wet.
Answer:
Given,
Speed of water = v
Solution:
Water comes from a fountain can be taken as projectile and the distance covered is maximum range of projectile i.e. θ = 45°.
Range of the particle (Rmax) = \(\frac{v^{2}}{g}\) sin 2θ = \(\frac{v^{2}}{g}\)
here, Rmax is radius of the area covered.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 11.
The following table gives the range of a particle when thrown on different planets. All the particles are thrown at the same angle with the horizontal and with the same initial speed. Arrange the planets in ascending order according to their acceleration due to gravity, (g value).
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Range = \(\frac{v^{2}}{g}\) sin 2θ ∴ g α \(\frac { 1 }{ range }\)
Ascending order of the planet with respect to their “g” is Mercury, Mars, Earth, Jupiter.

Question 12.
The resultant of two vectors A and B is perpendicular to vector A and its magnitude is equal to half of the magnitude of vector B. Then the angle between A and B is
(a) 30°
(b) 45°
(c) 150°
(d) 120°
Answer:
Given:
Resultant of \(\overrightarrow{\mathrm{A}}\) & \(\overrightarrow{\mathrm{B}}\) is perpendicular to \(\overrightarrow{\mathrm{A}}\) and magnitude of resultant (C) = \(\frac { 1 }{ 2 }\) \(\overrightarrow{\mathrm{B}}\) and α = 90°
Solution:
(i) Magnitude of resultant:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

(ii) direction of resultant:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

SamacheerKalvi.Guru

Question 13.
Compare the components for the following vector equations
(a) T\(\hat{j}\) -mg\(\hat{j}\) = ma\(\hat{j}\)
(b) \(\overrightarrow{\mathrm{T}}\) + \(\overrightarrow{\mathrm{F}}\) = \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)
(c) \(\overrightarrow{\mathrm{T}}\) – \(\overrightarrow{\mathrm{F}}\) = \(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{B}}\)
(d) T\(\hat{j}\) + mg\(\hat{j}\)= ma\(\hat{j}\)
Answer:
Components of the vectors
(a) T – mg = ma
(b) \(\overline{\mathrm{T}}_{x}+\overline{\mathrm{F}}_{x}\) = \(\overline{\mathrm{A}}_{x}+\overline{\mathrm{B}}_{x}\) (or) \(\overline{\mathrm{T}}_{y}+\overline{\mathrm{F}}_{y}=\overline{\mathrm{A}}_{y}+\overline{\mathrm{B}}_{y}\)
(c) \(\overline{\mathrm{T}}_{x}-\overline{\mathrm{F}}_{x}=\overline{\mathrm{A}}_{x}+\overline{\mathrm{B}}_{x}\)  (or) \(\overline{\mathrm{T}}_{y}-\overline{\mathrm{F}}_{y}=\overline{\mathrm{A}}_{y}+\overline{\mathrm{B}}_{y}\)
(d) T + mg = ma

Question 14.
Calculate the area of the triangle for which two of its sides are given by the vectors A = 5\(\hat{i}\) – 3\(\hat{j}\), B = 4\(\hat{i}\) + 6\(\hat{j}\) .
Answer:
Solution:
Area of the triangle = \(\frac { 1 }{ 2 }\) |\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{A}}\)|
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 15.
If Earth completes one revolution in 24 hours, what is the angular displacement made by Earth in one hour? Express your answer in both radian and degree.
Answer:
Given,
time period of earth = 24 hours
Solution:
Earth covers 360° in 24 hours
∴Angular displacement m 1 hour = \(\frac { 360° }{ 24 }\) = 15° (or) \(\frac { π }{ 12 }\)
Angular displacement in radian = \(\frac { 15° }{ 57.295° }\) = 0.262 rad

Question 16.
An object is thrown with initial speed 5 ms -1 with an angle of projection 30°. What is the height and range reached by the particle?
Answer:
Given,
Initial speed (u) = 5 ms-1
Angle of projection θ = 30°
Solution:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 17.
A foot – ball player hits the ball with speed 20 ms-1 with angle 30° with respect to horizontal direction as shown in the figure. The goal post is at a distance of 40 m from him. Find out whether ball reaches the goal post.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Given:
Initial speed (u) = 20 ms-1
Angle of projection (θ) = 30°
The distance of the goal post = 40 m
Solution:
Range of the projectile
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
The distance of goal post is 40 m. But the range of the ball is 35.35 m only. So ball will not reach the goal post.

Question 18.
If an object is thrown horizontally with an initial speed 10 ms -1 from the top of a building of height 100 m. What is the horizontal distance covered by the particle?
Answer:
Given,
Initial speed =10 ms-1
Height of the building (h) = 100 m
Range = ?
Solution:
Range of the object = R = \(u \sqrt{\frac{2 h}{g}}\) = 10\(\sqrt{\frac{200}{9.8}}\) = 45.1 m
R = 45 m.

SamacheerKalvi.Guru

Question 19.
An object is executing uniform circular motion with an angular speed of \(\frac { π }{ 12 }\) radian per second. At t = 0 the object starts at angle θ = 0. What is the angular displacement of the particle after 4 s?
Answer:
Given:
Angular speed = \(\frac { π }{ 12 }\) rad/ sec
Solution:
Angular speed = \(\frac { Angular displacement}{ time taken }\)
Angular displacement = \(\frac { π }{ 12 }\) x 4 = \(\frac { π }{ 12 }\) = 60°

Question 20.
Consider the x-axis as representing east, the v – axis as north and z – axis as vertically upwards. Give the vector representing each of the following points.
(a) 5 m north east and 2 m up
(b) 4 m south east and 3 m up
(c) 2 m north west and 4 m up
Answer:
Given,
Solution:
(a) Length along X – axis = 5 cos 45° = \(\frac{5}{\sqrt{2}}\)m
Length along Y- axis = 5 sin 45° = \(\frac{5}{\sqrt{2}}\)m
Length along Z – Axis = 2 m
In vector rotation = \(\frac{5}{\sqrt{2}}\)\(\hat{i}\) + \(\frac{5}{\sqrt{2}}\)\(\hat{j}\) + 2\(\hat{k}\) = \(\frac{5(\hat{i}+\hat{j})}{\sqrt{2}}\) + 2\(\hat{k}\)

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

(b) Length along X = 4 cos 45° = \(\frac{4}{\sqrt{2}}\)m
Length along Y = 4 sin 45° = \(\frac{4}{\sqrt{2}}\)m
Length along Z-axis = 3 m
In vector rotation = \(\frac{4}{\sqrt{2}}\)\(\hat{i}\) – \(\frac{4}{\sqrt{2}}\)\(\hat{i}\) + 3k = 4(\(\hat{i}\) – \(\hat{j}\)) \(\sqrt{2}+3 \hat{k}\)

(c) Length along X = – 2 cos 45° = \(\sqrt{2}+3 \hat{k}\) = \(\frac{2}{\sqrt{2}} m=\sqrt{2} m\)
Length along Y = 2 sin 45° = \(\frac{2}{\sqrt{2}} m=\sqrt{2} m\)
length along Z = 4 m
∴ In vector rotation = \(-\sqrt{2} \hat{i}+\sqrt{2} \hat{j}+4 \hat{k}\)

Question 21.
The Moon is orbiting the Earth approximately once in 27 days, what is the angle transformed by the Moon per day?
Answer:
Given,
period of moon = 27 days
Solution:
i.e. in 27 days moon covers 360°
In one day angle traversed by moon = \(\frac { 360° }{ 2H }\) = 13.3°

Question 22.
An object of mass m has angular acceleration a = 0.2 rad s2. What is the angular displacement covered by the object after 3 second? (Assume that the object started with angle zero with zero angular velocity).
Answer:
Given,
Angular acceleration = α = 0.2 rad s-2
Time = 3s
Initial velocity = 0
Solution:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Samacheer Kalvi 11th Physics Kinematics Additional Questions Solved

Samacheer Kalvi 11th Physics Kinematics Multiple Choice Questions 

Question 1.
The radius of the earth was measured by –
(a) Newton
(b) Eratosthenes
(c) Galileo
(d) Ptolemy
Answer:
(b) Eratosthenes

Question 2
The branch of mechanics which deals with the motion of objects without taking force into account is –
(a) kinetics
(b) dynamics
(c) kinematics
(d) statics
Answer:
(c) kinematics

Question 3.
If the coordinate axes (x, y, z) are drawn in anticlockwise direction then the co-ordinate system is known as –
(a) Cartesian coordinate system
(b) right handed coordinate system
(c) left handed coordinate system
(d) cylindrical coordinate system
Answer:
(b) right handed coordinate system

Question 4.
The dimension of point mass is –
(a) 0
(b) 1
(c) 2
(d) kg
Answer:
(a) 0

Question 5.
If an object is moving in a straight line then the motion is known as –
(a) linear motion
(b) circular motion
(c) curvilinear motion
(d) rotational motion
Answer:
(a) linear motion

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Question 6.
An athlete running on a straight track is an example for the whirling motion of a stone attached to’a string is a –
(a) linear motion
(b) circular motion
(c) curvilinear motion
(d) rotational motion
Answer:
(a) linear motion

Question 7.
The whirling motion of a stone attached to a string is a –
(a) linear motion
(b) circular motion
(c) curvilinear motion
(d) rotational motion
Answer:
(b) circular motion

Question 8.
Spinning of the earth about its own axis is known as –
(a) linear motion
(b) circular motion
(c) curvilinear motion
(d) rotational motion
Answer:
(d) rotational motion

Question 9.
If an object executes a to and fro motion about a fixed point, is an example for –
(a) rotational motion
(b) vibratory motion
(c) circular motion
(d) curvilinear motion
Answer:
(b) vibratory motion

Question 10.
Vibratory motion is also known as –
(a) circular motion
(b) rotational motion
(c) oscillatory motion
(d) spinning
Answer:
(c) oscillatory motion

Question 11.
The motion of satellite around the earth is an example for –
(a) circular motion
(b) rotational motion
(c) elliptical motion
(d) spinning
Answer:
(a) circular motion

Question 12.
An object falling freely under gravity close to earth is –
(a) one dimensional
(b) circular motion
(c) rotational motion
(d) spinning motion
Answer:
(a) one dimensional

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Question 13.
Motion of a coin on a carrom board is an example of –
(a) one dimensional motion
(b) two dimensional motion
(c) three dimensional motion
(d) none
Answer:
(b) two dimensional motion

Question 15.
A bird flying in the sky is an example of –
(a) one dimensional motion
(b) two dimensional motion
(c) three dimensional motion
(d) none
Answer:
(c) three dimensional motion

Question 16.
Example for scalar is –
(a) distance
(b) displacement
(c) velocity
(d) angular momentum
Answer:
(a) distance

Question 17.
Which of the following is not a scalar?
(a) Volume
(b) angular momentum
(c) Relative density
(d) time
Answer:
(b) angular momentum

Question 18.
Vector is having –
(a) only magnitude
(b) only direction
(c) bot magnitude and direction
(d) either magnitude or direction
Answer:
(c) both magnitude and direction

Question 19.
“norm” of the vector represents –
(a) only magnitude
(b) only direction
(c) both magnitude and direction
(d) either magnitude or direction
Answer:
(a) only magnitude

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Question 20.
If two vectors are having equal magnitude and same direction is known as –
(a) equal vectors
(b) col-linear vectors
(c) parallel vectors
(d) on it vector
Answer:
(a) equal vectors

Question 21.
The angle between two collinear vectors is / are –
(a) 0°
(b) 90°
(c) 180°
(d) 0° (or) 180°
Answer:
(d) 0° (or) 180°

Question 22.
The angle between parallel vectors is –
(a) 0°
(b) 90°
(c) 180°
(d) 0° (or) 180°
Answer:
(a) 0°

Question 23.
The angle between anti parallel vectors is –
(a) 0°
(b) 90°
(c) 180°
(d) 0° (or) 180°
Answer:
(c) 180°

Question 24.
Unit vector is –
(a) having magnitude one but no direction
(b) \(A \widehat{A}\)
(c) \(\frac{\widehat{A}}{A}\)
(d) |A|
Answer:
(c) \(\frac{\widehat{A}}{A}\)

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Question 25.
A unit vector is used to specify –
(a) only magnitude
(b) only direction
(c) either magnitude (or) direction
(d) absolute value
Answer:
(b) only direction

Question 26.
The angle between any two orthogonal unit vectors is –
(a) 0°
(b) 90°
(c) 180°
(d) 360°
Answer:
(b) 90°

Question 27.
If \(\hat{n}\) is a unit vector along the direction of \(\overrightarrow{\mathrm{A}}\), the \(\hat{n}\) is-
(a) \(\overrightarrow{\mathrm{A}}\) A
(b) n x A
(c) \(\overrightarrow{\mathrm{A}} / \mathrm{A}\)
(d \(\overrightarrow{\mathrm{A}}\) |A|
Answer:
(c) \(\overrightarrow{\mathrm{A}} / \mathrm{A}\)

Question 28.
The magnitude of a vector can not be-
(a) positive
(b) negative
(e) zero
(cl) 90
Answer:
(b) negative

Question 29.
If R = P + Q, then which of the following is true?
(a) P > Q
(b) Q >P
(c) P = Q
(d) R > P, Q
Answer:
(d) R > P, Q

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Question 30.
A force of 3 N and 4 N are acting perpendicular to an object, the resultant force is-
(a) 9 N
(b) 16 N
(c) 5 N
(d) 7 N
Answer:
(c) 5 N

Question 31.
Torque is a-
(a) scalar
(b) vector
(c) either scalar (or) vector
(d) none
Answer:
(6) vector

Question 32.
The resultant of \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\) acts along x – axis. If A = 2\(\hat{i}\) – 3 \(\hat{j}\) + 2\(\hat{k}\) then B is-
(a) -2\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)
(b) 3\(\hat{j}\) – 2\(\hat{k}\)
(c) -2\(\hat{i}\) -3 \(\hat{j}\)
(d) -2\(\hat{i}\) – 2\(\hat{k}\)
Answer:
(b) 3\(\hat{j}\) – 2\(\hat{k}\)

Question 33.
The angle between (\(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)) and (\(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{B}}\)) can be –
(a) only 0°
(b) only 90°
(c) between 0° and 90°
(d) between 0° and 180°
Answer:
(d) between 0° and 180°

Question 34.
If a vector \(\overrightarrow{\mathrm{A}}\) = 3\(\hat{i}\) + 2\(\hat{j}\) then what is 4 A-
(a) 12\(\hat{i}\) + 8\(\hat{j}\)
(b) 0.75\(\hat{i}\) + 0.5\(\hat{j}\)
(c) 3\(\hat{i}\) + 2\(\hat{j}\)
(d) 7\(\hat{i}\) + 6\(\hat{j}\)
Answer:
(a) 12\(\hat{i}\) + 8\(\hat{j}\)

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Question 35.
If P = mV then the direction of P along-
(a) m
(b) v
(c) both (a) and (b)
(d) neither m nor v
Answer:
(b) v

Question 36.
The scalar product \(\overrightarrow{\mathrm{A}}\). \(\overrightarrow{\mathrm{B}}\) is equal to-
(a) \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)
(b) AB sin θ
(c) AB cos θ
(d) \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)
Answer:
(c) AB cos θ

Question 37.
The scalar product \(\overrightarrow{\mathrm{A}}\).\(\overrightarrow{\mathrm{B}}\)is equal to-
(a) \(\overrightarrow{\mathrm{A}}\) +\(\overrightarrow{\mathrm{B}}\)
(b) \(\overrightarrow{\mathrm{A}}\). \(\overrightarrow{\mathrm{B}}\)
(c) AB sin θ
(d) (\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)
Answer:
(b) \(\overrightarrow{\mathrm{A}}\). \(\overrightarrow{\mathrm{B}}\)

Question 38.
The scalar product of two vectors will be maximum when θ is equal to –
(a) 0°
(b) 90°
(c) 180°
(d) 270°
Answer:
(a) 0°

Question 39.
The scalar product of two vectors will be minimum. When θ is equal to –
(a) 0°
(b) 45°
(c) 180°
(d) 60°
Answer:
(c) 180°

Question 40.
The vectors A and B to be mutually orthogonal when –
(a) \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\) = 0
(b) \(\overrightarrow{\mathrm{A}}\) –\(\overrightarrow{\mathrm{B}}\) = 0
(c) \(\overrightarrow{\mathrm{A}}\).\(\overrightarrow{\mathrm{B}}\) = 0
(d) \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) = 0
Answer:
(c) \(\overrightarrow{\mathrm{A}}\).\(\overrightarrow{\mathrm{B}}\) = 0

Question 41.
The magnitude of the vector is –
(a) A2
(b) \(\sqrt{\mathrm{A}^{2}}\)
(c) \(\sqrt{\mathrm{A}}\)
(d) \(\sqrt[3]{\mathrm{A}}\)
Answer:
(b) \(\sqrt{\mathrm{A}^{2}}\)

Question 42.
\(\hat{i}\) .\(\hat{j}\) is –
(a) 0
(b) I
(c) ∞
(d) none
Answer:
(a) 0

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Question 43.
If \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are two vectors, which are acting along x, y respectively, then \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) lies along –
(a) x
(b) y
(c) z
(d) none
Answer:
(c) z

Question 44.
The direction of \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) is given by-
(a) right hand screw rule
(b) right hand thumb rule
(c) both (a) and (b)
(d) neither (a) and (b)
Answer:
(c) both (a) and (b)

Question 45.
\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) is –
(a) AB cos θ
(b) AB sin θ
(c) AB tan θ
(d) AB sec θ
Answer:
(b) AB sin θ

Question 46.
\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) isequal to –
(a) \(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)
(b) \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)
(c) –\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)
(d) \(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{B}}\)
Answer:
(c) –\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)

Question 47.
The vector product of any two vectors gives a –
(a) vector
(b) scalar
(e) tensor
(d) col-linear
Answer:
(a) vector

Question 48.
|\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)| is equal to –
(a) -|\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)|
(b) |\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)|
(c) -|\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)|
(d) \( \frac{\overline{\mathrm{A}} \times \overline{\mathrm{B}}}{|\overline{\mathrm{A}} \times \overline{\mathrm{B}}|}\)
Answer:
(b) |\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)|

Question 49.
The vector product of two vectors will have maximum magnitude when θ is equal to –
(a) 0°
(b) 90°
(c) 180°
(d) 360°
Answer:
(b) 90°

Question 50.
The vector product of two non-zero vectors will be minimum when O is equal to –
(a) 0°
(b) 180°
(e) both (a) and (b)
(d) neither (a) nor (b)
Answer:
(e) both (a) and (b)

Question 51.
The product of a vector with itself is equal to –
(a) 0
(b) 1
(c) ∞
(d) A2
Answer:
(a) 0

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Question 52.
\(\hat{i}\) x \(\hat{i}\) is –
(a) 0
(b) 1
(c) ∞
(d) \(\hat{j}\)
Answer:
(a) 0

Question 53.
\(\hat{i}\) x \(\hat{j}\) is –
(a) \(\hat{i}\)
(b) \(\hat{j}\)
(c) \(\hat{k}\)
(d) \(\overrightarrow{\mathrm{z}}\)
Answer:
(c) \(\hat{k}\)

Question 54.
\(\hat{j}\) x \(\hat{i}\) is –
(a) –\(\hat{i}\)
(b) –\(\hat{j}\)
(c) –\(\hat{k}\)
(d) \(\overrightarrow{\mathrm{z}}\)
Answer:
(c) –\(\hat{k}\)

Question 55.
If two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) form adjacent sides of parallelogram, then the magnitude of |\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)| will give of parallelogram –
(a) length
(b) area
(c) volume
(d) diagonal
Answer:
(b) area

Question 56.
If \(\overrightarrow{\mathrm{P}}\) – \(\overrightarrow{\mathrm{Q}}\) then which of the following is incorrect. –
(a) \(\hat{P}\) = \(\hat{Q}\)
(b) |\(\hat{P}\)| = |\(\hat{Q}\)|
(c) P\(\hat{Q}\) = Q\(\hat{A}\)
(d) \(\hat{P}\) \(\hat{Q}\) = PQ
Answer:
(d) \(\hat{P}\) \(\hat{Q}\) = PQ

Question 57.
The momentum of a particle is \(\overrightarrow{\mathrm{P}}\) = cos θ \(\hat{i}\) + sin θ \(\hat{j}\) . The angle between momentum and the force acting on a body is –
(a) 0°
(b) 45°
(c) 90°
(d) 180°
Answer:
(c) 90°

Question 58.
A and B are two vectors, if A and B are perpendicular to each other then –
(a) \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) = 0
(b) \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) = l
(c) \(\overrightarrow{\mathrm{A}}\) \(\overrightarrow{\mathrm{B}}\) = 0
(d) \(\overrightarrow{\mathrm{A}}\) \(\overrightarrow{\mathrm{B}}\) = \(\overrightarrow{\mathrm{A}}\)\(\overrightarrow{\mathrm{B}}\)
Answer:
(c) \(\overrightarrow{\mathrm{A}}\) \(\overrightarrow{\mathrm{B}}\) = 0

Question 59.
The angle between two vectors -3\(\hat{i}\) + 6\(\hat{k}\) and 2\(\hat{i}\) + 3\(\hat{j}\) + \(\hat{k}\) is –
(a) 0°
(b) 45°
(c) 60°
(d) 90°
Answer:
(d) 90°

Question 60.
The radius vector is 2\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\) while linear momentum is 2\(\hat{i}\) + 3\(\hat{j}\) + \(\hat{k}\) Then the angular momentum is
(a) -2\(\hat{i}\) + 4\(\hat{k}\)
(b) 4\(\hat{i}\) – 8\(\hat{k}\)
(c) 2\(\hat{k}\) – 4\(\hat{j}\) + 2\(\hat{k}\)
(d) 4\(\hat{i}\) – 8\(\hat{j}\)
Answer:
(a) -2\(\hat{i}\) + 4\(\hat{k}\)

Question 61.
Which of the following cannot be a resultant of two vectors of magnitude 3 and 6?
(a) 3
(b) 6
(c) 10
(d) 7
Answer:
(c) 10

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Question 62.
Twelve forces each of magnitude 10 N acting on a body at an angle of 30° with other forces then their resultant is-
(a) 10 N
(b)120 N
(c) \(\frac{10}{\sqrt{3}}\)
(d) zero

Question 63.
Two forces are in the ratio of 3 : 4. The maximum and minimum of their resultants are in the ratio is –
(a) 4 : 3
(b) 3 : 4
(c) 7 : 1
(d) 1 : 7
Answer:
(c) 7 : 1

Question 64.
If | \(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) | = |\(\overrightarrow{\mathrm{P}}\) | + |\(\overrightarrow{\mathrm{Q}}\)|. The angle between the vectors \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) is –
(a) 0°
(b) 180°
(c) 60°
(d) 90°
Answer:
(a) 0°
|\(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) | = |\(\overrightarrow{\mathrm{P}}\) | + |\(\overrightarrow{\mathrm{Q}}\)|
Square on both sides and the resultant becomes
P2 + Q2 + 2PQ cos θ = P2 + Q2 + 2PQ cos θ = 1
θ = 0

Question 65.
If |\(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) = |\(\overrightarrow{\mathrm{P}}\) | — |\(\overrightarrow{\mathrm{P}}\)|, then the angle between the vectors \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\)
(a) 0°
(b) 90°
(c) 180°
(d) 360°
Answer:
(c) |\(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\)| = |\(\overrightarrow{\mathrm{P}}\) | |\(\overrightarrow{\mathrm{P}}\)| ‘
Square on both side, and the resultant becomes
P2 + Q2 + 2PQ cos θ = P2 + Q2 – 2PQ .
cos θ = -1
θ = 180°

Question 66.
If |\(\overrightarrow{\mathrm{P}}\) x \(\overrightarrow{\mathrm{Q}}\)| = |\(\overrightarrow{\mathrm{P}}\) . \(\overrightarrow{\mathrm{Q}}\)| then angle between \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) then angle between P and Q will be –
(a) 0°
(b) 30°
(c) 45°
(d) 60°
Answer:
(c) |\(\overrightarrow{\mathrm{P}}\) x \(\overrightarrow{\mathrm{Q}}\)| = |\(\overrightarrow{\mathrm{P}}\) . \(\overrightarrow{\mathrm{Q}}\)| Expand the terms
PQ sinθ = PQ cos θ
tan θ = 1
θ = 45°

Question 67.
If | \(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) | = |\(\overrightarrow{\mathrm{P}}\) | |\(\overrightarrow{\mathrm{Q}}\)|, then angle between \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) will be –
(a) 0°
(b) 45°
(c) 90°
(d) 180°
Answer:
(c) | \(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) | = |\(\overrightarrow{\mathrm{P}}\)| |\(\overrightarrow{\mathrm{Q}}\) |
Square on both side, and the resultants become,
P2 + Q2 + 2PQ cos 0 = P2 + Q2 – 2PQ cos θ 4PQ cos θ = 0
θ = 90°

Question 68.
If A and B are the sides of triangle, then area of triangle –
(a) \(\frac{1}{2}|\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}|\)
(b) \(\frac{1}{2}|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|\)
(c) AB sin θ
(d) AB cos θ
Answer:
(b) \(\frac{1}{2}|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|\)

Question 69.
A particle moves in a circular path of radius 2 cm. If a particle completes 3 rounds, then the distance and displacement of the particle are –
(a) 0 and 37.7
(b) 37.7 and 0
(c) 0 and 0
(d) 37.7 and 37.7
Answer:
(b) Radius = 2 cm
Circumference of the circle = 2nr = 4n cm
Distance covered in 3 rounds = 127r cm = 37.7 cm
Initial and final positions are same
∴ Displacement = 0

Question 70.
If rx and r2 are position vectors, then the displacement vector is –
(a) \(\vec{r}_{1} \times \vec{r}_{2}\)
(b) \(\vec{r}_{1} \cdot \vec{r}_{2} \)
(c) \(\vec{r}_{1}+\vec{r}_{2}\)
(d) \(\vec{r}_{2}+\vec{r}_{1} \)
Answer:
(d) \(\vec{r}_{2}+\vec{r}_{1} \)

Question 71.
The ratio of the displacement vector to the corresponding time interval is –
(a) average speed
(b) average velocity
(c) instantaneous speed
(d) instantaneous velocity
Answer:
(b) average velocity

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Question 72.
The ratio of total path length travelled by the particle in a time interval –
(a) average speed
(b) average velocity
(c) instantaneous speed
(d) instantaneous velocity
Answer:
(a) average speed

Question 73.
The product of mass and velocity of a particle is –
(a) acceleration
(b) force
(c) torque
(d) momentum
Answer:
(d) momentum

Question 74.
The area under the force, displacement curve is –
(a) potential energy
(b) work done.
(c) impulse
(d) acceleration
Answer:
(b) work done

Question 75.
The area under the force, time graph is –
(a) momentum
(b) force
(c) work done
(d) impulse
Answer:
(d) impulse

Question 76.
The unit of momentum is –
(a) kg ms-1
(b) kg ms-2
(c) kg m2s-1
(d) kg-1 m2 s-1
Answer:
(b) kg ms-2

Question 77.
The slope of the position – time graph will give –
(a) displacement
(b) velocity
(c) acceleration
(d) force
Answer:
(d) force

Question 78.
The area under velocity-time graph gives-
(a) positive
(b) negative
(c) either positive (or) negative
(d) zero
Answer:
(c) either positive (or) negative

Question 79.
The magnitude of distance is always-
(a) positive
(b) negative
(c) either positive (or) negative
(d) zero
Answer:
(a) positive

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Question 80.
If two objects A and B are moving along a straight line in the same direction with the velocities vA and vB respectively, then the relative velocity is-
(a) vA + vB
(b) vA – vB
(c) vA vB
(d) vA / vB
Answer:
(b) VA – VB

Question 81.
If two objects A and B are moving along a straight line in the opposite direction with the velocities VA and VB respectively, then relative velocity is-
(a) VA + VB
(b) VA – VB
(c) VA . VB
(d) VA / VB
Answer:
(a) VA + VB

Question 82.
If two objects moving with a velocities of VA and VB at an angle of 0 between them, the relative velocity is –
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 83.
A person moving horizontally with velocity \(\overrightarrow{\mathrm{V}_{m}}\) The relative velocity of rain with respect to the person is –
(a) \(\mathrm{V}_{\mathrm{R}}+\mathrm{V}_{\mathrm{m}}\)
(b) \(\sqrt{\mathrm{V}_{\mathrm{R}}+\mathrm{V}_{m}}\)
(c) \(\mathrm{V}_{\mathrm{R}}-\mathrm{V}_{m}\)
(d) \(\sqrt{\mathrm{v}_{\mathrm{R}}^{2}+\mathrm{V}_{m}^{2}}\)
Answer:
(d) \(\sqrt{\mathrm{v}_{\mathrm{R}}^{2}+\mathrm{V}_{m}^{2}}\)

Question 84.
A person moving horizontally with velocity \(\overrightarrow{\mathrm{V}_{m}}\) . Rain falls vertically with velocity \(\overrightarrow{\mathrm{V}_{R}}\) To save himself from the rain, he should hold an umbrella with vertical at an angle of –
(a) \(\tan ^{-1}\left(\frac{V_{R}}{V_{m}}\right)\)
(b) \(\tan ^{-1}\left(\frac{V_{m}}{V_{R}}\right)\)
(c) \(\tan \theta=\mathrm{V}_{m}+\mathrm{V}_{\mathrm{R}}\)
(d) \(\tan ^{-1}\left(\mathrm{V}_{\mathrm{R}}+\mathrm{V}_{m} / \mathrm{V}_{\mathrm{R}}-\mathrm{V}_{m}\right)\)
Answer:
(b) \(\tan ^{-1}\left(\frac{V_{m}}{V_{R}}\right)\)

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Question 85.
A car starting from rest, accelerates at a constant rate x for sometime after which it decelerates at a constant rate v to come to rest. If the total time elapsed is t, the maximum velocity attained by the car is given by –
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 86.
A car covers half of its journey with a speed of 10 ms-1 and the other half by 20 ms-1. The average speed of car during the total journey is –
(a) 70 ms-1
(b) 15 ms-1
(c) 13.33 ms-1
(d) 7.5 ms-1
Answer:
(c) Let x is the total distance
Time to cover 1st half = \(\frac{x / 2}{10}\)
Time to cover 2nd half = \(\frac{x / 2}{20}\)
Average speed =
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 87.
A swimmer can swim in still water at of 10 ms-1 While crossing a river his average speed is 6 ms-1. If he crosses the river in the shortest possible time, what is the speed of flow of water?
(a) 16 ms-1
(b) 4 ms-1
(c) 60 ms-1
(d) 8 ms-1
Answer:
(d) The resultant velocity of swimmer must be perpendicular to speed of water to cross the river in a shortest time
∴ \(v_{s}^{2}=v^{2}+v_{w}^{2}\)
\(v_{w}^{2}=v_{s}^{2}-v^{2}\) = 100 – 36 = 64
∴ V = 8 m/s-1

Question 88.
A 100 m long train is traveling from North to South at a speed of 30 ms-1. A bird is flying from South to North at a speed of 10-1. How long will the bird take to, cross the train?
(a) 3 s
(b) 2.5 s
(c) 10 s
(d) 5 s
Answer:
(b) Length of train = 100 m
Relative velocity = 30 + 10 = 40 ms-1
Time taken to cross the train (t) = \(\frac {distance}{ R.velocity }\) = \(\frac { 100 }{ 40 }\) = 2.5 s

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Question 89.
The first derivative of position vector with respect to time is –
(a) velocity
(b) acceleration
(c) force
(d) displacement
Answer:
(a) velocity

Question 90.
The second derivative of position vector with respect to time is –
(a) velocity
(b) acceleration
(c) force
(d) displacement
Answer:
(b) acceleration

Question 91.
The slope of displacement-time graph gives –
(a) velocity
(b) acceleration
(c) force
(d) displacement
Answer:
(a) velocity

Question 92.
The slope of velocity-time graph gives –
(a) velocity
(b) acceleration
(c) force
(d) displacement
Answer:
(b) acceleration

Question 93.
The position vector of a particle is \(\vec{r}\) = 4t2\(\hat{i}\) + 2t\(\hat{j}\) + 3t\(\hat{k}\) The acceleration of a particle is having only –
(a) X – component
(b) Y – component
(c) Z – component
(d) X – Y component
Answer:
(a) X – component
4t2\(\hat{i}\) + 2t\(\hat{j}\) + 3t\(\hat{k}\)
\(\vec{v}\) = \(\frac{d \vec{r}}{dt}\) = 8t\(\hat{i}\) + 2\(\hat{j}\)
a = \(\frac{d^{2} r}{d t^{2}}\) = 8\(\hat{i}\) a is having only X-component.

Question 94.
The position vector of a particle is \(\vec{r}\) = 4t2\(\hat{i}\) + 2t\(\hat{j}\) + 3t\(\hat{k}\). The speed of the particle at t = 5 s is –
(a) 42 ms-1
(b) 3s
(c) 3 ms-1
(d) 40 ms-1
Answer:
(a) 42 ms-1
\(\vec{r}\) = 4t2\(\hat{i}\) + 2t\(\hat{j}\) + 3t\(\hat{k}\)
Speed v = — = \(\frac{d \vec{r}}{dt}\) = 8t\(\hat{i}\) + 2\(\hat{j}\)
at t = 5 s v = 40 + 2 = 42

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Question 95.
An object is moving in a straight line with uniform acceleration a, the velocity-time relation is –
(a) u = v + at
(b) v = u + at
(c) v2 = u2 + a2t2
(d) v2 – u2 = at
Answer:
(b) v = u + at

Question 96.
An object is moving in a straight line with uniform acceleration, the displacement-time relation is –
(a) S = \(u t^{2}+\frac{1}{2} a t^{2}\)
(b) S = \(u t-\frac{1}{2} a t^{2}\)
(c) S = \(u t+\frac{1}{2} a t^{2}\)
(d) S = \(u t-a t^{2}\)
Answer:
(c) S = \(u t+\frac{1}{2} a t^{2}\)

Question 97.
An object is moving in a straight line with uniform acceleration, the velocity-displacement reflation is –
(a) V = u + 2as
(b) S = ut + -at
(c) V2 = u2 – 2as
(d) V2 = u2 + 2as
Answer:
(d) V2 = u2 + 2as

Question 98.
For free-falling body, its initial velocity is –
(a) 0
(b) 1
(c) ∞
(d) none
Answer:
(a) 0

Question 99.
An object falls from a height h (h<<R), the speed of the object when it reaches the ground is –
(a) \(\frac{1}{2} g t^{2}\)
(b) \(\sqrt{g t}\)
(c) gh
(d) \(\sqrt{2 g h}\)
Answer:
(d) \(\sqrt{2 g h}\)

Question 100.
An object falls from a height h (h<< R) the time taken by an object to reaches the ground is –
(a) \(\frac{1}{2} g t^{2}\)
(b) \(\sqrt{2 g h}\)
(c) \(\sqrt{\frac{2 h}{g}}\)
(d) \(\sqrt{\frac{2 g}{h}}\)
Answer:
(d) \(\sqrt{\frac{2 g}{h}}\)

Question 101.
In the absence of air resistance, horizontal velocity of the projectile is –
(a) always negative
(b) equal to ‘g’
(c) directly proportional to g
(d) a constant
Answer:
(d) a constant

Question 102.
In the horizontal projection, the range of the projectile is –
(a) \(\sqrt{\frac{2 h}{g}}\)
(b) \(u \sqrt{\frac{h}{g}}\)
(c) \(u \sqrt{\frac{2 h}{g}}\)
(d) \(u \sqrt{\frac{g}{2 h}}\)
Answer:
(c) \(u \sqrt{\frac{2 h}{g}}\)

SamacheerKalvi.Guru

Question 103.
In oblique projection, maximum height attained by the projectile is –
(a) \(\frac { t }{ u cos θ }\)
(b) \(\frac { u cos θ }{ 2g }\)
(c) \(\frac { 2g }{ u cos θ }\)
(d) \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
Answer:
(d) \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)

Question 104.
In oblique projection time of flight of a projectile is –
(a) \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
(b) \(\frac { 2u cos θ }{ g }\)
(c) \(\frac{u^{2} \sin 2 \theta}{g}\)
(d) \(\frac{u^{2}}{g}\)
Answer:
(b) \(\frac { 2u cos θ }{ g }\)

Question 105.
In oblique projection horizontal range of the projectile is –
(a) \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
(b) \(\frac { 2u cos θ }{ g }\)
(c) \(\frac{u^{2} \sin 2 \theta}{g}\)
(d) \(\frac{u^{2}}{g}\)
Answer:
(a) \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)

Question 106.
In oblique projection, maximum horizontal range of the projectile is –
(a) \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
(b) \(\frac { 2u cos θ }{ g }\)
(c) \(\frac{u^{2} \sin 2 \theta}{g}\)
(d) \(\frac{u^{2}}{g}\)
Answer:
(d) \(\frac{u^{2}}{g}\)

Question 107.
One radian is equal to –
(a) \(\frac {π}{ 180 }\) degree
(b) 60°
(c) 57.295°
(d) 53.925°
Answer:
(c) 57.295°

Question 108.
In relation between linear and angular velocity is –
(a) ω = vr
(b) ω = \(\frac {v }{ r }\)
(c) ω = \(\frac { r}{ v }\)
(d) ω = \(\frac { r }{ ω }\)
Answer:
(b) ω = \(\frac {v }{ r }\)

Question 109.
Centripetal acceleration is given by –
(a) \(\frac{v^{2}}{r}\)
(b) \(-\frac{v^{2}}{r}\)
(c) \(\frac{r}{v^{2}}\)
(d) \(-\frac{r}{v^{2}}\)
Answer:
(b) \(-\frac{v^{2}}{r}\)

Question 110.
In uniform circular motion –
(a) Speed changes but velocity constant
(b) Velocity changes but speed constant
(c) both speed and velocity are constant
(d) both speed and velocity are variable
Answer:
(b) Velocity changes but speed constant

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Question 111.
In non – uniform circular motion, the resultant acceleration is given by –
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 112.
In non – uniform circular motion, the resultant acceleration makes an angle with the radius vector is –
Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics Q112
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 113.
A compartment of an uniformly moving train is suddenly detached from the train and stops after covering some distance. The distance covered by the compartment and distance covered by the train in the given time –
(a) both will be equal
(b) second will be half of first
(c) first will be half of second
(d) none
Answer:
(c) first will be half of second

Question 114.
An object is dropped from rest. Its v – t graph is –
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 115.
When a ball hits the ground as free fall and renounces but less than its original height? Which is represented by –
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 116.
Which of the following graph represents the equation y = mx – C?
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

SamacheerKalvi.Guru

Question 117.
Which of the following graph represents the equation y = mx + C?
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 118.
Which of the following graph represents the equation y = mx?
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 119.
Which of the following graph represents the equation y = -mx + C?
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

SamacheerKalvi.Guru

Question 120.
Which of the following graph represents the equation y = kx2?
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 121.
X = -ky2 is represented by –
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 122.
X = ky2 is represented by –
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 123.
y = kx2 is represented by –
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

SamacheerKalvi.Guru

Question 124.
X °∝\(\frac { 1 }{ Y }\) (or) XY = constant is represented by –
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 125.
y = e-kx is represented by –
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 126.
Y = 1 – e-kx is represented by –
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 127.
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 is represented by –
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 128.
Let y =f(x) is a function. Its maxima (or) minima can be obtained by –
(a) y = 0
(b) f(x) = 0
(c) \(\frac {dy}{dx}\) = 0
(d) \(\frac{d^{2} y}{d x^{2}}\) = 0
Answer:
(c) \(\frac {dy}{dx}\) = 0

Question 129.
A particle at rest starts moving in a horizontal straight line with uniform acceleration. The ratio of the distance covered during the fourth and the third second is –
(a) \(\frac {4}{ 3 }\)
(b) \(\frac { 26 }{ 9 }\)
(e) \(\frac { 7}{ 5 }\)
(d) 2
Answer:
(c) The distance travelled during nth second –
Sn  = u + \(\frac { 1 }{ 2 }\) a (2n -1)
Distance travelled during 4th second S1 = \(\frac { 1 }{ 2 }\) (8 – 1)
Distance travelled during 3rd second S2 = \(\frac { 1 }{ 2 }\) a(6 – 1)
\(\frac{\mathrm{S}_{1}}{\mathrm{S}_{2}}\) = \(\frac { 7 }{ 5 }\)

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Question 130.
The distance travelled by a body, falling freely from rest in t = 1s, t = 2s and t = 3s are in the ratio of –
(a) 1 : 2 : 3
(h) 1 : 3 : 5
(c) 1 : 4 : 9
(d) 9 : 4 : 1
Answer:
(c) The distance travelled by a free falling body S = \(\frac { 1 }{ 2 }\) gt2
∴ S α t2
∴ S1 : S2  : S3 : 12 : 22 : 32 = 1 : 4 : 9.

Question 131.
The displacement of the particle along a straight line at time ¡ is given by X = a + ht + ct2 where a, b, c are constants. The acceleration of the particle is-
(a) a
(b) b
(c) c
(d) 2c
Answer:
(d) X = a + bt + ct2
\(\frac { dX }{ dt }\) = v = b + 2ct
Acceleration = \(\frac{d^{2} X}{d t^{2}}\) = 2c.

Question 132.
Two bullets are fired at an angle of θ and (90 – θ) to the horizontal with same speed. The ratio of their times of flight is –
(a) 1 : 1
(b) 1: tan θ
(c) tan θ : 1
(d) tan2 θ : 1
Answer:
(c) Time of flight tf =  \(\frac { 2x sinθ}{ 9 }\)
tf α sinθ
∴ \(\frac{t_{f_{1}}}{t_{f_{2}}}\) = \(\frac { sinθ }{sin (90 – θ) }\) = \(\frac { sinθ }{cos θ }\) = tanθ
\(t_{f_{1}}: t_{f_{2}}\) = tan θ : 1

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Question 133.
A particle moves along a circular path under the action of a force. The work done by the force is –
(a) positive and non zero
(b) zero
(c) negative and non-zero
(d) none
Answer:
(b) zero

Question 134.
For a particle, revolving in a circle with speed, the acceleration of the particle is –
(a) along the tangent
(b) along the radius
(c) along its circumference
(d) zero
Answer:
(b) along the radius

Question 135.
A gun fires two bullets with same velocity at 60° and 30° with horizontal. The bullets strike at the same horizontal distance. The ratio of maximum height for the two bullets is in the ratio of –
(a) 1 : 2
(b) 3 : 1
(c) 2 : 1
(d) 1 : 3
Answer:
(b) 3 : 1
Max height attained hmax = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
∴ hmax α sin2 θ i.e hmax α \(\frac { 1-cos2θ}{ 2 }\)
\(\frac{h_{\max 1}}{h_{\max } 2}\) = \(\frac{3 / 2}{1 / 2}\) = 3

Question 136.
A ball is thrown vertically upward. it is a speed of lo m/s. When it has reached one half of its maximum height. I-low high does the ball rise? (g = 10 ms-2)
(a) 5 m
(b) 7 m
(c) 10 m
(d) 12 m
Answer:
(c) 10 m

SamacheerKalvi.Guru

Question 137.
A car moves from X to Y with a uniform speed Vn  and returns to Y with a uniform speed Vd The average speed for this round trip is –
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Answer:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 138.
Two projectiles of same mass and with same velocity are thrown at an angle of 60° and 30° with the horizontal then which of the following will remain same?
(a) time of flight
(b) range of projectile
(c) maximum height reached
(d) all the above
Answer:
(b) range of projectile

Question 139.
A n object of mass 3 kg is at rest. Now a force of \(\overrightarrow{\mathrm{F}}\) = 6 t2\(\hat{i}\) + 4t\(\hat{j}\) is applied on the object, then the velocity of object at t = 3 second is –
(a) 18\(\hat{i}\) + 3 \(\hat{j}\)
(b) 18\(\hat{i}\) + 6\(\hat{j}\)
(c) 3 \(\hat{i}\) + 18\(\hat{j}\)
(d) 18 \(\hat{i}\) + 4\(\hat{j}\)
Answer:
(b) F = 6 t2\(\hat{i}\) + 4t\(\hat{j}\)
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 140.
The angle for which maximum height and horizontal range are same for a projectile is –
(a) 32°
(b) 48°
(c) 76°
(d) 84°
Answer:
(c) 76°
Hmax = horizontal range
\(\frac{u^{2} \sin ^{2} \theta}{2 g}\) = \(\frac{u^{2} \sin 2 \theta}{g}\)
\(\frac{\sin ^{2} \theta}{2}\) = 2 sin θ cos θ = sin θ = 4 cos θ tan θ = 4 θ = 76°

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Question 141.
A bullet is dropped from some height, when another bullet is fired horizontally from the same height. They will hit the ground –
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
(a) depends upon mass of bullet
(b) depends upon the observer
(c) one after another
(d) simultaneously
Answer:
(d) simultaneously

Question 142.
From this velocity – time graph, which of the following is correct?
(a) Constant acceleration
(b) Variable acceleration
(c) Constant velocity
(d) Variable acceleration
Answer:
(b) Variable acceleration

Question 143.
When a projectile is at its maximum height, the direction of its velocity and acceleration are –
(a) parallel to each other
(b) perpendicular to each other
(c) anti – parallel to each other
(d) depends on its speed
Answer:
(b) perpendicular to each other

Question 144.
At the highest point of oblique projection, which of the following is correct?
(a) velocity of the projectile is zero
(b) acceleration of the projectile is zero
(c) acceleration of the projectile is vertically downwards
(d) velocity of the projectile is vertically downwards
Answer:
(c) acceleration of the projectile Is vertically downwards

Question 145.
The range of the projectile depends –
(a) The angle of projection
(b) Velocity of projection
(c) g
(d) all the above
Answer:
(d) all the above

Question 146.
A constant force is acting on a particle and also acting perpendicular to the velocity of the particle. The particle describes the motion in a plane. Then –
(a) angular displacement is zero
(b) its velocity is zero
(c) it velocity is constant
(d) it moves in a circular path
Answer:
(d) it moves in a circular path

Question 147.
If a body moving in a circular path with uniform speed, then –
(a) the acceleration is directed towards its center
(b) velocity and acceleration are perpendicular to each other
(c) speed of the body is constant but its velocity is varying
(d) all the above
Answer:
(d) all the above

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Question 148.
A body is projected vertically upward with the velocity y = 3\(\hat{i}\) + 4\(\hat{j}\) ms-1. The maximum height attained by the body is (g 10 ms-2).
(a) 7 m
(b) 1.25 m
(c) 8 m
(d) 0.08 m
Answer:
(b) v = 3\(\hat{i}\) + 4\(\hat{j}\)
Hmax = \(\frac{v^{2} \sin ^{2} \theta}{2 g}\) = \(\frac{v^{2}}{2 g}\) [ θ = 90 ]
v = \(\sqrt{9+16}\)  = \(\sqrt{25}\)
v2 = 25
Hmax = \(\frac{25}{20}\)  = \(\frac { 5 }{ 4 }\) = 1.25 m

Samacheer Kalvi 11th Physics Kinematics Short Answer Questions – I (1 Mark)

Question 1.
What is frame of reference?
Answer:
In a coordinate system, the position of an object is described relative to it, then such a coordinate system is called as frame of reference.

Question 2.
What are the types of motion?
Answer:

  • Linear motion
  • Circular motion
  • Rotational motion
  • Vibratory motion.

Question 3.
What is linear motion? Give example.
Answer:
An object is said to be in linear motion if it moves in a straight line.
Example – an athlete running on a straight track.

Question 4.
What is circular motion? Give example.
Answer:
Circular motion is defined as a motion described by an object traversing a circular path.
Example – The whirling motion of a stone attached to a string.

Question 5.
What is rotational motion? Give example.
Answer:
During a motion every point in the object traverses a circular path about an axis except the points located on the axis, is called as rotational motion.
Example – Spinning of the earth about its own axis.

Question 6.
What is vibratory motion? Give example.
Answer:
If an object or particle executes a to and fro motion about a fixed point, it is said to be in vibratory motion.
Example – Vibration of a string on a guitar.

Question 7.
What is one dimensional motion? Give example.
Answer:
One dimensional motion is the motion of a particle moving along a straight line.
Example –  Motion of a train along a straight railway track.

Question 8.
What is two dimensional motion? Give example.
Answer:
If a particle moving along a curved path in a plane, then it is said to be in two dimensional motion.
Example – Motion of a coin on a carrom board.

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Question 9.
What is three dimensional motion? Give example.
Answer:
If a particle moving in used three dimensional space, then the particle is said to be in three dimensional motion.
E.g. A bird flying in the sky.

Question 10.
Write about the properties of components of vectors.
Answer:
If two vectors \(\overline{\mathrm{A}}\) and \(\overline{\mathrm{B}}\) are equal, then their individual components are also equal. then their individual components are also equal.
Let\(\overline{\mathrm{A}}\) = \(\overline{\mathrm{B}}\)
then Ax  \(\hat{i}\) + Ay \(\hat{j}\) + Az \(\hat{k}\) = Bx \(\hat{i}\) + By \(\hat{j}\) + Bz \(\hat{k}\)
i.e. Ax = Bx, Ay =  By  = Az = Bz

Question 11.
Give an example for scalar product of two vectors.
Answer:
The work done by a force \(\overrightarrow{\mathrm{F}}\) to move an object through a small displacement \(\overrightarrow{\mathrm{dr}}\) then
Work done W = \(\overrightarrow{\mathrm{F}}\) .\(\overrightarrow{\mathrm{dr}}\) (or) W = F dr cos θ

Question 12.
Give any three example for vector product of two vectors.
Answer:

  1. Torque \(\overrightarrow{\mathrm{t}}\) = \(\overrightarrow{\mathrm{r}}\) x \(\overrightarrow{\mathrm{F}}\). Where i is force and \(\overrightarrow{\mathrm{F}}\) is force and \(\overrightarrow{\mathrm{r}}\) position vector of a particle.
  2. Angular momentum \(\overrightarrow{\mathrm{L}}\) = \(\overrightarrow{\mathrm{r}}\) x \(\overrightarrow{\mathrm{P}}\) where \(\overrightarrow{\mathrm{P}}\) is the linear momentum.
  3. Linear velocity \(\overrightarrow{\mathrm{v}}\) = \(\overrightarrow{\mathrm{ω}}\) x \(\overrightarrow{\mathrm{r}}\) where \(\overrightarrow{\mathrm{ω}}\) is angular velocity.

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Question 13.
What is position vector?
Answer:
It is vector which denotes the position of a particle at any instant of time, with respect to some reference frame or coordinate system.
The position \(\overrightarrow{\mathrm{r}}\) vector of the particle at a point P is given by
\(\overrightarrow{\mathrm{r}}\) = x\(\hat{i}\) + y\(\hat{j}\) + z\(\hat{k}\)
where x, y and z are components of \(\overrightarrow{\mathrm{r}}\).

Question 14.
Write a note an momentum.
Answer:
Momentum of a particle is defined as product of mass with velocity. It is denoted as \(\overrightarrow{\mathrm{p}}\) Momentum is also a vector quantity
\(\overrightarrow{\mathrm{r}}\) = m\(\overrightarrow{\mathrm{v}}\)
The direction of momentum is also in the direction of velocity, and the magnitude of momentum is equal to product of mass and speed of the particle.
p = mv
In component form the momentum can be written as
px\(\hat{i}\) + py\(\hat{j}\)+ pz\(\hat{k}\) = mvx\(\hat{i}\) + mvy\(\hat{j}\) + mvz\(\hat{k}\)
Here,
px = x component of momentum and is equal to mvx
Px = y component of momentum and is equal to mvy
Px = z component of momentum and is equal to mvz

Question 15.
“Displacement vector is basically a position vector”. Comment on it.
Answer:
This statement is almost correct only. Because the displacement vector also gives the position of a point just like a position vector. The difference between these two vectors is p. The displacement vector gives the position of a point with respect to a point other than origin but position vector gives the position of a point with respect to origin.

SamacheerKalvi.Guru

Question 16.
Will two dimensional motion with an acceleration only in one dimension?
Answer:
Yes. In oblique projection, the acceleration is acting vertically downward but the object follows a parabolic path.

Question 17.
A foot ball is kicked by a player with certain angle to the horizontal. Is there any point at which velocity is perpendicular to its acceleration.
Answer:
Yes. At its maximum height in the parabolic path vertical velocity is zero but due to horizontal component, velocity acts along horizontally, but acceleration of the

Question 18.
Give any two examples for parallelogram law of vectors.
Answer:

  • the flight of a bird
  • working of a sling.

Question 19.
Why does rubber ball bounce greater heights on hills than in plains?
Answer:
The maximum height attained by the projectile is inversely proportional to acceleration due to gravity. At greater height, acceleration due to gravity will be lesser than plains. So ball can bounce higher in hills than in plains.

Question 20.
Is it possible for body to have variable velocity but constant speed? Give example.
Answer:
Yes, it is possible. In horizontal circular motion the speed of a particle is always constant but due to the variation in direction continuously, the velocity of a particle varies.

SamacheerKalvi.Guru

Question 21.
What is relative velocity?
Answer:
When two objects are moving with different velocities, then the velocity of one object with respect to another object is called relative velocity of an object with respect to another.

Question 22.
What is average acceleration?
Answer:
The average acceleration is defined as the ratio of change in velocity over the time interval
aavg = \(\frac{\Delta \overrightarrow{\mathrm{v}}}{\Delta t}\) It is a vector quantity.

Question 23.
Write a note an instantaneous acceleration.
Answer:
Instantaneous acceleration or acceleration of a particle at time ‘t’ is given by the ratio of change in velocity over ∆t, as ∆t approaches zero.
Acceleration Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
In other words, the acceleration of the particle at an instant t is equal to rate of change of velocity

(1) Acceleration is a vector quantity. Its SI unit is ms-2 and its dimensional formula is [M°L1 T-2]

(2) Acceleration is positive if its velocity is increasing, and is negative if the velocity is decreasing. The negative acceleration is called retardation or deceleration.

Question 24.
Write an acceleration in terms of its component?
(Or)
Show that the acceleration is the second derivative of position vector with respect to time.
Answer:
in terms of components, we can write,
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
are the components of instantaneous acceleration. Since each component of velocity is the derivative of the corresponding coordinate, we can express the components ax, ay, and az as
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Then the acceleration vector \(\overrightarrow{\mathrm{a}}\) it self is
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Thus acceleration is the second derivative of position vector with respect to time.

SamacheerKalvi.Guru

This is free online projectile motion calculator. enter velocity and time value then click on calculate and result will be instant displayed ,this calculator.

Question 25.
What are the examples of projectile motion?
Answer:

  1. An object dropped from window of a moving train.
  2. A bullet fired from a rifle.
  3. A ball thrown in any direction.
  4. A javelin or shot put thrown by an athlete.
  5. A jet of water issuing from a hole near the bottom of a water tank.

Question 26.
Explain projectile motion.
Answer:
A projectile moves under the combined effect of two velocities.

  • A uniform velocity in the horizontal direction, which will not change provided there is no air resistance.
  • A uniformly changing velocity (i.e., increasing or decreasing) in the vertical direction.

There are two types of projectile motion:

  • Projectile given an initial velocity in the horizontal direction (horizontal projection)
  • Projectile given an initial velocity at an angle to the horizontal (angular projection)

To study the motion of a projectile, let us assume that,

  • Air resistance is neglected.
  • The effect due to rotation of Earth and curvature of Earth is negligible.
  • The acceleration due to gravity is constant in magnitude and direction at all points of the motion of the projectile.

Question 27.
What is time of flight?
Answer:
The time taken for the projectile to complete its trajectory or time taken by the projectile to hit the ground is called time of flight.

Question 28.
Under what condition is the average velocity equal the instantaneous velocity?
Answer:
When the body is moving with uniform velocity.

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Question 29.
Draw Position time graph of two objects, A & B moving along a straight line, when their relative velocity is zero.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 30.
Suggest a situation in which an object is accelerated and have constant speed.
Answer:
Uniform Circular Motion.

Question 31.
Two balls of different masses are thrown vertically upward with same initial velocity. Maximum heights attained by them are h1 and h2 respectively what is h1/h2?
Answer:
Same height,
∴  h1/h2 = 1

Question 32.
A car moving with velocity of 50 kmh-1 on a straight road is ahead of a jeep moving with velocity 75 kmh-1 would the relative velocity be altered if jeep is ahead of car?
Answer:
No change.

Question 33.
Which of the two – linear velocity or the linear acceleration gives the direction of motion of a body?
Answer:
Linear velocity.

Question 34.
Will the displacement of a particle change on changing the position of origin of the coordinate system?
Answer:
Will not change.

Question 35.
If the instantaneous velocity of a particle is zero, will its instantaneous acceleration be necessarily zero?
Answer:
No. (highest point of vertical upward motion under gravity).

SamacheerKalvi.Guru

Question 36.
A projectile is fired with Kinetic energy 1 KJ. If the range is maximum, what is its Kinetic energy, at the highest point?
Answer:
Here \(\frac { 1 }{ 2 }\) mv2 =1kJ=1000 J, θ = 45°
At the highest point, K.E. = \(\frac { 1 }{ 2 }\) m(v cos 0)2 = \(\frac { 1 }{ 2 }\)\(\frac{m v^{2}}{2}\) = \(\frac {1000}{ 2 }\) = 500 J.

Question 37.
Write an example of zero vector.
Answer:
The velocity vectors of a stationary object is a zero vectors.

Question 38.
State the essential condition for the addition of vectors.
Answer:
They must represent the physical quantities of same native.

Question 39.
When is the magnitude of (\(\overline{\mathrm{A}}\) + \(\overline{\mathrm{B}}\)) equal to the magnitude of (\(\overline{\mathrm{A}}\) – \(\overline{\mathrm{B}}\))?
Answer:
When \(\overline{\mathrm{A}}\) is perpendicular to \(\overline{\mathrm{B}}\).

Question 40.
What is the maximum number of component into which a vector can be resolved?
Answer:
Infinite.

Question 41.
A body projected horizontally moves with the same horizontal velocity although it moves under gravity Why?
Answer:
Because horizontal component of gravity is zero along horizontal direction.

Question 42.
What is the angle between velocity and acceleration at the highest point of a projectile motion?
Answer:
90°.

SamacheerKalvi.Guru

Question 43.
When does

  • height attained by a projectile maximum?
  • horizontal range is maximum?

Answer:

  • Height is maximum at θ = 90
  • Range is maximum at θ = 45.

Question 44.
What is the angle between velocity vector and acceleration vector in uniform circular motion?
Answer:
90°.

Question 45.
A particle is in clockwise uniform circular motion the direction of its acceleration is radially inward. If sense of rotation or particle is anticlockwise then what is the direction of its acceleration?
Answer:
Radial in ward.

Question 46.
A train is moving on a straight track with acceleration a. A passenger drops a stone. What is the acceleration of stone with respect to passenger?
Answer:
\(\sqrt{a^{2}+g^{2}}\) where g = acceleration due to gravity.

Question 47.
What is the average value of acceleration vector in uniform circular motion over one cycle?
Answer:
Null vector.

Question 48.
Does a vector quantity depends upon frame of reference chosen?
Answer:
No.

SamacheerKalvi.Guru

Question 49.
What is the angular velocity of the hour hand of a clock?
Answer:
ω = \(\frac {2π}{ 12 }\) = \(\frac { π }{6}\) rad h-1

Question 50.
What is the source of centripetal acceleration for earth to go round the sun?
Answer:
Gravitation force of sun.

Question 51.
What is the angle between (\(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{A}}\) ) and (\(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{A}}\)) ?
Answer:
90°

Samacheer Kalvi 11th Physics Kinematics Short Answer Questions – II (2 Marks)

Question 1.
What are positive and negative acceleration in straight line motion?
Solution:
If speed of an object increases with time, its acceleration is positive. (Acceleration is in the direction of motion) and if speed of an object decreases with time its acceleration is negative (Acceleration is opposite to the direction of motion).

Question 2.
Can a body have zero velocity and still be accelerating? If yes gives any situation.
Solution:
Yes, at the highest point of vertical upward motion under gravity.

Question 3.
The displacement of a body is proportional to t3, where t is time elapsed. What is the nature of acceleration –  time graph of the body?
Solution:
As a α t3 ⇒ s = kt3
Velocity, V = \(\frac { ds }{ dt }\) = 3 kt3
Acceleration, a = \(\frac { dv }{ dt }\) = 3 kt3
i.e., a α t
⇒ motion is uniform, acceleration motion, a – t graph is straight-line.

SamacheerKalvi.Guru

Question 4.
Suggest a suitable physical situation for the following graph.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Solution:
A ball thrown up with some initial velocity rebounding from the floor with reduced speed after each hit.

Question 5.
An object is in uniform motion along a straight line, what will be position time graph for the motion of object, if (i) x0 = positive, v = negative is constant.
(i) x0 = positive, v = negative is |\(\vec{v}\) | constant.
(ii) both x0 and v are negative |\(\vec{v}\) | is constant.
(iii) x0 = negative, v = positive |\(\vec{v}\) | is constant.
(iv) both x0 and v are positive |\(\vec{v}\) | is constant where x0 is position at t = 0.
Solution:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 6.
A cyclist starts from centre O of a circular park of radius 1 km and moves along the path OPRQO as shown. If he maintains constant speed of 10 ms-1. What is his acceleration at point R in magnitude & direction?
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Solution:
Centripetal acceleration, ac = \(\frac{v^{2}}{r}\) = \(\frac{10^{2}}{1000}\) = 0.1 m/s2 along RO.

Question 7.
What will be the effect on horizontal range of a projectile when its initial velocity is doubled keeping angle of projection same?
Solution:
\(\frac{u^{2} \sin 2 \theta}{g}\) ⇒ R α u2
Range comes four times.

SamacheerKalvi.Guru

Question 8.
The greatest height to which a man can throw a stone is h. What will be the greatest distance upto which he can throw the stone?
Solution:
Maximum height:
H = \(\frac{u^{2} \sin ^{2} \theta}{g}\) ⇒ Hmax = \(\frac{u^{2}}{2g}\) = h (at θ = 90)
Maximum range Rmax = \(\frac{u^{2}}{g}\) = 2h

Question 9.
A person sitting in a train moving at constant velocity throws a ball vertically upwards. How will the ball appear to move to an observer.

  • Sitting inside the train
  • Standing outside the train

Solution:

  • Vertical straight line motion
  • Parabolic path.

Question 10.
A gunman always keep his gun slightly tilted above the line of sight while shooting. Why?
Solution:
Because bullet follow Parabolic trajectory under constant downward acceleration.

Question 11.
Is the acceleration of a particle in circular motion not always towards the center. Explain.
Solution:
No acceleration is towards the center only in case of uniform circular motion.

SamacheerKalvi.Guru

Samacheer Kalvi 11th Physics Kinematics Short Answer Questions – III (3 Marks)

Question 1.
Draw
(a) acceleration – time
(b) velocity – time
(c) Position – time graphs representing motion of an object under free fall. Neglect air resistance.
Solution:
The object falls with uniform acceleration equal to ‘g’
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 2.
The velocity time graph for a particle is shown in figure. Draw acceleration time graph from it.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Solution:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 3.
For an object projected upward with a velocity v0, which comes back to the same point after some time, draw
(i) Acceleration – time graph
(ii) Position – time graph
(iii) Velocity time graph
Solution:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 4.
The acceleration of a particle in ms2 is given by a = 3t2 + 2t + 2, where time t is in second. If the particle starts with a velocity v = 2 ms-1 at t = 0, then find the velocity at the end of 2s.
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics Q4

SamacheerKalvi.Guru

Question 5.
At what angle do the two forces (P + Q) and (P – Q) act so that the resultant is \(\sqrt{3 P^{2}+Q^{2}}\)?
Solution:
Use
R = \(\sqrt{3 P^{2}+Q^{2}}\)
R = \(\sqrt{3 \mathrm{P}^{2}+\mathrm{Q}^{2}}\)
A = P + Q
B = P – Q
solve, θ = 60°

Question 6.
A car moving along a straight highway with speed of 126 km h 1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop?
Solution:
Initial velocity of car,
u = 126 kmh-1 = 126 x \(\frac {5}{18}\) ms-1 = 35 ms-1 ………(i)
Since, the car finally comes to rest, v = 0
Distance covered, s = 200 m, a = ?, t = ?
v2 = u2 – 2as
or a = \(\frac{v^{2}-u^{2}}{2 s}\) ………..(ii)
substituting the values from eq. (i) in eq . (ii), we get
a = \(\frac{0-(35)^{2}}{2 \times 200}\) = \(\frac{0-(35)^{2}}{2 \times 200}\)
= \(-\frac{46}{16}\) ms-2 = -3.06 ms-2
Negative sign shows that acceleration in negative which is called retardation, i.e., car is uniformly retarded at – a = 3.06 ms-2.
To find t, let us use the relation
v = u + at
t = \(\frac {v-u}{ a }\)
use a = -3.06 ms-2, v = 0, u = 35 ms-1
∴ t = \(\frac {v-u}{ a }\) = \(\frac {0-35}{ -3.06 }\) = 11.44 s
∴ t = 11.44 sec

SamacheerKalvi.Guru

Samacheer Kalvi 11th Physics Kinematics Long Answer Questions
Question 1.
Explain the types of motion with example.
Answer:
(a) Linear motion:
An object is said to be in linear motion if it moves in a straight line.
Examples:

  • An athlete running on a straight track
  • AA particle falling vertically downwards to the Earth.

(b) Circular motion:
Circular motion is defined as a motion described by an object traversing a circular path.
Examples:

  • The whirling motion of a stone attached to a string.
  • The motion of a satellite around the Earth.
  • These two circular motions are shown in figure.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

(c) Rotational motion:
If any object moves in a rotational motion about an axis, the motion is called ‘rotation’. During rotation every point in the object transverses a circular path about an axis, (except the points located on the axis).
Examples:

  • Rotation of a disc about an axis through its center
  • Spinning of the Earth about its own axis.
  • These two rotational motions are shown in figure

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

(d) Vibratory motion:
If an object or particle executes a to-and-fro motion about a fixed point, it is said to be in vibratory motion. This is sometimes also called oscillatory motion.
Examples:

  • Vibration of a string on a guitar
  • Movement of a swing
  • These motions are shown in figure

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
Other types of motion like elliptical motion and helical motion are also possible.

Question 2.
What are the different types of vectors? ,
Answer:
1. Equal vectors:
Two vectors A and B are said to be equal when they have equal magnitude and same direction and represent the same physical quantity
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
(a) Collinear vectors:
Col-linear vectors are those which act along the same line. The angle between them can be 0° or 180°.

(i) Parallel vectors:
If two vectors A and B act in the same direction along the same line or on parallel lines, then the angle between them is 0°. Geometrical representation of parallel vectors.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

(ii) Anti-parallel vectors:
Two vectors A and B are said to be anti – parallel when they are in opposite directions along the same line or on parallel lines. Then the angle between them is 180°.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
2. Unit vector:
A vector divided by its magnitude is a unit vector. The unit vector for \(\overrightarrow{\mathrm{A}}\) is denoted by \(\widehat{A}\) . It has a magnitude equal to unity or one.
Since, \(\widehat{A}\) = \(\frac{\bar{A}}{A}\) we can write \(\overrightarrow{\mathrm{A}}\) = A\(\widehat{A}\)
Thus, we can say that the unit vector specifies only the direction of the vector quantity.

3. Orthogonal unit vectors:
Let \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) be three unit vectors which specify the directions along positive x-axis, positive y-axis and positive z-axis respectively. These three unit vectors are directed perpendicular to each other, the angle between any two of them is 90°.\(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) and are examples of orthogonal vectors. Two vectors which are perpendicular to each other are called orthogonal vectors as shown in the figure.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

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Question 3.
Explain the concept of relative velocity in one and two dimensional motion.
Answer:
When two objects A and B are moving with different velocities, then the velocity of one object A with respect to another object B is called relative velocity of object A with respect to B.

Case I:
Consider two objects A and B moving with uniform velocities VA and VB, as shown, along straight tracks in the same direction \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\), \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) with respect to ground.
The relative velocity of object A with respect to object B is \(\overrightarrow{\mathrm{V}}_{\mathrm{AB}}\) = \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) – \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\).
The relative velocity of object B with respect to object A is \(\overrightarrow{\mathrm{V}}_{\mathrm{BA}}\) = \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) – \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) Thus, if two objects are moving in the same direction, the magnitude of relative velocity of one object with respect to another is equal to the difference in magnitude of two velocities.

Case II.
Consider two objects A and B moving with uniform velocities \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) and \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) along the same straight tracks but opposite in direction.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
The relative velocity of an object A with respect to object B is –
\(\overrightarrow{\mathrm{V}}_{\mathrm{AB}}\) = \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) – (-\(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\)) = \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) + \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\)

The relative velocity of an object B with respect to object A is
\(\overrightarrow{\mathrm{V}}_{\mathrm{AB}}\) = –\(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) – \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) = – (\(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) + \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\))
Thus, if two objects are moving in opposite directions, the magnitude of relative velocity of one object with respect to other is equal to the sum of magnitude of their velocities.

Case III.
Consider the velocities \(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\) and \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\) at an angle θ between their directions. The relative velocity of A with respect to B, \(\overrightarrow{\mathrm{v}}_{\mathrm{AB}}\) = \(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\) – \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\)
Then, the magnitude and direction of \(\overrightarrow{\mathrm{v}}_{\mathrm{AB}}\) is given by \(\overrightarrow{\mathrm{v}}_{\mathrm{AB}}\) = \(\sqrt{\vec{v}_{\mathrm{A}}^{2}+\vec{v}_{\mathrm{B}}^{2}-2 v_{\mathrm{A}} v_{\mathrm{B}} \cos \theta}\) and tan β = \(\frac{v_{\mathrm{B}} \sin \theta}{v_{\mathrm{A}}-v_{\mathrm{B}} \cos \theta}\) (Here β is angle between (\(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\) and \(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\))
\(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\) – \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\) cos θ .

(i) When θ = 0, the bodies move along parallel straight lines in the same direction, We have \(\overrightarrow{\mathrm{v}}_{\mathrm{AB}}\) = (\(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\) – \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\)) in the direction of \(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\) . Obviously \(\overrightarrow{\mathrm{v}}_{\mathrm{BA}}\) = (\(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\) + \(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\)) in the direction of \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\).

(ii) When θ = 180°, the bodies move along parallel straight lines in opposite directions,
We have \(\overrightarrow{\mathrm{v}}_{\mathrm{AB}}\) = (\(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\)+ \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\)) in the direction of \(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\) . Similarly, vBA = (vB + vA) in the direction of \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\) .

(iii) If the two bodies are moving at right angles to each other, then 0 = 90°. The magnitude of the relative velocity of A with respect to B = \(\overrightarrow{\mathrm{v}}_{\mathrm{AB}}\) = \(\sqrt{v_{\mathrm{A}}^{2}+v_{\mathrm{B}}^{2}}\).

(iv) Consider a person moving horizontally with velocity \(\overrightarrow{\mathrm{V}}_{\mathrm{M}}\) . Let rain fall vertically with velocity \(\overrightarrow{\mathrm{V}}_{\mathrm{R}}\) . An umbrella is held to avoid the rain. Then the relative velocity of the rain with respect to the person is,
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
which has magnitude
\(\overrightarrow{\mathrm{V}}_{\mathrm{RM}}\) = \(\overrightarrow{\mathrm{V}}_{\mathrm{R}}\) – \(\overrightarrow{\mathrm{V}}_{\mathrm{M}}\)
And direction 0 = tan-1\(\left(\frac{V_{\mathrm{M}}}{\mathrm{V}_{\mathrm{R}}}\right)\) with the vertical as shown in figure.

SamacheerKalvi.Guru

Question 4.
Shows that the path of horizontal projectile is a parabola and derive an expression for
1. Time of flight
2. Horizontal range
3. resultant relative and any instant
4. speed of the projectile when it hits the ground?
Answer:
Consider a projectile, say a ball, thrown horizontally with an initial velocity \(\vec{u}\) from the top of a tower of height h. As the ball moves, it covers a horizontal distance due to its uniform horizontal velocity u, and a vertical downward distance because of constant acceleration due to gravity g. Thus, under the combined effect the ball moves along the path OPA. The motion is in a 2 – dimensional plane. Let the ball take time t to reach the ground at point A, Then the horizontal distance travelled by the ball is x(t) = x, and the vertical distance travelled is y(t) = y.
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
We can apply the kinematic equations along the x direction and y direction separately. Since this is two-dimensional motion, the velocity will have both horizontal component ux and vertical component uy.

Motion along horizontal direction:
The particle has zero acceleration along x direction. So, the initial velocity ux remains constant throughout the motion. The distance traveled by the projectile at a time t is given by the equation x = uxt +\(\frac { 1 }{ 2 }\) at2 . Since a = 0 along x direction, we have x = uxt ……….(1)

Motion along downward direction:
Here uy = 0 (initial velocity has no downward component), a = g (we choose the + ve y – axis in downward direction), and distance y at time t.
From equation, y = uxt +\(\frac { 1 }{ 2 }\) at2 we get
y = \(\frac { 1 }{ 2 }\) at2 …………..(2)

Substituting the value oft from equation (i) in equation (ii) we have
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
y = Kx2
where K = \(\frac{g}{2 u_{x}^{2}}\) is constant
Equation (iii) is the equation of a parabola. Thus, the path followed by the projectile is a parabola.

1. Time of Flight:
The time taken for the projectile to complete its trajectory or time taken by the projectile to hit the ground is called time of flight. Consider the example of a tower and projectile. Let h be the height of a tower. Let T be the time taken by the projectile to hit the ground, after being thrown horizontally from the tower.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
We know that sy = uyt + \(\frac { 1 }{ 2 }\) at2 for vertical motion. Here .sy = h, t = T, uy = 0 (i.e., no initial vertical velocity). Then h = \(\frac { 1 }{ 2 }\) gt2 or T = \(\sqrt{\frac{2 h}{g}}\) Thus, the time of flight for projectile motion depends on the height of the tower, but is independent of the horizontal velocity of projection. If one ball falls vertically and another ball is projected horizontally with some velocity, both the balls will reach the bottom at the same time. This is illustrated in the Figure

2. Horizontal range:
The horizontal distance covered by the projectile from the foot of the tower to the point where the projectile hits the ground is called horizontal range.
For horizontal motion, we have
sx = uxt + \(\frac { 1 }{ 2 }\) at2
Here,sx = R (range), ux = u, a = 0 (no horizontal acceleration) T is time of flight. Then horizontal range = uT
Since the time of flight T = \(\sqrt{\frac{2 h}{g}}\) we substitute this and we get the horizontal range of the particle as R = u \(\sqrt{\frac{2 h}{g}}\)
The above equation implies that the range R is directly proportional to the initial velocity u and inversely proportional to acceleration due to gravity g.

3. Resultant Velocity (Velocity of projectile at any time):
At any instant t, the projectile has velocity components along both x-axis and y-axis. The resultant of these two components gives the velocity of the projectile at that instant t, as shown in figure. The velocity component at any t along horizontal (x-axis)
is Vx = Ux + axt
Since, ux = u, ax = 0 , we get
ux = u ax = 0 we get
vx = u
The component of velocity along vertical direction (y – axis) is vy = uy + ayt
Since, uy= 0, ay = g, we get
Vy = gt
Hence the velocity of the particle at any instant is –
v = u\(\hat{i}\) + g\(\hat{j}\)
The speed of the particle at any instant t is given by
v = \(\sqrt{v_{x}^{2}+v_{y}^{2}}\)
v= \(\sqrt{u^{2}+g^{2} t^{2}}\)

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

4. Speed of the projectile when it hits the ground:
When the projectile hits the ground after initially thrown horizontally from the top of tower of height h, the time of flight is –
t = \(\sqrt{\frac{2 h}{g}}\)
The horizontal component velocity of the projectile remains the same i.e vx = u.
The vertical component velocity of the projectile at time T is
v = gT = g \(\sqrt{\frac{2 h}{g}}\) = \(\sqrt{\frac{2gh}\)
The speed of the particle when it reaches the ground is
v = \(\sqrt{u^{2}+2 g h}\).

SamacheerKalvi.Guru

Question 5.
Derive the relation between Tangential acceleration and angular acceleration.
Answer:
Consider an object moving along a circle of radius r. In a time ∆t, the object travels in an arc distance As as shown in figure. The corresponding angle subtended is ∆θ
The ∆s can be written in terms of ∆θ
∆s = r∆θ ………(i)
in a time ∆t, we have
\(\frac { ∆s }{ ∆t}\) = t \(\frac { ∆θ }{ ∆t}\) …………(ii)
¡n the limit ∆t – 0, the above equation becomes
\(\frac { ds }{ dt}\) = rω …………….(iii)
Here \(\frac { ds }{ dt}\) is linear speed (y) which is tangential to the circle and co is angular speed.
So equation (iii) becomes.
v r = rω …….(iv)
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics
which gives the relation between linear speed and angular speed.
Eq. (iv) is true only for circular motion. In general the relation between linear and angular velocity is given by
\(\vec{v}\) = \(\vec{\omega} \times \vec{r}\) ………..(v)
For circular motion eq. (y) reduces to eq. (iv) since \(\overrightarrow{\mathrm{ω}}\) and \(\overrightarrow{\mathrm{r}}\)are perpendicular to each other.
Differentiating the eq. (iv) with respect to time, we get (since r is constant)
\(\frac { dv }{ dt}\) = \(\frac { rdv }{ dt}\) = rα
Here \(\frac { dv }{ dt}\) Is the tangential acceleration and is denoted as at = \(\frac {dω}{ dt}\)is the angular acceleration
α. Then eq. (v) becomes
at = rα ………..(vii)

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Samacheer Kalvi 11th Physics Kinematics Numerical Questions

Question 1.
The V – t graphs of two objects make angle 30° and 60° with the time axis. Find the ratio of their accelerations.
Solution:
\(\frac{a_{1}}{a_{2}}\) = \(\frac{tan 30}{tan 60}\) = \(\frac{1 / \sqrt{3}}{\sqrt{3}}\) = \(\frac{1}{3}\) = 1 : 3.

Question 2.
When the angle between two vectors of equal magnitudes is 2n/?>, prove that the magnitude of the resultant is equal to either.
Solution:
R = \(\left(\mathrm{P}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ} \cos \theta\right)^{1 / 2}\) = \(\left(p^{2}+p^{2}+2 p \cdot p \cos \frac{2 \pi}{3}\right)\) = \(\left[2 p^{2}+2 p^{2}\left(\frac{-1}{2}\right)\right]^{1 / 2}\) = p.

Question 3.
If \(\overline{\mathrm{A}}\) = 3\(\hat{i}\) + 4\(\hat{j}\) and \(\overline{\mathrm{B}}\) = 7\(\hat{i}\) + 24 \(\hat{j}\), find a vector having the same magnitude as \(\overline{\mathrm{B}}\) and parallel to \(\overline{\mathrm{A}}\).
Solution:
\(|\overrightarrow{\mathrm{A}}|=\sqrt{3^{2}+4^{2}}=5\)
also \(|\overrightarrow{\mathrm{B}}|=\sqrt{7^{2}+24^{2}}=25\)
desired vector = \(|\overrightarrow{\mathrm{B}}|\) \(\widehat{A}\) = 25 x \(\frac{3 \hat{i}+4 \hat{j}}{5}\) = 5(3\(\hat{j}\) + 4\(\hat{j}\)) = 15 \(\hat{i}\) + 20\(\hat{j}\).

Question 4.
What is the vector sum of n coplanar forces, each of magnitude F, if each force makes an angle of \(\frac {2 π}{ n }\) with the preceding force?
Solution:
Resultant force is zero.

Question 5.
A car is moving along X- axis. As shown in figure it moves from O to P in 18 seconds and return from P to Q in 6 second. What are the average velocity and average speed of the car in going from

  • O to P
  • From O to P and hack to Q

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Solution:

  • O to P, Average velocity =20 ms-1
  • O to P and back to Q

Average velocity = 10 ms-1
Average speed = 20 ms-1

Question 6.
On a 60 km straight road, a bus travels the first 30 km with a uniform speed of 30 kmh-1 . How fast must the bus travel the next 30 km so as to have average speed of 40 kmh-1 for the entire trip?
Solution:
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

Question 7.
The displacement x of a particle varies with time as x = 4t2 – 15t + 25. Find the position, velocity and acceleration of the particle at t = O.
Solution:
Position, x = 25 m
Velocity = \(\frac { dx}{dt}\)8t – 15
t = 0, v = 0 – 15 = -15 m/s
acceleration, a = \(\frac { dx}{dt}\) = 8 ms-2

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Question 8.
A driver takes 0.20 second to apply the breaks (reaction time). If he is driving car at a speed of 54 kmh-1 and the breaks cause a deceleration of 6.0 ms-2. Find the distance travelled by car after he sees the need to put the breaks.
Solution:
(distance covered during 0.20 s) + (distance covered until rest)
= (15 x 0.25) + [18.75] = 21.75 m.

Question 9.
From the top of a tower 100 m in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with a velocity of 25 m/s. Find when and where the two balls will meet? (g = 9.8 m/s)
Solution:
For the ball dropped from the top
x = 4.9 t2…….(i)
For the ball thrown upwards
100 – x =25t – 4.9 t2 …….(ii)
From eq. (i) and (ii)
t = 4s  x = 78.4 m.

Question 10.
A ball thrown vertically upwards with a speed of 19.6 ms-1 from the top of a tower returns to the earth in 6s. Find the height of the tower, (g = 9.8 m/s)
Solution:
s = ut + \(\frac { 1 }{ 2 }\) at2
-h = 19. 6 x 6 + \(\frac { 1 }{ 2 }\) x (-9.8) x (6)2
h = 58.8 m.

Question 11.
Two town A and B are connected by a regular bus service with a bus leaving in either direction every T min. A man cycling with a speed of 20 kmh-1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed do the buses ply of the road?
Solution:
V =40 krnh-1 and T = 9 min.

SamacheerKalvi.Guru

Question 12.
A motorboat is racing towards north at 25 kmh-1 and the water current in that region is 10 kmh-1 in the direction of 60° east of south. Find the resultant velocity of the boat.
Solution:
V= 21.8 kmh-1
angle with north, θ = 23.4°.

Question 13.
An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft position 10 second apart is 30°, what is the speed of the aircraft?
Solution:
Speed = 182.2 ms-1

Question 14.
A boat is moving with a velocity (3\(\hat{i}\) -4\(\hat{j}\)) with respect to ground. The water in river is flowing with a velocity (-3\(\hat{i}\) – 4\(\hat{j}\)) with respect to ground. What is the relative velocity of boat with respect to river?
Solution:
\(\overrightarrow{\mathrm{V}}_{\mathrm{BW}}\)= \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\).

Question 15.
A hiker stands on the edge of a clift 490 m above the ground and throws a stone horizontally with an initial speed of 15 ms-1. Neglecting air resistance, find the time taken by the stone to reach the ground and the speed with which it hits the ground. (g 9.8 ms-2)
Solution:
time = 10 seconds
V = \(\sqrt{\mathrm{V}_{s}^{2}+\mathrm{V}_{Y}^{2}}\) = \(\sqrt{15^{2}+98^{2}}\) = 99.1 m/s-1

Question 16.
A bullet fired at an angle of 30° with the horizontal hits the ground 3 km away. By adjusting the angle of projection, can one hope to hit the target 5 km away ? Assume that the muzzle speed to be fixed and neglect air resistance.
Solution:
Maximum Range = 3.46 km
So it is not possible.

Question 17.
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 seconds, what is the magnitude and direction of acceleration of the stone?
Solution:
\(\frac { 88 }{ 25 }\) rad s-1, \(\frac { 2π}{ T}\) = \(\frac { 2πN}{t}\)
a = 991.2 cm s-2

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Question 18.
A cyclist is riding with a speed of 27 kmh-1 . As he approaches a circular turn on the road of radius 30 m-2, he applies brakes and reduces his
speed at the constant rate 0.5 ms-2. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
Solution:
ac = \(\frac{v^{2}}{r}\) = 0.7ms-2
at = 0.5 ms-2
a = \(\sqrt{a^{2}-a^{2}}\) = 0.86 ms-2
If O is the angle between the net acceleration and the velocity of the cyclist, then
0= tan-1 \(\left(\frac{a_{c}}{a_{\mathrm{T}}}\right)\) = tan-1 = 54°28′

Question 19.
If the magnitude of two vectors are 3 and 4 and their scalar product is 6, find angle between them and also find \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\)|
Solution:
\(\overrightarrow{\mathrm{A}}\).\(\overrightarrow{\mathrm{B}}\) = AB cos θ
|\(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\)| = AB sin θ
or 6 = (3 x 4) cos θ = 3 x 4 x 60°
or θ = 60°
= 3 x 4 x \(\frac{\sqrt{3}}{2}\) = \(6 \sqrt{3}\)

Question 20.
Find the value ofA so that the vector \(\overrightarrow{\mathrm{A}}\) = 2 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\) and \(\overrightarrow{\mathrm{B}}\) = 4\(\hat{i}\) -2 \(\hat{j}\) +2\(\hat{k}\) are perpendicular to each other.
Solution:
\(\overrightarrow{\mathrm{A}} \perp \overrightarrow{\mathrm{B}}\)
⇒ \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}\)
⇒ t = 3

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Question 21.
The velocity time graph of a particle is given by –
Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

  •  Calculate distance and displacement of particle from given v – t graph.
  • Specify the time for which particle undergone acceleration, retardation and moves with constant velocity.
  • Calculate acceleration, retardation from given v – t graph.
  • Draw acceleration-time graph of given v – t graph.

Solution:
(i) distance = area of ∆OAB + area of trapezium BCDE = 12 + 28 = 40 m
(ii) displacement area of ∆OAB – area of trapezium BCDE = 12 – 28 = – 16 m

  • times acc. \((0 \leq t \leq 4) \) and \((12\leq t \leq 16) \)
  • retardation \((4\leq t \leq 8) \)
  • constant velocity \((8\leq t \leq 12) \)
    Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

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Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics

Students can Download Physics Chapter 6 Optics Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Physics Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics

Samacheer Kalvi 12th Physics Optics Textual Evaluation Solved

Samacheer Kalvi 12th Physics Optics Multiple Choice Questions 

Question 1.
The speed of light in an isotropic medium depends on,
(a) its intensity
(b) its wavelength
(c) the nature of propagation
(d) the motion of the source w.r.to medium
Answer:
(b) its wavelength

Question 2.
A rod of length 10 cm lies along the principal axis of a concav e mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is, (AIPMT Main 2012)
(a) 2.5 cm
(b) 5cm
(c) 10 cm
(d) 15cm
Answer:
(b) 5cm
Hint:
By mirror formula, image distance of A
\(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\) ; \(\frac { { 1 } }{ { v }_{ A } } \) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)
\(\frac { { 1 } }{ { v }_{ A } } \) + \(\frac { 1 }{ u }\) + \(\frac { 1 }{ (-30) }\) = \(\frac { 1 }{ (-10) }\)
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-1
∴ vA = – 15 cm
Image distance of C, vc = – 20 cm
The length of image = |vA – vc|
= |-15 + 20| = 5 cm

The full form of ktm is Kraftfahrzeuge Trunkenpolz Mattighofen. It translates into English to mean “motor Vehicle”.

Question 3.
An object is placed in front of a convex mirror of focal length of/and the maximum and minimum distance of an object from the mirror such that the image formed is real and magnified. (IEE Main 2009)]
(a) 2ƒ and c
(b) c and ∞
(c) ƒ and O
(d) None of these
Answer:
(d) None of these
Hint:
There is no maximum minimum object distance for convex mirror to form real and inverted image.

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Question 4.
For light incident from air onto a slab of refractive index 2. Maximum possible angle of refraction is,
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
(a) 30°
Hint:
From Snell’s law, µ = \(\frac { sin i }{ sin r }\)
Now consider an angle of incident is 90°
\(\frac { sin 90° }{ 2 }\)
sin r = sin-1 (0.5)
r = 30°

Question 5.
If the velocity and wavelength of light in air is Va and λa and that in water is Va and λw, then the refractive index of water is,
(a) \(\frac { { V }_{ w } }{ { V }_{ a } } \)
(b) \(\frac { { V }_{ a } }{ { V }_{ w } } \)
(c) \(\frac { { λ }_{ w } }{ { λ }_{ a } } \)
(d) \(\frac { { V }_{ a }{ \lambda } }{ { V }_{ w }{ \lambda }_{ w } } \)
Answer:
(b) \(\frac { { V }_{ a } }{ { V }_{ w } } \)

Question 6.
Stars twinkle due to
(a) reflection
(b) total internal reflection
(c) refraction
(d) polarisation
Answer:
(c) refraction

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Question 7.
When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have refractive index,
(a) less than one
(b) less than that of glass
(c) greater than that of glass
(d) equal to that of glass
Answer:
(d) equal to that of glass
Hint:
According to len’s maker formula,
\(\frac { 1 }{ ƒ } \) = \(\left(\frac{n_{2}}{n_{1}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
Biconvex lens dipped in a liquid, acts as a plane sheet of glass, ƒ = ∞; \(\frac { 1 }{ ƒ } \) = 0
\(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \) -1 = 0;  n2 = n1

Question 8.
The radius of curvature of curved surface at a thin planoconvex lens is 10 cm and the refractive index is 1.5. If the plane surface is silvered, then the focal length will be,
(a) 5 cm
(b) 10 cm
(c) 15 cm
(d) 20 cm
Answer:
(b) 10 cm
Hint:
According to len’s maker formula,
\(\frac { 1 }{ ƒ }\) = (n – 1) \(\left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
R1 = ∞; R2 = -6
∴ ƒ = \(\frac { R }{ \left( n – 1 \right) } \) (R= 10 cm; n= 1.5
ƒ = \(\frac { 10}{ \left( 1.5 – 1 \right) } \) =20 cm

Question 9.
An air bubble in glass slab of refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness of the slab is,
(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) 16 cm
Answer:
(c) 12 cm
Hint:
Let d1 = 5 cm and d2 = 3 cm ; n = 1.5
Actual width is the sum of real depth from 2 sides
Thickness of slab = d1n + d2 n
= (5 x 1.5) +(3 x 1.5)= 12 cm

Question 10.
A ray of light travelling in a transparent medium of refractive index n falls, on a surface separating the medium from air at an angle of incidents of 45°. The ray can undergo total internal reflection for the following n,
(a) n= 1.25
(b) n = 1.33
(c) n= 1.4
(d) n= 1.5
Answer:
(d) n= 1.5
Hint:
For total internal reflection
sin i > sin c            where, i – angle of incidence
But, sin c = 1/n      c – critical angle
sin i > 1/n
n > \(\frac { 1 }{ sin i }\)
n > \(\frac { 1 }{ sin 45 }\) ; n > √2 ; n > 1.414

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Question 11.
A plane glass is placed over a various coloured letters (violet, green, yellow, red) The letter which appears to be raised more is,
(a) red
(b) yellow
(c) green
(d) violet
Answer:
(d) violet
Hint:
Letters appear to be raised depending upon the refractive index of the material. Since violet has a higher refractive index than red (the index increases with frequency), red will be the lowermost.

Question 12.
Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm approximately. The maximum distance at which these dots can be resolved by the eye is, [take wavelength of light, λ = 500 nm]
(a) 1 m
(b) 5 m
(c) 3 m
(d) 6m
Answer:
(b) 5 m
Hint:
Resolution limit sin θ = \(\frac { Y }{ d }\) = \(\frac { 1.22c }{ d }\)
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-2

Question 13.
In a Young’s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to,
(a) 2D
(b) \(\frac { D }{ 2 }\)
(c) √2D
(d) \(\frac { D }{ √2 }\)
Answer:
(a) 2D
Hint:
Young’s double -slite experiment is
β = \(\frac { λD }{ d}\) ; β’ = \(\frac { λD’ }{ d’}\) ; d’ = 2d
Same fringe space, β = β’
⇒ \(\frac { λD }{ d}\) = \(\frac { λD’ }{ d’}\) ; D’ = 2D

Question 14.
Two coherent monochromatic light beams of intensities I and 41 are superposed. The maximum and minimum possible intensities in the resulting beam are [IIT-JEE 1988]
(a) 5I and I
(b) 5I and 3I
(c) 9I and I
(d) 9I and 3I
Answer:
(c) 9I and I
Hint:
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-3

Question 15.
When light is incident on a soap film of thickness 5 x 10-5
cm, the wavelength of light reflected maximum in the visible region is 5320 A. Refractive index of the film will be,
(a) 1.22
(b) 1.33
(c) 1.51
(cl) 1.83
Answer:
(b) 1.33
Hint.
The condition for constructive interference, (for reflection)
2µ tcos r = (2n +1) \(\frac { λ }{ 2 }\) [∴ cos r = 1]
µ = \(\frac { \left( 2n+1 \right) \lambda }{ 4t } \)
For visible region, n = 2
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-4

Question 16.
First diffraction minimum due to a single slit of width 1.0 x 10-5 cm is at 30°. Then wavelength of light used is,
(a) 400 Å
(b) 500 Å
(c) 600 Å
Answer:
(b) 500 Å
Hint.
For diffraction minima, d sin θ = nλ
λ = \(\frac { dsin\theta }{ n } =\frac { 1\times { 10 }^{ -5 }\times { 10 }^{ -2 }\times sin30° }{ 1 } \) = 0.5 x 10-7
λ = 500 Å

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Question 17.
A ray of light strikes a glass plate at an angle 60°. If the reflected and refracted rays are perpendicular to each other, the refractive index of the glass is,
(a) √3
(b) \(\frac { 3 }{ 2 }\)
(c) \(\sqrt { \frac { 3 }{ 2 } } \)
(d) 2
Answer:
(a) √3
Hint.
Angle of refraction r = 60° ; Angle of incident i = 30°
sin i =n x sin r
n = \(\frac {sin 30°}{ sin 60°} \) = √3

Question 18.
One of the of Young’s double slits is covered with a glass plate as shown in figure. The position of central maximum will,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-5
(a) get shifted downwards
(b) get shifted upwards
(c) will remain the same
(d) data insufficient to conclude
Answer:
(b) get shifted upwards

Question 19.
Light transmitted by Nicol prism is,
(a) partiallypolarised
(b) unpolarised
(c) plane polarised
(d) elliptically polarised
Answer:
(c) plane polarised

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Question 20.
The transverse nature of light is shown in,
(a) interference
(b) diffraction
(c) scattering
(d) polarisation
Answer:
(d) polarisation

Samacheer Kalvi 12th Physics Optics Short Answer Questions

Question 1.
State the laws of reflection.
Answer:
(a) The incident ray, reflected ray and normal to the reflecting surface all are coplanar (ie. lie in the same plane).
(b) The angle of incidence i is equal to the angle of reflection r. i = r

Question 2.
What is angle of deviation due to reflection?
Answer:
The angle between the incident and deviated light ray is called angle of deviation of the light ray. It is written as, d= 180 – (i + r).
As, i = r in reflection, we can write angle of deviation ‘ in reflection at plane surface as, d = 180 – 2i

Question 3.
Give the characteristics of image formed by a plane mirror.
Answer:

  1. The image formed by a plane mirror is virtual, erect, and laterally inverted.
  2. The size of the image is equal to the size of the object.
  3. The image distance far behind the mirror is equal to the object distance in front of it.
  4.  If an object is placed between two plane mirrors inclined at an angle 0, then the number of images n formed is as, n = \(\left( \frac { 360 }{ \theta } -1 \right) \)

Question 4.
Derive the relation between ƒ and R for a spherical mirror.
Let C be the centre of curvature of the mirror. Consider a light ray parallel to the principal axis is incident on the mirror at M and passes through the principal focus F after reflection. The geometry of reflection of the incident ray is shown in figure. The line CM is the normal to the mirror at M. Let i be the angle of incidence and the same will be the angle of reflection. If MP is the perpendicular from M on the principal axis, then from the geometry, The angles ∠MCP = i and ∠MFP = 2i From right angle triangles ∆MCP and ∆MFP,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-6
\(\frac { 1 }{ f }\) = \(\left(\frac{n_{2}}{n_{1}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
tan i = \(\frac { PM }{ PC }\) and tan 2 i = \(\frac { PM }{ PF }\)
As the angles are small, tan i ≈ i = \(\frac { PM }{ PC }\) and tan 2 i = \(\frac { PM }{ PF }\)
Simplifying further, 2 = \(\frac { PM }{ PC }\) and tan = \(\frac { PM }{ PF }\) ;2PF = PC
PF is focal length/and PC is the radius of curvature R.
2 ƒ= R (or) ƒ = \(\frac { R }{ 2 }\)
ƒ = \(\frac { R }{ 2 }\) is the relation between ƒ and R.

Question 5.
What are the Cartesian sign conventions for a spherical mirror?
Answer:

  1. The Incident light is taken from left to right (i.e. object on the left of mirror).
  2. All the distances are measured from the pole of the mirror (pole is taken as origin).
  3. The distances measured to the right of pole along the principal axis are taken as positive.
  4. The distances measured to the left of pole along the principal axis are taken as negative.
  5. Heights measured in the upward perpendicular direction to the principal axis are taken as positive.
  6. Heights measured in the downward perpendicular direction to the principal axis, are taken as negative.

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Question 6.
What is optical path? Obtain the equation for optical path of a medium of thickness d and refractive index n.
Answer:
Optical path of a medium is defined as the distance d’ light travels in vacuum in the same time it travels a distance d in the medium.
Let us consider a medium of refractive index n and thickness d. Light travels with a speed v through the medium in a time t. Then we can write,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-7
v = \(\frac { d }{ t }\); rewritten as, t = \(\frac { d }{ v }\)
In the same time, light can cover a greater distance d’
in vacuum as it travels with greater speed c in vacuum.
Then we have,
c = v = \(\frac { d’ }{ t }\); rewritten as, t = \(\frac { d’ }{ c }\)
As the time taken in both the cases is the same, we can equate the time t as,
\(\frac { d’ }{ c }\) = \(\frac { d }{ v }\)
rewritten for the optical path d’ as d’ = \(\frac { c }{ v }\)d
As, \(\frac { c }{ v }\) = n ; The optical path d’ is, d’ = nd
As n is always greater than 1, the optical path d’ of the medium is always greater than d.

Question 7.
State the laws of refraction.
Answer:
Law of refraction is called Snell’s law.
Snell’s law states that,
(a) The incident ray, refracted ray and normal to the refracting surface are all coplanar (i.e. lie in the same plane).
(b) The ratio of angle of incident i in the first medium to the angle of reflection r in the second medium is equal to the ratio of refractive index of the second medium n2 to that of the refractive index of the first medium n1.
\(\frac { sin i }{ sin r }\) = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \).

Question 8.
What is angle of deviation due to refraction?
Answer:
The angle between the incident and deviated light is called angle of deviation. When light travels from rarer to denser medium it deviates towards normal. The angle of deviation in this case is, d = i – r

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Question 9.
What is principle of reversibility?
Answer:
The principle of reversibility states that light will follow exactly the same path if its direction of travel is reversed.

Question 10.
What is relative refractive index?
Answer:
Snell’s law, the term \(\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } \right) \) is called relative refractive index of second medium with respect to the first medium which is denoted as (n21). n21 = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \)

Question 11.
Obtain the equation for apparent depth.
Answer:
Light from the object O at the bottom of the tank passes from denser medium (water) to rarer medium (air) to reach our eyes. It deviates away from the normal in the rarer medium at the point of incidence B. The refractive index of the denser medium is n1 and rarer medium is n2. Here, n1 > n2. The angle of incidence in the denser medium is i and the angle of refraction in the rarer medium is r. The lines NN’ and OD are parallel. Thus angle ∠DIB is also r. The angles i and r are very small as the diverging light from O entering the eye is very narrow. The Snell’s law in product form for this refraction is,
n1 sin i = n2 sin r
As the angles i and r are small, we can approximate, sin i ≈ tan i;
n1 tan i = n2 tan i
In triangles ∆DOB and ∆DIB,
tan(i) = \(\frac { DB }{ DO }\) and tan(r) = \(\frac { DB }{ DI }\)
n1 = \(\frac { DB }{ DO }\) n2 = \(\frac { DB }{ DI }\)
DB is cancelled on both sides, DO is the actual depth d and DI is the apparent depth d’.
Rearranging the above equation for the apparent depth d’,
d’ = \(\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } \right) \)d
As the rarer medium is air and its refractive index n2 can be taken as 1, (n2 = 1). And the refractive index n1 of denser medium could then be taken as n, (n1 = n).
In that case, the equation for apparent depth becomes,
d = \(\frac { d }{ n }\)

Question 12.
Why do stars twinkle?
Answer:
The stars actually do not twinkle. They appear twinkling because of the movement of the atmospheric layers with varying refractive indices which is clearly seen in the night sky.

Question 13.
What is critical angle and total internal reflection?
Answer:
The angle of incidence in the denser medium for which the refracted ray graces the boundary is called critical angle ic.
The entire light is reflected back into the denser medium itself. This phenomenon is called total internal reflection.

Question 14.
Obtain the equation for critical angle.
Answer:
Snell’s law in the product form, equation for critical angle incidence becomes,
n1 sini ic = n2 sin 90°
n1 sini ic = n2 (∵ sin 90° = 1)
sini ic = \(\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } \right) \)
Here, n1 > n2
If the rarer medium is air, then its refractive index is 1 and can be taken as n itself, i.e. (n2 = 1) and (n1= n).
sini ic = \(\frac { 1 }{ n }\) (or) ic = sin-1 \(\left( \frac { 1 }{ n } \right) \)

Question 15.
Explain the reason for glittering of diamond.
Answer:
Diamond appears dazzling because the total internal reflection of light happens inside the diamond. The refractive index of only diamond is about 2.417. It is much larger than that for ordinary glass which is about only 1.5. The critical angle of diamond is about 24.4°. It is much less than that of glass. A skilled diamond cutter makes use of this larger range of angle of incidence (24.4° to 90° inside the diamond), to ensure that light entering the diamond is total internally reflected from the many cut faces before getting out. This gives a sparkling effect for diamond.

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Question 16.
What are mirage and looming?
Answer:
Mirage: Mirage takes place in hot regions. The light from distant objects appears to be reflected from ground. For mirage to form refractive index goes on increasing as we go up. Looming: Looming takes place in cold regions. The light from distant objects appears to be flying. For looming to form refractive index goes on decreasing.

Question 17.
Write a short notes on the prisms making use of total internal reflection.
Answer:
Prisms can be designed to reflect light by 90° or by 180° by making use of total internal reflection. The critical angle ic for the material of the prism must be less than 45°. This is true for both crown glass and flint glass. Prisms are also used to invert images without changing their size.

Question 18.
What is Snell’s window?
Answer:
When light entering the water from outside is seen from inside the water, the view is restricted to a particular angle equal to the critical angle ic. The restricted illuminated circular area is called Snell’s window.

Question 19.
Write a note on optical fibre.
Answer:
Transmitting signals through optical fibres is possible due to the phenomenon of total internal reflection. Optical fibres consists of inner part called core and outer part called cladding (or) sleeving. The refractive index of the material of the core must be higher than that of the cladding for total internal reflection to happen. Signal in the form of light is made to incident inside the core-cladding boundary at an angle greater than the critical angle. Hence, it undergoes repeated total internal reflections along the length of the fibre without undergoing any refraction.

Question 20.
Explain the working of an endoscope.
Answer:
An endoscope is an instrument used by doctors which has a bundle of optical fibres that are used to see inside a patient’s body. Endoscopes work on the phenomenon of total internal reflection. The optical fibres are inserted in to the body through mouth, nose or a special hole made in the body. Even operations could be carried out with the endoscope cable which has the necessary instruments attached at their ends.

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Question 21.
What are primary focus and secondary focus of convex lens?
Answer:
Primary focus: The primary focus F1 is defined as a point where an object should be placed to give parallel emergent rays to the principal axis.
Secondary focus: The secondary focus F2 is defined as a point where all the parallel rays travelling close to the principal axis converge to form an image on the principal axis.

Question 22.
What are the sign conventions followed for lenses?
Answer:
The sign conventions for thin lenses differ only in the signs followed for focal lengths.
(a) The sign of focal length is not decided on the direction of measurement of the focal length from the pole of the lens as they have two focal lengths, one to the left and another to the right (primary and secondary focal lengths on either side of the lens).
(b) The focal length of the thin lens is taken as positive for a converging lens and negative for a diverging lens.

Question 23.
Arrive at lens equation from lens maker’s formula.
Answer:
From refraction through a double convex lens, the relation between the object distance u, image distance v1 and radius of curvature R1 as
\(\frac { { \mu }_{ 2 } }{ { v }_{ 1 } } -\frac { { \mu }_{ 1 } }{ u } =\frac { { \mu }_{ 2 }-{ \mu }_{ 1 } }{ { R }_{ 1 } } \) …… (1)
The relation between the object distance image distance v1 and radius of curvature R2 can be
\(\frac { { \mu }_{ 1 } }{ { v } } -\frac { { { \mu }_{ 2 } } }{ v_{ 1 } } =\frac { { \mu }_{ 1 }-{ \mu }_{ 2 } }{ { R }_{ 2 } } \) …… (2)
Adding equation (1) and (2)
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-8
If the object is placed at infinity (µ = ∞), the image will be formed at the focus. i.e.v = ƒ
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-9
This is len’s maker’s formula. When the lens is placed in air µ1 = 1 and µ2 = µ.
Equation (4) becomes,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-10
From equation (3) and (4), we have \(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\)
=\(\frac { 1 }{ ƒ }\)
This is the len’s equation.

Question 24.
Obtain the equation for lateral magnification for thin lens.
Answer:
Lateral magnification in terms of u and ƒ.
The thin lens formula is
\(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\)=\(\frac { 1 }{ ƒ }\)
Multiplying on both sides by ‘u’
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-11
In terms of v and ƒ multiplying by v, we get
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-12
Hence, lateral magnification for thin lens,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-13

Question 25.
What is power of a lens?
Answer:
The power of a lens P is defined as the reciprocal of its focal length.
p = \(\frac { 1 }{ ƒ }\)
The unit of power is diopter D.

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Question 26.
Derive the equation for effective focal length for lenses in contact.
Answer:
Consider a two thin lenses in contact. In the absence of second lens L2, the first lens L1 will form a real image I’. Using thin lens formula.
\(\frac { 1 }{ { f }_{ 2 } } \)=\(\frac { 1 }{ v’ }\)–\(\frac { 1 }{ u }\) ….. (1)
The image I’ acts as a virtual object (u = v’) for the second lens L2 which finally forms its real image I at distance v. Thus
\(\frac { 1 }{ { f }_{ 2 } } \)=\(\frac { 1 }{ v }\)–\(\frac { 1 }{ v’ }\) ….. (2)
Adding equation (1) and (2) we get,
\(\frac { 1 }{ { f }_{ 1 } } \)+\(\frac { 1 }{ { f }_{ 2 } } \)=\(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\) ….. (3)
For the combination of thin lenses in contact, if ‘f’ is the equivalent focal length, then
\(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\)=\(\frac { 1 }{ ƒ }\) ….. (4)
From equations (3) and (4), the effective focal length for lenses in contact.
\(\frac { 1 }{ ƒ }\)=\(\frac { 1 }{ { f }_{ 1 } } \)+\(\frac { 1 }{ { f }_{ 2 } } \)

Question 27.
What is angle of minimum deviation?
Answer:
The minimum value of the angle of deviation suffered by a ray on passing through a prism is called the angle of minimum deviation.

Question 28.
What is dispersion?
Answer:
The phenomenon of spliting of white light into its component colours on passing through a refracting medium is called dispersion.

Question 29.
How are rainbows formed?
Answer:
Rainbow is formed by dispersion of sunlight into its constituent colours by raindrops which disperse sunlight by refraction and deviate the colours by total internal reflection.

Question 30.
What is Rayleigh’s scattering?
Answer:
The scattering of light by particles in a medium without a change in wavelength is called as Rayleigh’s scattering.

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Question 31.
Why does sky appear blue?
Answer:
Blue colour of the sky is due to scattering of sunlight by air molecules. According to Rayleigh’s law, intensity of scattered light, I ∝ \(\frac { 1 }{ { \lambda }^{ 4 } } \) So blue light of shorter wavelength is scattered much more than red light of larger wavelength. The blue component is proportionally more in light coming from different parts of the sky. That is why the sky appears blue.

Question 32.
What is the reason for reddish appearance of sky during sunset and sunrise?
Answer:
During sunrise or sunset, the sun is near the horizon. Sunlight has to travel a greater distance. So shorter waves of blue region are scattered away by the atmosphere. Red waves of longer wavelength are least scattered and reach the observer. So the sun appears red.

Question 33.
Why do clouds appear white?
Answer:
Clouds have large particles like dust and water droplets which scatter light of all colours almost equally. Hence clouds generally appear white.

Question 34.
What are the salient features of corpuscular theory of light?
Answer:

  • According this theory, light is emitted as tiny, massless (negligibly small mass) and perfectly elastic particles called corpuscles.
  • As the corpuscles are very small, the source of light does not suffer appreciable loss of mass even if it emits light for a long time.
  • On account of high speed, they are unaffected by the force of gravity and their path is a straight line in a medium of uniform refractive index.
  • The energy of light is the kinetic energy of these corpuscles. When these corpuscles impinge on the retina of the eye, the vision is produced.
  • The different size of the corpuscles is the reason for different colours of light.
  • When the corpuscles approach a surface between two media, they are either attracted or repelled.
  • The reflection of light is due to the repulsion of the corpuscles by the medium and refraction of light is due to the attraction of the corpuscles by the medium.

Question 35.
What is wave theory of light?
Answer:
Light is a disturbance from a source that travels as longitudinal mechanical waves through the either medium that was presumed to pervade all space as mechanical wave requires medium for its propagation. The wave theory could successfully explain phenomena of reflection, refraction, interference and diffraction of light.

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Question 36.
What is electromagnetic wave theory of light?
Answer:
Electromagnetic wave theory:
Maxwell (1864) proved that light is an electromagnetic wave which is transverse in nature carrying electromagnetic energy. He could also show that no medium is necessary for the propagation of electromagnetic waves. All the phenomenon of light could be successfully explained by this theory.

Question 37.
Write a short note on quantum theory of light.
Answer:
Albert Einstein (1905), endorsing the views of Max Plank (1900), was able to explain photoelectric effect in which light interacts with matter as photons to eject the electrons. A photon is a discrete packet of energy. Each photon has energy E of,
E = hv
Where, h is Plank’s constant (h = 6.625 x 10-34J s) and v is frequency of electromagnetic wave. As light has both wave as well as particle nature it is said to have dual nature. Thus, it is concluded that light propagates as a wave and interacts with matter as a particle.

Question 38.
What is a wave front?
Answer:
A wavefront is the locus of points which are in the same state or phase of vibration.

Question 39.
What is Huygens’ principle?
Answer:
According to Huygens principle, each point of the wavefront is the source of secondary wavelets emanating from these points spreading out in all directions with the speed of the wave. These are called as secondary wavelets.

Question 40.
What is interference of light?
Answer:
The phenomenon of addition or superposition of two light waves which produces increase in intensity at some points and decrease in intensity at some other points is called interference of light.

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Question 41.
What is phase of a wave?
Answer:
Phase is a particular point in time on the cycle of a waveform, measured as an angle in degrees.

Question 42.
Obtain the relation between phase difference and path difference.
Answer:
Phase difference (Φ):
It is the difference expressed in degrees or radians between two waves having same frequency and referenced to same point in time.
Path difference (δ):
It is the difference between the lengths of two paths of the two different having same frequency and travelling at same velocity. δ =\(\frac { \lambda }{ 2\pi } \) Φ

Question 43.
What are coherent sources?
Answer:
Two light sources are said to be coherent if they produce waves which have same phase or constant phase difference, same frequency or wavelength (monochromatic), same waveform and preferably same amplitude.

Question 44.
What is intensity division?
Answer:
Intensity’ or amplitude division: If we allow light to pass through a partially silvered mirror (beam splitter), both reflection and refraction take place simultaneously. As the two light beams are obtained from the same light source, the two divided light beams will be coherent beams. They will be either in-phase or at constant phase difference.

Question 45.
How does wavefront division provide coherent sources?
Answer:
Wavefront division is the most commonly used method for producing two coherent sources. A point source produces spherical wavefronts. All the points on the wavefront are at the same phase. If two points are chosen on the wavefront by using a double slit, the two points will act as coherent sources.

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Question 46.
How do source and images behave as coherent sources?
Answer:
Source and images: In this method a source and its image will act as a set of coherent source, because the source and its image will have waves in-phase or constant phase difference. The Instrument, Fresnel’s biprism uses two virtual sources as two coherent sources and the instrument, Lloyd’s mirror uses a source and its virtual image as two coherent sources.

Question 47.
What is bandwidth of interference pattern?
Answer:
The bandwidth (β) is defined as the distance between any two consecutive bright or dark fringes.

Question 48.
What is diffraction?
Answer:
Diffraction is bending of waves around sharp edges into the geometrically shadowed region.

Question 49.
Differentiate between Fresnel and Fraunhofer diffraction.
Answer:

S.No. Fresnel diffraction Fraunhofer diffraction
1. Spherical or cylindrical wavefront undergoes diffraction Plane wavefront undergoes diffraction
2. Light wave is from a source at finite distance Light wave is from a source at infinity
3. For laboratory conditions, convex lenses need not be used In laboratory conditions, convex lenses are to be used
4. difficult to observe and analyse Easy to observe and analyse

Question 50.
Discuss the special cases on first minimum in Fraunhofer diffraction.
Let us consider the condition for first minimum with (n = 1). a sin θ = λ
The first minimum has an angular spread of, sin θ = \(\frac { \lambda }{ a } \). Special cases to discuss on the condition.
1. When a < λ, the diffraction is not possible, because sin 0 can never be greater than 1.
2. When a ≥ λ, the diffraction is possible.

  • For a = λ, sin θ = 1 i.e, θ = 90°. That means the first minimum is at 90°. Hence, the central maximum spreads fully in to the geometrically shadowed region leading to bending of the diffracted light to 90°.
  • For a >> λ, sin θ << 1 i.e, the first minimum will fall within the width of the slit itself. The diffraction will not be noticed at all.

3. When a > λ and also comparable, say a = 2λ, sin θ = \(\frac { \lambda }{ a } \) = \(\frac { \lambda }{ { 2\lambda } } \) =\(\frac { 1 }{ 2 }\); then θ = 30°. These are practical cases where diffraction could be observed effectively.

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Question 51.
What is Fresnel’s distance? Obtain the equation for Fresnel’s distance.
Answer:
Fresnel’s distance is the distance up to which the ray optics is valid in terms of rectilinear propagation of light.
Fresnel’s distance z as, z = \(\frac { { a }^{ 2 } }{ 2\lambda } \).

Question 52.
Mention the differences between interference and diffraction.
Answer:

S.No. Interference Diffraction
1. Superposition of two waves Bending of waves around edges
2. Superposition of waves from two coherent sources. Superposition wavefronts emitted from various points of the same wavefront.
3. Equally spaced fringes. Unequally spaced fringes
4. Intensity of all the bright fringes is almost same Intensity falls rapidly for higher orders
5. Large number of fringes are obtained Less number of fringes are obtained

Question 53.
What is a diffraction grating?
Answer:
A diffraction grating is an optical component with a periodic structure that splits and diffracts light into several beams travelling in different directions.

Question 54.
What are resolution and resolving power?
Answer:
Optical resolution describes the ability of an imaging system to resolve detail in the object that is being imaged. Resolving power is the ability of an optical instrument to resolve or separate the image of two nearby point objects so that they can be distinctly seen. It is equal to the reciprocal of the limit of resolution of the optical instrument.

Question 55.
What is Rayleigh’s criterion?
Answer:
The images of two point objects are just resolved when the central maximum of the diffraction pattern of one falls over the first minimum of the diffraction pattern of the other.

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Question 56.
What is polarisation?
Answer:
The phenomenon of restricting the vibrations of light (electric or magnetic field vector) to a particular direction perpendicular to the direction of wave propagation motion is called polarization of light.

Question 57.
Differentiate between polarised and unpolarised light.
Answer:

S.No. Polarised light Unpolarised light
1. Consists of waves having their electric field vibrations in a single plane normal to the direction of ray. Consists of waves having their electric field vibrations equally distributed in all directions normal to the direction of ray.
2. Asymmetrical about the ray direction Symmetrical about the ray direction
3. It is obtained from unpolarised light with the help of polarisers Produced by conventional light sources.

Question 58.
Discuss polarisation by selective absorption.
Answer:
Selective absorption is the property of a material which transmits w’aves whose electric fields vibrate in a plane parallel to a certain direction of orientation and absorbs all other waves. The polaroids or polarisers are thin commercial sheets which make use of the property of selective absorption to produce an intense beam of plane polarised light. Selective absorption is also called as dichroism.

Question 59.
What are polariser and analyser?
Answer:
Polariser:
The Polaroid which plane polarises the unpolarised light passing through it is called a polariser.
Analyser:
The polaroid which is used to examine whether a beam of light is polarised or not is called an analyser.

Question 60.
What are plane polarised, unpolarised and partially polarised light?
Answer:
Plane polarised:
If the vibrations of a wave are present in only one direction in a plane perpendicular to the direction of propagation of the wave is said to be polarised or plane polarised light.

Unpolarised:
A transverse wave which has vibrations in all directions in a plane perpendicular to the direction of propagation is said to be unpolarised light.

Partially polarised light:
If the intensity of light varies between maximum and minimum for every’ rotation of 90° of the analyser, the light is said to be partially polarised light.

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Question 61.
State and obtain Malus’ law.
Answer:
When a beam of plane polarised light of intensity I0 is incident on an analyser, the light transmitted of intensity I from the analyser varies directly as the square of the cosine of the angle 0 between the transmission axis of polariser and analyser. This is known as Malus’ law. I = I0 cos2θ

Question 62.
List the uses of polaroids.
Answer:
Uses of polaroids:

  1. Polaroids are used in goggles and cameras to avoid glare of light.
  2. Polaroids are useful in three dimensional motion pictures i.e., in holography.
  3. Polaroids are used to improve contrast in old oil paintings.
  4. Polaroids are used in optical stress analysis.
  5. Polaroids are used as window glasses to control the intensity of incoming light.

Question 63.
State Brewster’s law.
Answer:
The law states that the tangent of the polarising angle for a transparent medium is equal to its refractive index, tan i = n. This relation is known as Brewster’s law.

Question 64.
What is angle of polarisation and obtain the equation for angle of polarisation.
Answer:
The angle of incidence at which a beam of unpolarised light falling on a transparent surface is reflected as a beam of plane polarised light is called polarising angle or Brewster’s angle. It is denoted by ip
ip = 90° – Rp

Question 65.
Discuss about pile of plates.
Answer:
The phenomenon of polarisation by reflection is used in the construction of pile of plates. It consists of a number of glass plates placed one over the other. The plates are inclined at an angle of 33.7° (90° – 56.3°) to the axis of the tube. A beam of unpolarised light is allowed to fall on the pile of plates along the axis of the tube. So, the angle of incidence of light will be at 56.3° which is the polarising angle for glass.

The vibrations perpendicular to the plane of incidence are reflected at each surface and those parallel to it are transmitted. The larger the number of surfaces, the greater is the intensity of the reflected plane polarised light. The pile of plates is used as a polarizer and also as an analyser.

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Question 66.
What is double refraction?
Answer:
When a ray of unpolarised light is incident on a calcite crystal, two refracted rays are produced. Hence, two images of a single object are formed. This phenomenon is called double refraction.

Question 67.
Mention the types of optically active crystals with example.
Answer:
Types of optically active crystals:
Uniaxial crystals:
Crystals like calcite, quartz, tourmaline and ice having only one optic axis are called uniaxial crystals.

Biaxial crystals:
Crystals like mica, topaz, selenite and aragonite having two optic axes are called biaxial crystals.

Question 68.
Discuss about Nicol prism.
Answer:
Nicol prism is an optical device incorporated in optical instruments both for producing and analysing plane polarised light. The construction of a Nicol prism is based on the phenomenon of Double refraction. One of the most common forms of the Nicol prism is made by taking a calcite crystal which is a double refracting crystal with its length three times its breadth.

It is cut into two halves along the diagonal so that their face angles are 72° and 108°. The two halves are joined i together by a layer of Canada balsam, a transparent cement.

Question 69.
How is polarisation of light obtained by scattering of light?
Answer:
The light from a clear blue portion of the sky shows a rise and fall of intensity when viewed through a polaroid which is rotated. This is because of sunlight, which has changed its I direction (having been scattered) on encountering the molecules of the earth’s atmosphere. The electric field of light interact with the electrons present in the air molecules.

Under the influence of the electric field of the incident wave the electrons in the molecules acquire components of motion in both these directions. We have an observer looking at 90° to the direction of the sun. Clearly, charges accelerating parallel do not radiate energy towards this observer since their acceleration has no transverse component. The radiation scattered by the molecule is therefore polarized perpendicular to the plane.

Question 70.
Discuss about simple microscope and obtain the equations for magnification for near point focusing and normal focusing.
Answer:
A simple microscope is a single magnifying (converging) lens of small focal length. The idea is to get an erect, magnified and virtual image of the object. For this the object is placed between F and P on one side of the lens and viewed from other side of the lens. There are two magnifications to be discussed for two kinds of focusing.

  1. Magnification in near point focusing m = 1 + \(\frac { D }{ f }\)
  2. Magnification in normal focusing (angular magnification), m = \(\frac { D }{ f }\)

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Question 71.
What are near point and normal focusing?
Answer:

  • Near point focusing:
    The image is formed at near point, i.e. 25 cm for normal eye. This distance is also called as least distance D of distinct vision. In this position, the eye feels comfortable but there is little strain on the eye.
  • Normal focusing:
    The image is formed at infinity. In this position the eye is most relaxed to view the image.

Question 72.
Why is oil immersed objective preferred in a microscope?
Answer:
It is best to use an oil-immersed objective at high magnification as the oil compensates for short focal lengths associated with larger magnifications.

Question 73.
What are the advantages and disadvantages of using a reflecting telescope?
Answer:
Advantages:

  • The main advantage is reflector telescope can escape from chromatic aberration because wavelength does not effect reflection.
  • The primary mirror is very stable because it is located at the back of the telescope and can be support in the back.
  • More cost effective than refractor of similar size.
  • Easier to make a high quality mirror than lens because mirror need to only concern with one side of the curvature.

Disadvantages:

  • Optical misalignment can occur quite easily.
  • Require frequent cleaning because the inside is expose to the atmosphere.
  • Secondary mirror can cause diffraction of original incoming light rays causing the “Christmas star effect” where a bright object have spikes.

Question 74.
What is the use of an erecting lens in a terrestrial telescope?
Answer:
A terrestrial telescope has an additional erecting lens to make the final image erect.

Question 75.
What is the use of collimator?
Answer:
The collimator is an arrangement to produce a parallel beam of light.

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Question 76.
What are the uses of spectrometer?
Answer:
The spectrometer is an optical instrument used to study the spectra of different sources of light and to measure the refractive indices of materials.

Question 77.
What is myopia? What is its remedy?
Answer:
Myopia (or) short sightedness:
It is a vision defect in which a person can see nearby objects clearly but cannot see the distant objects clearly beyond a certain point.

Remedy (correction):
A myopia eye is corrected by using a concave lens of focal length equal to the distance of the far point F from the eye.

Question 78.
What is hypermetropia? What is its remedy?
Answer:
Hypermetropia (or) Long sightedness: It is a vision defect in which a person can see the distant objects clearly but cannot see the nearby objects clearly.
Remedy (correction): A hypermetropic eye is corrected by using a convex lens of suitable focal length.

Question 79.
What is presbyopia?
Answer:
This defect is similar to hypermetropia i.e., a person having this defect cannot see nearby objects distinctly, but can see distant objects without any difficulty. This defect occurs in elderly persons (aged persons).

Question 80.
What is astigmatism?
Answer:
Astigmatism is the defect arising due to different curvatures along different planes in the eye lens. Astigmatic person cannot see all the directions equally well. The defect due to astigmatism is more serious than myopia and hyperopia.

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Samacheer Kalvi 12th Physics Optics Long Answer Questions

Question 1.
Derive the mirror equation and the equation for lateral magnification.
Answer:
The mirror equation:
1. The mirror equation establishes a relation among object distance u, image distance v and focal length/for a spherical mirror. An object AB is considered on the principal axis of a concave mirror beyond the center of curvature C’.
2. Let us consider three paraxial rays from point B on the object.
3. The first paraxial ray BD travelling parallel to principal axis is incident on the concave mirror at D, close to the pole P. After reflection the ray passes through the focus F. The second paraxial ray BP incident at the pole P is reflected along PBThe third paraxial ray BC passing through centre of curvature C, falls normally on the mirror at E is reflected back along the same path.
4. The three reflected rays intersect at the point B’. A perpendicular drawn as A’ B’ to the principal axis is the real, inverted image of the object AB.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-14
As per law of reflection, the angle of incidence ∠BPA is equal to the angle of reflection ∠B’PA’. The triangles ∆BPA and ∆B’PA’ are similar. Thus, from the rule of similar triangles,
\(\frac { A’B’ }{ AB }\) = \(\frac { PA’ }{ PA }\) …….. (1)
The other set of similar triangles are, ADPF and A BA.’ F. (PD is almost a straight vertical line)
\(\frac { A’B’ }{ PD }\) = \(\frac { AF’ }{ PF }\)
As, the distances PD = AB the above equation becomes,
\(\frac { A’B’ }{ AB }\) = \(\frac { AF’ }{ PF }\) ……… (2)
From equations (1) and (2) we can write,
\(\frac { PA’ }{ PA }\) = \(\frac { AF’ }{ PF }\)
As, A’ F = PA’ – PF, the above equation becomes,
\(\frac {PA’ }{ PA }\) = \(\frac { PA’-PF’P }{ PF }\) ……….. (3)
We can apply the sign conventions for the various distances in the above equation.
PA = – u, PA’ = -v, PF = -f
All the three distances are negative as per sign convention, because they are measured to the left of the pole. Now, the equation (3) becomes,
\(\frac { -v }{ -u }\) = \(\frac { -v\left( -f \right) }{ -f } \)
On further simplification,
\(\frac { -v }{ -u }\) = \(\frac { v-f }{ f }\); \(\frac { v }{ u }\)=\(\frac { v }{ f }\)=1
Dividing either side with v,
\(\frac { 1 }{ u }\)=\(\frac { 1 }{ f }\)=\(\frac { 1 }{ v }\)
After rearranging,
\(\frac { 1 }{ v }\)+\(\frac { 1 }{ u }\)=\(\frac { 1 }{ f }\) ……… (4)
The above equation (4) is called mirror equation.

Lateral magnification in spherical mirrors:
The lateral or transverse magnification is defined as the ratio of the height of the image to the height of the object. The height of the object and image are measured perpendicular to the principal axis.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics q1
m=\(\frac { h’ }{ h }\) ……..(5)
Applying proper sign conventions for equation (1),
\(\frac { A’B’ }{ AB }\) = \(\frac { PA’ }{ PA }\)
A’B’ = -h, AB = h, PA’ = -v, PA = -u
\(\frac { -h’ }{ h }\)=\(\frac { -v }{ -u }\)
On simplifying we get,
m=\(\frac { h’ }{ h }\)=-\(\frac { v }{ u }\) ………. (6)
Using mirror equation, we can further write the magnification as,
m=\(\frac { h’ }{ h }\)–\(\frac { f-v }{ f }\)=\(\frac { f }{ f-u }\) ………… (7)

Question 2.
Describe the Fizeau’s method to determine speed of light.
Answer:
Fizeau’s method to determine speed of light:
Apparatus:
The light from the source S was first allowed to fall on a partially silvered glass plate G kept at an angle of 45° to the incident light from the source. The light then was allowed to pass through a rotating toothed-wheel with N teeth and N cuts of equal widths whose speed of rotation could be varied through an external mechanism. The light passing through one cut in the wheel will get reflected by a mirror M kept at a long distance d, about 8 km from the toothed w’heel. If the toothed wheel was not rotating, the reflected light from the mirror would again pass through the same cut and reach the eyes of the observer through the partially silvered glass plate.

Working:
The angular speed of rotation of the toothed wheel was increased from zero to a value to until light passing through one cut would completely be blocked by the adjacent tooth. This is ensured by the disappearance of light while looking through the partially silvered glass plate.

Expression for speed of light:
The speed of light in air v is equal to the ratio of the distance the light travelled from the toothed wheel to the mirror and back 2d to the time taken t.
v = \(\frac { 2d }{ t }\) c ….. (1)
The distance d is a known value from the arrangement. The time taken t for the light to travel the distance to and fro is calculated from the angular speed co of the toothed wheel.
The angular speed ω of the toothed wheel when the light disappeared for the first time is,
ω = \(\frac { θ }{ t }\) ……. (2)
Here, 0 is the angle between the tooth and the slot which is rotated by the toothed wheel within that time t.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-15
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics q2
θ = \(\frac { 2π }{ 2N }\) = \(\frac { π }{ N }\)
Substituting for 0 in the equation (2) for ,
ω = \(\frac { \pi /n }{ 2 } \) = \(\frac { π }{ Nt }\)
Rewriting the above equation for t,
t = \(\frac { π }{ Nω }\) ……. (3)
Substituting t from equation (3) in equation (1), v = \(\frac { 2d }{ \pi /N\omega } \)
After rearranging,
v = \(\frac { 2dNω }{ π }\) …….. (4)
Fizeau had some difficulty to visually estimate the minimum intensity of the light when blocked by the adjacent tooth, and his value for speed of light was very close to the actual value.

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Question 3.
Obtain the equation for radius of illumination (or) Snell’s window.
Answer:
The radius of Snell’s window can be deduced with the illustration as shown in figure. Light is seen from a point A at a depth d. The Snell’s law in product form, equation n2 sin i = n2 sin r for the refraction happening at the point B on the boundary between the two media is,
n1 sin ic = n2 sin 90° ……. (1)
n1 sin ic= n2 ∵ sin 90° = 1
sin ic = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \) …….. (2)
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-16
From the right angle triangle ∆ABC,
sin ic = \(\frac { R }{ \sqrt { { d }^{ 2 }+{ R }^{ 2 } } } \) ……. (3)
Equating the above two equation (3) and equation (2).
\(\frac { R }{ \sqrt { { d }^{ 2 }+{ R }^{ 2 } } } \) = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \)
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-17
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-18
If the rarer medium outside is air, then, n2 = 1, and we can take n1 = n
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-19

Question 4.
Derive the equation for acceptance angle and numerical aperture, of optical fiber. Acceptance angle in optical fibre:
Answer:
To ensure the critical angle incidence in the core-cladding boundary inside the optical fibre, the light should be incident at a certain angle at the end of the optical fiber while entering in to it. This angle is called acceptance angle. It depends on the refractive indices of the core n1, cladding n2 and the outer medium n3. Assume the light is incident at an angle called acceptance angle i at the outer medium and core boundary at A.
The Snell’s law in the product form, equation for this refraction at the point A.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-20
n3 sin ia = n1 sin ra …(1)
To have the total internal reflection inside optical fibre, the angle of incidence at the core-cladding interface at B should be atleast critical angle ic. Snell’s law in the product form, equation for the refraction at point B is,
n1 sin ic = n2 sin 90° …(2)
n1 sin ic= n2 ∵ sin 90° =1
∴sin ic= \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \) …(3)
From the right angle triangle ∆ABC,
ic = 90°-ra
Now, equation (3) becomes, sin (90° – ra) = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \)
Using trigonometry’, cos ra = ra = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \) …….. (4)
sin ra = \(\sqrt { 1-{ cos }^{ 2 }{ r }_{ a } } \)
Substituting for cos ra
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-21
If outer medium is air, then n3 = 1. The acceptance angle ia becomes,
ra=sin-1\(\left( \sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \right) \) …….. (9)
Light can have any angle of incidence from 0 to ia with the normal at the end of the optical fibre forming a conical shape called acceptance cone. In the equation (6), the term (n3 sin ia) is called numerical aperture NA of the optical fibre.
NA =n3 sin ia \(\left( \sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \right) \) …….. (10)
If outer medium is air, then n3 = 1. The numerical aperture NA becomes,
NA = sin ia = \(\left( \sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \right) \) …….. (11)

Question 5.
Obtain the equation for lateral displacement of light passing through a glass slab.
Answer:
Consider a glass slab of thickness t and refractive index n is kept in air medium. The path of the light is ABCD and the refractions occur at two points B and C in the glass slab. The angles of incidence i and refraction r are measured with respect to the normal N1and N2 at the two points B and C respectively. The lateral displacement L is the perpendicular distance CE drawn between the path of light and the undeviated path of light at point C. In the right angle triangle ∆BCE,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-22
sin(i – r) = \(\frac { L }{ BC }\); BC= \(\frac { L }{ sin(i-r) } \) …… (1)
In the right angle triangle ∆BCF,
cos(r) = \(\frac { t }{ BC }\) ; BC= \(\frac { t }{ cos(r) } \) …… (2)
Equating equations (1) and (2)
\(\frac { L }{ sin(i-r) } \) = \(\frac { t }{ cos(r) } \)
After rearranging,
L = t\(\left( \frac { sin(i-r) }{ cos(r) } \right) \)
Lateral displacement depends upon the thickness of the slab. Thicker the slab, greater will be the lateral displacement. Greater the angle of incident, larger will be the lateral displacement.

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Question 6.
Derive the equation for refraction at single spherical surface.
Answer:
Equation for refraction at single spherical surface:
Let us consider two transparent media having refractive indices n1and n2 are separated by a spherical surface. Let C be the centre of curvature of the spherical surface. Let a point object O be in the medium n1. The line OC cuts the spherical surface at the pole P of the surface. As the rays considered are paraxial rays, the perpendicular dropped for the point of incidence to the principal axis is very close to the pole or passes through the pole itself.
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Light from O falls on the refracting surface at N. The normal drawn at the point of incidence passes through the centre of curvature C. As n2 > n1, light in the denser medium deviates towards the normal and meets the principal axis at I where the image is formed.
Snell’s law in product form for the refraction at the point N could be written as,
n1 sin i = n2 sin r …(1)
As the angles are small, sin of the angle could be approximated to the angle itself.
n1 i = n2 r …(2)
Let the angles,
∠NOP = α, ∠NCP = β, ∠NIP = γ
tan α = \(\frac { PN }{ PO }\);tan β = \(\frac { PN }{ PC }\);tan γ = \(\frac { PN }{ PI }\)
As these angles are small, tan of the angle could be approximated to the angle itself.
α = \(\frac { PN }{ PO }\); β = \(\frac { PN }{ PC }\); γ = \(\frac { PN }{ PI }\) …….. (3)
For the triangle, ∆ONC,
I = α + β …….. (4)
For the triangle, ∆INC,
β = r + γ (or) r = β – γ ……… (5)
Substituting for I and r from equations (4) and (5) in the equation (2).
n1 (α + β) = n2 (β – γ)
Rearranging,
n1 α + n2γ = (n2 – n1
Substituting for α, β and γ from equation (3)
n1 = (\(\frac { PN }{ PO }\)) + n2 = (\(\frac { PN }{ PI }\)) = (n2-n1) (\(\frac { PN }{ PC }\))
Further simplifying by cancelling PN,
\(\frac { { n }_{ 1 } }{ PO } \) + \(\frac { { n }_{ 2 } }{ PI } \) = \(\frac { { n }_{ 2 }-{ n }_{ 1 } }{ PC } \)
Following sign conventions, PO = -u, PI = +v and PC = +R in equation (6),
\(\frac { { n }_{ 2 } }{ -v } \) – \(\frac { { n }_{ 1 } }{ u } \) = \(\frac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \)
After rearranging, finally we get,
\(\frac { { n }_{ 2 } }{ v } \) – \(\frac { { n }_{ 1 } }{ u } \) = \(\frac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \) ……… (7)
Equation (7) gives the relation among the object distance n, image distance v, refractive indices of the two media (n1 and n2) and the radius of curvature R of the spherical surface. It holds for any spherical surface.
If the first medium is air then, n1 = 1 and the second medium is taken just as n2 = n, then the equation is reduced to,
\(\frac { n }{ v}\)–\(\frac { 1 }{ u}\) = \(\frac { \left( { n }-1 \right) }{ R } \) ………(8)

Question 7.
Obtain lens maker’s formula and mention its significance.
Answer:
Lens maker’s formula and lens equation:
Let us consider a thin lens made up of a medium of refractive index n2, is placed in a medium of refractive index n1. Let R1 and R2 be the radii of curvature of two spherical surfaces (1) and (2) respectively and P be the pole. Consider a point object O on the principal axis. The ray which falls very close to P, after refraction at the surface (1) forms image at I’. Before it does so, it is again refracted by the surface (2). Therefore the final image is formed at I. The general equation for the refraction at a spherical surface is given by
\(\frac { { n }_{ 2 } }{ v } \) – \(\frac { { n }_{ 1 } }{ u } \) = \(\frac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \)
For the refracting surface (1), the light goes from n1ton2.
\(\frac { { n }_{ 2 } }{ v’ } \) – \(\frac { { n }_{ 1 } }{ u } \) = \(\frac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R }_{ 1 } \) …….. (1)
For the refracting surface (2), the light goes from medium n2ton1.
\(\frac { { n }_{ 1 } }{ v } \) – \(\frac { { n }_{ 2 } }{ v’ } \) = \(\frac { \left( { n }_{ 1 }-{ n }_{ 2 } \right) }{ R }_{ 2 } \) …….. (2)
Adding the above two equation (1) and (2)
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-24
\(\frac { { n }_{ 1 } }{ v } \) – \(\frac { { n }_{ 1 } }{ u } \) = (n2 – n1) \(\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
Furhter simplifying and rearranging,
\(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\) = \(\left(\frac{n_{2}-n_{1}}{n_{1}}\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
\(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\) = \(\left(\frac{n_{2}}{n_{1}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) ……….. (3)
If the object is at infinity, the image is formed at the focus of the lens. Thus, for u = ∞, v= f. Then the equation becomes.
\(\frac { 1 }{ f }\)–\(\frac { 1 }{ ∞ }\) = \(\left(\frac{n_{2}}{n_{1}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
\(\frac { 1 }{ f }\)=\(\left(\frac{n_{2}}{n_{1}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) ………… (4)
If the refractive index of the lens is and it is placed in air, then n2 = n and n1 = 1 equation (4) becomes,
\(\frac { 1 }{ f }\) = (n-1) \(\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) ………. (5)
The above equation is called the lens maker’s formula, because it tells the lens manufactures what curvature is needed to make a lens of desired focal length with a material of particular refractive index. This formula holds good also for a concave lens. By comparing the equations (3) and (4) we can write,
\(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\) =\(\frac { 1 }{ f }\) …… (6)
This equation is known as lens equation which relates the object distance u and image distance v with the focal length f of the lens. This formula holds good for a any type of lens.

Question 8.
Derive the equation for thin lens and obtain its magnification.
Answer:
Lateral magnification in thin lens:
Let us consider an object 00′ of height h1 placed on the principal axis with its height perpendicular to the principal axis. The ray Op passing through the pole of the lens goes undeviated. The inverted real image II’ formed has a height h2.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-25
The lateral or transverse magnification m is defined as the ratio of the height of the image to that of the object.
m = \(\frac { II’ }{ OO’ }\) …….. (1)
From the two similar triangles ∆ POO’ and ∆ PII’, we can write,
\(\frac { H’ }{ OO’ }\) = \(\frac { PI }{ PO }\) ……..(2)
Applying sign convention,
\(\frac{-h_{2}}{h_{1}}\) = \(\frac { v }{ -u}\)
Substituting this in the equation (1) for magnification,
m = \(\frac{-h_{2}}{h_{1}}\) = \(\frac { v }{ -u}\)
After rearranging,
m = \(\frac{h_{2}}{h_{1}}\) = \(\frac { v }{ u}\) …….. (3)
The magnification is negative for real image and positive for virtual image. In the case of a concave lens, the magnification is always positive and less than one. We can also have the equations for magnification by combining the lens equation with the formula for magnification
as,
m = \(\frac{h_{2}}{h_{1}}=\frac{f}{f+u}\) (or) m = \(\frac{h_{2}}{h_{1}}=\frac{f-v}{f}\) ……… (4)

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Question 9.
Derive the equation for effective focal length for lenses in out of contact.
Answer:
Consider a two lenses of focal length f1 and f2 arranged coaxially but separated by a distance d can be considered. For a parallel ray that falls on the arrangement, the two lenses produce deviations δ1 and δ2 respectively and The net deviation δ is,
δ = δ1 + δ2 ……. (1)
From Angle of deviation in lens equation, δ = \(\frac { h }{ f }\)
δ1 = \(\frac{h_{1}}{f_{1}}\); δ2 = \(\frac{h_{2}}{f_{2}}\) δ = \(\frac{h_{1}}{f}\) ….. (2)
The equation (1) becomes,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-26
\(\frac{h_{1}}{f}\) = \(\frac{h_{1}}{f_{1}}\) + \(\frac{h_{2}}{f_{2}}\) ……. (3)
From the geometry,
h2 – h1 = P2C = CG
h2 – h1 = BG tan δ1 ≈ BG δ1
h2 – h1 = d\(\frac{h_{1}}{f_{1}}\)
h2 = h1 + d\(\frac{h_{1}}{f_{1}}\) …….. (4)
Substituting the above equation in Equation (3)
\(\frac{h_{1}}{f}=\frac{h_{1}}{f_{1}}+\frac{h_{1}}{f_{2}}+\frac{h_{1} d}{f_{1} f_{2}}\)
On further simplification,
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}+\frac{d}{f_{1} f_{2}}\) …….. (5)
The above equation could be used to find the equivalent focal length.

Question 10.
Derive the equation for angle of deviation produced by a prism and thus obtain the equation for refractive index of material of the prism.
Answer:
Angie of deviation produced by prism:
Let light ray PQ is incident on one of the refracting faces of the prism. The angles of incidence and refraction at the first face AB are i1 and r1. The path of the light inside the prism is QR. The angle of incidence and refraction at the second face AC is r2 and i2 respectively. RS is the ray emerging from p the second face. Angle i2 is also called angle of emergence. The angle between the direction of the incident ray PQ and the emergent ray RS is called the angle of deviation d.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-27
The two normals drawn at the point of incidence Q and emergence R are QN and RN. They meet at point N. The incident ray and the emergent ray meet at a point M.
The deviation d1 at the surface AB is,
angle ∠RQM = d1 = i1 – r1 …(1)
The deviation d2 at the surface AC is,
angle ∠QRM = d2 = i2 – r2 …(2)
Total angle of deviation d produced is,
d = d1 + d2 …(3)
Substituting for d1 and d2,
d=(i1 – r1) + (i2 – r2)
After rearranging,
d = (i1 – r1) + (i2 – r2) …(4)
In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°.
∠A + ∠QNR = 180° …(5)
From the triangle ∆QNR,
r1+ r2 ∠QNR = 180° …(6)
Comparing these two equations (5) and (6) we get,
r1+ r2 = A …(7)
Substituting this in equation (4) for angle of deviation,
d= i1+ i2 -A …(8)
Thus, the angle of deviation depends on the angle of incidence angle of emergence and the angle for the prism.
Refractive index of the material of the prism:
At minimum deviation,
i1 = i2 = i and r1 = r2 = r
Now, the equation (8) becomes,
D = i1 + i2-A (or) i = \(\frac { \left( A+D \right) }{ 2 } \)
The equation (7) becomes,
r1 + r2 = A ⇒ 2r = A (or) r = \(\frac { A }{ 2 }\)
Substituting i and r in Snell’s law,
n = \(\frac { sin i }{ sin r }\)
n = \(\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
The above equation is used to find the refractive index of the material of the prism.

Question 11.
What is dispersion? Obtain the equation for dispersive power of a medium.
Answer:
Dispersion. Dispersion is splitting of white light into its constituent colours.
Dispersive Power:
Consider a beam of white light passes through a prism; It gets dispersed into its constituent colours. Let δV, δR are the angles of deviation for violet and red light. Let nV and nR are the refractive indices for the violet and red light respectively.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-28
The refractive index of the material of a prism is given by the equation
n = \(\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
Here A is the angle of the prism and D is the angle of minimum deviation. If the angle of prism is small of the order of 10°, the prism is said to be a small angle prism. When rays of light pass through such prisms, the angle of deviation also becomes small. If A be the angle of a small angle prism and 5 the angle of deviation then the prism formula becomes.
n = \(\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin \left(\frac{A}{2}\right)}\) ……… (1)
For small angles of A and δm,
\(\sin \left(\frac{\mathrm{A}+\delta}{2}\right) \approx\left(\frac{\mathrm{A}+\delta}{2}\right)\) ……. (2)
\(\sin \left(\frac{\mathrm{A}}{2}\right) \approx\left(\frac{\mathrm{A}}{2}\right)\) …… (3)
∴ n = \(\frac{\left(\frac{A+\delta}{2}\right)}{\left(\frac{A}{2}\right)}=\frac{A+\delta}{A}=1+\frac{\delta}{A}\)
Further simplifying, \(\frac { δ }{ A }\) = n – 1
δ = (n – 1) A ……. (4)
When white light enters the prism, the deviation is different for different colours. Thus, the refractive index is also different for different colours.
For Violet light, δV = (nV – 1)A …(5)
For Red light, δR = (nR – 1) …(6)
As, angle of deviation for violet colour δV is greater than angle of deviation for red colour δR, the refractive index for violet colour nV is greater than the refractive index for red colour nR. Subtracting δV from δR we get,
δV – δR = (nV – nR)A ….. (7)
The term (δV – δR) is the angular separation between the two extreme colours (violet and red) in the spectrum is called the angular dispersion. Clearly, the angular dispersion produced by a prism depends upon.

  1. Angle of the prism
  2. Nature of the material of the prism.

If we take 8 is the angle of deviation for any middly ray (green or yellow) and n the corresponding refractive index. Then,
8 = (n – 1) A … (8)
Dispersive power (ω) is the ability of the material of the prism to cause dispersion. It is defined as the ratio of the angular dispersion for the extreme colours to the deviation for any mean colour. Dispersive power (ω),
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics q11
Substituting for (δVR)and (δ),
ω = \(\frac{\left(n_{\mathrm{V}}-n_{\mathrm{R}}\right)}{(n-1)}\) ……. (10)
Dispersive power is a dimensionless quality. It has no unit. Dispersive power is always positive. The dispersive power of a prism depends only on the nature of material of the prism and it is independent of the angle of the prism.

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Question 12.
Prove laws of reflection using Huygens’ principle.
Answer:
Proof for laws of reflection using Huygens’ Principle:
Let us consider a parallel beam of light, incident on a reflecting plane surface such as a plane mirror XY. The incident wavefront is AB and the reflected wavefront is A’B’ in the same medium. These wavefronts are perpendicular to the incident rays L, M and reflected rays L’, M’ respectively.

By the time point A of the incident wavefront touches the reflecting surface, the point B is yet to travel a distance BB’ to touch the reflecting surface a B’. When the point B falls on the reflecting surface at B’ , the point A would have reached A’. This is applicable to all the points on the wavefront.

Thus, the reflected wavefront A’B’ emanates as a plane wavefront. The two normals N and N’ are considered at the points where the rays L and M fall on the reflecting surface. As reflection happens in the same medium, the speed of light is same before and after the reflection. Hence, the time taken for the ray to travel from B to B’ is the same as the time taken for the ray to travel from A to A’. Thus, the distance BB’ is equal to the distance AA’; (AA’= BB’).
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(i) The incident rays, the reflected rays and the normal are in the same plane.
(ii) Angle of incidence,
∠i = ∠NAL = 90° – ∠NAB = ∠BAB’
Angle of reflection,
∠r = ∠N’B’M’ = 90° – ∠N’B’A’ = ∠A’B’A
For the two right angle triangles, A ABB’ and A B’A’A, the right angles, ∠B and ∠A’ are equal, (∠B and ∠A’ = 90°); the two sides, AA’ and BB’ are equal, (AA’ = BB’); the side AB’ is the common. Thus, the two triangles are congruent. As per the property of congruency, the two angles, ∠BAB’ and ∠A’B’A must also be equal. i = r
Hence, the laws of reflection are proved.

Question 13.
Prove laws of refraction using Huygens’ principle.
Answer:
Let us consider a parallel beam of light is incident on a refracting plane surface XY such as a glass surface. The incident wavefront AB is in rarer medium (1) and the refracted wavefront A’B’ is in denser medium (2). These wavefronts are perpendicular to the incident rays L, M and refracted rays L’, M’ respectively. By the time the point A of the incident wavefront touches the refracting surface, the point B is yet to travel a distance BB’ to touch the refracting surface at B’
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-30
When the point B falls on the refracting surface at B’, the point A’ would have reached A in the other medium. This is applicable to all the points on the wavefront. Thus, the refracted wavefront A’B’ emanates as a plane wavefront.
The two normals N and N’ are considered at the points where the rays L and M fall on the refracting surface. As refraction happens from rarer medium (1) to denser medium (2), the speed of light is v1 and v2 before and after refraction and v1 is greater than v2 (v1 > v2). But, the time taken t for the ray to travel from B to B’ is the same as the time taken for the ray to
travel from A to A’.
t = \(\frac { BB’ }{ { v }_{ 1 } } \) = \(\frac { AA’ }{ { v }_{ 2 } } \) (or) \(\frac { BB’ }{ AA’ }\) = \(\frac { { v }_{ 1 } }{ { v }_{ 2 } } \)
(i) The incident rays, the refracted rays and the normal are in the same plane.
(ii) Angle of incidence,
i = ∠NAL = 90° – ∠NAB = ∠BAB’
Angle of refraction,
r =∠N’B’M’ = 90° – ∠N’B’A’ = ∠A’B’A’
For the two right angle triangles ∆ ABB’ and ∆ B’A’A,
\(\frac { sin i }{ sin r }\) = \(\frac{B B^{\prime} / A B^{\prime}}{A A^{\prime} / A B^{\prime}}\) =\(\frac { BB’ }{ AA’ }\) =\(\frac{v_{1}}{v_{2}}=\frac{c / v_{2}}{c / v_{1}}\)
Here, c is speed of light in vacuum. The ratio c/v is the constant, called refractive index of the medium. The refractive index of medium (1) is, c/v1 = n1 and that of medium (2) is, c/v2 = n2.
\(\frac { sin i }{ sin r }\) = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \)
In product form,
n1 sin i = n2 sin r
Hence, the laws of refraction are proved.

Question 14.
Obtain the equation for resultant intensify’ due to interference of light.
Answer:
Let us consider two light waves from the two sources S1 and S2 meeting at a point P. The wave from S1 at an instant t at P is,
y1 = a1 sin ωt … (1)
The wave form S2 at an instant t at P is,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-31
y2 = a2 sin (ωt + Φ) … (2)
The two waves have different amplitudes a1 and a2, same angular frequency ω, and a phase difference of Φ between them. The resultant displacement will be given by,
y = y1 + y2 = a1 sin ωt + a1 sin2 (ωt + Φ) … (3)
The simplification of the above equation by using trigonometric identities gives the equation,
y = A sin (ωt + θ) … (4)
Where, A = \(\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \phi}\) …. (5)
θ = tan-1 \(\frac{a_{2} \sin \phi}{a_{1}+a_{2} \cos \phi}\) …. (6)
The resultant amplitude is maximum,
Amax = \(\sqrt{\left(a_{1}+a_{2}\right)^{2}}\) ; when Φ = 0, ±2π, ±4π …., …… (7)
The resultant amplitude is minimum,
Amax = \(\sqrt{\left(a_{1}-a_{2}\right)^{2}}\) ; when Φ = ±π, ±3,π ±5π …., …… (8)
The intensity of light is proportional to square of amplitude,
I ∝ A2 ….. (9)
Now, equation (5) becomes,
I ∝ I1+I 2+2 \(\sqrt { { I }_{ 1 }{ I }_{ 2 } } \) cos Φ …… (10)
In equation (10) if the phase difference, Φ = 0, ±2π, ±4π …., it corresponds to the condition for maximum intensity of light called as constructive interference. The resultant maximum intensity is,
Imax ∝ ( a1 + a2)2 ∝ I1 + I2 +2 \(\sqrt { { I }_{ 1 }{ I }_{ 2 } } \) …… (11)
In equation (10) if the phase difference, Φ = ±π, ±3,π ±5π …., it corresponds to the condition for minimum intensity of light called destructive interference. The resultant minimum intensity is,
Imax ∝( a1 – a2)2 ∝ I1 + I2 -2 \(\sqrt { { I }_{ 1 }{ I }_{ 2 } } \) ……. (12)

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Question 15.
Explain the Young’s double slit experimental setup and obtain the equation for path difference.
Answer:
I Experimental setup:

1. Wavefronts from S1 and S2 spread out and overlapping takes place to the right side of double slit. When a screen is placed at a distance of about 1 meter s from the slits, alternate bright and dark fringes which are equally spaced appear on the screen. These are called interference fringes or bands.
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2. Using an eyepiece the fringes can be seen directly. At the center point O on the screen, waves from S1 and S2 travel equal distances and arrive in-phase. These two waves constructively interfere and bright fringe is observed at O. This is called central bright fringe.

3. The fringes disappear and there is uniform illumination on the screen when one of the slits is covered. This shows clearly that the bands are due to interference.

II Equation for path difference:

1. Let d be the distance between the double slits S1 and S2 which act as coherent sources of wavelength λ. A screen is placed parallel to the double slit at a distance D from it. The mid-point of S1 and S2 is C and the midpoint of the screen O is equidistant from S1 and S2. P is any point at a distance y from O.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-33

2. The waves from S1 and S2 meet at P either inphase or out-of-phase depending upon the path difference between the two waves.
3. The path difference δ between the light waves from S1 and S2 to the point P is,
δ = S2 P – S1
4. A perpendicular is dropped from the point S1 to the line S2 P at M to find the path difference more precisely.
δ = S2 P – MP = S2 M
The angular position of the point P from C is θ. ∠OCP = θ.
From the geometry, the angles ∠OCP and ∠S2 S1M are equal. ∠OCP = ∠S2 S1 M = θ
In right angle triangle ∆S1 S2 M, the path difference, S2 M = d sin θ
δ = d sin θ
If the angle 0 is small, sin θ ≈ tan θ ≈ θ. From the right angle triangle ∆OCP, tan θ = \(\frac { y }{ D }\)
The path difference, δ = \(\frac { dy }{ D }\)

Question 16.
Obtain the equation for bandwidth in Young’s double slit experiment.
Answer:
Condition for bright fringe (or) maxima
The condition for the constructive interference or the point P to be have a bright fringe is,
Path difference, δ = nλ
where, n = 0, 1, 2,. . .
∴ \(\frac { dy }{ D }\) = nλ
y = n \(\frac { λD }{ d }\) (or) yn = n \(\frac { λD }{ d }\)
This is the condition for the point P to be a bright fringe. The distance is the distance of the nth bright fringe from the point O.
Condition for dark fringe (or) minima
The condition for the destructive interference or the point P to be have a dark fringe is,
Path difference, δ = (2n – 1) \(\frac { λ }{ 2 }\)
where, n = 1, 2, 3 …
∴ \(\frac { dy }{ D }\) = (2n – 1) \(\frac { λ }{ 2 }\)
y = \(\frac { (2n-1) }{ 2 }\) \(\frac { λD }{ d }\) (or) yn = \(\frac { (2n-1) }{ 2 }\) \(\frac { λD }{ d }\)
Equation for bandwidth
The bandwidth (β) is defined as the distance between any two consecutive bright or dark fringes.
The distance between (n +1)th and nth consecutive bright fringes from O is given by,
β = y(n +1) – yn = \(\left((n+1) \frac{\lambda \mathrm{D}}{d}\right)-\left(n \frac{\lambda \mathrm{D}}{d}\right)\)
β = \(\frac { λD }{ d }\)
Similarly, the distance between (n +1)th and nth consecutive dark fringes from O is given by,
β = y(n+1) – yn = \(\left(\frac{(2(n+1)-1)}{2} \frac{\lambda D}{d}\right)-\left(\frac{(2 n-1)}{2} \frac{\lambda D}{d}\right)\)
β = \(\frac { λD }{ d }\)
The above equation show that the bright and dark fringes are of same width equally spaced on either side of central bright fringe.

Question 17.
Obtain the equations for constructive and destructive interference for transmitted and reflected waves in thin films.
Answer:
Interference in thin films:
Let us consider a thin film of transparent material of refractive index p (not to confuse with order of fringe n) and thickness d. A parallel beam of light is incident on the film at an angle i. The wave is divided into two parts at the upper surface, one is reflected and the other is refracted.

The refracted part, which enters into the film, again gets divided at the lower surface into two parts; one is transmitted out of the film and the other is reflected back in to the film. Reflected as well as refracted waves are sent by the film as multiple reflections take place inside the film. The interference is produced by both the reflected and transmitted light.
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For transmitted light:
The light transmitted may interfere to produce a resultant intensity. Let us consider the path difference between the two light waves transmitted from B and D. The two waves moved together and remained in phase up to B where splitting occured. The extra path travelled by the wave transmitted from D is the path inside the film, BC + CD.
If we approximate the incidence to be nearly normal (i = 0), then the points B and D are very close to each other. The extra distance travelled by the wave is approximately twice thickness of the film, BC + CD = 2d. As this extra path is traversed in a medium of refractive index µ, the optical path difference is, δ = 2µd.
The condition for constructive interference in transmitted ray is,
2µd = nλ
Similarly, the condition for destructive interference in transmitted ray is,
2µd = (2n – 1) \(\frac { λ }{ 2 }\)
For reflected light:
It is experimentally and theoretically proved that a wave while travelling in a rarer medium and getting reflected by a denser medium, undergoes a phase change of π. Hence, an additional path difference of \(\frac { λ }{ 2 }\) should be considered.

Let us consider the path difference between the light waves reflected by the upper surface at A and the other wave coming out at C after passing through the film. The additional path travelled by wave coming out from C is the path inside the film, AB + BC. For nearly normal incidence this distance could be approximated as, AB + BC = 2d. As this extra path is travelled in the medium of refractive index p, the optical path difference is, δ = 2µd.
The condition for constructive interference for reflected ray is,
2µd + \(\frac { λ }{ 2 }\) = nλ (or) 2µd = (2n – 1) \(\frac { λ }{ 2 }\)
The additional path difference \(\frac { λ }{ 2 }\) is due to the phase change of n in rarer to denser reflection taking place at A. The condition for destructive interference for reflected ray is,
2µd + \(\frac { λ }{ 2 }\) = (2n + 1) \(\frac { λ }{ 2 }\) (or) 2µd = nλ

Question 18.
Discuss diffraction at single slit and obtain the condition for nth minimum.
Answer:
Diffraction at single slit:
Let a parallel beam of light fall normally on a single slit AB of width. The diffracted beam falls on a screen kept at a distance. The center of the slit is C. A straight line through C perpendicular to the plane of slit meets the center of the screen at O. We would like to find the intensity at any point P on the screen. The lines joining P to the different points on the slit can be treated as parallel lines, making an angle 9 with the normal CO.

All the waves start parallel to each other from different points of the slit and interfere at point P and other points to give the resultant intensities. The point P is in the geometrically shadowed region, up to which the central maximum is spread due to diffraction. We need to give the condition for the point P to be of various minima.

The basic idea is to divide the slit into much smaller even number of parts. Then, add their contributions at P with the proper path difference to show that destructive interference takes place at that point to make it minimum. To explain maximum, the slit is divided into odd number of parts.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-35
Condition for P to be first minimum:
Let us divide the slit AB into two half’s AC and CB. Now the width of AC is (a/2). We have different points on the slit which are separated by the same width (here a/2) called corresponding points.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-36
The path difference of light waves from different corresponding points meeting at point P and interfere destructively to make it first minimum. The path difference 8 between waves from these corresponding points
is, δ = \(\frac { a }{ 2 }\) sin θ
The condition for P to be first minimum, \(\frac { a }{ 2 }\) sin θ = \(\frac { λ }{ 2 }\)
a sin θ = λ (first minimum) ….. (1)
Condition for P to be first minimum:
Let us divide the slit AB into two half’s AC and CB. Now the width of AC is (a/2). We have different points on the slit which are separated by the same width (here a/2) called corresponding points.
The path difference of light waves from different corresponding points meeting at point P and interfere destructively to make it first minimum.
The path difference δ between waves from these corresponding points δ = \(\frac { a }{ 4 }\) sin θ.
The condition for P to be first minimum, \(\frac { a }{ 4 }\) sin θ = \(\frac { λ }{ 2 }\)
a sin θ = 2λ (second minimum) …(2)
Condition for P to be third order minimum:
The same way the slit is divided into six equal parts to explain the condition for P to be third
minimum is, \(\frac { a }{ 6 }\) sin θ = \(\frac { λ }{ 2 }\)
a sin θ = 3λ (third minimum) …(3)
Condition for P to be nth order minimum:
Dividing the slit into 2n number of (even number of) equal parts makes the light produced by one of the corresponding points to be cancelled by its counterpart. Thus, the condition for nth
order minimum is, \(\frac { a }{ 2n }\) sin θ = \(\frac { λ }{ 2 }\)
a sin θ = nλ (nth minimum)

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Question 19.
Discuss the diffraction at a grating and obtain the condition for the mth maximum.
Answer:
A plane transmission grating is represented by AB. Let a plane wavefront of monochromatic light with wavelength λ be incident normally on the grating. As the slits size is comparable to that of wavelength, the incident light diffracts at the grating.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-37
A diffraction pattern is obtained on the screen when the diffracted waves are focused on a screen using a convex lens. Let us consider a point P at an angle 0 with the normal drawn from the center of the grating to the screen. The path difference 5 between the diffracted waves from one pair of corresponding points is,
δ = (a + b) sin θ
This path difference is the same for any pair of corresponding points. The point P will be bright, when
δ = m λ where m = 0, 1 , 2, 3

Combining the above two equations, we get,
(a + b) sin θ = m λ
Here, m is called order of diffraction.

Condition for zero order maximum, m = 0
For (a + b) sin θ = 0, the position, θ = 0. sin θ = 0 and m = 0. This is called zero order diffraction or central maximum.

Condition for first order maximum, m = 1
If (a + b) sin θ1 = λ, the diffracted light meet at an angle θ1 to the incident direction and the first order maximum is obtained.

Condition for second order maximum, m = 2
Similarly, (a + b) sin θ2 = 2λ, forms the second order maximum at the angular position θ2.

Condition for higher order maximum
On either side of central maxima different higher orders of diffraction maxima are formed at different angular positions.
If we take, N = \(\frac { 1 }{ a+b }\)

Then, N gives the number of grating elements or rulings drawn per unit width of the grating. Normally, this number N is specified on the grating itself. Now, the equation becomes,
\(\frac { 1 }{ N }\) sin θ = mλ, (or) sin θ = Nmλ

Question 20.
Discuss the experiment to determine the wavelength of monochromatic light using diffraction grating.
Answer:
Experiment to determine the wavelength of monochromatic light:
The wavelength of a spectral line can be very accurately determined with the help of a diffraction grating and a spectrometer. Initially all the preliminary adjustments of the spectrometer are made. The slit of collimator is illuminated by a monochromatic light, whose wavelength is to be determined. The telescope is brought in line with collimator to view the image of the slit. The given plane transmission grating is then mounted on the prism table with its plane perpendicular to the incident beam of light coming from the collimator.

The telescope is turned to one side until the first order diffraction image of the slit coincides with the vertical cross wire of the eye piece. The reading of the position of the telescope is noted. Similarly the first order diffraction image on the other side is made to coincide with the vertical cross wire and corresponding reading is noted. The difference between two positions gives 2θ. Half of its value gives θ, the diffraction angle for first order maximum. The wavelength of light is calculated from the equation,
λ = \(\frac { sin θ }{ N m}\)
Here, N is the number of rulings per metre in the grating and m is the order of the diffraction image.

Question 21.
Discuss the experiment to determine the wavelength of different colours using diffraction grating.
Answer:
Determination of wavelength of different colours:
When white light is used, the diffraction pattern consists of a white central maximum and on both sides continuous coloured diffraction patterns are formed. The central maximum is white as all the colours meet here constructively with no path difference.

As θ increases, the path difference, (a + b) sin θ, passes through condition for maxima of diffraction of different orders for all colours from violet to red. It produces a spectrum of diffraction pattern from violet to red on either side of central maximum. By measuring the angle at which these colours appear for various orders of diffraction, the wavelength of different colours could be calculated using the formula,
λ = \(\frac { sin θ }{ N m}\)
Here, N is the number of rulings per metre in the grating and m is the order of the diffraction image.

Question 22.
Obtain the equation for resolving power of optical instrument.
Answer:
Resolution:
The effect of diffraction has an adverse impact in the image formation by the optical instruments such as microscope and telescope. For a single rectangular slit, the half angle θ subtended by the spread of central maximum (or position of first minimum) is given by the ‘ relation,
a sin θ = λ … (1)
Similar to a rectangular slit, when a circular aperture or opening (like a lens or the iris of our eye) forms an image of a point object, the image formed will not be a point but a diffraction pattern of concentric circles that becomes fainter while moving away from the center. These are known as Airy’s discs. The circle of central maximum has the half angular spread given by the equation,
a sin θ = 1.22λ … (2)
Here, the numerical value 1.22 comes for central maximum formed by circular apertures. This involves higher level mathematics which is avoided in this discussion.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-38
For small angles, sin θ ≈ θ
a θ = 1.22λ … (3)
Rewriting further, θ = \(\frac { 1.22λ }{ a}\) and \(\frac { { r }_{ 0 } }{ f } \) = \(\frac { 1.22λ }{ a}\)
r0 = \(\frac { 1.22λf }{ a}\) …. (4)

When two point sources close to each another form image on the screen, the diffraction pattern of one point source can overlap with another and produce a blurred image. To obtain a good image of the two sources, the two point sources must be resolved i.e., the point sources must be imaged in such a way that their images are sufficiently far apart that their diffraction patterns do not overlap.

According to Rayleigh’s criterion, for two point objects to be just resolved, the minimum distance between their diffraction images must be in such a way that the central maximum of one coincides with the first minimum of the other and vice versa. Such an image is said to be just resolved image of the object. The Rayleigh’s criterion is said to be limit of resolution.

According to Rayleigh’s criterion the two point sources are said to be just resolved when the distance between the two maxima is at least r0. The angular resolution has a unit in radian (rad) and it is given by the equation.
θ = \(\frac { 1.22λf}{ a}\)
It shows that the first order diffraction angle must be as small as possible for greater resolution. This further shows that for better resolution, the wavelength of light used must be as small as possible and the size of the aperture of the instrument used must be as large as possible. The equation (4) is used to calculate spacial resolution. The inverse of resolution is called resolving power. This implies, smaller the resolution, greater is the resolving power of the instrument. The ability of an optical instrument to • separate or distinguish small or closely adjacent objects through the image formation is said to be resolving power of the instrument.

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Question 23.
Discuss about simple microscope and obtain the equations for magnification for near point focusing and normal focusing.
Answer:
Simple microscope:
A simple microscope is a single magnifying (converging) lens of small focal length. The idea is to get an erect, magnified and virtual image of the object. For this the object is placed between F and P on one side of the lens and viewed from other side of the lens. There are two magnifications to be discussed for two kinds of focusing.

  • Near point focusing :
    The image is formed at near point, i.e. 25 cm for normal eye. This distance is also called as least distance D of distinct vision. In this position, the eye feels comfortable but there is little strain on the eye.
  • Normal focusing :
    The image is formed at infinity. In this position the eye is most relaxed to view the image.

Magnification in near point focusing:
Object distance u is less than f. The image distance is the near point D. The magnification m is given by the relation,
m = \(\frac { v}{ u}\)
With the help of lens equation, \(\frac { 1}{ v}\) – \(\frac { 1}{ u}\) \(\frac { 1}{ f}\) the magnification can further be writen as,
m = 1- \(\frac { v}{ f}\)
Substituting for v with sign convention, v = -Derivem
m = 1 + \(\frac { D}{ f}\)
This is the magnification for near point focusing
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-39
Magnification in normal focusing (angular magnification):
We will now find the magnification for the image formed at infinity. If we take the ratio of height of image to height of object \(\left(m=\frac{h^{\prime}}{h}\right)\) to find the magnification, we will not get a practical relation as the image will also be of infinite size when the image is formed at infinity. Hence, we can practically use the angular magnification. The angular magnification is defined as the ratio of angle 0j subtended by the image with aided eye to the angle 90 subtended by the object with unaided eye.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-40
m = \(\frac{\theta_{i}}{\theta_{0}}\)
For unaided eye, tan θ0 ≈ θ0 = \(\frac { h }{ D }\)
For aided eye, tan θi ≈ θi = \(\frac { h }{ f }\)
The angular magnification is, m = \(\frac{\theta_{i}}{\theta_{0}}\) = \(\frac { h/f }{ h/d }\)
m = \(\frac { d }{ f }\)
This is the magnification for normal focusing. The magnification for normal focusing is one less than that for near point focusing.

Question 24.
Explain about compound microscope and obtain the equation for magnification. Compound microscope:
Answer:
The lens near the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens which is the eyepiece. Eyepiece serves as a simple microscope that produces finally an enlarged and virtual image. The first inverted image formed by the objective is to be adjusted close to, but within the focal plane of the eyepiece so that the final image is formed nearly at infinity or at the near point. The final image is inverted with respect to the original object. We can obtain the magnification for a compound microscope.
Magnification of compound microscope :
From the ray diagram, the linear magnification due to the objective is,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-41
m0 = \(\frac { h’ }{ f}\)
From the figure, tan β = \(\frac { h }{ { f }_{ 0 } } \) =\(\frac { h’ }{ L’}\),then
\(\frac { h’ }{ h}\) = \(\frac { L }{ { f }_{ 0 } } \); m0 =\(\frac { L }{ { f }_{ 0 } } \)
Here, the distance L is between the first focal point of the eyepiece to the second focal point of the objective. This is called the tube length
L of the microscope as f0 and fe are comparatively smaller than L. If the final image is formed at P (near point focusing), the magnification me of the eyepiece is,
me = 1 + \(\frac { D }{ { f }_{ e } } \)
The total magnification m in near point focusing is,
m = m0me = \(\left(\frac{L}{f_{o}}\right)\left(1+\frac{D}{f_{e}}\right)\)
If the final image is formed at infinity (normal focusing), the magnification me of the eyepiece is,
me = \(\frac { D }{ { f }_{ e } } \)
The total magnification m in normal focusing is,
m = m0me = \(\left(\frac{L}{f_{o}}\right)\left(\frac{D}{f_{e}}\right)\)

Question 25.
Obtain the equation for resolving power of microscope.
Answer:
Resolving power of microscope:
The diagram related to the calculation of resolution of microscope. A microscope is used to see the details of the object under observation. The ability of microscope depends not only in magnifying the object but also in resolving two points on the object separated by a small distance dmin.
Smaller the value of dmin better will be the resolving power of the microscope.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-42
The radius of central maxima is, r0 = \(\frac { 1.22λv }{ a }\)
In the place of focal length f we have the image distance v. If the difference between the two points on the object to be resolved is dmin, then the magnification m is, m = \(\frac{r_{o}}{d_{\min }}\)
dmin = \(\frac{r_{o}}{m}\) =\(\frac{1.22 \lambda v}{a m}\) = \(\frac{1.22 \lambda v}{a(v / u)}\) = \(\frac{1.22 \lambda u}{a}\) [∴m = v/u] [∴ [u ≈ ƒ]
On the object side, 2tan β ≈ 2sin β = \(\frac { a }{ f }\)
dmin = \(\frac{1.22 \lambda}{2 sin β }\) ∴ [a = ƒ2 sinβ]
To further reduce the value of dmin the optical path of the light is increased by immersing the objective of the microscope in to a bath containing oil of refractive index n.
dmin = \(\frac{1.22 \lambda}{2n sin β }\)
Such an objective is called oil immersed objective. The term n sin p is called numerical aperture NA.
dmin = \(\frac{1.22 \lambda}{2(NA) }\)

Question 26.
Discuss about astronomical telescope.
Answer:
Astronomical telescope:
An astronomical telescope is used to get the magnification of distant astronomical objects like stars, planets, moon etc. the image formed by astronomical telescope will be inverted. It has an objective of long focal length and a much larger aperture than the eyepiece. Light from a distant object enters the objective and a real image is formed in the tube at its second focal point. The eyepiece magnifies this image producing a final inverted image.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-43
Magnification of astronomical telescope:
The magnification m is the ratio of the angle β subtended at the eye by the final image to the angle a which the object subtends at the lens or the eye
m = \(\frac { β }{ α }\)
From the diagram, m = \(\frac{h / f_{e}}{h / f_{0}}\)
m = \(\frac { { f }_{ 0 } }{ { f }_{ e } } \)
The length of the telescope is approximately, L = f0 + fe

Question 27.
Mention different parts of spectrometer and explain the preliminary adjustments. Spectrometer:
Answer:
The spectrometer is an optical instrument used to study the spectra of different sources of light and to measure the refractive indices of materials. It consists of basically three parts. They are (i) collimator, (ii) prism table and (iii) Telescope.
Adjustments of the spectrometer:
The following adjustments must be made before doing the experiment using spectrometer.

(a) Adjustment of the eyepiece:
The telescope is turned towards an illuminated surface and the eyepiece is moved to and fro until the cross wires are clearly seen.

(b) Adjustment of the telescope:
The telescope is adjusted to receive parallel rays by turning it towards a distant object and adjusting the distance between the objective lens and the eyepiece to get a clear image on the cross wire.

(c) Adjustment of the collimator:
The telescope is brought along the axial line with the collimator. The slit of the collimator is illuminated by a source of light. The distance between the slit and the lens of the collimator is adjusted until a clear image of the slit is seen at the cross wire of the telescope. Since the telescope is already adjusted for parallel rays, a well-defined image of the slit can be formed, only when the light rays emerging from the collimator are parallel.

(d) Levelling the prism table:
The prism table is adjusted or levelled to be in horizontal position by means of levelling screws and a spirit level.

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Question 28.
Explain the experimental determination of material of the prism using spectrometer. Determination of refractive index of material of the prism:
Answer:
The preliminary adjustments of the telescope, collimator and the prism table of the spectrometer are made. The refractive index of the prism can be determined by knowing the angle of the prism and the angle of minimum deviation.

(i) Angle of the prism (A):
The prism is placed on the prism table with its refracting edge facing the collimator. The slit is illuminated by a sodium light (monochromotic light). The parallel rays coming from the collimator fall on the two faces AB and AC. The telescope is rotated to the position T1 until the image of the slit formed by the reflection at the face AB is made to coincide with the vertical cross wire of the telescope.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-44
The readings of the verniers are noted. The telescope is then rotated to the position T2 where the image of the slit formed by the reflection at the face AC coincides with the vertical cross wire. The readings are again noted.
The difference between these two readings gives the angle rotated by the telescope, which is twice the angle of the prism. Half of this value gives the angle of the prism A.

(ii) Angle of minimum deviation (D):
The prism is placed on the prism table so that the light from the collimator falls on a refracting face, and the refracted image is observed through the telescope. The prism table is now rotated so that the angle of deviation decreases. A stage comes when the image stops for a moment and if we rotate the prism table further in the same direction, the image is seen to recede and the angle of deviation increases. The vertical cross wire of the telescope is made to coincide with the image of the slit where it turns back. This gives the minimum deviation position.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-45
The readings of the verniers are noted. Now, the prism is removed and the telescope is turned to receive the direct ray and the vertical cross wire is made to coincide with the image. The readings of the verniers are noted. The difference between the two readings gives the angle of minimum deviation D. The refractive index of the material of the prism n is calculated using the formula,
n = \(\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
The refractive index of a liquid may be determined in the same way using a hollow glass prism filled with the given liquid.

Samacheer Kalvi 12th Physics Optics Conceptual Questions

Question 1.
Why are dish antennas curved?
Answer:
Dish antenna is curved so as it can receive parallel signal rays coming from same direction. These parallel signal rays reflect from parabolic dish, and gathered at main antenna part. This increases directivity of antenna, and gives sufficient amplitude signal.

Question 2.
What type of lens is formed by a bubble inside water?
Answer:
Air bubble has spherical surface and is surrounded by medium (water) of higher refractive index. When light passes from water to air it gets diverged. So air bubble behaves as a concave lens.

Question 3.
It is possible for two lenses to produce zero power?
Answer:
Yes. It is possible for two lenses to produce zero power. Both the surfaces of lenses are equally curved, i.e. R1 = R2 and hence
Power (P) = (µ – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)=0\)

Question 4.
Why does sky look blue and clouds look white?
Answer:
Blue colour of the sky is due to scattering of sunlight by air molecules. According to Rayleigh’s law, intensity of scattered light, I ∝ \(\frac { 1 }{ { \lambda }^{ 4 } } \). So blue light of shorter wavelength is scattered much more than red light of larger wavelength. The blue component is proportionally more in light coming from different parts of the sky. That is why the sky appear blue. Clouds have large particles like dust and water droplets which scatter light of all colours almost equally. Hence clouds generally appear white.

Question 5.
Why is yellow light preferred to during fog?
Answer:
Yellow light has longer wavelength than green, blue or violet components of white lights. As scattered intensity, I ∝ \(\frac { 1 }{ { \lambda }^{ 4 } } \). so yellow colour is least scattered and produces sufficient illumination.

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Question 6.
Two independent monochromatic sources cannot act as coherent sources, why?
Answer:
Two independent sources of light cannot be coherent. This is because light is emitted by individual atoms. When they return to ground state. Even the smallest source of light contains billions of atoms which obviously cannot emit light waves in the same phase.

Question 7.
Does diffraction take place at the Young’s double slit?
Answer:
Both diffraction and interference in the double slit experiment. The wavefront is diffracted as it passes through each of the slits. The diffraction causes the wavefronts to spread out as if they were coming from light sources located at the slits. These two wavefronts overlap, and interference occurs.

Question 8.
Is there any difference between colored light obtained from prism and colours of soap bubble?
Answer:
Yes. there is a difference between colored light obtained from the prism is the phenomenon of‘dispersion of light’ and colored light obtained from the soap bubble is the phenomenon of ‘interference of light’.

Question 9.
A small disc is placed in the path of the light from distance source. Will the center of the shadow be bright or dark?
Answer:
When a tiny circular small disc is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the disc because, wave diffracted from the edge of the circular disc interface constructively at the centre of the shadow, which produces bright spot.

Question 10.
When a wave undergoes reflection at a denser medium, what happens to its phase?
Answer:
When a wave undergoes a reflection at a denser medium then it’s crest reflected as trough and vice versa. So, its phase changes at 180°.

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Samacheer Kalvi 12th Physics Optics Numerical Problems

Question 1.
An object is placed at a certain distance from a convex lens of focal length 20 cm. Find the distance of the object if the image obtained is magnified 4 times.
Solution:
ƒ = – 20 cm
v = – 4u
According to lens formula
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ (-20) }\) = \(\frac { 1 }{ (-4u) }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ (-20) }\) = \(\left[-\frac{1}{4}+1\right]\)
= \(\frac { 1 }{ u }\) \(\frac { 3 }{ 4 }\)
u = \(\frac { 3 × 20 }{ 4 }\) = -15 cm.

Question 2.
A compound microscope has a magnification of 30. The focal length of eye piece is 5 cm. Assuming the final image to be at least distance of distinct vision, find the magnification produced by the objective.
Solution:
Magnification of compound microscope, M = 30
Focal length, f = 5 cm
Least distance of distinct vision, D = 25 cm
Now, M = M0 x Me
= M0 x \(\left[1+\frac{\mathrm{D}}{f_{e}}\right]\)
30 = M0 x \(\left[1+\frac{25}{5}\right]\)
M0 = \(\frac { 30 }{ 6 }\) = 5

Question 3.
An object is placed in front of a concave mirror of focal length 20 cm. The image formed is three times the size of the object. Calculate two possible distances of the object from the mirror.
Solution:
m = + 3 and m = – 3; f = – 20 cm
When, m = 3
m = \(\frac { v }{ u }\) = 3
v = – 3 u
From mirror equation,
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ -20 }\) = \(\frac { 1 }{ -3u }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ u }\) \(\left[-\frac{1}{3}+1\right]\)
\(\frac { 1 }{ -20 }\) = \(\frac { 2 }{ 3u }\)
u = –\(\frac { 40 }{ 3 }\) cm
When, m = – 3 ⇒ v = 3u
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ -20 }\) = \(\frac { 1 }{ 3u }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ -20 }\) = \(\frac { 4 }{ 3u }\)
u = \(\frac { 4×20 }{ 3 }\)
u = –\(\frac { 80 }{ 3 }\) cm

Question 4.
A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm.What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
Solution:
Actual depth of the bulb in water
d1 = 80 cm = 0.8 m
Refractive index of water, p = 1.33
The given situation is shown in the figure.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-46
Where, i = Angle of incidence
r = Angle of refraction = 90°
Since the bulb is a point source, the emergent light can be considered as a circle AT
R= \(\frac { AC }{ 2 }\) = AO = OB
Using snell’s law, the relation for the refractive index of water is sin
µ = \(\frac { sin r }{ sin i }\)
i = sin-1 (0.75) = 48.75°
Using the given figure, we have the relation
tan i = \(\frac { OC }{ BC }\) = \(\frac { R }{ { d }_{ 1 } } \)
R = tan i x d1 = tan 48.75° x 0.8
R = 0.91 m
Area of the surface of water = πR²
= 3.14 x (0.91)² = 2.61 m²
Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61 m².

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Question 5.
A thin converging glass lens made of glass with refractive index 1.5 has a power of + 5.0 D. When this lens is immersed in a liquid of refractive index n, it acts as a divergent lens of focal length 100 cm. What must be the value of n?
Solution:
According to Lens maker’s formula, the focal length for a convex lens placed in air can be obtained as
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-47
\(\frac { 0.5 }{ 5 }\) = \(\frac { 1.5 }{ n }\) -1
0.9 = \(\frac { 1.5 }{ n }\)
n = \(\frac { 1.5 }{ 0.9 }\) = \(\frac { 5 }{ 3 }\) -1

Question 6.
If the distance D between an object and screen is greater than 4 times the focal length of a convex lens, then there are two positions of the lens for which images are formed on the screen. This method is called conjugate foci method. If d is the distance between the two positions of the lens, obtain the equation for focal length of the convex lens.
Solution:
Let us fix the position of object and place the screen to get the enlarged image first. Also, let us fix the position of screen where we get the enlarged image.
Let D be the distance between object and screen. Let us mark the position of lens dv Then let us move the lens away from the object to get a diminished image. Let this position of lens be d2. Let d be the distance between the lens position d1 and d2. Let V be the distance b/w image and lens. Let ‘u’ be the distance between object and lens.
From mirror equation,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-48
\(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)
Let us replace v by substituting v = D – u
\(\frac { 1 }{ D-u }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)
we get the equation, u² -Du + fD = 0
the quadratic equation for above equation,
u = \(\frac{\mathrm{D} \pm \sqrt{\mathrm{D}^{2}-4 f \mathrm{D}}}{2}\)
When D = 4f, we get only position of lens to get image.
This corresponds to placing the object at 2f and getting the image at 2f on the otherside. Hence, for displacement method we need D > 4 f. When this condition is satisfied we get
u1 = \(\frac{\mathrm{D}-\sqrt{\mathrm{D}^{2}-4 f \mathrm{D}}}{2}\) ; corresponding v1 – D – u2 = \(\frac{\mathrm{D}+\sqrt{\mathrm{D}^{2}-4 f \mathrm{D}}}{2}\) after changing the location
u1 =\(\frac{\mathrm{D}+\sqrt{\mathrm{D}^{2}-4f \mathrm{D}}}{2}\) ; corresponding v2 – D – u2 = \(\frac{\mathrm{D}-\sqrt{\mathrm{D}^{2}-4 f \mathrm{D}}}{2}\)
now the displacement d = v1 – u1 = u1 – v1 = \(\sqrt{D^{2}-4 f D}\)
Hence we get focal length, ƒ = \(\frac{D^{2}-d^{2}}{4 D}\)

Question 7.
A beam of light of wavelength 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. What is the distance between the first dark fringe on either side of the central bright fringe?
Solution:
For first minimum (n = 1) on either side of central maximum.
sin θ = \(\frac { λ }{ a }\)
Where a is width of the slit Since 0 is very small, (sin θ ≈ θ)
θ = \(\frac { λ }{ a }\) …… (1)
sin θ ≈ θ = \(\frac { x }{ 2D }\) …… (2)
Where, x is distance of first dark fringe from central maximum.
D is distance between slit and screen.
From equation (1) and (2)
\(\frac { x }{ 2D }\) = \(\frac { λ }{ a }\)
x = \(\frac { 2Dλ }{ a }\) = \(\frac { 2\times 2\times 600\times { 10 }^{ -9 } }{ 1\times { 10 }^{ -3 } } \)
= 2400 x 10-6 = 2.4 x 10-3m
x = 2.4 mm

Question 8.
In Young’s double slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelength λ0 = 750 nm and λ = 900 nm. What is the minimum distance from the common central bright fringe on a screen 2 m from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other?
Solution:
Now from the question we can infer that
D = 2; d = 2
n1 λ1 = n2 λ2
\(\frac{n_{1}}{n_{2}}=\frac{\lambda_{2}}{\lambda_{1}}\) =\(\frac { 900 }{ 750 }\) \(\frac { 1 }{ 2 }\)
Thus, we have
\(\frac{n_{1}}{n_{1}}=\frac{\lambda_{2}}{\lambda_{1}}\) =\(\frac { 6 }{ 5 }\)
5th and 6th fringes will coincide respectively. The minimum distance is given as
Xmin = \(\frac{n_{2} \lambda_{2} \mathrm{D}}{d}\) = \(\frac{5 \times 900 \times 10^{-9} \times 2}{2 \times 10^{-3}}\)
= 4500 x 10-6 = 4.5 x 10-3m
Xmin = 4.5 mm

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Question 9.
In Young’s double slit experiment, 62 fringes are seen in visible region for sodium light of wavelength 5893 Å. If violet light of wavelength 4359 Å is used in place of sodium light, then what is the number of fringes seen?
Solution:
From young’s double slit experiment,
λ1 = 5893 Å ; λ2 = 4359 Å
\(\frac{n_{1} \lambda_{1} D}{d}\) = \(\frac{n_{2} \lambda_{2} \mathrm{D}}{d}\)
The above condition is total extent of fringes is constant for both wavelengths.
\(\frac{62 \times 5893 \times 10^{-10} \times D}{d}\) = \(\frac{n_{2} \times 4359 \times 10^{-10} \times D}{d}\)
n2 = \(\frac{62 \times 5893}{4359}\) = \(\frac{365366}{4359}\) = 83.8
n2 = 84

Question 10.
A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. What is the focal length of the eyepiece.
Solution:
Magnifying Power, m = 100
Focal length of the objective,f0 = 0.5 cm
Tube length, l = 6.5 cm
Since the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eyepiece.
v0e= 6.5 cm ….. (1)
The magnifying power for normal adjustment is given by
m = \(\left(\frac{v_{o}}{u_{o}}\right) \times \frac{\mathrm{D}}{f_{e}}\)
= – \(\left[1-\frac{v_{o}}{f_{\mathrm{o}}}\right] \frac{\mathrm{D}}{f_{e}}\)
100 = – \(\left[1-\frac{v_{o}}{0.5}\right] \times \frac{25}{f_{e}}\)
2v0 -4ƒe= 1
On solving equations (1) and (2), we get
v0 = 4.5 cm and ƒe = 2 cm
Thus, the focal length of the eyepiece is 2 cm.

Samacheer Kalvi 12th Physics Optics Additional Questions

Samacheer Kalvi 12th Physics Optics Multiple Choice Questions

Question 1.
When a ray of light enters a glass slab from air
(b) its wavelength increases
(c) its frequency increases
(d) neither its wavelength nor its frequency changes
Answer:
(a) its wavelength decreases
Hint:
Wavelength, λ = \(\frac { Velocity }{ Frequency }\) = \(\frac { u }{ v }\)
When light travels from air to glass, frequency v remains unchanged, velocity u decreases and hence wavelength X also decreases.

Question 2.
A source emits sound of frequency 600 Hz inside water. The frequency heard in air (velocity of sound in water = 1500 m/s, velocity of sound in air = 300 m/s) will be
(a) 300 Hz
(b) 120 Hz
(c) 600 Hz
(d) 6000 Hz
Answer:
(c) 600 Hz
Hint:
Frequency does not change when sound travels from one medium to another
∴ Frequency of sound in air = Frequency of sound in water = 600 Hz

Question 3.
Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is 60°). In the position of minimum deviation, the angle of refraction will be
(a) 30° for both the colours
(b) greater for the violet colour
(c) greater for the violet colour
(d) equal but not 30° for both the colours
Answer:
(a) 30° for both the colours
Hint:
For any prism, r1 = r2 = A
In the position of minimum deviation for any wavelength,
r1 = r2 = \(\frac { A }{ 2 }\) = \(\frac { 60° }{ θ }\) = 30°

Question 4.
To get three images of a single object, one should have two plane mirrors at an angle of
(a) 60°
(b) 90°
(c) 120°
(d) 30°
Answer:
(b) 90°
Hint:
The number of images formed,
n = \(\frac { 360° }{ θ }\) – 1 or 3 = \(\frac { 360° }{ θ }\) -1 or θ = 90°

Question 5.
Which of the following is used in optical fibres?
(a) Total internal reflection
(b) Diffraction
(c) Refraction
(d) Scattering
Answer:
(a) Total internal reflection
Hint:
The working of optical fibres is based on total internal reflection.

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Question 6.
Two lenses of power – 15 D and +15D are in contact with each other. The focal length of the combination is
(a) + 10 cm
(b) -20 cm
(c) – 10 cm
(d) + 20cm
Answer:
(c) – 10 cm
Hint P = P1 + P2 = -15 + 5= -10 D
F = \(\frac { 1 }{ P }\) = \(\frac { 1 }{ -10 }\) m = -10 cm

Question 7.
The refractive index of glass is 1.520 for red light and 1.525 for blue light. Let δ1 and δ2 be angles of minimum deviation for red and blue light respectively in a prism of this glass, then
(a) δ1, can be less than or greater than δ2 depending upon the values of δ1 and δ2
(b) δ1 > δ2
(c) δ1 < δ2
(d) δ1 = δ2
Answer:
(c) δ1 < δ2
Hint:
δ1 = (μR – 1)A, δ2 = (μB – 1)A
As, μR < μB ∴ δ1 < δ2

Question 8.
Time image formed by an objective of a compound microscope is
(a) virtual and diminished
(b) real and diminished
(c) real and enlarged
(d) virtual and enlarged
Answer:
(c) real and enlarged
Hint
The image formed by the objective of a compound microscope is real and enlarged.

Question 9.
An astronomical telescope has a large aperture to,
(a) reduce spherical aberration
(b) have high resolution
(c) increase span of observation
(d) have low dispersion
Answer:
(b) have high resolution

Question 10.
Two plane mirrors are inclined to each other at an angle of 60°. A point object is placed in between them. The total number of images produced by both the mirror is
(a) 2
(b) 4
(c) 5
( d) 6
Answer:
(c) 5
Hint:
Number of images formed, θ = \(\frac { 360° }{ θ }\) – 1 = \(\frac { 360 }{ 60 }\) = 1 = 5.

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Question 11.
A boy 1.5 m tall with his eye level at 1.38 m stands before a mirror fixed on a wall. The minimum length of mirror required to view the complete image of boy is
(a) 0.75 m
(b) 0.06 m
(c) 0.69 m
(d) 0.12 m
Answer:
(a) 0.75 m
Hint
Minimum length of mirror required \(\frac { 1 }{ 2 }\) x Height of boy = \(\frac { 1 }{ 2 }\) x 1.5 = 0.75 m.

Question 12.
A pencil of light rays falls on a plane mirror and forms a real image, so the incident rays are
(a) parallel
(b) diverging
(c) converging
(d) statement is false
Answer:
(c) converging
Hint:
When converging rays fall on a plane mirror, they get reflected to a point d in front of the mirror forming a real image.

Question 13.
For a real object, which of the following can produce a real image?
(a) plane mirror
(b) concave lens
(c) convex lens
(d) concave mirror
Answer:
(d) concave mirror
Hint:
Only concave mirror produces real image provided the object is not placed between its focus and pole.

Question 14.
Which mirror is to be used to obtain a parallel beam of light from a small lamp?
(a) Plane mirror
(b) Convex mirror
(c) Concave mirror
(d) None of the above
Answer:
(c) Concave mirror
Hint:
When the small lamp is placed at the focus of the concave mirror, the reflected light is a parallel beam.

Question 15.
When a plane electromagnetic wave enters a glass slab, then which of the following will not change?
(a) Wavelength
(b) Frequency
(c) Speed
(d) Amplitude
Answer:
(b) Frequency
Only the frequency of the electromagnetic wave remains unchanged.

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Question 16.
If wavelength of light in air is 2400 x 10-10 m, then what will be the wavelength of light in glass (μ = 1.5)?
(a) 1600 Å
(b) 7200 Å
(c) 1080 Å
(d) None of these
Answer:
(a) 1600 Å
Hint:
μ = \(\frac { { \lambda }_{ a } }{ { \lambda }_{ g } } \) ⇒λλg = \(\frac { { \lambda }_{ a } }{ μ } \) = \(\frac{2400 \times 10^{-10} \mathrm{m}}{1.5}\) = 1600 Å

Question 17.
Why is refractive index in a transparent medium greater than one?
(a) Because the speed of light in vacuum is always less than speed in a transparent medium.
(b) Because the speed of light in vacuum is always greater than the speed in a transparent medium.
(c) Frequency of wave changes when it Gasses medium.
(d) None of the above.
Answer:
(b) Because the speed of light in vacuum is always greater than the speed in a transparent medium.
Hint:
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-49
As c > v, μ > 1

Question 18.
The wavelength of sodium light in air is 5890 Å. The velocity of light in air is 3 x 108 ms-1. The wavelength of light in a glass of refractive index 1.6 would be close to
(a) 5890 Å
(b) 3681 Å
(c) 9424 Å
(d) 15078 Å
Answer:
(b) 3681 Å
Hint:
μ = \(\frac { { \lambda }_{ a } }{ { \lambda }_{ g } } \) ⇒λg= \(\frac { { \lambda }_{ a } }{ μ } \) = \(\frac { 5890 Å }{ 1.6 }\) = 3681 Å

Question 19.
A glass slab (μ = 1.5) of thickness 6 cm is placed over a paper. Qhat is the shift in the letters?
(a) 4 cm
(b) 2 cm
(c) 1 cm
(d) None of these
Answer:
(b) 2 cm
Hint:
Normal shift, x = \(t\left(1-\frac{1}{\mu}\right)=6\left(1-\frac{1}{1.5}\right)\) cm = 2cm

Question 20.
Light traveling from a transparent medium to air undergoes total internal reflection at an angle of incident of 45°. Then refractive index of the medium may be
(a) 1.5
(b) 1.3
(c) 1.1
(d) \(\frac { 1 }{ √2 }\)
Answer:
(a) 1.5
Hint:
μ = \(\frac{1}{\sin i_{c}}\) = \(\frac{1}{\sin 45^{\circ}}\) = √2 = 1.414 ≈ 1.5.

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Question 21.
A point source of light is placed 4 m below the surface of water of refractive index 5/3. The minimum diameter of a disc which should be placed over the source, on the surface of water to cut-off all light coming out of water is
(a) infinite
(b) 6 cm
(c) 4 cm
(d) 3 cm
Answer:
(b) 6 cm
Hint:
r = \(\frac{h}{\sqrt{\mu^{2}-1}}\) = \(\frac{4}{\sqrt{\left(\frac{5}{3}\right)^{2}-1}}\) = 3 cm

Question 22.
In optical fibres, propagation of light is due to
(a) diffraction
(b) total internal reflection
(c) reflection
(d) refraction
Answer:
(b) total internal reflection
Hint:
In optical fiberes, light propagates due to tatal internal reflection.

Question 23.
Sparkling of diamond is due to
(a) reflection
(b) dispersion
(c) total internal reflection
(d) high refractive index of diamond
Answer:
(c) total internal reflection

Question 24.
For a given lens, the magnification was found to be twice as large as when the object was 0.15m distant from it as when the distance was 0.2 m. The focal length of the lens is
(a) 1.5 m
(b) 0.20 m
(c) 0.10 m
(d) 0.05 m
Asnwer:
(c) 0.10 m
Hint:
Here m1 = 2m2
\(\frac { f }{ f-0.15 }\) = 2 \(\frac { f }{ f-0.20 }\)
2f – 0.30 = f- 0.20 ; f = 0.10 m

Question 25.
Two lenses of focal lengths f1 and f2 are kept in contact coaxially. The resultant power of combination will be
(a) \(\frac { { f }_{ 1 }{ f }_{ 2 } }{ { { f }_{ 1 }-f }_{ 2 } } \)
(b) \(\frac { { f }_{ 1 }{ +f }_{ 2 } }{ { { f }_{ 1 }f }_{ 2 } } \)
(c) \({ f }_{ 1 }{ +f }_{ 2 }\)
(d) \(\frac { { f }_{ 1 } }{ { f }_{ 2 } } +\frac { { f }_{ 2 } }{ { f }_{ 1 } } \)
Answer:
(b) \(\frac { { f }_{ 1 }{ +f }_{ 2 } }{ { { f }_{ 1 }f }_{ 2 } } \)
Hint:
P = \(\frac { 1 }{ { f }_{ 1 } } +\frac { 1 }{ { f }_{ 2 } } \) = \(\frac { { f }_{ 1 }{ +f }_{ 2 } }{ { { f }_{ 1 }f }_{ 2 } } \).

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Question 26.
Two lenses of power 3D and -ID are kept in contact. What is focal length and nature of combined lens?
(a) 50 cm, convex
(b) 200 cm, convex
(c) 50 cm, concave
(d) 200 cm, concave
Answer:
(a) 50 cm, convex
Hint:
P = P1 + P2 = 3 – 1 = 2D
F= \(\frac { 1 }{ P }\) = \(\frac { 1 }{ 2 }\) m = 50 cm

Question 27.
If two thin lenses are kept coaxially together, then their power is proportional (R1,R2) being the radii of curved surfaces) to
(a) R1 + R2
(b) \(\left[\frac{\mathbf{R}_{1}+\mathbf{R}_{2}}{\mathbf{R}_{1} \mathbf{R}_{2}}\right]\)
(c) \(\left[\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1} \mathrm{R}_{2}}\right]\)
(d) None of these
Answer:
(b) \(\left[\frac{\mathbf{R}_{1}+\mathbf{R}_{2}}{\mathbf{R}_{1} \mathbf{R}_{2}}\right]\)
Hint:
P = \(\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{2}{\mathrm{R}_{1}}+\frac{2}{\mathrm{R}_{2}}\) = 2\(\left[\frac{\mathbf{R}_{1}+\mathbf{R}_{2}}{\mathbf{R}_{1} \mathbf{R}_{2}}\right]\)
or p ∝ \(\left[\frac{\mathbf{R}_{1}+\mathbf{R}_{2}}{\mathbf{R}_{1} \mathbf{R}_{2}}\right]\)

Question 28.
A ray incident at 15° on one refracting surface of a prism of angle 60°, suffers a deviation of 55°. What is the angle of emergance?
(a) 95°
(b) 45°
(c) 30°
(d) none of these
Answer:
(d) none of these
Hint:
A + δ = i + each
60° + 55° = 15° + each e = 115 -15 =100°

Question 29.
Dispersion of light is caused due to
(a) Wavelength
(b) intensity of light
(c) density of medium
(d) none of these
Answer:
(a) Wavelength
Hint:
Dispersion is due to the dependence of the speed of a wave on its wavelength in any medium.

Question 30.
White light is incident on one of the refracting surfaces of a prism of angle 50°. If the refractive indices for red and blue colours are 1.641 and 1.659 respectively, the angular separation between these two colours when they emerge out of the prism is
(a) 0.9°
(b) 0.09°
(c) 1.8°
(d) 1.2°
Answer:
(b) 0.09°
Hint:
Angular dispersion,
δB – δR = (µB – µR) A
= (1.659 – 1.641) x 5° = 0.09°

Question 31.
The sky would appear red instead of blue if
(a) atmospheric particles scatter blue light more than red light
(b) atmospheric particles scatter all colours equally
(c) atmospheric particle scatter red light more than blue light
(d) the sun was much hotter
Answer:
(c) atmospheric particle scatter red light more than blue light

Question 32.
A setting sun appears to be at an altitude higher than it really is. This is because of
(a) absorption of light
(b) reflection of light
(c) refraction of light
(d) dispersion of light
Answer:
(c) refraction of light
Hint:
This is due to refraction of light by the earth’s atmosphere.

Question 33.
The reddish appearance of rising and setting sun is due to
(a) reflection of light
(b) diffraction of light
(c) scattering of light
(d) interference of light
Answer:
(c) scattering of light
Hint:
The reddish appearance of the rising and the setting sun is due to scattering of light.

Question 34.
In the formation of a rainbow, the light from the sun on water droplets undergoes
(a) dispersion only
(b) only total internal reflection
(c) dispersion and total internal reflection
(d) none of the above
Answer:
(c) dispersion and total internal reflection
Hint:
Rainow is formed due to dispersion of sunlight by raindrops which also deviate the colours by total internal reflection.

Question 35.
The angular magnification of a simple microscope can be increased by increasing
(a) focal length of lens
(b) size of object
(c) aperture of lens
(d) power of lens
Answer:
(b) size of object

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Question 36.
For compound microscope f0 = 1 cm, fe = 2.5 cm. An object is placed at distance 1.2 cm from objective lens. What should be length of microscope for normal adjustment?
(a) 8.5 cm
(b) 8.3 cm
(c) 6.5 cm
(d) 6.3 cm
Answer:
(a) 8.5 cm
Hint:
In the normal adjustment of a compound microscope, L = υ0 + υe = \(\frac{υ_{0} f_{e}}{υ_{0}+f_{e}}\) + fe = \(\frac { 1.2×1 }{ -1.2+1 }\) + 2.5 = 6 + 25 = 8.5 cm

Question 37.
Magnifying power of an astronomical telescope for normal vision with usual notation is
(a) -f0 / fe
(b) -f0 x fe
(c) -f0 / f0
(d) -f0 + fe
Answer:
(a) -f0 / fe
Hint:
In normal adjustment of the telescope, m = -f0 / fe

Question 38.
F1 and F2 are focal length of objective and eyepiece respectively of the telescope. The angular magnification for the given telescope is equal to
(a) \(\frac {F_{1}}{F_{2}}\)
(b) \(\frac { { F }_{ 2 } }{ { F }_{ 1 } } \)
(c) \(\frac { { F }_{ 1 }{ F }_{ 2 } }{ { F }_{ 1 }+{ F }_{ 2 } } \)
(d) \(\frac { { F }_{ 1 }+{ F }_{ 2 } }{ { F }_{ 1 }{ F }_{ 2 } } \)
Answer:
(a) \(\frac { { F }_{ 1 } }{ { F }_{ 2 } } \)
Hint:
In normal adjustment of the telescope, \(\left| m \right| \) = \(\frac { { f }_{ 0 } }{ { f }_{ e } } \) = \(\frac { { F }_{ 1 } }{ { F }_{ 2 } } \)

Question 39.
Focal length of objective and eyepiece of telescope are 200 cm and 4 cm respectively. What is length of telescope for normal adjustment?
(a) 196 cm
(b) 204 cm
(c) 250 cm
(d) 225 cm
Answer:
(b) 204 cm
Hint L = \(-{ f }_{ 0 }+{ f }_{ e }\) = 200 + 4 = 204 cm

Question 40.
For normal vision, what is minimum distance of object from eye?
(a) 30 cm
(b) 25 cm
(c) Infinite
(d) 40 cm
Answer:
(b) 25 cm
Hint:
For normal eye, the least distance of distinct vision is 25 cm.

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Question 41.
The focal length of the objective and eyepiece of a telescope are respectively 100 cm and 2 cm. The moon subtends angle of 0.5° ; the angle subtended by the moon’s image will be
(a) 10°
(b) 250
(e) 100°
(d) 75°
Answer:
b) 250
Hint ;
m = \(\frac { β }{ α }\) β = \(\frac { { f }_{ 0 } }{ { f }_{ e } } \) ; α = \(\frac { 100 }{ 2 }\) x 0.5° = 25°

Question 42.
A person cannot clearly see distance more than 40 cm. He is advised to use lens of power,
(a) – 2.5 D
(b) 2.5 D
(c) – 6.25 D
(d) 1.5 D
Answer:
(a) – 2.5 D
Hint;
For the remedial lens, u = ∞,
v = – 40 cm = – 0.40 m
∴ \(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) – \(\frac { 1 }{ u }\) = \(\frac { 1 }{ -0.40 }\) – \(\frac { 1 }{ ∞ }\) = -2.5 ⇒ P = 2.5D

Question 43.
The light gathering power of a camera lens depends on
(a) its diameter only
(b) ratio of diameter and focal length
(c) product of focal length and diameter
(d) wavelength of the light used
Answer:
(a) its diameter only
Hint:
The light gathering power of a camera lens is proportional to its area or to the square of its diameter.

Question 44.
Amount of light entering into the camera depends upon
(a) focal length of objective lens
(b) product of focal length and diameter of the objective lens
(c) distance of object from camera
(d) aperture setting of the camera
Answer:
(d) aperture setting of the camera
Hint:
The amount of light entering into the camera depends upon the aperture setting of the camera.

Question 45.
Line spectrum can be obtained from
(a) sun
(b) candle
(c) mercury vapour lamp
(d) electric bulb
Answer:
(c) mercury vapour lamp

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Question 46.
The Production of band spectra is caused by
(a) atomic nuclei
(b) hot metals
(c) molecules
(d) electrons
Answer:
(c) molecules

Question 47.
Two mirrors are kept at 60° to each other and a body is placed at middle. The total number of images formed is
(a) six
(b) four
(c) five
(d) three
Answer:
(a) six
Hint:
Number of images formed, n = \(\frac { 360 }{ θ }\) -1 = \(\frac { 360 }{ 60 }\) -1 = 5

Question 48.
A point source kept at a distance of 1000 m has a illumination I. To change the illumination to 161, the new distance should become
(a) 250 m
(b) 500 m
(c) 750 m
(d) 800 m
Answer:
(a) 250 m
Hint:
\(\frac{I_{2}}{I_{1}}=\left(\frac{r_{1}}{r_{2}}\right)^{2}\)
\(\frac{16 \mathrm{I}_{1}}{\mathrm{I}_{1}}=\frac{(1000)^{2}}{r_{2}^{2}}\) ; r2 \(\frac { 1000 }{ 4 }\) = 250 cm

Question 49.
A concave mirror of focal length 15 cm forms an image having twice the linear dimensions of the object. The position of the object, when the image is virtual will be
(a) 22.5 cm
(b) 7.5 cm
(c) 30 cm
(d) 45 cm
Answer:
(b) 7.5 cm
Hint:
For virtual images, m = \(\frac { -v }{ u }\) = + 2 or v = – 2u
As \(\frac { 1 }{ u }\) + \(\frac { 1 }{ v }\) = \(\frac { 1 }{ f }\)
∴ \(\frac { 1 }{ u }\) – \(\frac { 1 }{ 2u }\) = \(\frac { 1 }{ -15 }\) or \(\frac { 1 }{ 2u }\) = \(\frac { 1 }{ -15 }\) ⇒ u = -7.5 cm

Question 50.
when a ray of light enters a glass slab, then
(a) its frequency and velocity change
(b) only frequency changes
(c) its frequency and wavelength change
(d) its frequency does not change
Answer:
(d) its frequency does not change
Hint:
When a ray of light enters a glass slab, its velocity and wavelength change while frequency does not change.

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Question 51.
A light wave of frequency u and wavelength A travels from air to glass. Then,
(a) y changes
(b) y does not change, A changes
(c) A does not change
(d) y and A change
Answer:
(b) y does not change, A changes
Hint:
Same reasoning as in the above question.

Question 52.
In refraction, light waves are bent on passing from one medium to the second medium, because in the second medium.
(a) the frequency is different
(b) the coefficient of elasticity is different
(c) the speed is different
(d) the amplitude is smaller
Answer:
(c) the speed is different
Hint:
Speed of light in second medium is different than that in first medium.

Question 53.
A ray of light having wavelength 720 nm enters in a glass of refractive index 1.5. The wavelength of the ray within the glass will be
(a) 360 nm
(b) 480 nm
(e) 720 nm
(d) 1080 nm
Answer:
(b) 480 nm
Hint:
\({ \lambda }_{ g }\) = \(\frac { { \lambda }_{ 0 } }{ \mu } \) = \(\frac { 790nm }{ 1.5 }\) = 480nm

Question 54.
Brilliance of a diamond is due to
(a) shape
(b) cutting
(c) reflection
(d) total internal reflection
Answer:
(d) total internal reflection
Hint:
Brilliance of a diamond is due to total internal reflection of light.

Question 55.
An endoscope is employed by a physician to view the internal parts of a body organ. If is based on the principle of
(a) refraction
(b) reflection
(c) total internal reflection
(d) dispersion
Answer:
(c) total internal reflection
Hint:
An endoscope is made of optical fibres which work on the principle of total internal reflection.

SamacheerKalvi.Guru

Question 56.
‘Mirage’ is a phenomenon due to
(a) reflection of light
(b) refraction of light
(c) total internal reflection of light
(d) diffraction of light Hint Mirage occurs due to total internal reflection of light.
Answer:
(c) total internal reflection of light

Question 57.
Two lenses of power +12D and -2D are combined together. What is their equivalent focal length?
(a) 10 cm
(b) 12.5 cm
(c) 16.6 cm
(d) 8.33 cm
Answer:
(a) 10 cm
Hint:
P = P1 + P2 = + 12 – 2 = 10 D
F = \(\frac { 1 }{ P }\) = \(\frac { 1 }{ 10 }\)m = 10 cm

Question 58.
If two lenses of power + 1.5 D and + 1.0 D are placed in contact, then the effective power of combination will be
(a) 2.5 D
(b) 1.5 D
(c) 0.5 D
(d) 3.25 D
Answer:
(a) 2.5 D
Hint:
P = P1 + P2 = + 1.5 + 1.0 = + 2.5D

Question 59.
The angle of a prism is 6° and its refractive index for green light is 1.5. If a green ray passes through it, the deviation will be
(a) 30°
(b) 15°
(c) 3°
(d) 0°
Answer:
(c) 3°
Hint:
5 = (p – 1) A = (1.5 – 1) x 6 = 3°

Question 60.
Sky appears to be blue in clear atmosphere due to light’s
(a) diffraction
(b) dispersion
(c) scattering
(d) polarisation
Answer:
(c) scattering
Hint:
Sky appears blue due to scattering of light by atmospheric modecules.

SamacheerKalvi.Guru

Question 61.
One can not see through fog, because
(a) fog absorbs the light
(b) light suffers total reflection at droplets
(c) refractive index of the fog is infinity
(d) light is scattered by the droplets
Answer:
(d) light is scattered by the droplets

Question 62.
Fraunhofer lines of the solar system is an example of
(a) emission lines spectrum
(b) emission band spectrum
(c) continuous emission spectrum
(d) line absorption spectrum
Answer:
(d) line absorption spectrum
Hint:
Fraunhofer lines is an example of line absorption spectrum.

Question 63.
A person using a lens as a sample microscope sees an
(a) inverted virtual image
(b) inverted real magnified image
(c) upright virtual image
(d) upright real magnified image
Answer:
(d) upright real magnified image
Hint:
A person sees an upright virtual image in a simple microscope.

Question 64.
Four lenses of focal length + 10 cm, + 50 cm, + 100 cm and + 200 cm are available for making an astronomical telescope. To produce the largest magnification, the focal length of the eyepiece should be
(a) + 10 cm
(b) + 50 cm
(c) + 100 cm
(d) + 200 cm
Answer:
(a) + 10 cm
Hint:
To produce the largest magnification, the eyepiece should have minimum focal length.

Question 65.
The camera lens has an aperture of f and the exposure time is 1/60 s. What will be the new exposure time if the aperture become 1.4f?
(a) \(\frac { 1 }{ 42 }\) s
(b) \(\frac { 1 }{ 56 }\) s
(c) \(\frac { 1 }{ 72 }\) s
(d) \(\frac { 1 }{ 31 }\) s
Answer:
(d) \(\frac { 1 }{ 31 }\) s
Hint:
Time of exposure ∝ (f – number)2
\(\frac{t}{\left(\frac{1}{60}\right)}=\left(\frac{1.4}{1}\right)^{2}\) ⇒ t = \(\frac { 1.4×1.4 }{ 60 }\) ≈ \(\frac { 1 }{ 31 }\) s.

SamacheerKalvi.Guru

Question 66.
For a person near point of vision is 100 cm. Then the power of lens he must wear so as have normal vision, should be
(a) + ID
(b) – ID
(c) + 3D
(d) – 3D
Answer:
(c) + 3D
Hint:
f = \(\frac { yD }{ y-D }\) = \(\frac { 100×25 }{ 100-25 }\) = \(\frac { 100 }{ 3 }\) cm = \(\frac { 1 }{ 3 }\) cm ; P = \(\frac { 1 }{ f }\) = +3d

Question 67.
Ray optics is valid, when characteristic dimension ions are
(a) much smaller than the wavelength of light
(b) much larger than the wavelength of light
(c) of the same order as the wavelength of light
(d) of the order of one millimetre
Hint:
Ray optics is valid, when characteristic dimensions are much larger than the wavelength of
light.

Question 68.
A tall man of height 6 feet, want to see his full image. Then required minimum length of the mirrorwillbe
(a) 12 feet
(b) 3 feet
(e) 6 feet
(d) any length
Hint:
inimum height of mirror required for seeing full image Height of the man = 3 feet.

Question 69.
The refractive index of water is 1.33. What will be the speed of light in water?
(a) 3 x 108 ms-1
(b) 2.26 x 108 ms-1
(c) 4 x 108 ms-1
(d) 1.33 x 108 ms-1
Answer:
(b) 2.26 x 108 ms-1
Hint:
As, μ = \(\frac { c }{ v }\) ⇒ v = \(\frac { c }{ μ }\) = \(\frac { 3\times { 10 }^{ 8 } }{ 1.33 } \) = 2.26 x 108 ms-1

Question 70.
A beam of monochromatic light is refracted from vacuum into a medium of refractive index
1.5. The wavelength of refracted light will be
(a) same
(b) dependent on intensity of refracted light
(c) larger
(d) smaller
Answer:
(d) smaller
Hint:
As light enters the medium, its wavelength decreases and becomes equal to \(\frac { λ }{ μ }\).

SamacheerKalvi.Guru

Question 71.
Optical fibers are based on
(a) total internal reflection
(b) less scattering
(c) refraGtion
(d) less absorption coefficient
Answer:
(a) total internal reflection

Question 72.
A convex lens is dipped in a liquid, whose refractive index is equal to the refractive index of the lens. Then, its focal length will
(a) become zero
(b) becomes infinite
(e) remain unchanged
(d) become small, but non-zero
Answer:
(b) becomes infinite
Hint:
\(\frac { 1 }{ { f }_{ 1 } } \) = \(\left(\frac{\mu_{g}}{\mu_{e}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) = (1 – 1) \(\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) = 0 ; f1 = ∞

Question 73.
A convex lens and a concave lens, each having same focal length of 25 cm, are put in contact to form a combination of lenses. The power of the combination (in diopter) is
(a) zero
(b) 25
(c) 50
(d) infinite
Answer:
(a) zero
Hint:
\(\frac { 1 }{ F }\) = \(\frac { 1 }{ { f }_{ 1 } } \) + \(\frac { 1 }{ { f }_{ 2 } } \) = \(\frac { 1 }{ +25 }\) + \(\frac { 1 }{ -25 }\) = 0 ; P = \(\frac { 1 }{ F }\) = 0

Question 74.
The focal length of a converging lens is measured for violet, green and red colours. If is fV, fG and fR respectively. We will get
(a) fV = fG
(b) fG = fR
(c) fV < fR
(d) fV > fR
Answer:
(c) fV < fR
Hint:
\(\frac { 1 }{ f }\) = (μ – 1) \(\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) i.e., f ∝ \(\frac { 1 }{ μ – 1 }\)
As μV > μR, so, fV < fR

Question 75.
Rainbow is formed due to combination of
(a) refraction and scattering
(b) refraction and absorption
(c) dispersion and total internal reflection
(d dispersion and focusing
Answer:
(c) dispersion and total internal reflection
Hint:
Rainbow is formed due to dispersion and total internal reflection of sunlight by raindrops.

SamacheerKalvi.Guru

Question 76.
The blue colour of the sky is due to the phenomenon of
(a) scattering
(b) dispersion
(c) reflection
(d) refraction
Answer:
(a) scattering
Hint:
The blue colour of the sky is due to the scattering of sunlight by atmospheric molecules.

Question 77.
An astronomical telescope often fold angular magnification has a length of 44 cm. The focal length of the object is
(a) 4 cm
(b) 40 cm
(c) 44 cm
(d) 440 cm
Answer:
(b) 40 cm
Hint:
Here, f0 + fe = 44 cm
m =\(\frac { { f }_{ 0 } }{ { f }_{ e } } \) (or) f0 = 10 fe
∴ 10 fe + 10 fe = 44 cm (or) fe = 4 cm
Hence, f0 = 10 x 4 (or) f0 = 40 cm

Question 78.
Exposure time of a camera lens at the \(\frac { f }{ 2.8 }\) Setting is \(\frac { 1 }{ 200 }\) second. The correct time of exposure at \(\frac { f }{ 5.6 }\) is
(a) 0.20 second
(b) 0.40 second
(c) 0.02 second
(d) 0.04 second
Answer:
(c) 0.02 second
Hint:
Time of exposure, t ∝ (f- number)2
∴ \(\frac{t}{\left(\frac{1}{200}\right)}=\left(\frac{5.6}{2.8}\right)^{2}\) = 4 ⇒ t = 0.02s

Question 79.
Which of the following is not due to total internal reflection?
(a) Working of optical fibre
(b) Difference between apparent and real depth of a pond
(c) Mirage on hot summer day
(d) Brilliance of diamond
Answer:
(b) Difference between apparent and real depth of a pond
Hint:
Difference between apparent and real depth of a pond is due to refraction of light and the other three phenomena involve total internal reflection.

Question 80.
An object is placed at a distance of 0.5 m infront of a plane mirror. The distance between object and image will be
(a) 0.25 m
(b) 0.5 m
(c) 1.0 m
(d) 2.0 m
Answer:
(c) 1.0 m
Hint:
Distance between object and image = 0.5 + 0.5 = 1.0 m.

SamacheerKalvi.Guru

Question 81.
An observer moves towards a stationary plane mirror at a speed of 4 ms-1 with what speed – will his image move towards him?
(a) 2 ms-1
(b) 4 ms-1
(c) 8 ms-1
(d) the image will stay at rest
Answer:
(c) 8 ms-1
Hint:
Speed of the image towards the observer = 2 x 4 = 8 ms-1

Question 82.
If two mirrors are kept at 60° to each other and a body is placed at the middle, then total number of images formed is
(a) six
(b) four
(c) five
(d) three
Answer:
(c) five
Hint:
Number of images formed = \(\frac { 360 }{ θ }\) -1 = \(\frac { 360 }{ 60 }\) -1 = 5

Question 83.
If an object is placed at 10 cm infront of a concave mirror of focal length 15 cm. The magnification of image is
(a) -1.5
(b) 1.5
(c) -3
(d) 3
Answer:
d) 3
Hint:
m = \(\frac { f }{ f-u }\) = \(\frac { -15 }{ -15(-10) }\) = +3

Question 84.
An object of length 2.5 cm is placed at the principal axis of a concave mirror at a distance 1.5f. The image height is
(a) + 5 m
(b) -5 cm
(c) – 10 cm
(d) + 1 cm
Answer:
(b) -5 cm
Hint:
m = \(\frac { f }{ f-u }\) = \(\frac { f }{ f-1.5f}\) = -2cm
Height image = m x height of object -2 x 2.5 = – 5 cm

Question 85.
Which of the following mirror is used by a dentist to examine a small cavity?
(a) Concave mirror
(b) Convex mirror
(c) Combination of (a) and (b)
(d) None of these
Answer:
(a) Concave mirror
Hint:
A concave mirror, because it forms erect and enlarged image when held closer to the cavity.

SamacheerKalvi.Guru

Question 86.
When a ray of light enters from one medium to another, then which of the following does not change?
(a) Frequency
(b) Wavelength
(c) Speed
(d) Amplitude
Answer:
(a) Frequency
Hint:
Only frequency remains unchanged.

Question 87.
When light travels from one medium to the other medium of which the refractive index is different, then which of the following will change?
(a) Frequency, wavelength and velocity
(6) Frequency and wavelength
(c) Frequency and velocity
(d) Wavelength and velocity
Answer:
(d) Wavelength and velocity
Hint:
Wavelength and velocity will change while frequency remains unchanged.

Question 88.
The time taken by the light to cross a glass of thickness 4 mm and refractive index (μ = 3), will be
(a) 4 x 10-11 sec
(b) 16 x 10-11 sec
(c) 8 x 10-11 sec
(d) 24 x 10-10 sec
Answer:
(a) 4 x 10-11 sec
Hint:
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-50
\(\frac { d }{ c/a }\) = \(\frac { μd }{ c }\) = \(\frac { 3×4×10^{-3} }{ 3×10^{8} }\) =  4 × 10-11 s

Question 89.
The critical angle of a medium with respect to air is 45°. The refractive index of medium is
(a) 1.41
(b) 1.2
(c) 1.5
(d) 2
Answer:
(a) 1.41
Hint:
μ = \(\frac{1}{\sin i_{c}}\) = \(\frac { 1 }{ sin 45° }\) = \(\frac { 1 }{ 1/√2 }\) ≈ 1.41

Question 90.
If the critical angle for total internal reflection from a medium to vacuum is 30°, then velocity of light in the medium is
(a) 6 x 108 m/sec
(b) 2 x 108 m/sec
(c) 3 x 108 m/sec
(d) 1.5 x 108 m/sec
Answer:
(c) 3 x 108 m/sec
Hint:
μ = \(\frac{1}{\sin i_{c}}\) = \(\frac { c }{ v }\)

SamacheerKalvi.Guru

Question 91.
When a ray of light enter from one medium to another, its velocity is doubled. The critical angle for the ray for two internal reflection will be
(a) 30°
(b) 60°
(c) 90°
(d) Information is incomplete
Answer:
(a) 30°
Hint:
μ = \(\frac{1}{\sin i_{c}}\) = \(\frac { { v }_{ 1 } }{ { v }_{ 2 } } \) =\(\frac { { 2v }_{ 2 } }{ { v }_{ 2 } } \) = 2 ; sin ic = \(\frac { 1 }{ 2 }\) ∴ic = 30°

Question 92.
A driver at a depth 12 m inside water (μ = 4/3) see the sky in a cone of semi-vertical angle is
(a) sin-1\(\left( \frac { 4 }{ 3 } \right) \)
(b) tan-1\(\left( \frac { 4 }{ 3 } \right) \)
(c) sin-1\(\left( \frac { 3 }{ 4 } \right) \)
(d) 90°
Answer:
(c) sin-1\(\left( \frac { 3 }{ 4 } \right) \)
Hint:
Required semi vertical angle = Critical angle ic = sin-1 \(\frac { 1 }{ μ }\) = sin-1\(\left( \frac { 3 }{ 4 } \right) \)

Question 93.
The principle behind optical fibres is
(a) total internal reflection
(b) total external reflection
(c) both (a) and (b)
(d) diffraction
Answer:
(a) total internal reflection
Hint:
Optical fibres work on the principle of total internal reflection.

Question 94.
Air bubble in water behaves as
(а) some times concave, sometimes convex lens
(b) concave lens
(c) convex lens
(d) always refracting surface
Answer:
(b) concave lens

Question 95.
A convex lens of 40 cm focal length is combined with a concave lens of focal length 25 cm. The power of combination is
(a) -1.5 D
(b) – 6.5 D
(c) + 6.6 D
(d) + 6.5 D
Answer:
(a) -1.5 D
Hint:
P = P1 + P2 = \(\frac { 1 }{ { f }_{ 1 } } \) + \(\frac { 1 }{ { f }_{ 2 } } \) = \(\frac { 100 }{ 40 }\) + \(\frac { 100 }{ -25 }\) = -1.5d.

SamacheerKalvi.Guru

Question 96.
Two thin lenses, one of focal length + 60 cm and the other of focal length – 20 cm are put in contact, the combined focal length is,
(a) 15 cm
(b) – 15 cm
(c) – 30 cm
(d) 30 cm
Answer:
(c) – 30 cm
Hint:
F = \(\frac { { f }_{ 1 }{ f }_{ 2 } }{ { f }_{ 1 }{ +f }_{ 2 } } \) = \(\frac { 60 ×(-20) }{ 60-20 }\) = 30 cm

Question 97.
How does refractive index (μ) of a material vary with respect to wavelength (λ). (A and B are constants).
(a) μ = A + \(\frac { B }{ { \lambda }^{ 2 } } \)
(b) μ = A + Bλ2
(c) μ = A + \(\frac { B }{ λ }\)
(d) μ = A + Bλ
Answer:
(a) μ = A + \(\frac { B }{ { \lambda }^{ 2 } } \)
Hint:
According to cauchy’s relation, μ = A + \(\frac { B }{ { \lambda }^{ 2 } } \)

Question 98.
a prism of a refracting angle 60° is made with a material of refractive index p. For a certain wavelength of light, the angle of minimum deviation is 30°. For this wavelength, the value of p of material is
(a) 1.820
(b) 1.414
(c) 1.503
(d) 1.231
Answer:
(b) 1.414
Hint:
μ = \(\frac{\sin \left(\frac{60^{\circ}+30^{\circ}}{2}\right)}{\sin 30^{\circ}}\) = \(\frac { sin 45° }{ sin 30° }\) = \(\frac { 1 }{ √2 }\) x \(\frac { 2 }{ 1 }\) √2 = 1.414

Question 99.
Refractive index of red and violet light are 1.52 and 1.54 respectively. If the angle of prism is 10°, the angular dispersion will be
(a) 0.02°
(b) 0.20°
(c) 3.06°
(d) 30.6°
Answer:
(b) 0.20°
Hint:
Angular dispersion = δV – δR = A(μV – μV) = 10° (1.54 – 1.52) = 0.20°

Question 100.
In a simple microscope, if the final image is located at 25 cm from the eye placed close to the lens, then magnifying power is
(a) \(\frac { 25 }{ f }\)
(b) 1 + \(\frac { 25 }{ f }\)
(c) \(\frac { f }{ 25 }\)
(d) \(\frac { f }{ 25 }\) + 1
Answer:
(b) 1 + \(\frac { 25 }{ f }\)
Hint:
When the final image is formed at the least distance of distinct vision in a simple microscope,
m = 1 + \(\frac { 25 }{ f }\)

SamacheerKalvi.Guru

Question 101.
Magnification at least distance of distinct vision of a simple microscope of focal length 5 cm is
(a) 2
(b) 5
(c) 4
(d) 6
Answer:
(d) 6
Hint:
m = 1 + \(\frac { 25 }{ f }\) = 1 + \(\frac { 25 }{ 5 }\) = 6

Question 102.
Magnification of a compound microscope is 30. Focal length of eyepiece is 5 cm and the image is formed at a distance of distinct vision of 25 cm. The magnification of the objective
lens is
(a) 6
(b) 5
(c) 7.5
(d) 10
Answer:
(b) 5
Hint:
me = \(\frac { v }{ u}\) = \(\frac { D }{ { u }_{ c } } \) = 1 + \(\frac { D }{ { f }_{ e } } \) = 1 + \(\frac { 25 }{ 5 }\) = 6
For the compound microscope, m = m0 x me ⇒ 30 = m0 x 6 (or) m0 = 5

Question 103.
The astronomical microscope consists of objective and eyepiece. The focal length of the objective is
(a) equal to that of the eyepiece
(b) shorter than that of the eyepiece
(c) greater than that of the eyepiece
(d) five times shorter than that of eyepiece
Answer:
(c) greater than that of the eyepiece
Hint:
For producing large magnification,f0 > fe

Question 104.
The number of lenses in terrestrial telescope is
(a) 2
(b) 4
(c) 3
(d) 6
Answer:
(c) 3
Hint:
A terrestrial telescope consists of three lenses: objective, erecting lens and eyepiece.

Question 105.
An achromatic combination of lenses is formed by joining
(a) 2 convex lens
(b) 1 convex, 1 concave lens
(c) 2 concave lenses
(d) 1 convex and 1 plane mirror
Answer:
(b) 1 convex, 1 concave lens
Hint:
An achromatic doublet should satisfy the condition \(\frac { { w }_{ 1 } }{ { f }_{ 1 } } \) + \(\frac { { w }_{ 2 } }{ { f }_{ 2 } } \) = 0.

SamacheerKalvi.Guru

Question 106.
The amount of light received by a camera depends upon
(a) diameter only
(b) ratio of focal length and diameter
(c) product of focal length and diameter
(d) only one of the focal length
Answer:
(b) ratio of focal length and diameter
Hint:
The amount of light received by a camera depends on the ratio of the focal length and diameter of the aperture.

Question 107.
Myopia is corrected by using a
(a) cylindrical lens
(b) bifocal lens
(c) convex lens
(d) concave lens
Answer:
(d) concave lens

Question 108.
The critical angle for total internal reflection in diamond is 24.5°. The angle refractive index of diamond is
(a) 2.41
(b) 1.41
(c) 2.59
(d) 1.59
Answer:
(a) 2.41
Hint:
μ = \(\frac { 1 }{ { sin i }_{ c } } \) = \(\frac { 1 }{ sin 24.5° }\) = 2.41

Question 109.
When a glass lens with μ = 1.47 is immersed in a trough of liquid, it looks to be disappeared. The liquid in the trough could be
(a) water
(b) kerosene
(c) glycerine
(d) alcohol
Answer:
(c) glycerine
Hint:
Glass lens will disappear, if µl = µg.

Question 110.
In optical fibres, the refractive index of the core is
(a) greater than that of the cladding
(b) equal to that of the cladding
(c) smaller than that of the cladding
(d) independent of that of the cladding
Answer:
(a) greater than that of the cladding
Hint:
In optical fibres, refractive index of core material > refractive index of the cladding.

SamacheerKalvi.Guru

Question 111.
For a wavelength of light ‘λ’ and scattering object of size ‘a’, all wavelength are scattered nearly equally, if
(a) a = λ
(b) a >> λ
(c) a << λ
(d) a ≥ λ
Answer: (b) a >> λ
Hint:
For a >> λ, the scattering power is not selective.

Question 112.
Two coherent monochromatic light beams of intensities I and 4I are supperposed. The maximum and minimum possible intensities in the resulting beams are
(a) 5I and I
(b) 9I and I
(c) 5I and 3I
(d) 9I and 3I
Answer:
(b) 9I and I
Hint:
Imax = \((\sqrt{\mathrm{I}_{1}}+\sqrt{\mathrm{I}_{2}})^{2}\) = \((\sqrt{41}+\sqrt{1})^{2}\) = 9I
Imax = \((\sqrt{\mathrm{I}_{1}}-\sqrt{\mathrm{I}_{2}})^{2}\) = \((\sqrt{41}-\sqrt{1})^{2}\) = I

Question 113.
screen is doubled. The fringe width is
(a) unchanged
(b) halved
(c) doubled
(d) quadrupled
Answer:
(d) quadrupled
Hint:
β = \(\frac { λD }{ d }\) ; β = \(\frac { λ ×2D }{ D/2 }\) = 4β

Question 114.
In a young’s double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen, when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by
(a) 12
(b) 18
(c) 24
(d) 30
Answer:
(b) 18
Hint:
n1λ1= n2λ2 ⇒12 x 600 = n2 x 400 or n2 = 18

Question 115.
Consider ffaunhoffer diffraction pattern obtained with a single slit illuminated at normal incident. At the angular position of the first diffraction minimum the phase difference between the wavelets from the opposite edges of the slits is
(a) \(\frac { π }{ 4 }\)
(b) \(\frac { π }{ 2 }\)
(c) 2π
(d) π
Answer:
(c) 2π
Hint:
At the angular position of first minimum wavelets from opposite edges of the slits have a path difference of X and a phase difference of 2π radian.

SamacheerKalvi.Guru

Question 116.
A beam of light of wavelength 600 nm from a distant source falls on a single slit 1.00 mm wide and the resulting diffraction pattern is observed on a screen 2m away. The distance between the first dark fringes on either side of the central bright fringe is
(a) 1.2 cm
(b) 1.2 mm
(c) 2.4 cm
(d) 2.4 mm
Answer:
(c) 2.4 cm
Hint:
Distance between the first dark fringes on either side = width of central maximum
= \(\frac { 2Dλ }{ d }\) = \(\frac{2 \times 2 \times 600 \times 10^{-9}}{1.00 \times 10^{-3}}\) m = 2.4 x 10-3 m = 2.4mm

Question 117.
A young’s double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is.
(a) hyperbola
(b) circle
(c) straight line
(d) parabola
Answer:
(a) hyperbola
Hint:
In young’s double slit experiment, the fringes obtained are hyperbolic in shape. But in a small interference pattern, the fringes appear straight.

Question 118.
The initial shape of the wavefront of the beam is
(a) planar
(b) convex
(c) concave
(d) convex near the axis and concave near the periphery
Answer:
(a) planar
Hint:
As the beam is initially parallel, the shape of wavefront is planar.

Question 119.
The angle of incident at which reflected light is totally polarised for reflection from air to glass (refractive index μ) is
(a) sin-1 (μ)
(b) sin-1 \(\left( \frac { 1 }{ μ } \right) \)
(c) tan-1 \(\left( \frac { 1 }{ μ } \right) \)
(d) tan-1 (μ)
Answer:
(d) tan-1 (μ)
According to Brewster’s law,
μ = tan ip ∴ ip = tan-1 (μ)

Question 120.
According to Huygen’s principle, light is a form of
(a) particle
(b) rays
(c) wave
(d) none of the above
Answer:
(c) wave
Hint:
According to Huygen’s principle, light travels in the form of a longitudinal wave.

SamacheerKalvi.Guru

Question 121.
Which one of the following phenomena is not explained by Huygen’s construction of wavefront?
(a) refraction
(b) reflection
(c) diffraction
(d) origin of spectra
Answer:
(d) origin of spectra
Hint:
Huygen’s construction of wavefront cannot explain origin of spectra which can be explained on the basis of quantum theory.

Samacheer Kalvi 12th Physics Optics Additional Problems

Question 1.
Light from a point source in air falls on a convex spherical glass surface (n = 1.5, radius of curvature = 20 cm). The distance of light source from the glass surface is 100 cm. At
What position is the image formed?
Solution:
n1 = 1; n2 = 1.5
u = 100cm ; R = + 20 cm
(R is + Ve for a convex refracting surfce)
As \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 2 }\)
\(\frac{n_{2}}{v}-\frac{n_{1}}{u}\) = \(\frac{n_{2}-n_{1}}{R}\)
\(\frac{1.5}{v}+\frac{1}{100}\) = \(\frac{1.5-1}{20}=\frac{1}{40}\)
\(\frac { 3 }{ 2v }\) = \(\frac{1}{40}-\frac{1}{100}=\frac{5-2}{200}=\frac{3}{200}\)
\(\frac { 1 }{ v }\) = \(\frac { 1 }{ 100 }\)
v = 100 cm
Thus the image is formed at a distance of 100 cm from the glass surface in the direction of incident light.

Question 2.
Find the value of critical angle for a material of refractive index √3 .
Solution:
Here, n = √3
sin ic = \(\frac { 1 }{ n }\) = \(\frac { 1 }{ √3 }\) = \(\frac { √3 }{ 3 }\)
∴ critical angle, ic = 35.3°

Question 3.
The radius of curvature of each face of biconcave lens, made of glass of refractive index 1.5 is 30 cm. Calculate the focal length of the lens in air.
Solution:
Here n = 1.5 ; R1 = – 30 cm ; R2 = 30 cm
Using len’s maker’s formula,
\(\frac { 1 }{ f }\) = (n – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
= (1.5-1)\(\left[\frac{1}{-30}-\frac{1}{30}\right]\) = 0.5 × \(\left(\frac{-2}{30}\right)\)
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ 30 }\)
f = -30

Question 4.
The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. If focal length is 12 cm. What is the refractive index of glass?
Solution:
f = +12 cm ; R1 = 10 cm ; R2 = – 15 cm ; n = ?
As, \(\frac { 1 }{ f }\) = (n – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
\(\frac { 1 }{ 12 }\) = (n – 1) \(\left(\frac{1}{10}+\frac{1}{15}\right)\) = (n – 1) x latex]\frac { 5 }{ 30 }[/latex]
(n – 1) = \(\frac { 6 }{ 12 }\) = 0.5
n = 0.5 +1
n = 1.5

Question 5.
A double convex lens made of glass of refractive index 1.5 has its both surfaces of equal radii of curvature of 20 cm each. An object of 5 cm height is placed at a distance of 10 cm from the lens. Find the position, nature and size of the image.
Solution:
Here n = 1.5 ; R1= +20 cm ; R2 = – 20 cm
Using lens maker’s formula,
\(\frac { 1 }{ f }\) = (n – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
(1.5-1) = \(\left[\frac{1}{20}-\frac{1}{-20}\right]\) 0.5 x \(\left(\frac{2}{20}\right)\)
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ 20 }\)
f = 20 cm
Now , u = -10 cm andf = + 20 cm
From thin lens formula, \(\frac { 1 }{ v }\) = \(\frac { 1 }{ f }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ 20 }\) – \(\frac { 1 }{ 10 }\) = \(\frac { 1 }{ 20 }\)
∴ v = -20 cm
Magnification, m = \(\frac{h_{2}}{h_{1}}\)= \(\frac { v }{ u }\)
\(\frac{h_{2}}{5cm}\) = \(\frac { -20 }{ -10 }\)
h2
Hence a virtual and erect image of height 10 cm is formed at a distance of 20 cm from the lens on the same side as the the object.

SamacheerKalvi.Guru

Question 6.
The image obtained with a convex lens is erect and its length is four times the length of the object. If the focal length of the lens is 20 cm, calculate the object and image distances.
Solution.
Heref= 20 cm, m = + 4 for a virtual image.
To calculate u, we have
m = \(\frac { f }{ u+f }\)

Question 7.
The radius of curvature of each surface of a convex lens of refractive index 1.5 is 40 cm. Calculate its power.
Solution:
Here, n = 105 ; R1 = + 40cm = 0.40 m
R2 = – 40cm = – 0.40 m
Power (p) = \(\frac { 1 }{ f }\) = (n – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
= (1.5 – 1)\(\left[\frac{1}{0.40}-\frac{1}{(-0.40)}\right]\) = 0.5 x \(\frac { 2 }{ 0.40 }\)
p = 2.5 D

Question 8.
A ray of light incident on an equilateral glass prism shows minimum deviation of 30°. Calculate the speed of light through the prism.
Solution:
Here, A = 60° ; D = 30°
Refractive index, n = \(\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}\) = \(\frac{\sin \left(\frac{60+30}{2}\right)}{\sin \left(\frac{60}{2}\right)}\)
n = \(\frac{\sin 45^{\circ}}{\sin 30^{\circ}}\) = \(\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\) = √2
n = 1.414
Velocity or light in glass, v = \(\frac { c }{ n }\) = \(\frac{3 \times 10^{8}}{1.414}\)
v = 2.12 x 108 ms-1

Question 9.
Two sources of intensity I and 41 are used in an interference experiment. Find the intensity at points where the waves from two sources superimpose with a phase
(i) Zero
(ii) \(\frac { π }{ 2 }\)
(iii) π
Solution:
The resultant intensity at a point where phase difference is Φ is
IR+I1+ I2 2\(+\sqrt{\mathrm{I}_{1} \mathrm{I}_{2}}\) cos Φ
As I1 + I and I2 = 4I, therefore,
IR = I + 4I + 2\(\sqrt{I.4I}\) cos Φ
IR = 5I + 4I cos Φ
(i) When Φ = 0 ; IR = 5I + 4I cos 0 = 9I
(ii) When Φ = \(\frac { π }{ 2 }\) ; IR = 5I + 4I cos \(\frac { π }{ 2 }\) = 5I
(iii) When Φ = π ; IR = 5I + 4I cos π = 51 – 41 = I
Φ = 0 ; IR = 9I
Φ = \(\frac { π }{ 2 }\) ; IR = 5I
Φ = π ; IR = I

Question 10.
Assume that light of wavelength 600 A is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch?
Solution:.
The limit of resolution of a telescope, dθ = \(\frac { 1.22λ }{ D }\)
D = 100 inch = 254 cm
[∴ 1 inch = 2.54 cm]
λ = 6000 Å = 6000 x 10-10 m/s
dθ = \(\frac{1.22 \times 6000 \times 10^{-10}}{254 \times 10^{-2}}\) = 2.9 x 10-7
dθ = 2.9 x 10-7 rad.

SamacheerKalvi.Guru

Question 11.
Two polarising sheet have their polarising directions parallel so that the intensity of the transmitted light is maximum. Through what angle must the either sheet be turned if the intensity is to drop by one-half?
Solution:
Here, I = \(\frac{I_{o}}{2}\)
Using Malus law,,
I = Io cos2 θ
\(\frac{I_{o}}{2}\) = Io cos2 θ
cos θ ± \(\frac { 1 }{ √2 }\)
θ = ± 45°, ± 135°

Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics Read More »

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Students can Download Physics Chapter 4 Work, Energy and Power Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Physics Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Samacheer Kalvi 11th Physics Work, Energy and Power Textual Questions Solved

Samacheer Kalvi 11th Physics Work, Energy and Power Multiple Choice Questions
Question 1.
A uniform force of (\(2 \hat{i}+\hat{j}\)) + N acts on a particle of mass 1 kg. The particle displaces from position \((3 \hat{j}+\hat{k})\) m to \((5 \hat{i}+3 \hat{j})\) m. Th e work done by the force on the particle is
[AIPMT model 2013]
(a) 9 J
(b) 6 J
(c) 10 J
(d) 12 J
Answer:
(c) 10 J

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 2.
A ball of mass 1 kg and another of mass 2 kg are dropped from a tall building whose height is 80 m. After, a fall of 40 m each towards Earth, their respective kinetic energies will be in the ratio of [AIPMT model 2004]
(a) \(\sqrt{2}\) : 1
(b) 1 : \(\sqrt{2}\)
(c) 2 : 1
(d) 1 : 23
Answer:
(d) 1 : 23

Question 3.
A body of mass 1 kg is thrown upwards with a velocity 20 m s-1. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction?
(Take g = 10 ms-2) [AIPMT 2009]
(a) 20 J
(b) 30 J
(c) 40 J
(d) 10 J
Answer:
(a) 20 J

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 4.
An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water of the jet. What is the rate at which kinetic energy is imparted to water ? [AIPMT 2009]
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 1
Answer:
(a) \(\frac{1}{2} m v^{2}\)

Question 5.
A body of mass 4 m is lying in xv-plane at rest. It suddenly explodes into three pieces. Two pieces each of mass m move perpendicular to each other with equal speed v the total kinetic energy generated due to explosion is [AIPMT 2014]
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 7
Answer:
(b) \(\frac{3}{2} m v^{2}\)

The potential energy calculator to find how much energy is stored in an object raised off the ground.

Question 6.
The potential energy of a system increases, if work is done
(a) by the system against a conservative force
(b) by the system against a non-conservative force
(c) upon the system by a conservative force
(d) upon the system by a non-conservative force
Answer:
(a) by the system against a conservative force

Question 7.
What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 8
Answer:
(c) \(\sqrt{5 g R}\)

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 8.
The work done by the conservative force for a closed path is
(a) always negative
(b) zero
(c) always positive
(d) not defined
Answer:
(b) zero

Question 9.
If the linear momentum of the obj ect is increased by 0.1 %, then the kinetic energy is increased by
(a) 0.1%
(b) 0.2%
(c) 0.4%
(d) 0.01%
Answer:
(b) 0.2%

Question 10.
If the potential energy of the particle is Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 7031, then force experienced by the particle is
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 16
Answer:
(c) F = -βx

Question 11.
A wind-powered generator converts wind energy into electric energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to
(a) v
(b) v2
(c) v3
(d) v4
Answer:
(c) v4

Question 12.
Two equal masses m1 and m2 are moving along the same straight line with velocities 5 ms-1 and -9 ms-1 respectively. If the collision is elastic, then calculate the velocities after the collision of Wj and m2, respectively
(a) -4 ms-1 and 10 ms-1
(b) 10 ms-1 and 0 ms-1
(c) -9 ms-1 and 5 ms-1
(d) 5 ms-1 and 1 ms-1
Answer:
(c) -9 ms-1 and 5 ms-1

Question 13.
A particle is placed at the origin and a force F = kx is acting on it (where k is a positive constant). If U(0) = 0, the graph of U(x) versus x will be (where U is the potential energy function) [IIT 2004]
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 20
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 21

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 14.
A particle which is constrained to move along x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x) = -kx + ax3. Here, k and a are positive constants. For x ≥ 0, the functional form of the potential energy U(x) of the particle is [IIT 2002]
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 22
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 23

Question 15.
A spring of force constant k is cut into two pieces such that one piece is double the length of the other. Then, the long piece will have a force constant of
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 24
Answer:
(b) \(\frac{3}{2} k\)

Samacheer Kalvi 11th Physics Work, Energy and Power Short Answer Questions

Question 1.
Explain how the definition of work in physics is different from general perception.
Answer:
The term work is used in diverse contexts in daily life. It refers to both physical as well as mental work. In fact, any activity can generally be called as work. But in Physics, the.term work is treated as a physical quantity with a precise definition. Work is said to be done by the force when the force applied on a body displaces it.

Question 2.
Write the various types of potential energy. Explain the formulae.
Answer:
(a) U = mgh
U – Gravitational potential energy
m – Mass of the object,
g – acceleration due to gravity
h – Height from the ground,
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 26
u – Elastic potential energy
k – String constant; x-displacement.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 27
U – electrostatic potential energy
\(\varepsilon_{0}\) = absolute permittivity
q1, q2 – electric charges

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 3.
Write the differences between conservative and non-conservative forces. Give two examples each.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 28

Question 4.
Explain the characteristics of elastic and inelastic collision.
Answer:
In any collision process, the total linear momentum and total energy are always conserved whereas the total kinetic energy need not be conserved always. Some part of the initial kinetic energy is transformed to other forms of energy. This is because, the impact of collisions and deformation occurring due to collisions may in general, produce heat, sound, light etc. By taking these effects into account, we classify the types of collisions as follows:
(a) Elastic collision
(b) Inelastic collision
(a) Elastic collision: In a collision, the total initial kinetic energy of the bodies (before collision) is equal to the total final kinetic energy of the bodies (after collision) then, it is called as elastic collision, i.e.,
Total kinetic energy before collision = Total kinetic energy after collision
(b) Inelastic collision: In a collision, the total initial kinetic energy of the bodies (before collision) is not equal to the total final kinetic energy of the bodies (after collision) then, it is called as inelastic collision, i.e.,
Total kinetic energy before collision ≠ Total kinetic energy after collision
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 261
Even though kinetic energy is not conserved but the total energy is conserved. This is because the total energy contains the kinetic energy term and also a term ∆Q, which includes all the losses that take place during collision. Note that loss in kinetic energy during collision is transformed to another form of energy like sound, thermal, etc. Further, if the two colliding bodies stick together after collision such collisions are known as completely inelastic collision or perfectly inelastic collision. Such a collision is found very often. For example when a clay putty is thrown on a moving vehicle, the clay putty (or Bubblegum) sticks to the moving vehicle and they move together with the same velocity.

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 5.
Define the following
(a) Coefficient of restitution
(b) Power
(c) Law of conservation of energy
(d) Loss of kinetic energy in inelastic collision.
Answer:
(a) The ratio of velocity of separation after collision to the velocity of approach before collision
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 281
(b) Power is defined as the rate of work done or energy delivered
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 29
Its unit is watt.
(c) The law of conservation of energy states that energy can neither be created nor destroyed. It may be transformed from one form to another but the total energy of an isolated system remains constant.
(d) In perfectly inelastic collision, the loss in kinetic energy during collision is transformed to another form of energy like sound, thermal, heat, light etc. Let KEi be the total kinetic energy before collision and KEf be the total kinetic energy after collision.
Total kinetic energy before collision,
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 30
Total kinetic energy after Collision,
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 31
Then the loss of kinetic energy is
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 32

Samacheer Kalvi 11th Physics Work, Energy and Power Long Answer Questions

Question 1.
Explain with graphs the difference between work done by a constant force and by a variable force.
Answer:
Work done by a constant force: When a constant force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation,
dW = (F cos θ) dr ..(1)
The total work done in producing a displacement from initial position ri to final position rf is,
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 33
The graphical representation of the work done by a constant force is shown in figure given below. The area under the graph shows the work done by the constant force.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 34
Work done by a variable force: When the component of a variable force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation
dW = F cos θ dr [F cos θ is the component of the variable force F]
where, F and θ are variables. The total work done for a displacement from initial position ri to final position rf is given by the relation,
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 35
A graphical representation of the work done by a variable force is shown in figure given below. The area under the graph is the work done by the variable force.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 36

Question 2.
State and explain work energy principle. Mention any three examples for it.
Answer:
(i) If the work done by the force on the body is positive then its kinetic energy increases.
(ii) If the work done by the force on the body is negative then its kinetic energy decreases.
(iii) If there is no work done by the force on the body then there is no change in its kinetic energy, which means that the body has moved at constant speed provided its mass remains constant.
(iv) When a particle moves with constant speed in a circle, there is no change in the kinetic energy of the particle. So according to work energy principle, the work done by centripetal force is zero.

Question 3.
Arrive at an expression for power and velocity. Give some examples for the same.
Answer:
The work done by a force \(\overrightarrow{\mathrm{F}}\) for a displacement \(d \vec{r}\) is
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 37
Left hand side of the equation (i) can be written as
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 38
Since, velocity is Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 39. Right hand side of the equation (i) can be written as dt
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 391
Substituting equation (ii) and equation (iii) in equation (i), we get
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 40
This relation is true for any arbitrary value of dt. This implies that the term within the bracket must be equal to zero, i.e.,
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 41
Hence power \(\mathrm{P}=\overrightarrow{\mathrm{F}} \cdot \vec{v}\)

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 4.
Arrive at an expression for elastic collision in one dimension and discuss various cases.
Answer:
Consider two elastic bodies of masses m1 and m2 moving in a straight line (along positive x direction) on a frictionless horizontal surface as shown in figure given below.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 42
In order to have collision, we assume that the mass m] moves faster than mass m2 i.e., u1 > u2. For elastic collision, the total linear momentum and kinetic energies of the two bodies before and after collision must remain the same.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 43
From the law of conservation of linear momentum,
Total momentum before collision (pi) = Total momentum after collision (pf)
m1u1 + m2u2 = m1v1 + m2v2 …(i)
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 44
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 45
This means that for any elastic head on collision, the relative speed of the two elastic bodies after the collision has the same magnitude as before collision but in opposite direction. Further note that this result is independent of mass.
Rewriting the above equation for v1 and v2,
v1 = v2 + u2 – u2 …(vi)
Or v2 = u1 + v1 – u2 …(vii)
To find the final velocities v1 and v2:
Substituting equation (vii) in equation (ii) gives the velocity of as m1 as
m1 (u1 – v1) = m2(u1 + v1 – u2 – u2)
m1 (u1 – y1) = m2 (u1 + + v1  – 2u2)
m1u1 – m1v1 = m2u1 + m2v1 + 2m2u2
m1u1 – m2u1 + 2m2u2 = m1v1 + m2v1
(m1– m2) u1 + 2m2u2 = (m1 + m2) v1
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 46
Similarly, by substituting (vi) in equation (ii) or substituting equation (viii) in equation (vii), we get the final velocity of m2 as
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 47
Case 1: When bodies has the same mass i.e., m1 = m2,
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 48
The equations (x) and (xi) show that in one dimensional elastic collision, when two bodies of equal mass collide after the collision their velocities are exchanged.
Case 2: When bodies have the same mass i.e., m1 = m2 and second body (usually called target) is at rest (u2 = 0),
By substituting m1 = m2 = and u2 = 0 in equations (viii) and equations (ix) we get,
from equation (viii) ⇒ v1 = 0 …(xii)
from equation (ix) ⇒ v2 = u1 ….. (xiii)
Equations (xii) and (xiii) show that when the first body comes to rest the second body moves with the initial velocity of the first body.
Case 3: The first body is very much lighter than the second body
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 49
Dividing numerator and denominator of equation (viii) by m2, we get
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 50
Similarly the numerator and denominator of equation (ix) by m2, we get
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 501
The equation (xiv) implies that the first body which is lighter returns back (rebounds) in the opposite direction with the same initial velocity as it has a negative sign. The equation (xv) implies that the second body which is heavier in mass continues to remain at rest even after collision. For example, if a ball is thrown at a fixed wall, the ball will bounce back from the wall with the same velocity with which it was thrown but in opposite direction.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 51
Dividing numerator and denominator of equation (xiii) by m1, we get
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 52
The equation (xvi) implies that the first body which is heavier continues to move with the same initial velocity. The equation (xvii) suggests that the second body which is lighter will move with twice the initial velocity of the first body. It means that the lighter body is thrown away from the point of collision.

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 5.
What is inelastic collision? In which way it is different from elastic collision. Mention few examples in day to day life for inelastic collision.
Answer:
Inelastic collision: In a collision, the total initial kinetic energy of the bodies (before collision) is not equal to the total final kinetic energy of the bodies (after collision) then, it is called as inelastic collision, i.e.,
Total kinetic energy before collision ≠ Total kinetic energy after collision
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 53
Even though kinetic energy is not conserved but the total energy is conserved. This is because the total energy contains the kinetic energy term and also a term ∆Q, which includes all the losses that take place during collision. Note that loss in kinetic energy during collision is transformed to another form of energy like sound, thermal, etc. Further, if the two colliding bodies stick together after collision such collisions are known as completely inelastic collision or perfectly inelastic collision. Such a collision is found very often. For example when a clay putty is thrown on a moving vehicle, the clay putty (or Bubblegum) sticks to the moving vehicle and they move together with the same velocity.
Difference between Elastic & in elastic collision
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 54

Samacheer Kalvi 11th Physics Numerical Problems

Question 1.
Calculate the work done by a force of 30N in lifting a load of 2 Kg to a height of 10m(g = 10 ms-1)
Answer:
Given: F = 30 N, load (m) = 2 kg; height = 10 m, g = 10 ms-2
Gravitational force F = mg = 30 N
The distance moved h = 10 m
Work done on the object W = Fh = 30 × 10 = 300 J.

Question 2.
A ball with a velocity of 5 ms-1 impinges at angle of 60° with the vertical on a smooth horizontal plane. If the coefficient of restitution is 0.5, find the velocity and direction after the impact.
Answer:
Given: Velocity of ball: 5 ms-1
Angle of inclination with vertical: 60°
Coefficient of restitution = 0.5.
Note: Let the angle reflection is θ’ and the speed after collision is v’. The floor exerts a force on the ball along the normal during the collision. There is no force
parallel to the surface. Thus, the parallel component of the velocity of the ball remains unchanged. This gives
v’ sin θ’ = v sin θ …… (i)
Vertical component with respect to floor = v’ cos θ’ (velocity of separation)
Velocity of approach = v cos θ
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 60
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 601
from (i) and (ii)
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 61
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 62

Question 3.
A bob of mass m is attached to one end of the rod of negligible mass and length r, the other end of which is pivoted freely at a fixed center O as shown in the figure.
What initial speed must be given to the object to reach the top of the circle?
(Hint: Use law of conservation of energy). Is this speed. less or greater than speed obtained in the section 4.2.9?
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 63
Answer:
To get the vertical speed given to the object to reach the top of the circle, law of conservation of energy can be used at a points (1) and (2)
Total energy at 1 = Total energy at 2
∴ Potential energy at point 1 = 0
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 64
from eqn (i)
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 641
In this case bob of mass m is connected with a rod of negligible mass, so the velocity of bob at highest point can be equal to zero i.e. v2 = 0
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 65
The speed of bob obtained here is lesser than the speed obtained in section 4.2.9. It is only because of string is replaced by a massless rod here.

Question 4.
Two different unknown masses A and B collide. A is initially at rest when B has a speed v. After collision B has a speed v/2 and moves at right angles to its original direction of motion. Find the direction in which A moves after collision.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 66

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 5.
A bullet of mass 20 g strikes a pendulum of mass 5 kg. The centre of mass of pendulum rises a vertical distance of 10 cm. If the bullet gets embedded into the pendulum, calculate its initial speed.
Answer:
Given: m1 = 20 g = 20 × 10-3 kg; m2 = 5 kg; s = 10 × 10-2 m.
Let the speed of the bullet be v. The common velocity of bullet and pendulum bob is V. According to law of conservation of linear momentum.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 67
The bob with bullet go up with a deceleration of g = 9.8 ms-2. Bob and bullet come to rest at a height of 10 × 10-2 m.
from III rd equation of motion
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 68

Samacheer Kalvi 11th Physics Conceptual Questions

Question 1.
A spring which in initially in un-stretched condition, is first stretched by a length x and again by a further length x. The work done in the first case W1 is one third of the work done in second case W2. True or false?
Answer:
The amount of work done to stretching distance x
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 691
Total work done in stretching the spring through a distance 2x is
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 701
Extra work required to stretch the additional x distance is
W = W2 – W1 = 4W1 – W1 = 3W1
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 712
Hence it is true

Question 2.
Which is conserved in inelastic collision? Total energy (or) Kinetic energy?
Answer:
In inelastic collision total energy is only conserved but kinetic energy is not conserved. A part of kinetic energy is converted into some other form of energy such as sound, heat energy.
Note: The linear momentum is also conserved.

Question 3.
Is there any net work done by external forces on a car moving with a constant speed along a straight road?
Answer:
If the car moves at constant speed, then there is no change in its kinetic energy. It implies that if there is no change in kinetic energy then there is no work done by the force on the body provided its mass remains constant.

Question 4.
A car starts from rest and moves on a surface with uniform acceleration. Draw the graph of kinetic energy versus displacement. What information you can get from that graph?
Answer:
A car starts from rest and moves with uniform acceleration. The graph between kinetic energy and displacement, is a straight line.
The slope of KE and displacement graph gives net force acting on the car to keep the car with uniform acceleration.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 72

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 5.
A charged particle moves towards another charged particle. Under what conditions the total momentum and the total energy of the system conserved?
Answer:
Coulomb force is acting in between the charged particles Internal force is a conservative force. If no external forces act or the work done by external forces is zero then the mechanical energy of the system and also total linear momentum also remains constant.

Samacheer Kalvi 11th Physics Work, Energy and Power Additional Questions Solved

Samacheer Kalvi 11th Physics Multiple Choice Questions

Question 1.
Thrust and linear momentum
(a) Thrust and linear momentum
(b) Work and energy
(c) Work and power
(d) Power and energy
Answer:
(b) Work and energy

Question 2.
The rate of work done is called as
(a) energy
(b) power
(c) force
(d) mechanical energy
Answer:
(b) power

Question 3.
Unit of work done
(a) Nm
(b) joule
(c) either a or b
(d) none
Answer:
(c) either a or b

Question 4.
Dimensional formula for work done is
(a) MLT-1
(b) ML2T2
(c) M-1L-1T2
(d) ML2T-2
Answer:
(d) ML2T-2

Question 5.
When a body moves on a horizontal direction, the amount of work done by the gravitational force is
(a) positive
(b) negative
(c) zero
(d) infinity
Answer:
(c) zero

Question 6.
The amount of work done by centripetal force on the object moving in a circular path is
(a) zero
(b) infinity
(c) positive
(d) negative
Answer:
(a) zero

Question 7.
The work done by the goal keeper catches the ball coming towards him by applying a force is
(a) positive
(b) negative
(c) zero
(d) infinity
Answer:
(b) negative

Question 8.
If the angle between force and displacement is acute then the work done is
(a) positive
(b) negative
(c) zero
(d) maximum
Answer:
(a) positive

Question 9.
If the force and displacement are perpendicular to each other, then the work done is
(a) positive
(b) negative
(c) zero
(d) maximum
Answer:
(c) zero

Question 10.
If the angle between force and displacement is obtuse, then the work done is
(a) positive
(b) negative
(c) zero
(d) minimum
Answer:
(b) negative

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 11.
The area covered under force and displacement graph is
(a) work done
(b) acceleration
(c) power
(d) kinetic energy
Answer:
(a) work done

Question 12.
The capacity to do work is
(a) force
(b) energy
(c) work done
(d) power
Answer:
(b) energy

Question 13.
The energy possessed by a body due to its motion is called as
(a) potential energy
(b) kinetic energy
(c) mechanical energy
(d) none
Answer:
(b) kinetic energy

Question 14.
The energy possessed by the body by virtue of its position is called as
(a) potential energy
(b) kinetic energy
(c) mechanical energy
(d) none
Answer:
(a) potential energy

Question 15.
1 erg is equivalent to
(a) 10-7 J
(b) 1.6 × 10-19 J
(c) 4.186 J
(d) 3.6 × 10-6 J
Answer:
(a) 10-7 J

Question 16.
1 electron volt is equivalent to
(a) 10-7 J
(b) 1.6 × 10-19 J
(c) 4.186 J
(d) 3.6 × 10-6 J
Answer:
(b) 1.6 × 10-19 J

Question 17.
1 kilowatt hour is equivalent to
(a) 10-7 J
(b) 1.6 × 10-19 J
(c) 4.186 J
(d) 3.6 × 10-6 J
Answer:
(d) 3.6 × 10-6 J

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 18.
1 calorie is equivalent to
(a) 10-7 J
(b) 1.6 × 10-19 J
(c) 4.186 J
(d) 3.6 × 106 J
Answer:
(c) 4.186 J

Question 19.
The amount of work done by a moving body depends on the
(a) mass of the body
(b) velocity
(c) both (a) and (b)
(d) time
Answer:
(c) both (a) and (b)

Question 20.
The kinetic energy of a body is given by
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 301
Answer:
(a) \(\frac{1}{2} m v^{2}\)

Question 21.
Kinetic energy of the body is always
(a) zero
(b) infinity
(c) negative
(d) positive
Answer:
(d) positive

Question 22.
If the work done by the force on the body is positive then its kinetic energy
(a) increases
(b) decreases
(c) zero
(d) either increases or decreases
Answer:
(a) increases

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 23.
If p is the momentum of the particle then its kinetic energy is
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 312
Answer:
(c) \(\frac{\mathbf{p}^{2}}{2 \mathbf{m}}\)

Question 24.
If two objects of masses m1 and m2 (m1 > m2) are moving with the same momentum then the kinetic energy will be greater for
(a) m1
(b) m2
(c) m1 or m2
(d) both will have equal kinetic energy
Answer:
(b) m2

Question 25.
For a given momentum, the kinetic energy is proportional to
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 321
Answer:
(b) \(\frac{1}{\mathrm{m}}\)

Question 26.
Elastic potential energy possessed by a spring is
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 331
Answer:
(c) \(\frac{1}{2}\)kx2

Question 27.
Potential energy stored in the spring depends on
(a) spring constant
(b) mass
(c) gravity
(d) length
Answer:
(b) mass

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 28.
Two springs of spring constants k1 and k2 (k1 > k2). If they are stretched by the same force then (u1, u2 are potential energy of the springs) is
(a) u1 > u2
(b) u2 > u1
(c) u1 = u2
(d) u1 ≥ u2
Answer:
(b) u2 > u1

Question 29.
Conservative force is
(a) electrostatic force
(b) magnetic force
(c) gravitational force
(d) all the above
Answer:
(d) all the above

Question 30.
Non conservative force is
(a) frictional force
(b) viscous force
(c) air resistance
(d) all the above
Answer:
(d) all the above

Question 31.
If the work done is completely recoverable, then the force is
(a) conservative
(b) non-conservative
(c) both (a) and (b)
(d) frictional in nature
Answer:
(b) non-conservative

Question 32.
The work done by the conservative forces in a cycle is
(a) zero
(b) one
(c) infinity
(d) having negative value
Answer:
(a) zero

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 33.
Negative gradient of potential energy gives
(a) conservative force
(b) non conservative force
(c) kinetic energy
(d) frictional force
Answer:
(a) conservative force

Question 34.
When a particle moving in a vertical circle, the variable is/are
(a) velocity of the particle
(b) tension of the string
(c) both (a) and (b)
(d) mass of the particle
Answer:
(c) both (a) and (b)

Question 35.
Which of the following is zero at the highest point in vertical circular motion?
(a) velocity of the particle
(b) tension of the spring
(c) potential energy
(d) none
Answer:
(a) velocity of the particle

Question 36.
The body must have a speed at highest point in vertical circular motion to stay in the circular path
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 702
Answer:
(a) \(\geq \sqrt{\mathbf{g r}}\)

Question 37.
The body must have a minimum speed of lowermost point in vertical circular motion to complete the circle
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 713
Answer:
(c) \(\geq \sqrt{5 \mathrm{gr}}\)

Question 38.
The rate of work done is
(a) energy
(b) force
(c) power
(d) energy flow
Answer:
(c) power

Question 39.
The unit of power is
(a) J
(b) W
(c) J s-1
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 40.
One horse power (1 hp) is
(a) 476 W
(b) 674 W
(c) 746 W
(d) 764 W
Answer:
(c) 746 W

Question 41.
The dimension of power is
(a) ML2T-2
(b) ML2T-3
(c) ML-2T2
(d) ML-2T3
Answer:
(b) ML2T-3

Question 42.
kWh is the practical unit of
(a) energy
(b) power
(c) electrical energy
(d) none
Answer:
(a) energy

Question 43.
If a force F is applied on a body and the body moves with velocity v, the power will be
(a) F.V
(b) F/V
(c) FV2
(d) FW2
Answer:
(a) F.V

Question 44.
A body of mass m is thrown vertically upward with a velocity v. The height at which the kinetic energy of the body is one third of its initial value is given by
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 75
Answer:
(c) \(\frac{v^{2}}{6 g}\)
Solution:
Initial Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 76. The loss in K.E will be the gain in potential energy
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 767

Question 45.
A body of mass 5 kg is initially at rest. By applying a force of 20 N at an angle of 60° with horizontal the body is moved to a distance of 4 m. The kinetic energy acquired by the body is
(a) 80 J
(b) 60 J
(c) 40 J
(d) 17.2 J
Answer:
(c) 40 J
Solution:
The work done is equal to its kinetic energy
∴ K.E gained = Fs cos θ = 20 × 4 cos 60° = 40 J.

Question 46.
A bullet is fired normally on an immovable wooden plank of thickness 2 m. It loses 20% of its kinetic energy in penetrating a thickness 0.2 m of the plank. The distance penetrated by the bullet inside the wooden plank is
(a) 0.2 m
(b) 0.8 m
(c) 1 m
(d) 1.5 m
Answer:
(c) 1 m
Solution:
The wood offers a constant retardation. If the bullet loses 20% of its kinetic energy by penetrating 0.2m. it can penetrate further into 4 × 0.2 = 0.8 m with the remaining kinetic energy. So the total distance penetrated by the bullet is 0.2 + 0.8 = 1 m.

Question 47.
Which of the following quantity is conserved in all collision process?
(a) kinetic energy
(b) linear momentum
(c) both (a) and (b)
(d) none.
Answer:
(b) linear momentum

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 48.
The kinetic energy is conserved in
(a) elastic collision
(b) inelastic collision
(c) both (a) and (b)
(d) none
Answer:
(a) Elastic collision

Question 49.
The kinetic energy is not conserved in
(a) Elastic collision
(b) In elastic collision
(c) both (a) and (b)
(d) none
Answer:
(b) In elastic collision

Question 50.
In inelastic collision, which is conserved
(a) linear momentum
(b) total energy
(c) both (a) and (b)
(d) none
Answer:
(c) both (a) and (b)

Question 51.
If the two colliding bodies stick together after collision such collisions are
(a) elastic collision
(b) inelastic collision
(c) perfectly inelastic collision
(d) head on collision
Answer:
(c) perfectly inelastic collision

Question 52.
When bubblegum is thrown on a moving vehicle, it sticks is an example for
(a) elastic collision
(b) inelastic collision
(c) perfectly inelastic collision
(d) none
Answer:
(c) perfectly inelastic collision

Question 53.
Elastic collision is due to
(a) conservative force
(b) non conservative force
(c) gravitational force
(d) electrostatic force
Answer:
(b) non conservative force

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 54.
Inelastic collision is due to
(a) conservative force
(b) non conservative force
(c) gravitational force
(d) electrostatic force
Answer:
(b) non conservative force

Question 55.
If the velocity of separation is equal to the velocity of approach, then the collision is
(a) conservative force
(b) non conservative force
(c) gravitational force
(d) electrostatic force
Answer:
(a) conservative force

Question 56.
For elastic collision, coefficient of restitution is
(a) 0
(b) 1
(c) 0 < e < 1
(d) ∞
Answer:
(b) 1

Question 57.
For inelastic collision co-efficient of restitution is
(a) 0
(b) 1
(c) 0 < e < 1
(d) ∞
Answer:
(c) 0 < e < 1

Question 58.
For perfectly inelastic collision, coefficient of restitution is
(a) 0
(b) 1
(c) 0 < e < 1
(d) ∞
Answer:
(a) 0

Question 59.
The ratio of velocities of equal masses in an inelastic collision with one of the masses is stationary is
60. A box is dragged across a surface by a rope which makes an angle 45° with the horizontal. The tension in the rope is 100 N when the box is dragged 10 m. The work done is
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 80
Answer:
(a) \(\frac{1-e}{1+e}\)

Question 60.
A box is dragged across a surface by a rope which makes an angle 45° with the horizontal. The
tension in the rope is 100 N when the box is dragged 10 m. The work done is
(a) 707.1 J
(b) 607.1 J
(c) 1414.2 J
(d) 900 J
Answer:
(a) 707.1 J
Solution:
The component of force acting along the surface is T cos θ
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 800
∴ Work done = T cos θ × x
= 10o cos 45° × 10
= 707.1 J

Question 61.
A position dependent force F = (7 – 2x + 3x2) N acts on a small body of mass 2 kg and displaces it from x = 0 to x = 5 m. Work done is
(a) 35 J
(b) 70 J
(c) 135 J
(d) 270 J
Answer:
(c) 135 J
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 81

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 62.
In gravitational field, the work done in moving a body from one point into another depends on
(a) initial and final positions
(b) distance between them
(c) actual distance covered
(d) velocity of motion
Answer:
(c) initial and final positions

Question 63.
A particle of mass “m” moving with velocity v strikes a particle of mass “2m” at rest and sticks to it. The speed of the combined mass is
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 82
Answer:
(c) \(\frac{v}{3}\)
Solution:
According to conservation of linear momentum
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 83

Question 64.
A force of (\(10 \hat{i}-3 \hat{j}+6 \hat{k}\)) N acts on a body of 5 kg and displaces it from (\(6 \hat{i}+5 \hat{j}-3 \hat{k}\)) to (\(10 \hat{i}-2 \hat{j}+7 k\)) m. The work done is
(a) 100 J
(b) 0
(c) 121 J
(d) none of these
Answer:
(c) 121 J
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 84

Question 65.
A 9 kg mass and 4 kg mass are moving with equal kinetic energies. The ratio of their momentum is
(a) 1 : 1
(b) 3 : 2
(c) 2 : 3
(d) 9 : 4.
Answer:
(b) 3 : 2
Solution:
Given that K.E are equal
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 85

Question 66.
If momentum of a body increases by 25% its kinetic energy will increase by
(a) 25%
(b) 50%
(c) 125%
(d) 56.25%
Answer:
(d) 56.25%
Solution:
Let momentum of p1 = 100% momentum of p2 = 125%.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 86

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 67.
A missile fired from a launcher explodes in mid air, its total
(a) kinetic energy increases
(b) momentum increases
(c) kinetic energy decreases
(d) momentum decreases
Answer:
(a) kinetic energy increases

Question 68.
A bullet hits and gets embedded in a wooden block resting on a horizontal friction less surface. Which of the following is conserved?
(a) momentum alone
(b) kinetic energy alone
(c) both momentum and kinetic energy
(d) no quantity is conserved
Answer:
(a) momentum alone

Question 69.
Two balls of equal masses moving with velocities 10 m/s and -7 m/s respectively collide elastically. Their velocities after collision will be
(a) 3 ms-1 and 17 ms-1
(b) -7 ms-1 and 10 ms-1
(c) 10 ms-1 and -7 ms-1
(d) 3 ms-1 and -70 ms-1
Answer:
(b) -7 ms-1 and 10 ms-1

Question 70.
A spring of negligible mass having a force constant of 10 Nm-1 is compressed by a force to a distance of 4 cm. A block of mass 900 g is free to leave the top of the spring. If the spring is released, the speed of the block is
(a) 11.3 ms-1
(b) 13.3 × 101 ms-1
(c) 13.3 × 10-2 ms-1
(d) 13.3 × 10-3 ms-1
Answer:
(c) 13.3 × 10-2 ms-1
Solution:
We know that, the potential energy of the spring = \(\frac{1}{2}\)kx2. Here the potential energy of the spring is converted into kinetic energy of the block.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 90

Question 71.
A particle falls from a height ftona fixed horizontal plate and rebounds. If e is the coefficient ” of restitution, the total distance travelled by the particle on rebounding when it stops is
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 91
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 912
S = h + 2e2h + 2e4h + 2e6h + …..
S = h + 2h (e2 + e4 + e6 +…)
By using binomixal expansion we can write it as
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 92

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 72.
If the force F acting on a body as a function of x then the work done in moving a body from x = 1 m to x = 3m is
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 923
(a) 6 J
(b) 4 J
(c) 2.5 J
(d) 1 J
Answer:
(b) 4 J

Question 73.
A boy “A” of mass 50 kg climbs up a staircase in 10 s. Another boy “B” of mass 60 kg climbs up a Same staircase in 15s. The ratio of the power developed by the boys “A” and “B” is
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 93
Answer:
(a) \(\frac{5}{4}\)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 94

Samacheer Kalvi 11th Physics Short Answer Questions

Question 1.
Define work, energy, power.
Answer:
Work: Work is said to be done by the force when the force applied on a body displaces it.
Energy: Energy is defined as the ability to do work.
Power: The rate of work done is called power.

Question 2.
Discuss the possibilities of work done to be zero.
Answer:
Work done is zero in the following cases.
(i) When the force is zero (F = 0). For example, ,a body moving on a horizontal smooth frictionless surface will continue to do so as no force (not even friction) is acting along the plane. (This is an ideal situation.)
(ii) When the displacement is zero (dr = 0). For example, when force is applied on a rigid wall it does not produce any displacement. Hence, the work done is zero as shown in figure.
(iii) When the force and displacement are perpendicular (0 = 90°) to each other, when a body moves on a horizontal direction, the gravitational force (mg) does not work on the body, since it acts at right angles to the displacement as shown in Figure (b). In circular motion the centripetal force does not do work on the object moving on a circle as it is always perpendicular to the displacement as shown in Figure (c).
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 95

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 3.
Derive the relation between momentum and kinetic energy.
Answer:
Consider an object of mass m moving with a velocity v. Then its linear momentum is
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 96
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 97
Multiplying both the numerator and denominator of equation (i) by mass m
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 98
where | \(\vec{p}\) | is the magnitude of the momentum. The magnitude of the linear momentum can be obtained by
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 99
Note that if kinetic energy and mass are given, only the magnitude of the momentum can be calculated but not the direction of momentum. It is because the kinetic energy and mass are scalars.

Question 4.
How can an object move with zero acceleration (constant velocity) when the external force is acting on the object?
Answer:
It is possible when there is another force which acts exactly opposite to the external applied force. They both cancel each other and the resulting net force becomes zero, hence the object moves with zero acceleration.

Question 5.
Why should the object be moved at constant velocity when we define potential energy?
Answer:
If the object does not move at constant velocity, then it will have different velocities at the initial and final locations. According to work-kinetic energy theorem, the external force will impart some extra kinetic energy. But we associate potential energy to the forces like gravitational force, spring force and coulomb force. So the external agency should not impart any kinetic energy when the object is taken from initial to final location.

Question 6.
Derive an expression for potential energy near the surface of the earth.
Answer:
The gravitational potential energy (U) at some height h is equal to the amount of work required to take the object from ground to that height h with constant velocity. Let us consider a body of mass m being moved from ground to the height h against the gravitational force as shown.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 100
The gravitational force \(\overrightarrow{\mathrm{F}}_{g}\) acting on the body is, \(\overrightarrow{\mathrm{F}}_{g}=-m g \hat{j}\) (as Gravitational potential energy the force is in y direction, unit vector \(\hat{j}\) is used). Here, negative sign implies that the force is acting vertically downwards. In order to move the body without acceleration (or with constant velocity), an external applied force \(\overrightarrow{\mathrm{F}}_{a}\), equal in magnitude but opposite to that of gravitational force \(\overrightarrow{\mathrm{F}}_{g}\) has to be applied on the body i.e., \(\overrightarrow{\mathrm{F}}_{a}=-\overrightarrow{\mathrm{F}}_{g}\).
This implies that \(\overrightarrow{\mathrm{F}}_{a}=+m g \hat{j}\). The positive sign implies that the applied force is in vertically upward direction. Hence, when the body is lifted up its velocity remains unchanged and thus its kinetic energy also remains constant.
The gravitational potential energy (U) at some height h is equal to the amount of work required to take the object from the ground to that height h.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 101
Since the displacement and the applied force are in the same upward direction, the angle between them, θ = 0°. Hence, cos 0° = 1 and | \(\overrightarrow{\mathrm{F}}_{a}\) | = mg and | \(d \vec{r}\) | = dr.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 102

Question 7.
Explain force displacement graph for a spring.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 7030
Since the restoring spring force and displacement are linearly related as F = – kx, and are opposite in direction, the graph between F and x is a straight line with dwelling only in the second and fourth quadrant as shown in Figure. The elastic potential energy can be easily calculated by drawing a F – x graph. The shaded area (triangle) is the work done by the spring force.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 103

Question 8.
Explain the potential energy – displacement graph for a spring.
Answer:
A compressed or extended spring will transfer its stored potential energy into kinetic energy of the mass attached to the spring. The potential energy-displacement graph is shown in Figure.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 105
In a frictionless environment, the energy gets transferred from kinetic to potential and potential to kinetic repeatedly such that the total energy of the system remains constant. At the mean position,
∆KE = ∆U

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 9.
Define unit of power.
Answer:
The unit of power is watt. One watt is defined as the power when one joule of work is done in one second.

Question 10.
Define average power and instantaneous power.
Answer:
The average power is defined as the ratio of the total work done to the total time taken.
Pav = total work done/total time taken The instantaneous power is defined as the power delivered at an instant
pinst = dw/dt

Question 11.
Define elastic and inelastic collision.
Answer:
In any collision, if the total kinetic energy of the bodies before collision is equal to the total final kinetic energy of the bodies after collision then it is called as elastic collision.
In a collision the total initial kinetic energy of the bodies before collision is not equal to the . total final kinetic energy of the bodies after collision. Then it is called as inelastic collision.

Question 12.
What will happen to the potential energy of the system.
If (i) Two same charged particles are brought towards each other
(ii) Two oppositely charged particles are brought towards each other.
Answer:
(i) When the same charged particles are brought towards each other, the potential energy of the system will increase. Because work has to be done against the force of repulsion. This work done only stored as potential energy.
(ii) When two oppositely charged particles are brought towards each other, the potential energy of the system will decrease. Because work is done by the force of attraction between the charged particles.

Question 13.
Define the conservative and non-conservative forces. Give examples of each.
Answer:
Conservative force : e.g., Gravitational force, electrostatic force.
Non-Conservative force : e.g., forces of friction, viscosity.

Question 14.
A light body and a heavy body have same linear momentum. Which one has greater K.E ?
Answer:
Lighter body has more K.E. as K.E. = \(\frac{p^{2}}{2 m}\) and for constant p, K.E. \(\propto \frac{1}{m}\)

Question 15.
The momentum of the body is doubled, what % does its K.E change?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 120

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 16.
A body is moving along a circular path. How much work is done by the centripetal force?
Answer:
W = FS cos 90° = 0.

Question 17.
Which spring has greater value of spring constant – a hard spring or a delicate spring?
Answer:
Hard spring.

Question 18.
Two bodies stick together after collision. What type of collision is in between these two bodies? .
Answer:
Inelastic collision.

Question 19.
State the two conditions under which a force does not work.
Answer:

  1. Displacement is zero or it is perpendicular to force.
  2. Conservative force moves a body over a closed path.

Question 20.
How will the momentum of a body changes if its K.E. is doubled?
Answer:
Momentum becomes \(\sqrt{2}\) times.

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 21.
K.E. of a body is increased by 300 %. Find the % increase in its momentum.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 122

Question 22.
A light and a heavy body have same K.E., which of the two have more momentum and why?
Answer:
Heavier body.

Question 23.
Does the P.E. of a spring decreases or increases when it is compressed or stretched?
Answer:
Increases because W.D. on it when it increases is compressed or stretched.

Question 24.
Name a process in which momentum changes but K.E. does not.
Answer:
Uniform circular motion.

Question 25.
What happens to the P.E. of a bubble when it rises in water?
Answer:
Decreases.

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 26.
A body is moving at constant speed over a frictionless surface. What is the work done by the weight of the body?
Answer:
W = 0.

Question 27.
Define spring constant of a spring.
Answer:
It is the restoring force set up in a string per unit extension.

Samacheer Kalvi 11th Physics Short Answer Questions 2 Marks

Question 28.
How much work is done by a coolie walking on a horizontal platform with a load on his head? Explain.
Answer:
W = 0 as his displacement is along the horizontal direction and in order to balance the load on his head, he applies a force on it in the upward direction equal to its weight. Thus angle between force and displacement is zero.

Question 29.
Mountain roads rarely go straight up the slope, but wind up gradually. Why?
Answer:
If roads go straight up then angle of slope 0 would be large so frictional force f = µ mg cos θ would be less and the vehicles may slip. Also greater power would be required.

Question 30.
A truck and a car moving with the same K.E. on a straight road. Their engines are simultaneously switched off which one will stop at a lesser distance?
Answer:
By Work – Energy Theorem,
Loss in K.E. = W.D. against the force × distance of friction
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 125
∴ Truck will stop in a lesser distance because of greater mass.

Question 31.
Is it necessary that work done in the motion of a body over a closed loop is zero for every force in nature? Why?
Answer:
No. W.D. is zero only in case of a conservative force.

Question 32.
How high must a body be lifted to gain an amount of P.E. equal to the K.E. it has when moving at speed 20 ms-1. (The value of acceleration due to gravity at a place is 9.8 ms-2).
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 132

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 33.
Give an example in which a force does work on a body but fails to change its K.E.
Answer:
When a body is pulled on a rough, horizontal surface with constant velocity. Work is done on the body but K.E. remains unchanged.

Question 34.
A bob is pulled sideway so that string becomes parallel to horizontal and released. Length of the pendulum is 2 m. If due to air resistance loss of energy is 10%, what is the speed with which the bob arrived at the lowest point.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 133

Question 35.
Two springs A and B are identical except that A is harder than B (KA > KB) if these are stretched by the equal force. In which spring will more work be done?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 134
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 135

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 36.
Find the work done if a particle moves from position r1 = to a position \((3 \hat{i}+2 \hat{j}-6 \hat{k})\) to a position \(\vec{r}_{2}=(14 \hat{i}+13 \hat{j}-9 \hat{k})\) under the effect of force \(\overrightarrow{\mathrm{F}}=(4 \hat{i}+\hat{j}+3 \hat{k}) \mathrm{N}\)
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 138

Question 37.
Spring A and B are identical except that A is stiffer than B, i.e., force constant kA > kB. In which spring is more work expended if they are stretched by the same amount?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 139

Question 38.
A ball at rest is dropped from a height of 12 m. It loses 25% of its kinetic energy in striking the ground, find the height to which it bounces. How do you account for the loss in kinetic energy?
Answer:
If ball bounces to height h’, then
mgh’ = 75% of mgh
∴ h’ = 0.75 h = 9 m.

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 39.
Which of the two kilowatt hour or electron volt is a bigger unit of energy and by what factor?
Answer:
kwh is a bigger unit of energy.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 140

Question 40.
A spring of force constant K is cut into two equal pieces. Calculate force constant of each part.
Answer:
Force constant of each half becomes twice the force constant of the original spring.

Samacheer Kalvi 11th Physics Short Answer Questions 3 Marks

Question 41.
A car of mass 2000 kg is lifted up a distance of 30 m by a crane in 1 min. A second crane does the same job in 2 min. Do the cranes consume the same or different amounts of fuel? What is the power supplied by each crane? Neglect Power dissipation against friction.
Answer:
t1 = 1 min = 60 s, t2 = 2 min = 120 s
W = Fs = mgs = 5.88 × 105 J
As both cranes do same amount of work so both consume same amount of fuel.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 152

Question 42.
20 J work is required to stretch a spring through 0.1m. Find the force constant of the spring. If the spring is further stretched through 0.1 m, calculate work done.
Answer:
P.E. of spring when stretched through a distance 01m,
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 153
when spring is further stretched through 01m, then P.E. will be :
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 154

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 43.
A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, how much electric power is consumed by the pump. The efficiency of the pump is 30%.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 155

Question 44.
A ball bounces to 80% of its original height. Calculate the mechanical energy lost in each bounce.
Answer:
Let Initial P.E. = mgh
P.E. after first bounce = mg × 80% of h = 0.80 mgh
P.E. lost in each bounce = 0.20 mgh
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 156

Samacheer Kalvi 11th Physics Long Answer Questions

Question 1.
Obtain an expression for the critical vertical of a body revolving in a vertical circle
Answer:
Imagine that a body of mass (m) attached to one end of a massless and inextensible string executes circular motion in a vertical plane with the other end of the string fixed. The length of the string becomes the radius (r) of the circular path (See figure).
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 160
Let us discuss the motion of the body by taking the free body diagram (FBD) at a position where the position vector (\(\vec{r}\)) makes an angle θ with the vertically downward direction and the instantaneous velocity is as shown in Figure.
There are two forces acting on the mass.
1. Gravitational force which acts downward
2. Tension along the string.
Applying Newton’s second law on the mass, in the tangential direction,
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 161
The circle can be divided into four sections A, B, C, D for better understanding of the motion. The four important facts to be understood from the two equations are as follows:
(i) The mass is having tangential acceleration (g sin θ) for all values of θ (except θ = 0°), it is clear that this vertical circular motion is not a uniform circular motion.
(ii) From the equations (ii) and (i) it is understood that as the magnitude of velocity is not a constant in the course of motion, the tension in the string is also not constant.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 162
Hence velocity cannot vanish, even when the tension vanishes.
These points are to be kept in mind while solving problems related to motion in vertical circle.
To start with let us consider only two positions, say the lowest point 1 and the highest point 2 as shown in Figure for further analysis. Let the velocity of the body at the lowest point 1 be \(\vec{v}_{1}\), at the highest point 2 be \(\vec{v}_{2}\) and \(\vec{v}\) at any other point. The direction of velocity is tangential to the circular path at all points. Let \(\overrightarrow{\mathrm{T}}_{1}\) be the tension in the string at the lowest point and \(\overrightarrow{\mathrm{T}}_{2}\) be , the tension at the highest point and \(\overrightarrow{\mathrm{T}}\) be the tension at any other point. Tension at each point acts towards the center. The tensions and velocities at these two points can be found by applying the law of conservation of energy.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 163
For the lowest point (1)
When the body is at the lowest point 1, the gravitational force \(m \vec{g}\) which acts on the body (vertically downwards) and another one is the tension \(\overrightarrow{\mathrm{T}}_{1}\), acting vertically upwards, i.e. towards the center. From the equation (ii), we get
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 164
For the highest point (2)
At the highest point 2, both the gravitational force mg on the body and the tension T2 act downwards, i.e. towards the center again.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 165
From equations (iv) and (ii), it is understood that T1 > T2. The difference in tension T1 – T2 is obtained by subtracting equation (iv) from equation (ii).
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 166
The term Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 7032 can be found easily by applying law of conservation of energy at point 1 and also at point 2.
Note: The tension will not do any work on the mass as the tension and the direction of motion is always perpendicular.
The gravitational force is doing work on the mass, as it is a conservative force the total energy of the mass is conserved throughout the motion.
Total energy at point 1 (E1) is same as the total energy at a point 2 (E2)
E1 = E2
Potential energy at point 1, U1 = 0 (by taking reference as point 1)
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 167
Similarly, Potential energy at point 2, U2 = mg (2r) (h is 2r from point 1)
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 168
From the law of conservation of energy given in equation (vi), we get
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 169
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 170
Substituting equation (vii) in equation (iv) we get,
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 171
Therefore, the difference in tension is
T1 – T2 = 6 mg …(viii)
Minimum speed at the highest point (2)
The body must have a minimum speed at point 2 otherwise, the string will slack before reaching point 2 and the body will not loop the circle. To find this minimum speed let us take the tension T2 = 0 in equation (iv).
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 172
The body must have a speed at point 2, \(v_{2} \geq \sqrt{g r}\) to stay in the circular path.
Maximum speed at the lowest point 1
To have this minimum speed (\(v_{2}=\sqrt{g r}\)) at point 2, the body must have minimum speed also at point 1.
By making use of equation (vii) we can find the minimum speed at point 1.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 173
Substituting equation (ix) in (vii),
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 174
The body must have a speed at point 1, \(v_{1} \geq \sqrt{5 g r}\) to stay in the circular path.
From equations (ix) and (x), it is clear that the minimum speed at the lowest point 1 should be v 5 times more than the minimum speed at the highest point 2, so that the body loops without leaving the circle.

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 2.
Obtain the expressions for the velocities of the two bodies after collision in the case of one dimensional elastic collision and discuss the special cases.
Answer:
Consider two elastic bodies of masses m1 and m2 moving in a straight line (along positive x direction) on a frictionless horizontal surface as shown in figure.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 180
In order to have collision, we assume that the mass m1 moves faster than mass m2 i.e., u1 > u2.
For elastic collision, the total linear momentum and kinetic energies of the two bodies before and after collision must remain the same
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 181
From the law of conservation of linear momentum,
Total momentum before collision (pi) = Total momentum after collision (pf)
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 182
For elastic collision,
Total kinetic energy before collision KEi = Total kinetic energy after collision KFf
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 183
After simplifying and rearranging the terms,
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 184
Using the formula a2 – b2 = (a + b) (a – b), we can rewrite the above equation as
m1(u1 + v1)(u1 – v1) = m2 (v2 + u2) (v2 – u2) …(iv)
Dividing equation (iv) by (ii) gives,
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 185
This means that for any elastic head on collision, the relative speed of the two elastic bodies after the collision has the same magnitude as before collision but in opposite direction. Further note that this result is independent of mass.
Rewriting the above equation for V1 and v2,
v1 = v2 + u2 – u1
Or v2 = u1 + v1 – u1
To find the final velocities v1 and v2 :
Substituting equation (vii) in equation (ii) gives the velocity of m1 as
m1 (u1 – v1 ) = m2 (u1 + v1 – u2 – u2)
m1u1 – m1v1 = m2(u1 + v1 – 2u2)
m1u1 + 2m2u2 = m1v1 + m2v1
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 186
Similarly, by substituting (vi) in equation (ii) or substituting equation (viii) in equation (vii), we get the final velocity of m2 as
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 1901
Case 1: When bodies has the same mass i.e., m1 = m2,
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 191
The equations (x) and (xi) show that in one dimensional elastic collision, when two bodies of equal mass collide after the collision their velocities are exchanged.
Case 2: When bodies have the same mass i.e., m1 = m2 and second body (usually called target) is at rest (u2 = 0),
By substituting m1m2 = and u2 = 0 in equations (viii) and equations (ix) we get, from equation
(viii) ⇒ V1 = 0 …(xii)
from equation (ix) ⇒ v2 = u1 … (xiii)
Equations (xii) and (xiii) show that when the first body comes to rest the second body moves with the initial velocity of the first body.
Case 3: The first body is very much lighter than the second body
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 192
Similarly, Dividing numerator and denominator of equation (ix) by m2, we get
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 193
v2 = 0
The equation (xiv) implies that the first body which is lighter returns back (rebounds) in the opposite direction with the same initial velocity as it has a negative sign. The equation (xv) implies that the second body which is heavier in mass continues to remain at rest even after collision. For example, if a ball is thrown at a fixed wall, the ball will bounce back from the wall with the same velocity with which it was thrown but in opposite direction.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 194
Similarly,
Dividing numerator and denominator of equation (xiii) by m1, we get
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 195
The equation (xvi) implies that the first body which is heavier continues to move with the same initial velocity. The equation (xvii) suggests that the second body which is lighter will move with twice the initial velocity of the first body. It means that the lighter body is thrown away from the point of collision.

Samacheer Kalvi 11th Physics Numerical Questions

Question 1.
A body is moving along z-axis of a coordinate system under the effect of a constant force F = Find the work done by the force in moving the body a distance of 2 m along z-axis.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 196
\(\mathrm{W}=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{S}}=2 \mathrm{J}\)

Question 2.
Water is pumped out of a well 10 m deep by means of a pump rated 10 KW. Find the efficiency of the motor if 4200 kg of water is pumped out every minute. Take g = 10 m/s2.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 197

Question 3.
A railway carriage of mass 9000 kg moving with a speed of 36 kmph collides with a stationary carriage of same mass. After the collision, the carriages get coupled and move together. What is their common speed after collision? What type of collision is this?
Answer:
m1 = 9000 kg, u1 = 36 km/h = 10 m/s
m2 = 9000 kg, u2 = 0, v = v1 = v2 = ?
By conservation of momentum:
m1u1 + m2u2 = (m1 + m2)v
∴ v = 5 m/s
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 198
As total K.E. after collision < Total K.E. before collision
∴ collision is inelastic

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 4.
In lifting a 10 kg weight to a height of 2m, 230 J energy is spent. Calculate the acceleration with which it was raised.
Answer:
W = mgh + mah = m(g + a)h
∴ a = 1.5 m/s2.

Question 5.
A bullet of mass 0.02 kg is moving with a speed of 10 ms-1. It can penetrate 10 cm of a wooden block, and comes to rest. If the thickness of the target would be 6 cm only, find the K.E. of the bullet when it comes out.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 199
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 200

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 6.
A man pulls a lawn roller through a distance of 20 m with a force of 20 kg weight. If he applies the force at an angle of 60° with the ground, calculate the power developed if he takes 1 min in doing so.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 201

Question 7.
A body of mass 0.3 kg is taken up an inclined plane to length 10 m and height 5 m and then allowed to slide down to the bottom again. The coefficient of friction between the body and the plane is 0.15. What is the
(i) work done by the gravitational force over the round trip?
(ii) work done by the applied force over the upward journey?
(iii) work done by frictional force over the round trip?
(iv) kinetic energy of the body at the end of the trip?
How is the answer to (iv) related to the first three answers?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 202
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 203
(i) W = FS = – mg sin θ × h = -14.7 J is the W.D. by gravitational force in moving plane.
W’ = FS = + mg sin θ × h = 14.7 J is the W.D. by gravitational force in moving the body down the inclined plane.
∴ Total W.D. round the trip, W1 = W + W’ = 0
(ii) Force needed to move the body up the inclined plane,
F = mg sin θ + fk = mg sin θ + µkR = mg sin θ + µk mg cos θ
∴ W.D. by force over the upward journey is
W2 = F × l = mg (sin θ + µk cos θ)l = 18.5 J
(iii) W.D. by frictional force over the round trip,
W3 = -fk(l + l) = -2fkl = -2µkcos θ l = -7.6 J
(iv) K.E. of the body at the end of round trip
= W.D. by net force in moving the body down the inclined plane
= (mg sin θ – µkcos θ) l
= 10.9 J
⇒ K.E. of body = net W.D. on the body.

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 8.
Two identical 5 kg blocks are moving with same speed of 2 ms-1 towards each other along a frictionless horizontal surface. The two blocks collide, stick together and come to rest. Consider the two blocks as a system. Calculate work done by
(i) external forces and
(ii) Internal forces.
Answer:
Here no external forces are acting on the system so :
\(\overrightarrow{\mathrm{F}}_{\mathrm{ext}}=0 \Rightarrow \mathrm{W}_{\mathrm{ext}}=0\)
According to work-energy theorem :
Total W.D. = Change in K.E.
or Wext + = Final K.E. – Initial K.E.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 2012

Question 9.
A truck of mass 1000 kg accelerates uniformly from rest to a velocity of 15 ms-1 in 5 seconds. Calculate
(i) its acceleration,
(ii) its gain in K.E.,
(iii) average power of the engine during this period, neglect friction.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 2013

Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Question 10.
An elevator which can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 ms-1. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power.
Answer:
Downward force on the elevator is :
F = mg + f = 22000 N
∴ Power supplied by motor to balance this force is :
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 2022

Question 11.
To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass 1000 kg moving with a speed 18.0 kmh-1 on a smooth road and colliding with a horizontally mounted spring of spring constant 6.25 × 10-3 Nm-1. What is the maximum compression of the spring?
Answer:
At maximum compression xm, the K.E. of the car is converted entirely into the P.E. of the spring.
Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power 2031

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Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Students can Download Physics Chapter 3 Laws of Motion Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Physics Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Samacheer Kalvi 11th Physics Chapter 3 Laws of Motion Textual Questions Solved

Samacheer Kalvi 11th Physics Laws of Motion Multiple Choice Questions

Question 1.
When a car takes a sudden left turn in the curved road, passengers are pushed towards the right due to
(a) inertia of direction
(b) inertia of motion
(c) inertia of rest
(d) absence of inertia
Answer:
(a) inertia of direction

Question 2.
An object of mass m held against a vertical wall by applying horizontal force F as shown in the figure. The minimum value of the force F is [IIT JEE 1994]
(a) Less than mg
(b) Equal to mg
(c) Greater than mg
(d) Cannot determine
Answer:
(c) Greater than mg

Question 3.
A vehicle is moving along the positive x direction, if sudden brake is applied, then
(a) frictional force acting on the vehicle is along negative x direction
(b) frictional force acting on the vehicle is along positive x direction
(c) no frictional force acts on the vehicle
(d) frictional force acts in downward direction
Answer:
(a) frictional force acting on the vehicle is along negative x direction

SamacheerKalvi.Guru

The normal force formula is defined as the force that any surface exerts on any other object.

Question 4.
A book is at rest on the table which exerts a normal force on the book. If this force is considered as reaction force, what is the action force according to Newton’s third law?
(a) Gravitational force exerted by Earth on the book
(b) Gravitational force exerted by the book on Earth
(c) Normal force exerted by the book on the table
(d) None of the above
Answer:
(c) Normal force exerted by the book on the table

Question 5.
Two masses m1 and m2 are experiencing the same force where m1 < m2 The ratio of their acceleration \(\frac{a_{1}}{a_{2}}\) is –
(a) 1
(b) less than 1
(c) greater than 1
(d) all the three cases
Answer:
(c) greater than 1

Question 6.
Choose appropriate free body diagram for the particle experiencing net acceleration along negative y direction. (Each arrow mark represents the force acting on the system).
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Question 7.
A particle of mass m sliding on the smooth double inclined plane (shown in figure) will experience –
(a) greater acceleration along the path AB
(b) greater acceleration along the path AC
(c) same acceleration in both the paths
(d) no acceleration in both the paths
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Answer:
(a) greater acceleration along the path AC

SamacheerKalvi.Guru

Question 8.
Two blocks of masses m and 2m are placed on a smooth horizontal surface as shown. In the first case only a force F1  is applied from the left. Later only a force F2  is applied from the right. If the force acting at the interface of the two blocks in the two cases is same, then F1 : F2 is [Physics Olympiad 2016]
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Answer:
(c) 2 : 1

Question 9.
Force acting on the particle moving with constant speed is –
(a) always zero
(b) need not be zero
(c) always non zero
(d) cannot be concluded
Answer:
(b) need not be zero

Question 10.
An object of mass m begins to move on the plane inclined at an angle 0. The coefficient of static friction of inclined surface is lay. The maximum static friction experienced by the mass is –
(a) mg
(b) µs mg
(c) µs mg sin θ
(d) µs mg cos θ
Answer:
(d) µs mg cos θ

Question 11.
When the object is moving at constant velocity on the rough surface –
(a) net force on the object is zero
(b) no force acts on the object
(c) only external force acts on the object
(d) only kinetic friction acts on the object
Answer:
(a) net force on the object is zero

Question 12.
When an object is at rest on the inclined rough surface –
(a) static and kinetic frictions acting on the object is zero
(b) static friction is zero but kinetic friction is not zero
(c) static friction is not zero and kinetic friction is zero
(d) static and kinetic frictions are not zero
Answer:
(c) static friction is not zero and kinetic friction is zero

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Question 13.
The centrifugal force appears to exist –
(a) only in inertial frames
(b) only in rotating frames
(c) in any accelerated frame
(d) both in inertial and non-inertial frames
Answer:
(b) only in rotating frames

Question 14.
Choose the correct statement from the following –
(a) Centrifugal and centripetal forces are action reaction pairs
(b) Centripetal forces is a natural force
(c) Centrifugal force arises from gravitational force
(d) Centripetal force acts towards the center and centrifugal force appears to act away from the center in a circular motion
Answer:
(d) Centripetal force acts towards the center and centrifugal force appears to act away from the center in a circular motion

Question 15.
If a person moving from pole to equator, the centrifugal force acting on him –
(a) increases
(b) decreases
(c) remains the same
(d) increases and then decreases
Answer:
(a) increases

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions

Question 1.
Explain the concept of inertia. Write two examples each for inertia of motion, inertia of rest and inertia of direction.
Answer:
The inability of objects to move on its own or change its state of motion is called inertia. Inertia means resistance to change its state. There are three types of inertia:
1. Inertia of rest:
The inability of an object to change its state of rest is called inertia of rest.
Example:

  • When a stationary bus starts to move, the passengers experience a sudden backward push.
  • A book lying on the table will remain at rest until it is moved by some external agencies.

2. Inertia of motion:
The inability of an object to change its state of uniform speed (constant speed) on its own is called inertia of motion.
Example:

  • When the bus is in motion, and if the brake is applied suddenly, passengers move forward and hit against the front seat.
  • An athlete running is a race will continue to run even after reaching the finishing point.

3. Inertia of direction:
The inability of an object to change its direction of motion on its own is called inertia of direction.
Example:

  • When a stone attached to a string is in whirling motion, and if the string is cut suddenly, the stone will not continue to move in circular motion but moves tangential to the circle.
  • When a bus moving along a straight line takes a turn to the right. The passengers are thrown towards left.

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Question 2.
State Newton’s second law.
Answer:
The force acting on an object is equal to the rate of change of its momentum –
\(\overline{\mathrm{F}}\) = \(\frac{d \bar{p}}{d t}\)

Question 3.
Define one newton.
Answer:
One newton is defined as the force which acts on 1 kg of mass to give an acceleration 1 ms-2 in the direction of the force.

Question 4.
Show that impulse is the change of momentum.
Answer:
According to Newton’s Second Law
F = \(\frac {dp}{dt}\) i.e. dp = Fdt
Integrate it over a time interval from ti  to tf
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Pi → initial momentum of the object at ti
Pf → Final momentum of the object at tf
Pf – Pi = ∆p = change in momentum during the time interval ∆t.
\(\int_{t_{i}}^{t_{f}} \mathrm{F} \cdot d t=\mathrm{J}\) is called the impulse.
If the force is constant over the time interval ∆t, then
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Hence the proof.

Question 5.
Using free body diagram, show that it is easy to pull an object than to push it.
Answer:
When a body is pushed at an arbitrary angle θ [0 to \(\frac {π}{2}\)], the applied force F can be resolved into two components as F sin 0 parallel to the surface and F cos 0 perpendicular to the surface as shown in figure. The total downward force acting on the body is mg + F cos θ. It implies that the normal force acting on the body increases. Since there is no acceleration along the vertical direction the normal force N is equal to
Npush = mg + F cos θ …………(1)
As a result the maximal static friction also increases and is equal to
\(f_{S}^{\max }\) = \(\mu_{r} \mathrm{N}_{\mathrm{push}}\) = µs(mg + F cos θ) ……(2)
Equation (2) shows that a greater force needs to be applied to push the object into motion.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

When an object is pulled at an angle θ, the applied force is resolved into two components as shown in figure. The total downward force acting on the object is –
Npull = mg – F cos θ ………….(3)
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Equation (3) shows that the normal force is less than – Npush. From equations (1) and (3), it is easier to pull an object than to push to make it move.

SamacheerKalvi.Guru

Question 6.
Explain various types of friction. Suggest a few methods to reduce friction.
Answer:
There are two types of Friction:
(1) Static Friction:
Static friction is the force which opposes the initiation of motion of an object on the surface. The magnitude of static frictional force fs lies between
\(0 \leq f_{s} \leq \mu_{s} \mathrm{N}\)
where, µs – coefficient of static friction
N – Normal force

(2) Kinetic friction:
The frictional force exerted by the surface when an object slides is called as kinetic friction. Also called as sliding friction or dynamic friction,
fk – µkN
where µk – the coefficient of kinetic friction
N – Normal force exerted by the surface on the object

Methods to reduce friction:
Friction can be reduced

  • By using lubricants
  • By using Ball bearings
  • By polishing
  • By streamlining

Question 7.
What is the meaning by ‘pseudo force’?
Answer:
Pseudo force is an apparent force which has no origin. It arises due to the non-inertial nature of the frame considered.

Question 8.
State the empirical laws of static and kinetic friction.
Answer:
The empirical laws of friction are:

  • Friction is independent of surface of contact.
  • Coefficient of kinetic friction is less than coefficient of static friction.
  • The direction of frictional force is always opposite to the motion of one body over the other.
  • Frictional force always acts on the object parallel to the surface on which the objet is placed,
  • The magnitude of frictional force between any two bodies in contact is directly proportional to the normal reaction between them.

Question 9.
State Newton’s third law.
Answer:
Newton’s third law states that for every action there is an equal and opposite reaction.

Question 10.
What are inertial frames?
Answer:
Inertial frame is the one in which if there is no force on the object, the object moves at constant velocity.

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Question 11.
Under what condition will a car skid on a leveled circular road?
Answer:
On a leveled circular road, if the static friction is not able to provide enough centripetal ’force to turn, the vehicle will start to skid
\(\mu_{s}<\frac{v^{2}}{r g}\)

Samacheer Kalvi 11th Physics Laws of Motion Long Answer Questions

Question 1.
Prove the law of conservation of linear momentum. Use it to find the recoil velocity of a gun when a bullet is fired from it.
Answer:
In nature, conservation laws play a very important role. The dynamics of motion of bodies can be analysed very effectively using conservation laws. There are three conservation laws in mechanics. Conservation of total energy, conservation of total linear momentum, and conservation of angular momentum. By combining Newton’s second and third laws, we can derive the law of conservation of total linear momentum. When two particles interact with each other, they exert equal and opposite forces on each other.

The particle 1 exerts force \(\overrightarrow{\mathrm{F}}_{12}\) on particle 2 and particle 2 exerts an exactly equal and opposite force \(\overrightarrow{\mathrm{F}}_{12}\) on particle 1 according to Newton’s third law.
\(\overrightarrow{\mathrm{F}}_{12}\) = –\(\overrightarrow{\mathrm{F}}_{12}\) ……..(1)
In terms of momentum of particles, the force on each particle (Newton’s second law) can be written as –
\(\overrightarrow{\mathrm{F}}_{12}\) = \(\frac{d \vec{p}_{1}}{d t}\) and \(\overrightarrow{\mathrm{F}}_{21}\) = \(\frac{d \vec{p}_{2}}{d t}\) ………(2)
Here \(\vec{p}_{1}\) is the momentum of particle 1 which changes due to the force \(\overrightarrow{\mathrm{F}}_{12}\) exerted by particle 2. Further Po is the momentum of particle \(\vec{p}_{2}\) This changes due to \(\overrightarrow{\mathrm{F}}_{21}\) exerted by particle 1.
Substitute equation (2) in equation (1)
\(\frac{d \vec{p}_{1}}{d t}\) = – \(\frac{d \vec{p}_{2}}{d t}\) …………(3)
\(\frac{d \vec{p}_{1}}{d t}\) + \(\frac{d \vec{p}_{2}}{d t}\) = 0 ………(4)
\(\frac {d}{dt}\)(\(\vec{p}_{1}\) + \(\vec{p}_{2}\)) = 0
It implies that \(\vec{p}_{1}\) + \(\vec{p}_{2}\) = constant vector (always).
\(\vec{p}_{1}\) + \(\vec{p}_{2}\) is the total linear momentum of the two particles (\(\vec{p}_{tot}\) = \(\vec{p}_{1}\) + \(\vec{p}_{2}\)).It is also called as total linear momentum of the system. Here, the two particles constitute the system. From this result, the law of conservation of linear momentum can be stated as follows.

If there are no external forces acting on the system, then the total linear momentum of the system (\(\vec{p}_{tot}\)) is always a constant vector. In other words, the total linear momentum of the system is conserved in time. Here the word ‘conserve’ means that \(\vec{p}_{1}\) and \(\vec{p}_{2}\) can vary, in such a way that \(\vec{p}_{1}\) + \(\vec{p}_{2}\) is a constant vector.

The forces \(\overrightarrow{\mathrm{F}}_{12}\) and \(\overrightarrow{\mathrm{F}}_{21}\) are called the internal forces of the system, because they act only between the two particles. There is no external force acting on the two particles from outside. In such a case the total linear momentum of the system is a constant vector or is conserved.

Meaning of law of conservation of momentum:
1. The Law of conservation of linear momentum is a vector law. It implies that both the magnitude and direction of total linear momentum are constant. In some cases, this total momentum can also be zero.

2. To analyse the motion of a particle, we can either use Newton’s second law or the law of conservation of linear momentum. Newton’s second law requires us to specify the forces involved in the process. This is difficult to specify in real situations. But conservation of linear momentum does not require any force involved in the process. It is convenient and hence important.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

For example, when two particles collide, the forces exerted by these two particles on each other is difficult to specify. But it is easier to apply conservation of linear momentum during the collision process.

Examples:
Consider the firing of a gun. Here the system is Gun+bullet. Initially the gun and bullet are at rest, hence the total linear momentum of the system is zero. Let \(\vec{p}_{1}\) be the momentum of the bullet and \(\vec{p}_{2}\) the momentum of the gun before firing. Since initially both are at rest,

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Total momentum before firing the gun is zero, \(\vec{p}_{1}\) + \(\vec{p}_{2}\) = 0.
According to the law of conservation of linear momentum, total linear momentum has to be zero after the firing also.

When the gun is fired, a force is exerted by the gun on the bullet in forward direction. Now the momentum of the bullet changes from \(\vec{p}_{1}\) to \(\vec{p}_{1}\) To conserve the total linear momentum of the system, the momentum of the gun must also change from \(\vec{p}_{2}\) to \(\vec{p}_{2}^{\prime}\). Due to the conservation of linear momentum, \(\vec{p}_{1}\) + \(\vec{p}_{2}^{\prime}\) = 0.

It implies that \(\vec{p}_{1}^{\prime}\)= \(\vec{p}_{2}^{\prime}\), the momentum of the gun is exactly equal, but in the opposite direction to the momentum of the bullet. This is the reason after firing, the gun suddenly moves backward with the momentum (-\(\vec{p}_{2}^{\prime}\)). It is called ‘recoil momentum’. Th is is an example of conservation of total linear momentum.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Question 2.
What are concurrent forces? State Lami’s theorem.
Answer:
Concurrent force:
A collection of forces is said to be concurrent, if the lines of forces act at a common point.

Lami’s Theorem:
If a system of three concurrent and coplanar forces is in equilibrium, then Lami’s theorem states that the magnitude of each force of the system is proportional to sine of the angle between the other two forces.
i.e. |\(\overrightarrow{\mathrm{F}}_{1}\)|∝ sin α, |\(\overrightarrow{\mathrm{F}}_{2}\)| ∝ sin β, |\(\overrightarrow{\mathrm{F}}_{2}\)| ∝ sin γ,

Question 3.
Explain the motion of blocks connected by a string in

  1. Vertical motion
  2. Horizontal motion.

Answer:
When objects are connected by strings and When objects are connected by strings and a force F is applied either vertically or horizontally or along an inclined plane, it produces a tension T in the string, which affects the acceleration to an extent. Let us discuss various cases for the same.

Case 1:
Vertical motion:
Consider two blocks of masses m1 and m2 (m1> m2) connected by a light and in-extensible string that passes over a pulley as shown in Figure.

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Let the tension in the string be T and acceleration a. When the system is released, both the blocks start moving, m2 vertically upward and mk, downward with same acceleration a. The gravitational force m1g on mass m1 is used in lifting the mass m2. The upward direction is chosen as y direction. The free body diagrams of both masses are shown in Figure.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Applying Newton’s second law for mass m2 T \(\hat{j}\) – m2 g \(\hat{j}\) = m2 a \(\hat{j}\) The left hand side of the above equation is the total force that acts on m2 and the right hand side is the product of mass and acceleration of m2 in y direction.
By comparing the components on both sides, we get
T = m2 g = m2 a ……….(1)
Similarly, applying Newton’s second law for mass m2
T \(\hat{j}\) – m1 g\(\hat{j}\) = m1a\(\hat{j}\)
As mass m1 moves downward (-\(\hat{j}\)), its acceleration is along (-\(\hat{j}\))
By comparing the components on both sides, we get
T = m1g = -m1a
m1g – T = m1a ………..(2)
Adding equations (1) and (2), we get
m1g – m2g = m1a + m2a
(m1 – m2)g = (m1 + m2)a …………(3)
From equation (3), the acceleration of both the masses is –
a = (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ………..(4)
If both the masses are equal (m1 = m2), from equation (4)
a = 0
This shows that if the masses are equal, there is no acceleration and the system as a whole will be at rest.
To find the tension acting on the string, substitute the acceleration from the equation (4) into the equation (1).
T = m2g = m2(\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))
T = m2g + m2 (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ……….(5)
By taking m2g common in the RHS of equation (5)
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Equation (4) gives only magnitude of acceleration.
For mass m1, the acceleration vector is given by \(\vec{a}\) = –\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\)\(\hat{j}\)
For mass m2, the acceleration vector is given by \(\vec{a}\) = \(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\) \(\hat{j}\)

Case 2:
Horizontal motion:
In this case, mass m2 is kept on a horizontal table and mass m1, is hanging through a small pulley as shown in figure. Assume that there is no friction on the surface
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
As both the blocks are connected to the un stretchable string, if m1 moves with an acceleration a downward then m2 also moves with the same acceleration a horizontally.
The forces acting on mass m2 are

  • Downward gravitational force (m2g)
  • Upward normal force (N) exerted by the surface
  • Horizontal tension (T) exerted by the string

The forces acting on mass m1 are

  • Downward gravitational force (m1g)
  • Tension (T) acting upwards

The free body diagrams for both the masses is shown in figure.

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Applying Newton’s second law for m1
T\(\hat{i}\) – m1g\(\hat{j}\) = -m1a\(\hat{j}\) (alongy direction)
By comparing the components on both sides of the above equation,
T – m1g = -m1a …………(1)
Applying Newton’s second law for m2
Ti = m1ai (along x direction)
By comparing the components on both sides of above equation,
T = m2a ………….(2)
There is no acceleration along y direction for m2.
N\(\hat{j}\) – m2g\(\hat{j}\) = 0
By comparing the components on both sides of the above equation
N – m2g = 0
N = m2g ……….(3)
By substituting equation (2) in equation (1), we can find the tension T
m2a – m1g = -m1a
m2a + m1a = m1g
a = \(\frac{m_{1}}{m_{1}+m_{2}}\)g …………(4)
Tension in the string can be obtained by substituting equation (4) in equation (2)
T = \(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\)g ………..(5)

Comparing motion in both cases, it is clear that the tension in the string for horizontal motion is half of the tension for vertical motion for same set of masses and strings. This result has an important application in industries. The ropes used in conveyor belts (horizontal motion) work for longer duration than those of cranes and lifts (vertical motion).

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Question 4.
Briefly explain the origin of friction. Show that in an inclined plane, angle of friction is pqual to angle of repose.
Answer:
If a very gentle force in the horizontal direction is given to an object at rest on the table it does not move. It is because of the opposing force exerted by the surface on the object which resists its motion. This force is called the frictional force. During the time of Newton and Galileo, frictional force was considered as one of the natural forces like gravitational force. But in the twentieth century, the understanding on atoms, electron and protons has changed the perspective.

The frictional force is actually the electromagnetic force between the atoms on the two surfaces. Even well polished surfaces have irregularities on the surface at the microscopic level. The component of force parallel to the inclined plane (mg sin θ) tries to move the object down. The component of force perpendicular to the inclined plane (mg cos θ) is balanced by the Normal force (N).
N = mg cos θ ………(1)

When the object just begins to move, the static friction attains its maximum value
fs = \(f_{s}^{\max }\) = µs N
This friction also satisfies the relation
\(f_{s}^{\max }\) = µs mg sin θ ……….(2)
Equating the right hand side of equations (1) and (2),
(\(f_{s}^{\max }\))/N = sin θ / cos θ
From the definition of angle of friction, we also know that
tan θ = µs ………..(3)
in which θ is the angle of friction.
Thus the angle of repose is the same as angle of friction. But the difference is that the angle of repose refers to inclined surfaces and the angle of friction is applicable to any type of surface.

Question 5.
State Newton’s three laws and discuss their significance.
Answer:
First Law:
Every object continues to be in the state of rest or of uniform motion (constant velocity) unless there is external force acting on it.

Second Law:
The force acting on an object is equal to the rate of change of its momentum

Third Law:
For every action there is an equal and opposite reaction.

Significance of Newton’s Laws:
1. Newton’s laws are vector laws. The equation \(\overline{\mathrm{F}}\) = m\(\overline{\mathrm{a}}\) is a vector equation and essentially it is equal to three scalar equations. In Cartesian coordinates, this equation can be written as Fx\(\hat{i}\) + Fy\(\hat{j}\) + Fz\(\hat{j}\) = max\(\hat{i}\) + may\(\hat{j}\) + maz\(\hat{j}\)
By comparing both sides, the three scalar equations are

Fx = max The acceleration along the x-direction depends only on the component of force acting along the x – direction.
Fy = may The acceleration along the y direction depends only on the component of force acting along the y – direction.
Fz = maz The acceleration along the z direction depends only on the component of force acting along the z – direction.
From the above equations, we can infer that the force acting along y direction cannot alter the acceleration along x direction. In the same way, Fz cannot affect ay and ax. This understanding is essential for solving problems.

2. The acceleration experienced by the body at time t depends on the force which acts on the body at that instant of time. It does not depend on the force which acted on the body before the time t. This can be expressed as
\(\overline{\mathrm{F}}\)(t) = m\(\overline{\mathrm{a}}\)(t)
Acceleration of the object does not depend on the previous history of the force. For example, when a spin bowler or a fast bowler throws the ball to the batsman, once the ball leaves the hand of the bowler, it experiences only gravitational force and air frictional force. The acceleration of the ball is independent of how the ball was bowled (with a lower or a higher speed).

3. In general, the direction of a force may be different from the direction of motion. Though in some cases, the object may move in the same direction as the direction of the force, it is not always true. A few examples are given below.

Case 1:
Force and motion in the same direction:
When an apple falls towards the Earth, the direction of motion (direction of velocity) of the apple and that of force are in the same downward direction as shown in the Figure.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Case 2:
Force and motion not in the same direction:
The Moon experiences a force towards the Earth. But it actually moves in elliptical orbit. In this case, the direction of the force is different from the direction of motion as shown in Figure.

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Case 3:
Force and motion in opposite direction:
If an object is thrown vertically upward, the direction of motion is upward, but gravitational force is downward as shown in the Figure.

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Case 4:
Zero net force, but there is motion:
When a raindrop gets detached from the cloud it experiences both downward gravitational force and upward air drag force. As it descends towards the Earth, the upward air drag force increases and after a certain time, the upward air drag force cancels the downward gravity. From then on the raindrop moves at constant velocity till it touches the surface of the Earth. Hence the raindrop comes with zero net force, therefore with zero acceleration but with non-zero terminal velocity. It is shown in the Figure

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

4. If multiple forces \(\overrightarrow{\mathrm{F}}_{1}\), \(\overrightarrow{\mathrm{F}}_{2}\), \(\overrightarrow{\mathrm{F}}_{3}\),……. \(\overrightarrow{\mathrm{F}}_{n}\) act on the same body, then the total force (\(\overrightarrow{\mathrm{F}}_{net}\)) is equivalent to the vectorial sum of the individual forces. Their net force provides the acceleration.
\(\overrightarrow{\mathrm{F}}_{net}\) = \(\overrightarrow{\mathrm{F}}_{1}\) + \(\overrightarrow{\mathrm{F}}_{2}\) + \(\overrightarrow{\mathrm{F}}_{3}\) + ……… + \(\overrightarrow{\mathrm{F}}_{n}\)
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Newton’s second law for this case is –
\(\overrightarrow{\mathrm{F}}_{net}\) = m\(\overline{\mathrm{a}}\)
In this case the direction of acceleration is in the direction of net force.
Example:
Bow and arrow

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

5. Newton’s second law can also be written in the following form.
Since the acceleration is the second derivative of position vector of the body \(\left(\vec{a}=\frac{d^{2} \vec{r}}{d t^{2}}\right)\)
the force on the body is –
\(\overline{\mathrm{F}}\) = m\(\frac{d^{2} \vec{r}}{d t^{2}}\)
From this expression, we can infer that Newton’s second law is basically a second order ordinary differential equation and whenever the second derivative of position vector is not zero, there must be a force acting on the body.

6. If no force acts on the body then Newton’s second law, m = \(\frac{d \vec{v}}{d t}\) = 0
It implies that \(\overline{\mathrm{v}}\) = constant. It is essentially Newton’s first law. It implies that the second law is consistent with the first law. However, it should not be thought of as the reduction of second law to the first when no force acts on the object. Newton’s first and second laws are independent laws. They can internally be consistent with each other but cannot be derived from each other.

7. Newton’s second law is cause and effect relation. Force is the cause and acceleration is the effect. Conventionally, the effect should be written on the left and cause on the right hand side of the equation. So the correct way of writing Newton’s second law is –
m\(\overline{\mathrm{a}}\) = \(\overline{\mathrm{F}}\) or \(\frac{d \vec{p}}{d t}\) = \(\overline{\mathrm{F}}\)

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Question 6.
Explain the similarities and differences of centripetal and centrifugal forces.
Answer:
Salient features of centripetal and centrifugal forces.
Centripetal Force:

  • It is a real force which is exerted on the body by the external agencies like gravitational force, tension in the string, normal force etc.
  • Acts in both inertial and non-inertial frames
  • It acts towards the axis of rotation or center of the circle in circular motion
    \(\left|\overrightarrow{\mathrm{F}}_{\mathrm{C}_{\mathrm{P}}}\right|\) = mω2r = \(\frac{m v^{2}}{r}\)
  • Real force and has real effects
  • Origin of centripetal force is interaction between two objects.
  • In inertial frames centripetal force has to be included when free body diagrams are drawn.

Centrifugal Force:

  • It is a pseudo force or fictitious force which cannot arise from gravitational force, tension force, normal force etc.
  • Acts only in rotating frames (non-inertial frame)
  • It acts outwards from the axis of rotation or radially outwards from the center of the circular motion
    \(\left|\overrightarrow{\mathrm{I}}_{\mathrm{C}_{\mathrm{f}}}\right|\) = mω2r = \(\frac{m v^{2}}{r}\)
  • Pseudo force but has real effects
  • Origin of centrifugal force is inertia. It does not arise from interaction. In an inertial frame the object’s inertial motion appears as centrifugal force in the rotating frame.
  • In inertial frames there is no centrifugal force. In rotating frames, both centripetal and centrifugal force have to be included when free body diagrams are drawn.

Question 7.
Briefly explain ‘centrifugal force’ with suitable examples.
Answer:
To use Newton’s first and second laws in the rotational frame of reference, we need to include a Pseudo force called centrifugal force. This centrifugal force appears to act on the object with respect to rotating frames.

Circular motion can be analysed from two different frames of reference. One is the inertial frame (which is either at rest or in uniform motion) where Newton’s laws are obeyed. The other is the rotating frame of reference which is a non – inertial frame of reference as it is accelerating.

When we examine the circular motion from these frames of reference the situations are entirely different. To use Newton’s first and second laws in the rotational frame of reference, we need to include a pseudo force called ‘centrifugal force’. This ‘centrifugal force’ appears to act on the object with respect to rotating frames. To understand the concept of centrifugal force, we can take a specific case and discuss as done below.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Free body diagram of a particle including the centrifugal force Consider the case of a whirling motion of a stone tied to a string. Assume that the stone has angular velocity ω in the inertial frame (at rest). If the motion of the stone is observed from a frame which is also rotating along with the stone with same angular velocity ω then, the stone appears to be at rest.

This implies that in addition to the inward centripetal force – mω2r there must be an equal and opposite force that acts on the stone outward with value +mω2r . So the total force acting on the stone in a rotating frame is equal to zero (-mω2r + mω2 r = 0). This outward force +mω2r is called the centrifugal force. The word ‘centrifugal’ means ‘flee from center’.

Note that the ‘centrifugal force’ appears to act on the particle, only when we analyse the motion from a rotating frame. With respect to an inertial frame there is only centripetal force which is given by the tension in the rstring. For this reason centrifugal force is called as a ‘pseudo force’. A pseudo force has no origin. It arises due to the non inertial nature of the frame considered. When circular motion problems are solved from a rotating frame of reference, while drawing free body diagram of a particle, the centrifugal force should necessarily be included as shown in the figure.

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Question 8.
Briefly explain ‘rolling friction’.
Answer:
The invention of the wheel plays a crucial role in human civilization. One of the important applications is suitcases with rolling on coasters. Rolling wheels makes it easier than carrying luggage. When an object moves on a surface, essentially it is sliding on it. But wheels move on the surface through rolling motion. In rolling motion when a wheel moves on a surface, the point of contact with surface is always at rest. Since Rolling and kinetic friction the point of contact is at rest, there is no relative motion between the wheel and surface.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Hence the frictional fore is very less. At the same time if an object moves without a wheel, there is a relative motion between the object and the surface. As a result frictional force is larger. This makes it difficult to move the object. The figure shows the difference between rolling and kinetic friction. Ideally in pure rolling, motion of the point of contact with the surface should be at rest, but in practice it is not so.

Due to the elastic nature of the surface at the point of contact there will be some deformation on the object at this point on the wheel or surface as shown in figure. Due to this deformation, there will be minimal friction between wheel and surface. It is called ‘rolling friction. In fact, rolling friction’ is much smaller than kinetic friction.

Question 9.
Describe the method of measuring angle of repose.
Answer:
When objects are connected by strings and When objects are connected by strings and a force F is applied either vertically or horizontally or along an inclined plane, it produces a tension T in the string, which affects the acceleration to an extent. Let us discuss various cases for the same.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Case 1:
Vertical motion:
Consider two blocks of masses m1 and m2 (m1> m2) connected by a light and in extensible string that passes over a pulley as shown in Figure.

Let the tension in the string be T and acceleration a. When the system is released, both the blocks start moving, m2 vertically upward and mk, downward with same acceleration a. The gravitational force m1g on mass m1 is used in lifting the mass m2. The upward direction is chosen as y direction. The free body diagrams of both masses are shown in Figure.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Applying Newton’s second law for mass m2
T \(\hat{j}\) – m2g\(\hat{j}\) = m2a\(\hat{j}\) The left hand side of the above equation is the total force that acts on m2 and the right hand side is the product of mass and acceleration of m2 in y direction.
By comparing the components on both sides, we get
T = m2g = m2a ……….(1)

Similarly, applying Newton’s second law for mass m2
T \(\hat{j}\) – m1g\(\hat{j}\) = m1a\(\hat{j}\)
As mass mj moves downward (-\(\hat{j}\)), its acceleration is along (-\(\hat{j}\))
By comparing the components on both sides, we get
T = m1g = -m1a
m1g – T = m1a ………..(2)

Adding equations (1) and (2), we get
m1g – m2g = m1a + m2a
(m1 – m2)g = (m1 + m2)a …………(3)
From equation (3), the acceleration of both the masses is –
a = (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ………..(4)
If both the masses are equal (m1 = m2), from equation (4)
a = 0
This shows that if the masses are equal, there is no acceleration and the system as a whole will be at rest.
To find the tension acting on the string, substitute the acceleration from the equation (4) into the equation (1).
T = m2g = m2(\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))
T = m2g + m2 (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ……….(5)
By taking m2g common in the RHS of equation (5)
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Equation (4) gives only magnitude of acceleration
For mass m1, the acceleration vector is given by \(\vec{a}\) = –\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\)\(\hat{j}\)
For mass m2, the acceleration vector is given by \(\vec{a}\) = \(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\) \(\hat{j}\)

Case 2:
Horizontal motion:
In this case, mass m2 is kept on a horizontal table and mass m1, is hanging through a small pulley as shown in figure. Assume that there is no friction on the surface
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
As both the blocks are connected to the un stretchable string, if m1 moves with an acceleration a downward then m2 also moves with the same acceleration a horizontally.
The forces acting on mass m2 are

  1. Downward gravitational force (m2g)
  2. Upward normal force (N) exerted by the surface
  3. Horizontal tension (T) exerted by the string

The forces acting on mass m1 are

  1. Downward gravitational force (m1g)
  2. Tension (T) acting upwards

The free body diagrams for both the masses is shown in figure 2.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Applying Newton’s second law for m1
T\(\hat{i}\) – m1g\(\hat{j}\) = -m1a\(\hat{j}\) (alongy direction)
By comparing the components on both sides of the above equation,
T – m1g = -m1a …………(1)
Applying Newton’s second law for m2
T\(\hat{i}\) = m1a\(\hat{i}\) (along x direction)
By comparing the components on both sides of above equation,
T = m2a ………….(2)
There is no acceleration along y direction for m2.
N\(\hat{j}\) – m2g\(\hat{j}\) = 0
By comparing the components on both sides of the above equation
N – m2g = 0
N = m2g ……….(3)
By substituting equation (2) in equation (1), we can find the tension T
m2a – m1g = -m1a
m2a + m1a = m1g
a = \(\frac{m_{1}}{m_{1}+m_{2}}\)g …………(4)
Tension in the string can be obtained by substituting equation (4) in equation (2)
T = \(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\)g ………..(5)

Comparing motion in both cases, it is clear that the tension in the string for horizontal motion is half of the tension for vertical motion for same set of masses and strings. This result has an important application in industries. The ropes used in conveyor belts (horizontal motion) work for longer duration than those of cranes and lifts (vertical motion).

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Question 10.
Explain the need for banking of tracks.
Answer:
In a leveled circular road, skidding mainly depends on the coefficient of static friction py. The coefficient of static friction depends on the nature of the surface which has a maximum limiting value. To avoid this problem, usually the outer edge of the road is slightly raised compared to inner edge. This is called banking of roads or tracks. This introduces an inclination, and the angle is called banking angle. “Let the surface of the road make angle 9 with horizontal surface. Then the normal force makes the same angle 9 with the vertical. When the car takes a turn.Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

there are two forces acting on the car:
(a) Gravitational force mg (downwards)
(b) Normal force N (perpendicular to surface)
We can resolve the normal force into two components N cos θ and N sin θ. The component balances the downward gravitational force ‘mg’ and component will provide the necessary centripetal acceleration. By using Newton second law.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

The banking angle 0 and radius of curvature of the road or track determines the safe speed of the car at the turning. If the speed of car exceeds this safe speed, then it starts to skid outward but frictional force comes into effect and provides an additional centripetal force to prevent the outward skidding.

At the same time, if the speed of the car is little lesser than safe speed, it starts to skid inward and frictional force comes into effect, which reduces centripetal force to prevent inward skidding. However if the speed of the vehicle is sufficiently greater than the correct speed, then frictional force cannot stop the car from skidding.

Question 11.
Calculate the centripetal acceleration of Moon towards the Earth.
Answer:
The centripetal acceleration is given by a = \(\frac{v^{2}}{r}\) This expression explicitly depends on Moon’s speed which is nontrivial. We can work with the formula
ω2Rm = am
am is centripetal acceleration of the Moon due to Earth’s gravity, ω is angular velocity
Rm is the distance between Earth and the Moon, which is 60 times the radius of the Earth.
Rm = 60R = 60 x 6.4 x 106 = 384 x 106 m
As we know the angular velocity ω = \(\frac { 2π}{ T }\) and T = 27.3 days = 27.3 x 24 x 60 x 60 second = 2.358 x 106 sec.
By substituting these values in the formula for acceleration
a6 = \(\frac{\left(4 \pi^{2}\right)\left(384 \times 10^{6}\right)}{\left(2.358 \times 10^{8}\right)^{2}}\) = 0.00272 ms-2

Samacheer Kalvi 11th Physics Laws of Motion Conceptual Questions

Question 1.
Why it is not possible to push a car from inside?
Answer:
While trying to push a car from outside, he pushes the ground backwards at an angle. The ground offers an equal reaction in the opposite direction, so car can be moved. But the person sits inside means car and the person becomes a single system, and the force given will be a internal force. According to Newton’s third law, total internal force acting on the system is zero and it cannot accelerate the system.

Question 2.
There is a limit beyond which the polishing of a surface increases frictional resistance rather than decreasing it why?
Answer:
Friction is a contact force. Friction is directly proportional to area of contact. In the normal surfaces there are bumps to interlock the surfaces in contact. But the surfaces are polished beyond certain limit. The area of contact will be increased and the molecules come closer to each other. It increases electrostatic force between the molecules. As a result it increases friction.

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Question 3.
Can a single isolated force exist in nature? Explain your answer.
Answer:
No. According to Newton’s third law, for every action, there is an equal and opposite reaction. So, whatever case we consider, if there is an action there is always a reaction. So it is impossible.

Question 4.
Why does a parachute descend slowly?
Answer:
A parachute descends slowly because the surface area of parachute is large so that air gives more resistance when it descends down.

Question 5.
When walking on ice one should take short steps. Why?
Answer:
Let R represent the reaction offered by the ground. The vertical component R cos θ will balance the weight of the person and the horizontal component R sin θ will help the person to walk forward.
Now, normal reaction = R cos θ
Friction force = R sin θ
Coefficient of friction, µ = \(\frac {R sin θ}{R cos θ }\) = tan θ
In a long step, θ is more. So tan θ is more. But p has a fixed value. So, there is danger of slipping in a long step.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Question 6.
When a person walks on a surface, the frictional force exerted by the surface on the person is opposite to the direction of motion. True or false?
Answer:
False. In frictional force exerted by the surface on the person is in the direction of his motion. Frictional force acts as an external force to move the person. When the person trying to move, he gives a push to ground on the backward direction and by Newton’s third law he is pushed by the ground in the forward direction. Hence frictional force acts along the direction of motion.

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Question 7.
Can the coefficient of friction be more than one?
Answer:
Yes. The coefficient of friction can be more than one in some cases such as silicone rubber. Coefficient of friction is the ratio of frictional force to normal force, i.e. F = μR. If p is greater than one means frictional force is greater than normal force. But in general case the value is usually between 0 and 1.

Question 8.
Can we predict the direction of motion of a body from the direction of force on it?
Answer:
Yes. The direction of motion is always opposite to the force of kinetic friction. By using the principle of equilibrium, the direction of force of static friction can be determined. When the object is in equilibrium, the frictional force must point in the direction which results as a net force is zero.

Question 9.
The momentum of a system of particles is always conserved. True or false?
Answer:
True. The total momentum of a system of particles is always constant i.e. conserved. When no external force acts on it.

Samacheer Kalvi 11th Physics Laws of Motion Numerical Problems

Question 1.
A force of 50 N act on the object of mass 20 kg. shown in the figure. Calculate the acceleration of the object in x and y directions.
Answer:
Given F = 50 N and m = 20 kg
(1) component of force along x – direction
Fx = F cos θ
= 50 x cos 30° = 43.30 N
ax = \(\frac{F_{x}}{m}\) = \(\frac {43.30}{20}\) =2.165 ms-2

(2) Component of force along y – direction
Fy = F sin θ = 50 sin 30° = 25 N
ay = \(\frac{F_{y}}{m}\) = \(\frac {25}{20}\) = 1.25 ms-2

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Question 2.
A spider of mass 50 g is hanging on a string of a cob web as shown in the figure. What is the tension in the string?
Answer:
Given m = 50 g, T = ?
Tension in the string T = mg
= 50 x 10-2 x 9.8 = 0.49 N

Question 3.
What is the reading shown in spring balance?
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Answer:
When a spring balance hung on a rigid support and load is attached at its other end, the weight of the load exerts a force on the rigid support in turn support exerts equal and opposite force on that load, so that balance will be stretched. This is the principle of spring balance. Flence the answer is 4 kg.
Given: m = 2 kg, 0 = 30°.
Resolve the weight into its component as mg sin θ and mg cos θ.
Here mg sin θ acts parallel to the surface
∴ W = mg sin θ
= 2 x 9.8 x sin 30° = 2 x 9.8 x \(\frac {1}{2}\) = 9.8 N

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Question 4.
The physics books are stacked on each other in the sequence: +1 volumes 1 and 2; +2 volumes 1 and 2 on a table
(a) Identify the forces acting on each book and draw the free body diagram.
(b) Identify the forces exerted by each book on the other.
Answer:
Let m1, m2, m3, m4, are the masses of +1 volume I and II and +2 volumes I & II
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

(a) Force on book m4

  • Downward gravitational force acting downward (m3g)
  • Upward normal force (N3) exerted by book of mass m3
    Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion |

(b) Force on book m3

  • Downward gravitational force (m3g)
  • Downward force exerted by m4 (N4)
  • Upward force exerted by m2 (N2)
    Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

(c) Force on book m2

  • Downward gravitational force (m2g)
  • Downward force exerted by m3 (N3)
  • Upward force exerted by m1 (CN1)
    Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

(d) Force on book m1

  • Downward gravitational force exerted by earth (m1g)
  • Downward force exerted by m2 (N2)
  • Upward force exerted by the table (Ntable)
    Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Question 5.
A bob attached to the string oscillates back and forth. Resolve the forces acting on the bob in to components. What is the acceleration experienced by the bob at an angle θ.
Answer:
The gravitational force (mg) acting downward can be resolved into two components as mg cos θ and mg sin θ
T – tension exerted by the string.
Tangential force FT = maT = mg sin θ
∴ Tangential acceleration aT = g sin θ
Centripetal force Fc = mac = T – mg cos θ
ac = \(\frac { T – mg cos θ }{ m }\)
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Question 6.
Two masses m1 and m2 are connected with a string passing over a friction-less pulley fixed at the comer of the table as shown in the figure. The coefficient of static friction of mass m1 with the table is µs Calculate the minimum mass m3 that may be placed on m1to prevent it from sliding. Check if m1 = 15 kg, m2 = 10 kg, m3 = 25 and µs = 0.2.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Solution:
Let m3 is the mass added on m1
Maximal static friction
\(f_{s}^{\max }\) = µsN = µs (m1 + m3 )g
Here
N = (m1 + m3 )g
Tension acting on string = T = m2 g
Equate (1) and (2)
µs(m1 + m3) = m2g
µsm1 + µsm3 = m2
m3 = \(f_{s}^{\max }\) – m1

(ii) Given,
m1 = 15 kg, m2 = 10 kg : m3 = 25 kg and µs = 0.2
m3 = \(f_{s}^{\max }\) – m1
m3 = \(\frac {10}{ 0.2 }\) – 15 = 50 – 15 = 35 kg
The minimum mass m3 = 35 kg has to be placed on ml to prevent it from sliding. But here m3 = 25 kg only.
The combined masses (m1 + m3) will slide.

Question 7.
Calculate the acceleration of the bicycle of mass 25 kg as shown in Figures 1 and 2.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Answer:
Given:
Mass of bicycle m = 25 kg
Fig. I:
Net force acting in the forward direction, F = 500 – 400 = 100 N
acceleration a = \(\frac { F}{ m }\) = \(\frac { 100 }{25}\) = 4 ms-2

Fig. II:
Net force acting on bicycle F = 400 – 400 = 0
∴ acceleration a = \(\frac { F}{ m }\) = \(\frac { 0}{25}\) = 0

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Question 8.
Apply Lami’s theorem on sling shot and calculate the tension in each string?
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Answer:
Given F = 50 N, θ = 30°
Here T is resolved into its components as T sin θ and T cos θ as shown.
According to Lami’s theorem,
\(\frac { F}{sin θ}\) = \(\frac { T}{sin (180 – θ)}\) = \(\frac { T}{sin (180 – θ)}\)
\(\frac { F}{sin θ}\) = \(\frac { T}{sin θ}\)
\(\frac { F}{2 sin θ cos θ}\) = \(\frac { T}{sin θ}\) [T = \(\frac { T}{2 cos θ}\) ]
T = \(\frac {T}{2 cos θ}\) = \(\frac { 50}{ 2 cos 30}\) = 28.868 N

Question 9.
A football player kicks a 0.8 kg ball and imparts it a velocity 12 ms-1. The contact between the foot and ball is only for one – sixtieth of a second. Find the average kicking force.
Answer:
Given,
Mass of the ball = 0.8 kg
Final velocity (V) =12 ms-1 and time t =\(\frac {1}{60}\) s
Initial velocity = 0
We know the average kicking force
F = ma = \(\frac {m(v – u)}{t}\) = \(\frac{0.8(12-0)}{\left(\frac{1}{60}\right)}\)
F = 576 N

Question 10.
A stone of mass 2 kg is attached to a string of length 1 meter. The string can withstand maximum tension 200 N. What is the maximum speed that stone can have during the whirling motion?
Solution:
Given,
Mass of a stone = 2 kg,
length of a string = 1 m
Maximum tension = 200 N
The force acting on a stone in the whirling motion is centripetal force. Which is provided by tension of the string.
Tmax = Fmax = \(\frac{m \mathrm{V}_{\mathrm{max}}^{2}}{r}\)
200 = \(v_{\max }^{2}\) = 100
\(v_{\max }^{2}\) = 10 ms-1

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Question 11.
Imagine that the gravitational force between Earth and Moon is provided by an invisible string that exists between the Moon and Earth. What is the tension that exists in this invisible string due’ to Earth’s centripetal force? (Mass of the Moon = 7.34 x 1022 kg, Distance between Moon and Earth = 3.84 x 108 m).

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Solution:
Given,
Mass of the moon = 7.34 x 1022 kg
Distance between moon and earth = 3.84 x 108 m
Centripetal force = F = \(\frac{m \mathrm{V}^{2}}{r}\) = \(\frac{7.34 \times 10^{22} \times\left(1.023 \times 10^{3}\right)^{2}}{3.84 \times 10^{8}}\) = 2 x 1020

Question 12.
Two bodies of masses 15 kg and 10 kg are connected with light string kept on a smooth surface. A horizontal force F = 500 N is applied to a 15 kg as shown in the figure. Calculate the tension acting in the string.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Answer:
Given,
m1  = 15 kg, m2  = 10 kg, F = 500 N
Tension acting in the string T = \(\frac{m_{2}}{m_{1}+m_{2}}\) F
T = \(\frac {10}{25}\) x 500 = 200 N

Question 13.
People often say “For every action there is an equivalent opposite reaction”. Here they meant ‘action of a human’. Is it correct to apply Newton’s third law to human actions? What is meant by ‘action’ in Newton third law? Give your arguments based on Newton’s laws.
Answer:
Newton’s third law is applicable to only human’s physical actions which involves physical force. Third law is not applicable to human’s psychological actions or thoughts.

Question 14.
A car takes a turn with velocity 50 ms-1 on the circular road of radius of curvature To m. Calculate the centrifugal force experienced by a person of mass 60 kg inside the car?
Answer:
Given,
Mass of a person = 60 kg
Velocity of the car = 50 ms-1
Radius of curvature = 10 m
Centrifugal force F = \(\frac{m \mathrm{V}^{2}}{r}\) = \(\frac{60 \times(50)^{2}}{10}\) = 15,000 N

Question 15.
A long stick rests on the surface. A person standing 10 m away from the stick. With what minimum speed an object of mass 0.5 kg should he thrown so that it hits the stick. (Assume the coefficient of kinetic friction is 0.7).
Answer:
Given,
Distance (s) = 10 m
Mass of the object (m) = 0.5 kg
Coefficient of kinetic friction (µ) = 0.7
Work done in moving a body in horizontal surface ω = µR x s = µmg x s
This work done is equal to initial kinetic energy of the object
\(\frac{1}{2} m v^{2}\) = µ mg s
\(\left|v^{2}\right|\) = 2 µgs = 2 x 0.7 x 9.8 x 10
v2 = 14 x 9.8 = 137. 2
v = 11. 71 ms-1

Samacheer Kalvi 11th Physics Laws of Motion Additional Questions Solved

Samacheer Kalvi 11th Physics Laws of Motion Multiple Choice Questions

Question 1.
The concept “force causes motion” was given by –
(a) Galileo
(b) Aristotle
(c) Newton
(d) Joule
Answer:
(b) Aristotle

Question 2.
Who decoupled the motion and force?
(a) Galileo
(b) Aristotle
(c) Newton
(d) Joule
Answer:
(a) Galileo

Question 3.
The inability of objects to move on its own or change its state of motion is called as –
(a) force
(b) momentum
(c) inertia
(d) impulse
Answer:
(c) inertia

Question 4.
Inertia means –
(a) inability
(b) resistance to change its state
(c) movement
(d) inertial frame
Answer:
(b) resistance to change its state

Question 5.
When a bus starts to move from rest, the passengers experience a sudden backward push is an example for –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(c) Inertia of rest

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Question 6.
If the brake is applied in the moving bus suddenly, passengers move forward is an example for –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(a) Inertia of motion

Question 7.
In whirling motion, if the string is cut suddenly, the stone moves tangential to circle is an –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(b) Inertia of direction

Question 8.
Newtons laws are applicable in –
(a) Inertial frame
(b) non inertial frame
(c) in any frame
(d) none
Answer:
(a) Inertial frame

Question 9.
The accelerated train is an example for –
(a) inertial frame
(b) non-inertial frame
(c) both (a) and (b)
(d) none of the above
Answer:
(b) non-inertial frame

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Question 10.
Rate of change of momentum of an object is equal to –
(a) acceleration
(b) work done
(c) force
(d) impulse
Answer:
(c) force

Question 11.
The product of mass and velocity is –
(a) force
(b) impulse
(c) momentum
(d) acceleration
Answer:
(c) momentum

Question 12.
Unit of momentum –
(a) kg ms-2
(b) kg ms-1
(c) MLT-2
(d) MLT-1
Answer:
(b) kg ms-1

Question 13.
According to Newton’s third law –
(a) F12 = F21
(*) F12 = -F21
(c) F12 + F21 = 0
(d) F12 x F21 = 0
Answer:
(a) F12 = F21

Question 14.
According to Newton’s third law –
(a) \(\overrightarrow{\mathrm{F}_{12}}=\overrightarrow{\mathrm{F}_{21}}\)
(b) \(\overrightarrow{\mathrm{F}_{12}}=-\overrightarrow{\mathrm{F}_{21}}\)
(c) \(\mathrm{F}_{12}+\mathrm{F}_{21}\) = 0
(d) \(\mathrm{F}_{12}x\mathrm{F}_{21}\) = 0
Answer:
(b) \(\overrightarrow{\mathrm{F}_{12}}=-\overrightarrow{\mathrm{F}_{21}}\)

Question 15.
The law which is valid in both inertial and non-inertial frame is –
(a) Newton’s first law
(b) Newton’s second law
(c) Newton’s third law
(d) none
Answer:
(c) Newton’s third law

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Question 16.
When a force is applied on a body, it can change –
(a) velocity
(b) momentum
(c) direction of motion
(d) all the above
Answer:
(d) all the above

Question 17.
The rate of change of velocity is 1 ms-2 when a force is applied on the body of mass 75 gm the force is –
(a) 75 N
(b) 0.75 N
(c) 0.075 N
(d) 0.75 x 10-3 N
Answer:
(c) Force is given by
F = m a
= 75 gm x 1 cm s-2 = 75 x 10-3 x 1 = 75 x 10-3 = 0.075 N

Question 18.
The action and reaction forces acting on –
(a) same body
(b) different bodies
(c) either same or different bodies
(d) none of the above
Answer:
(b) different bodies

Question 19.
Newton’s first law of motion gives the concept of –
(a) velocity
(b) energy
(c) momentum
(d) Inertia
Answer:
(d) Inertia

Question 20.
Inertia of a body has direct dependence on –
(a) velocity
(b) area
(c) mass
(d) volume
Answer:
(c) mass

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Question 21.
If a car and a scooter have the same momentum, then which one is having greater speed?
(a) scooter
(b) car
(c) both have same velocity
(d) data insufficient
Answer:
(a) scooter

Question 22.
Newton’s second law gives –
(a) \(\overrightarrow{\mathrm{F}} \propto \frac{d \overrightarrow{\mathrm{P}}}{\mathrm{dt}}\)
(b) \(\overrightarrow{\mathrm{F}}=\frac{d \overrightarrow{\mathrm{P}}}{\mathrm{dt}}\)
(c) \(\overrightarrow{\mathrm{F}}=m \vec{a}\)
(d) all the above
Answer:
(d) all the above

Question 23.
1 dyne is –
(a) 105N
(b) 10-5N
(c) 1N
(d) 10-3N
Answer:
(b) 10-5N

Question 24.
If same force is acting on two masses m1 and m2, and the accelerations of two bodies are a1 and a2 respectively, then –
(a) \(\frac{a_{2}}{a_{1}}=\frac{m_{2}}{m_{1}}\)
(b) \(\frac{a_{1}}{a_{2}}=\frac{m_{1}}{m_{2}}\)
(c) \(\frac{a_{1}}{a_{2}}=\frac{m_{2}}{m_{1}}\)
(d) m1 a1 + m2a2 = 0
Answer:
(c) \(\frac{a_{1}}{a_{2}}=\frac{m_{2}}{m_{1}}\)

Question 25.
If a force \(\overline{\mathrm{F}}\) = 3\(\hat{i}\) – 4\(\hat{j}\) N produces an acceleration of 10 ms-2 on a body, then the mass of a body is –
(a) 10 kg
(b) 9 kg
(c) 0.9 kg
(d) 0.5 kg
Answer:
\(\overline{\mathrm{F}}\) = 3\(\hat{i}\) – 4\(\hat{j}\)
Magnitude:
|\(\overline{\mathrm{F}}\)| = \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5N
F = ma
⇒ m = \(\frac{|\mathrm{F}|}{a}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\) = 0.5 kg

Question 26.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms-1. How long does the body take to stop?
(a) 0.75 s
(b) 1.33 s
(c) 6 s
(d) 35 s
Answer:
Acceleration a = \(\frac{-F}{m}\) = \(\frac{50}{20}\) = – 2.5 ms-2
u = l5 ms-1
v = 0
t = ?
v = u + at
0 = 15 – 2.5t
t = \(\frac{15}{2.5}\) = 6s

Question 27.
Rain drops come down with –
(a) zero acceleration and non zero velocity
(b) zero velocity with non zero acceleration
(c) zero acceleration and non zero net force
(d) none
Answer:
(a) zero acceleration and non zero velocity

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Question 28.
If force is the cause then the effect is –
(a) mass
(b) potential energy
(c) acceleration
(d) Inertia
Answer:
(c) acceleration

Question 29.
In free body diagram, the object is represented by a –
(a) line
(b) arrow
(c) circle
(d) point
Answer:
(d) point

Question 30.
When an object of mass m slides on a friction less surface inclined at an angle 0, then normal force exerted by the surface is –
(a) g cos θ
(b) mg cos θ
(c) g sin θ
(d) mg tan θ
Answer:
(b) mg cos θ

Question 31.
The acceleration of the sliding object in an inclined plane –
(a) g cos θ
(b) mg cos θ
(c) g sin θ
(d) mg sin θ
Answer:
(c) g sin θ

Question 32.
The speed of an object sliding in an inclined plane at the bottom is –
(a) mg cos θ
(b) \(\sqrt{2 s g sin θ}\)
(c) \(\sqrt{2 s g cos θ}\)
(d) \(\sqrt{2 s g tan θ}\)
Answer:
(b) \(\sqrt{2 s g sin θ}\)

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Question 33.
The acceleration of two bodies of mass m1 and m2 in contact on a horizontal surface is –
(a) \(a=\frac{\mathbf{F}}{m_{1}}\)
(b) \(a=\frac{F}{m_{2}}\)
(c)  \(a=\frac{\mathrm{F}}{m_{1}+m_{2}}\)
(d) \(a=\frac{\mathrm{F}}{m_{1} m_{2}}\)
Answer:
(c)  \(a=\frac{\mathrm{F}}{m_{1}+m_{2}}\)

Question 34.
Two blocks of masses m1 and m2 (m1 > m2) in contact with each other on frictionless, horizontal surface. If a horizontal force F is given on m1, set into motion with acceleration a, then reaction force on mass m1 by m2, is –
(a) \(\frac{\mathrm{F} m_{1}}{m_{1}+m_{2}}\)
(b) \(\frac{m_{1} m_{2}}{\mathrm{F} m_{1}}\)
(c) \(\frac{m_{1} m_{2}}{\mathrm{F} m_{2}}\)
(d) \(\frac{\mathrm{F} m_{2}}{m_{1}+m_{2}}\)
Answer:
(d) \(\frac{\mathrm{F} m_{2}}{m_{1}+m_{2}}\)

Question 35.
If two masses m1 and m2 (m1 > m2) tied to string moving over a frictionless pulley, then acceleration of masses –
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g
(b) \(\frac{m_{1}+m_{2}}{\left(m_{1}-m_{2}\right)}\) g
(c) \(\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}\) g
(d) \(\frac{m_{1} m_{2}}{2 m_{1} m_{2}}\) g
Answer:
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g

Question 36.
if two masses m1 and m2 (m1 > m2)tied to string moving over a frictionless pulley, then acceleration of masses –
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g
(b) \(\frac{m_{1}+m_{2}}{\left(m_{1}-m_{2}\right)}\) g
(c) \(\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}\) g
(d) \(\frac{m_{1} m_{2}}{2 m_{1} m_{2}}\) g
Answer:
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g

Question 37.
Three massses is in contact as shown. If force F is applied to mass m1, the acceleration of three masses is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion q37
Answer:
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)

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Question 38.
Three masses in contact is as shown above. If force F is applied to mass m1 then the contact force acting on mass m2 is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)

Question 39.
Three masses is contact as shown. It force F is applied to mass m1, then the contact force acting on mass m3 is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)

Question 40.
Two masses connected with a string. When a force F is applied on mass m2. The acceleration produced is –
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion q40
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}}\)
(b) \(\frac{\mathbf{F}}{m_{1}-m_{2}}\)
(c) \(\frac{m_{1}+m_{2}}{\mathrm{F}}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}}\)

Question 41.
Two masses connected with a string. When a force F is applied on mass m2. The force acting on m1 is –
(a) \(\frac{m_{1} \mathrm{F}}{m_{1}+m_{2}}\)
(b) \(\frac{m_{2} \mathrm{F}}{m_{1}+m_{2}}\)
(c) \(\frac{m_{1}+m_{2}}{m_{1}} \mathbf{F}\)
(d) \(\frac{m_{1}+m_{2}}{m_{2}} \mathbf{F}\)
Answer:
(b) \(\frac{m_{2} \mathrm{F}}{m_{1}+m_{2}}\)

Question 42.
If a block of mass m lying on a frictionless inclined plane of length L height h and angle of inclination θ, then the velocity at its bottom is –
(a) g sin θ
(b) g cos θ
(c) \(\sqrt{2 g h}\)
(d) \(\sqrt{2 a sin θ}\)
Answer:
(c) \(\sqrt{2 g h}\)

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Question 43.
If a block of mass m lying on a frictionless inclined plane of length L, height h and angle of inclination θ, then the time take taken to reach the bottom is –
(a) g sing θ
(b) sin θ \(\sqrt{\frac{2 h}{g}}\)
(c) sin θ \(\sqrt{\frac{g}{h}}\)
(d) \(\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}\)
Answer:
(d) \(\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}\)

Question 44.
A rocket works on the principle of conservation of –
(a) energy
(b) mass
(c) angular momentum
(d) linear momentum
Answer:
(b) mass

Question 45.
A bomb at rest explodes. The total momentum of all its fragments is –
(a) zero
(b) infinity
(c) always 1
(d) always greater then 1
Answer:
(a) zero

Question 46.
A block of mass m1 is pulled along a horizontal friction-less surface by a rope of mass m2 If a force F is given at its free end. The net force acting on the block is –
(a) \(\frac{m_{1} \mathrm{F}}{m_{1}-m_{2}}\)
(b) F
(c) \(\frac{m_{2} \mathrm{F}}{\left(m_{1}+m_{2}\right)}\)
(d) \(\frac{m_{1} \mathrm{F}}{\left(m_{1}+m_{2}\right)}\)
Answer:
(b) F

Question 47.
A block of mass m is pulled along a horizontal surface by a rope. The tension in the rope will be same at all the points –
(a) if the rope is accelerated
(b) if the rope is mass less
(c) always
(d) none of the above
Answer:
(b) if the rope is mass less

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Question 48.
The lines of forces act at a common point is called as –
(a) concurrent forces
(b) co-planar forces
(c) equilibrium
(d) resultant
Answer:
(a) concurrent forces

Question 49.
If the lines of forces act in the same plane, they can be –
(a) concurrent forces
(b) coplanar forces
(c) either concurrent force or coplanar forces
(d) Lami’s force
Answer:
(d) concurrent forces

Question 50.
Lami’s theorem is applicable only when the system of forces are is –
(a) same plane
(b) different plane
(c) equilibrium
(d) none of the above
Answer:
(c) equilibrium

Question 51.
Due to the action of internal forces of the system, the total linear momentum of the system is –
(a) a variable
(b) a constant
(c) always zero
(d) always infinity
Answer:
(c) always zero

Question 52.
The velocity with which a gun suddenly moves backward after firing is –
(a) linear velocity
(b) positive velocity
(c) recoil velocity
(d) v1 + v2
Answer:
(c) recoil velocity

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Question 53.
If a very large force acts on an object for a very short duration, then the force is called as –
(a) Newtonian force
(b) impulsive force
(c) concurrent force
(d) coplanar force
Answer:
(A) impulsive force

Question 54.
The unit of impulse is –
(a) Nm
(b) Ns
(c) Nm2
(d) Ns-2
Answer:
(b) Ns

Question 55.
The force which always opposes the relative motion between an object and the surface where it is placed is –
(a) concurrent force
(b) frictional force
(c) impulsive force
(d) coplanar force
Answer:
(b) frictional force

Question 56.
The force which opposes the initiation of motion of an object on the surface is –
(a) static friction
(b) kinetic friction
(c) friction
(d) zero
Answer:
(d) static friction

Question 57.
When the object is at rest, the resultant of gravitational force and upward normal force is –
(a) Static force
(b) zero
(c) one
(d) infinity
Answer:
(b) zero

Question 58.
The magnitude of static frictional force d lies between –
(a) 0 ≤ f ≤ µsN
(b) 0 ≥f ≥ µsN
(c) 0 and 1
(d) 0 and minimal static frictional force.
Answer:
(a) 0 ≤ f ≤ µsN

Question 59.
The unit of co-efficient of static friction is –
(a) N
(b) N m
(c) N s
(d) no unit
Answer:
(d) no unit

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Question 60.
If the object is at rest and no external force is applied on the object, the static friction acting on the object is –
(a) µsN
(b) zero
(c) one
(d) infinity
Answer:
(d) no unit

Question 61.
When object begins to slide, the static friction acting on the object attains –
(a) zero
(b) minimum
(c) maximum
(d) infinity
Answer:
(c) maximum

Question 62.
The static friction does not depend upon –
(a) the area of contact
(b) normal force
(c) the magnitude of applied force
(d) none of the above
Answer:
(a) the area of contact

Question 63.
Which of the following pairs of materials has minimum amount of coefficient of static friction is –
(a) Glass and glass
(b) wood and wood
(c) ice and ice
(d) steel and steel
Answer:
(c) ice and ice

Question 64.
Kinetic friction is also called as –
(a) sliding friction
(b) dynamic friction
(c) both (a) and (b)
(d) static friction
Answer:
(c) both (a) and (b)

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Question 65.
The unit of coefficient of kinetic friction is/has –
(a) Nm
(b) Ns
(c) Nm2
(d) no unit
Answer:
(d) no unit

Question 66.
The nature of materials in mutual contact decides –
(a) µs
(b) µk
(c) µs or µk
(d) none
Answer:
(c) µs or µk

Question 67.
Coefficient of kinetic friction is less than –
(a) O
(b) one
(c) µs
(d) µsN
Answer:
(c) µs

Question 68.
The static friction –
(a) increases linearly
(b) is constant
(c) zero
(d) varies parabolically
Answer:
(a) increases linearly

Question 69.
The kinetic friction –
(a) increases linearly
(b) is constant
(c) zero
(d) varies parabolically
Answer:
(b) is constant

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Question 70.
Kinetic friction is independent of –
(a) nature of materials
(b) temperature of the surface
(c) applied force
(d) none of the above
Answer:
(c) applied force

Question 71.
The angle between the normal force and the resultant force of normal force and maximum frictional force is –
(a) angle of friction
(b) angle of repose
(c) angle of inclination
(d) none of the above
Answer:
(a) angle of friction

Question 72.
The angle friction θ is given by –
(a) tan µs
(b) tan-1 µs
(c) \(\frac{f S^{\mathrm{max}}}{N}\)
(d) sin-1 µs
Answer:
(b) tan-1 µs

Question 73.
The angle of inclined plane with the horizontal such that an object placed on it begins to slide is –
(a) angle of friction
(b) angle of repose
(c) angle of response
(d) angle of retardation
Answer:
(b) angle of repose

Question 74.
Comparatively, which of the following has lesser value than others?
(a) static friction
(b) kinetic friction
(c) Rolling friction
(d) skiping friction
Answer:
(c) Rolling friction

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Question 75.
The origin of friction is –
(a) electrostatic interaction
(b) electromagnetic interaction magnetic
(c) photon interaction
(d) interaction
Answer:
(b) electromagnetic interaction

Question 76.
Friction can be reduced by –
(a) polishing
(b) lubricating
(c) using ball bearings
(d) all the above
Answer:
(c) using ball bearings

Question 77.
For a particle revolving in a circular path, the acceleration of the particle is –
(a) along the tangent
(b) along the radius
(c) along the circumference of the circle
(d) zero
Answer:
(b) along the radius

Question 78.
A particle moves along a circular path under the action of a force. The work done by the force is –
(a) Positive and non zero
(b) zero
(c) Negative and non zero
(d) none of the above
Answer:
(b) zero

Question 79.
A bullet hits and gets embedded in a solid block resting on a horizontal frictionless table. Which of the following is conserved?
(a) Momentum and kinetic energy
(b) kinetic energy alone
(c) Momentum alone
(d) potential energy alone
Answer:
(c) Momentum alone

Question 80.
The origin of the centripetal force can be –
(a) gravitational force
(b) frictional force
(c) coulomb force
(d) all the above
Answer:
(d) all the above

Question 81.
Centripetal acceleration is –
(a) \(\frac{m v^{2}}{r}\)
(b) \(\frac{v^{2}}{r}\)
(c) r v2
(d) rω
Answer:
(b) \(\frac{v^{2}}{r}\)

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Question 82.
Centripetal acceleration is –
(a) \(\frac{m v^{2}}{r}\)
(b) r ω2
(c) rv2
(d) rω
Answer:
(c) rω2

Question 83.
The centripetal force is –
(a) \(\frac{m v^{2}}{r}\)
(b) rω2
(c) both (a) and (b)
(d) none
Answer:
(c) both (a) and (b)

Question 84.
When a car is moving on a circular track the centripetal force is due to –
(a) gravitational force
(b) frictional force
(c) magnetic force
(d) elastic force
Answer:
(b) frictional force

Question 85.
If the road is horizontal then the normal force and gravitational force are –
(a) equal and along the same direction
(b) equal and opposite
(c) unequal and along the same direction
(d) unequal and opposite
Answer:
(b) equal and opposite

Question 86.
The velocity of a car for safe turn on leveled circular road –
(a) \(v \leq \sqrt{\mu_{s} r g}\)
(b) \(v \geq \sqrt{\mu_{s} r g}\)
(c) \(v=\sqrt{\mu_{s} rg}\)
(d) \(v \leq \mu_{s} rg\)
Answer:
(a) \(v \leq \sqrt{\mu_{s} r g}\)

Question 87.
In a leveled circular road, skidding mainly depends on –
(a) µs
(b) µk
(c) acceleration
(d) none
Answer:
(a) µs

Question 88.
The speed of a car to move on the banked road so that it will have safe turn is –
(a) µsrg
(b) \(\sqrt{r g \tan \theta}\)
(c) rg tan θ
(d) r2g tan θ
Answer:
(b) \(\sqrt{r g \tan \theta}\)

Question 89.
Centrifugal force is a –
(a) pseudo force
(b) real force
(c) forced acting towards center
(d) none of the above
Answer:
(a) pseudo force

Question 90.
Origin of centrifugal force is due to –
(a) interaction between two
(b) inertia
(c) electromagnetic interaction
(d) inertial frame
Answer:
(b) inertia

Question 91.
Centripetal force acts in –
(a) inertial frame
(b) non inertial frame
(c) both (a) and (h)
(d) linear motion
Answer:
(c) both (a) and (b)

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Question 92.
Centrifugal force acts in –
(a) inertial frame
(b) non inertial frame
(c) both (a) and (b)
(d) linear motion
Answer:
(b) non inertial frame

Question 93.
A cricket ball of mass loo g moving with a velocity of 20 ms-1 is brought to rest by a player in 0.05s the impulse of the ball is –
(a) 5 Ns
(b) – 2 Ns
(c) – 2.5 Ns
(d) zero
Answer:
(b) – 2 Ns
mass = 0.1 kg
Initial velocity t = 20 ms-1
Final velocity y = 0
Change in momentum in impulse = m(v – u) = 0.1(0 – 20) = – 2 Ns

Question 94.
If a stone tied at the one end of a string of length 0.5 m is whirled in a horizontal circle with a constant speed 6 ms-1  then the acceleration of the shone is –
(a) 12 ms-2
(b) 36 ms-2
(c) 2π2 ms-2
(d) 72 ms-2
Answer:
(d) Centripetal acceleration = \(\frac{v^{2}}{r}\) = \(\frac{6^{2}}{0.5}\) = \(\frac{36}{0.5}\) = 72 ms-2

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Question 95.
A block of mass 3 kg is at rest on a rough inclined plane with angle of inclination 30° with horizontal. If .is 0.7, then the frictional force is –
(a) 17.82 N
(b) 1.81 N
(c) 3.63 N
(d) 2.1 N
Answer:
(a) Frictional force = µmg cos θ = 0.7 x 3 x 9.8 cos 30° = 17.82 N

Question 96.
Two masses 2 kg and 4 kg are tied at the ends of a mass less string and which is passing over a friction-less pulley. The tension in the string is –
(a) 3.68 N
(b) 78.4 N
(c) 26 N
(d) 13.26 N
Answer:
(c) Tension in the string T = \(\frac{2 m_{1} m_{2}}{\left(m_{1}+m_{2}\right)}\)g
T = \(\frac{2 x 2 x 4}{2 + 4}\) x 9.8 = \(\frac{16}{6}\) x 9.8 = 26.13 N

Question 97.
A bomb of 10 kg at rest explodes into two pieces of mass 4 kg and 6 kg. if the velocity of 4 kg mass is 6 ms-1 then the velocity of 6 kg is –
(a) – 4 ms-1
(b) – 6 ms-1
(c) – 24 ms-1
(d) – 2.2 ms-1
Answer:
(a) According to law of conservation of momentum
m1v1 + m2v2 = 0
v2 = –\(\frac{m_{1} v_{1}}{m_{2}}\) = \(\frac{4 x 6}{6}\) = -4ms-1

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Question 98.
A body is subjected under three concurrent forces and it is in equilibrium. The resultant of any two forces is –
(a) coplanar with the third force
(b) is equal and opposite to third force
(c) both (a) and (b)
(d) none of the above
Answer:
(c) both (a) and (b)

Question 99.
An impulse is applied to a moving object with the force at an angle of 20° with respect to velocity vector. The angle between the impulse vector and the change in momentum vector is –
(a) 0°
(b) 30°
(c) 60°
(d) 120°
Answer:
(a) Impulse and change in momentum are in same direction. So the angle is zero.

Question 100.
A bullet of mass m and velocity v1 is fired into a large block of wood of mass M. The final velocity of the system is-
(a) \(\frac{v_{1}}{m+\mathrm{M}}\)
(b) \(\frac{m v_{1}}{m+\mathrm{M}}\)
(c) \(\frac{m+m}{m} v_{1}\)
(d) \(\frac{m+m}{m-M} v_{1}\)
Answer:
(b) \(\frac{m v_{1}}{m+\mathrm{M}}\)

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Question 101.
A block of mass 2 kg is placed on the floor. The co – efficient of static friction is 0.4. The force of friction between the block and floor is –
(a) 2.8 N
(b) 7.8 N
(c) 2 N
(d) zero
Answer:
(b) The force required to move = = µR = µmg = 0.4 x 2 x 9.8 = 7.84 N

Question 102.
A truck weighing 1000 kg is moving with velocity of 50 km/h on smooth horizontal roads. A mass of 250 kg is dropped into it. The velocity with which it moves now is –
(a) 12.5 km/h
(b) 20 km/h
(c) 40 km/h
(d) 50 km/h
Answer:
(c) According to law of conservation of linear momentum
m2 v2 = (m1 + m2)v2
v2 = \(\frac{m_{1} v_{1}}{m_{1}+m_{2}}\) = \(\frac{1000 \times 50}{1250}\) = 40 km/h

Question 103.
A body of mass loo g is sliding from an inclined plane of inclination 30°. if u = 1.7, then the frictional force experienced is –
(a) \(\frac{3.4}{\sqrt{3}}\)N
(b) 1.47 N
(c) \(\frac{\sqrt{3}}{3.4}\)N
(d) 1.38 N
Answer:
(b) Frictional force F = µ mg cos θ = 1.7 x 0.1 x 10 cos 30°= \(\frac{1.7}{2}\) x \(\sqrt{3}\) = 1.47 N

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions (1 Mark)

Question 1.
A passenger sitting in a car at rest, pushes the car from within. The car doesn’t move, why?
Answer:
For motion, there should be external force.

Question 2.
Give the magnitude and directions of the net force acting on a rain drop falling with a constant speed.
Answer:
as \(\overline{\mathrm{a}}\) = 0 so \(\overline{\mathrm{F}}\) = 0.

Question 3.
Why the passengers in a moving car are thrown outwards when it suddenly takes a turn?
Answer:
Due to inertia of direction.

Question 4.
You accelerate your car forward. What is the direction of the frictional force on a package resting on the floor of the car?
Answer:
The package in the accelerated car (a non inertial frame) experiences a Pseudo force in a direction opposite to that of the motion of the car. The frictional force on the package which acts opposite to this pseudo force is thus in the same direction (forward) as that of the car.

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Question 5.
What is the purpose of using shockers in a car?
Answer:
To decrease the impact of force by increasing the time for which force acts.

Question 6.
Why are types made of rubber not of steel?
Answer:
Since coefficient of friction between rubber and road is less than the coefficient of friction between steel and road.

Question 7.
Wheels are made circular. Why?
Answer:
Rolling friction is less than sliding friction.

Question 8.
If a ball is thrown up in a moving train, it comes back to the thrower’s hands. Why?
Answer:
Both during its upward and downward motion, the ball continues to move inertia of motion with the same horizontal velocity as the train. In this period, the ball covers the same horizontal distance as the train and so it comes back to the thrower’s hand.

Question 9.
Calculate the force acting on a body which changes the momentum of the body at the rate of 1 kg-m/s2 .
Answer:
As F = rate change of momentum
F = 1 kg-m/s2 = 1N

SamacheerKalvi.Guru

Question 10.
On a rainy day skidding takes place along a curved path. Why?
Answer:
As the friction between the types and road reduces on a rainy day.

Question 11.
Why does a gun recoils when a bullet is being fired?
Answer:
To conserve momentum.

Question 12.
Why is it difficult to catch a cricket ball than a tennis ball even when both are moving with the same velocity?
Answer:
Being heavier, cricket ball has higher rate of change of momentum during motion so more force sumed.

Question 13.
The distance travelled by a moving body is directly proportional to time. Is any external force acting on it?
Answer:
As s ∝ t, so acceleration a = 0, therefore, no external force is acting on the body.

Question 14.
Calculate the impulse necessary to stop a 1500 kg car moving at a speed of 25 ms-1.
Answer:
Use formula I = change in momentum = m(v – u) (Impulse – 37500 Ns)

Question 15.
Lubricants are used between the two parts of a machine. Why?
Answer:
To reduce friction and so to reduce wear and tear.

Question 16.
What provides the centripetal force to a car taking a turn on a level road?
Answer:
Force of friction between the type and road provides centripetal force.

Question 17.
A body is acted upon by a number of external forces. Can it remain at rest?
Answer:
Yes, if the external forces acting on the body can be represented in magnitude and direction by the sides of a closed polygon taken in the same order.

Question 18.
Bodies of larger mass need greater initial effort to put them in motion. Why?
Answer:
As F = ma so for given a, more force will be required to put a large mass in motion.

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Question 19.
An athlete runs a certain distance before taking a long jump Why?
Answer:
So that inertia of motion may help him in his muscular efforts to take a longer jump.

Question 20.
Action and reaction forces do not balance each other. Why?
Answer:
As they acts on different bodies.

Question 21.
The wheels of vehicles are provided with mudguards. Why?
Answer:
When the wheel rotates at a high speed, the mud sticking to the wheel flies off tangentially, this is due to inertia of direction. If order that the flying mud does not spoil the clothes of passer by the wheels are provided with mudguards.

Question 22.
China wares are wrapped in straw paper before packing. Why?
Answer:
The straw paper between the China ware increases the Time of experiencing the jerk during transportation. Hence impact of force reduces on China wares.

SamacheerKalvi.Guru

Question 23.
Why is it difficult to walk on a sand?
Answer:
Less reaction force.

Question 24.
The outer edge of a curved road is generally raised over the inner edge Why?
Answer:
In addition to the frictional force, a component of reaction force also provides centripetal force.

Question 25.
Explain why the water doesn’t fall even at the top of the circle when the bucket full of water is upside down rotating in a vertical circle?
Answer:
Weight of the water and bucket is used up in providing the necessary centripetal force at the top of the circle.

Question 26.
Why does a speedy motor cyclist bends towards the center of a circular path while taking a turn on it?
Answer:
So that in addition of the frictional force, the horizontal component of the normal reaction also provides the necessary centripetal forces.

Question 27.
An impulse is applied to a moving object with a force at an angle of 20° wr.t. velocity vector, what is the angle between the impulse vector and change in momentum vector ?
Answer:
Impulse and change in momentum are along the same direction. Therefore angle between these two vectors is zero.

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions (2 Marks)

Question 28.
A man getting out of a moving bus runs in the same direction for a certain distance. Comment.
Answer:
Due to inertia of motion.

Question 29.
If the net force acting upon the particle is zero, show that its linear momentum remains constant.
Answer:
As F x \(\frac {dp}{dt}\)
when F = 0, \(\frac {dp}{dt}\) = 0 so P = constant

SamacheerKalvi.Guru

Question 30.
A force of 36 dynes is inclined to the horizontal at an angle of 60°. Find the acceleration in a mass of 18 g that moves in a horizontal direction.
Answer:
F = 36 dyne at an angle of 60°
Fx = F cos 60° = 18 dyne
Fx = max
So ax = \(\frac{F_{x}}{m}\) = 1 cm /s2

Question 31.
The motion of a particle of mass m is described by h = ut + \(\frac {1}{2}\) gt2. Find the force acting on particle.
Answer:
a = ut + \(\frac {1}{2}\) gt2
find a by differentiating h twice w.r.t.
a = g
As F = ma so F = mg (answer)

Question 32.
A particle of mass 0.3 kg is subjected to a force of F = -kx with k= 15 Nmr-1. What will be its initial acceleration if it is released from a point 20 cm away from the origin?
Answer:
As F = ma so F = -kx = ma
a = \(\frac {-kx}{m}\)
for x = 20 cm, ⇒ a = -10 m/s2.

SamacheerKalvi.Guru

Question 33.
A 50 g bullet is fired from a 10 kg gun with a speed of 500 ms-1. What is the speed of the recoil of the gun?
Answer:
Initial momentum = 0
Using conservation of linear momentum mv + MV = 0
V = \(\frac {-mv}{M}\) ⇒ V = 2.5 m/s

Question 34.
Smooth block is released at rest on a 45° incline and then slides a distance d. If the time taken of slide on rough incline is n times as large as that to slide than on a smooth incline. Show that coefficient of friction, µ = \(\left(1-\frac{1}{n^{2}}\right)\)
Answer:
When there is no friction, the block slides down the inclined plane with acceleration. a = g sin θ
when there is friction, the downward acceleration of the block is a’ = g (sin θ – µ cos θ)
As the block Slides a distance d in each case so
d = \(\frac {1}{2}\) at2 = \(\frac {1}{2}\) a’t’2
\(\frac{a}{a^{\prime}}=\frac{t^{\prime 2}}{t^{2}}=\frac{(n t)^{2}}{t^{2}}\) = n2
or \(\frac {g sin θ}{g(sin θ – µ cos θ)}\) = n2
Solving, we get (Using θ = 45°)
µ = 1 – \(\frac{1}{n^{2}}\)

SamacheerKalvi.Guru

Question 35.
A spring balance is attached to the ceiling of a lift. When the lift is at rest spring balance reads 49 N of a body hang on it. If the lift moves:

  1. Downward
  2. upward, with an acceleration of 5 ms2
  3. with a constant velocity.

What will be the reading of the balance in each case?
Answer:
1.  R = m(g – a) = 49 N
so = m = \(\frac {49}{9.8}\) = 5 kg
R = 5 (9.8 – 5)
R = 24 N

2. R = m(g + a)
R = 5 (9.8 + 5)
R = 74 N

3.  as a = 0 so R = mg = 49 N

Question 36.
A bob of mass 0.1 kg hung from the ceiling of room by a string 2 m long is oscillating. At its mean position the speed of a bob is 1 ms-1. What is the trajectory of the ‘oscillating bob if the string is cut when the bob is –

  1. At the mean position
  2. At its extreme position.

Answer:

  1. Parabolic
  2. vertically downwards

Question 37.
A block placed on a rough horizontal surface is pulled by a horizontal force F. Let f be the force applied by the rough surface on the block. Plot a graph of f versus F.
Answer:

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Unto point A, f = F (50 Long as block is stationary) beyond A, when F increases, block starts moving f remains constant.

Question 38.
A mass of 2 kg is suspended with thread AB. Thread CD of the same type is attached to the other end of 2 kg mass.

  • Lower end of the lower thread is pulled gradually, hander and hander is the downward direction so as to apply force on AB Which of the thread will break & why?
  • If the lower thread is pulled with a jerk, what happens?

Answer:

  • Thread AB breaks down
  • CD will break.

SamacheerKalvi.Guru

Question 39.
A block of mass M is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is p and the acceleration due to gravity is g, calculate the minimum force required to be applied by the finger to held the block against the wall?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
For the block not to fall f = Mg
But f = µR = µF so
µF = Mg
F = \(\frac {Mg}{µ}\)

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions (3 Marks) & Numericals

Question 40.
A block of mass 500 g is at rest on a horizontal table. What steady force is required to give the block a velocity of 200 cm s-2 in 4 s?
Answer:
Use F – ma
a = \(\frac {v – u}{ t }\) = \(\frac {200- 0}{ 4 }\) = 50 cm/s2
F = 500 x 50 = 25,000 dyne.

Question 41.
A force of 98 N is just required to move a mass of 45 kg on a rough horizontal surface. Find the coefficient of friction and angle of friction?
Answer:
F = 48 N,R = 45 x 9.8 = 441 N
µ = \(\frac {F’}{ R}\) = 0.22
Angle of friction θ = tan-1 0.22 = 12°24′

Question 42.
Calculate the force required to move a train of 2000 quintal up on an incline plane of 1 in 50 with an acceleration of 2 ms-2. The force of friction per quintal is 0.5 N.
Answer:
Force of friction = 0.5 N per quintal
f = 0.5 x 2000 = 1000 N
m = 2000 quintals = 2000 x 100 kg
sin θ = \(\frac {1}{50}\), a – 2 m/s2
In moving up an inclined plane, force required against gravity
mg sin θ = 39200 N
And force required to produce acceleration = ma
= 2000 x 100 x 2 = 40,0000 N
Total force required = 1000 + 39,200 + 40,0000 = 440200 N.

SamacheerKalvi.Guru

Question 43.
A force of 100 N gives a mass m1, an acceleration of 10 ms-2 and of 20 ms-2 to a mass m2.
What acceleration must be given to it if both the masses are tied together?
Answer:
Suppose, a = acceleration produced if m1 and m2 are tied together,
F = 100 N
Let a1 and a2 be the acceleration produced in m1 and m2 respectively.
∴ a1 and a2 = 20ms-2 (given)
Again m1 = \(\frac{\mathrm{F}}{a_{1}}\) and m2 = \(\frac{\mathrm{F}}{a_{2}}\)
⇒ m1 = \(\frac {100}{10}\) = 10kg
and m2 = \(\frac {100}{20}\) = 5kg
∴ m1 + m2 = 10 + 5 = 15
so, a = \(\frac{\mathbf{F}}{m_{1}+m_{2}}\) = \(\frac {100}{15}\) = \(\frac {20}{3}\) = 6.67 ms2

Question 44.
The pulley arrangement of figure are identical. The mass of the rope is negligible. In (a) mass m is lifted up by attaching a mass (2m) to the other end of the rope. In (b), m is lifted up by pulling the other end of the rope with a constant downward force F = 2 mg. In which case, the acceleration of m is more?
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Answer:
Case (a):
a = \(\frac {2m – m}{2m + m}\) g = a = \(\frac {g}{3}\)
Case (b):
FBD of mass m
ma’ = T – mg
ma’ = 2 mg – mg
⇒ ma’ = mg
a’ = g
So in case (b) acceleration of m is more.

Question 45.
Figure shows the position-time graph of a particle of mass 4 kg. What is the
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
(a) Force on the particle for t < 0, t > 4s, 0 < t < 4s?
(b) Impulse at t = 0 and t = 4s?
(Consider one dimensional motion only)
Answer:
(a) For t < 0. No force as Particles is at rest. For t > 4s, No force again particle comes at rest.
For 0 < t < 4s, as slope of OA is constant so velocity constant i.e., a = 0, so force must be zero.

(b) Impulse at t = 0
Impulse = change in momentum
I = m(v – w) = 4(0 – 0.75) = 3 kg ms-1
Impulse at t = 4s
1 = m(v – u) = 4 (0 – 0.75) = -3 kg ms-1

SamacheerKalvi.Guru

Question 46.
What is the acceleration of the block and trolley system as the figure, if the coefficient of kinetic friction between the trolley and the surface is 0.04? Also Calculate friction in the string: Take g = 10 m/s2, mass of the string is negligible.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Answer:
Free body diagram of the block
30 – T = 3a
Free body diagram of the trolley
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
T – fk = 20 a ………….(2)
where fk = µk= 0.04 x 20 x 10 = 8 N
Solving (i) & (ii), a = 0.96 m/s2 and T = 27.2 N

Question 47.
Three blocks of masses ml = 10 kg, m2 = 20 kg are connected by strings on smooth horizontal surface and pulled by a force of 60 N. Find the acceleration of the system and frictions in the string.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Solution:
All the blocks more with common acceleration a under the force F = 60 N.
F = (m1 + m2 + m3)a
a = \(\frac{\mathrm{F}}{\left(m_{1}+m_{2}+m_{3}\right)}\) = 1 m/s2
to determine, T1 →Free body diagram of m1.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
T1 = m1a = 10 x 1 = 10 N
to determine, T2 →Free body diagram of m3
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
F – T2 = m3a
Solving, we get T2 = 30 N

SamacheerKalvi.Guru

Question 48.
The rear side of a truck is open and a box of 40 kg mass is placed 5m away from the open end. The coefficient of friction between the box and the surface below it is 0.15 on a straight road, the truck starts from rest and accelerates with 2 m/s2. At what distance from the starting point does the box fall off the truck ? (ignore the size of the box)
Answer:
Force on the box due to accelerated motion of the truck
F = ma = 40 x 2 = 80 N (in forward direction)
Reaction on the box, F’ = F = 80 N (in backward direction)
Force of limiting friction, f = µR = 0 .15 x 40 x 10 = 60 N
Net force on the box in backward direction is P = F’ f = 80 – 60 = 20 N
Backward acceleration in the box = a= \(\frac {p}{m}\) = \(\frac {20}{40}\) = 0.5 ms-2
t = time taken by the box to travel s = 5 m and falls off the truck, then from
s = ut + \(\frac {1}{2}\) at2
5 = 0 x t + \(\frac {1}{2}\) x 0.5 x t2
t = 4.47
If the truck travels a distance x during this time
then x = 0 x 4.34 +\(\frac {1}{2}\) x 2 x (4.471)2
x = 19.98 m

Question 49.
A block slides down as incline of 30° with the horizontal. Starting from rest, it covers 8 m in the first 2 seconds. Find the coefficient of static friction.
Use s = ut + \(\frac {1}{2}\) at2
a = \(\frac{2 s}{t^{2}}\) at2 as u = 0
µ = \(\frac{g sin θ – a}{g Cos θ }\)
Putting the value and solving, µ = 0.11

Question 50.
A helicopter of mass 2000 kg rises with a vertical acceleration of 15 m/s2 . The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the:
(a) Force on the floor of the helicopter by the crew and passenger.
(b) Action of the rotor of the helicopter on the surrounding air
(c) Force on the helicopter due to the surrounding air (g = 10 m/s2 )
Answer:
(a) Force on the floor of the helicopter by the crew and passengers
= apparent weight of crew and passengers
= 500(10+ 15)
=12500 N

(b) Action of rotor of helicopter on surrounding air is Obviously vertically downwards, because helicopter rises on account of reaction of this force. Thus force of action
= (2000 + 500) (10 + 15)
= 2500 x 25
= 62,500 N

(c) Force on the helicopter due to surrounding air is obviously a reaction. As action and reaction are equal and opposite, therefore
Force of reaction F’ = 62,500 vertically upwards.

SamacheerKalvi.Guru

Question 51.
A rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is (µ). Let the mass of the box be m.

  1.  At what angle of inclination θ of the plane to the horizontal will the box just start to slide down the plane ?
  2. What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to a > θ.
  3. What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed ?
  4. What is the force needed to be applied upwards along the plane to pk kg f make the box move up the plane with acceleration a ?

Answer:
1. When the box just starts sliding
µ = tanθ
or 0 = tan-1 µ

2. Force acting on the box down the plane
= mg (sin a – µ cos a)

3. Force needed mg (sin a + µ cos a)

4. Force needed = mg (sin a + µ cos a) + ma.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Question 52.
Two masses of 5 kg and 3 kg are suspended with help of mass less in extensible string as shown. Calculate T1 and T2 when system is going upwards with acceleration m/s2. (Use g 9.8 m/s2)
Answer:
According Newton’s second law of motion
(1) T1 – (m1 + m2)g = (m1 + m2)a
T1 = (m1 + m2)(a + g) = (5 + 3) (2 + 9.8)
T1 = 94.4 N

(2) T2 – m2g = m2a
T2 = m2 (a + g)
T2 = 3(2 + 9.8)
T2 = 35.4 N

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Question 53.
There are few forces acting at a Point P produced by strings as shown, which is at rest. Find the forces F1 & F1
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Answer:
Using Resolution of forces IN and 2N and then applying laws of vector addition. Calculate for F1 & F1.
F1 = \(\frac{1}{\sqrt{2}}\) N, F2 = \(\frac{3}{\sqrt{2}}\)N

Question 54.
A hunter has a machine gun that can fire 50g bullets with a velocity of 150 ms A 60 kg tiger springs at him with a velocity of 10 ms-1. How many bullets must the hunter fire into the target so as to stop him in his track?
Answer:
Given m = mass of bullet = 50 gm = 0.50 kg
M = mass of tiger = 60 kg
v = Velocity of bullet – 150 m/s
V = Velocity of tiger = – 10 m/s
(v It is coming from opposite direction n = no. of bullets fired per second at the tiger so as to stop it.)
Pi = 0, before firing ……..(i)
Pf = n (mv) + MV …………(ii)
∴ From the law of conservation of momentum,
Pi = Pf
⇒ 0 = n (mv) + MV
n = \(\frac{MV}{mv}\) = \(\frac{-60 \times(-10)}{0.05 \times 150}\) = 80

SamacheerKalvi.Guru

Question 55.
Two blocks of mass 2 kg and 5 kg are connected by an ideal string passing over a pulley. The block of mass 2 kg is free to slide on a surface inclined at an angle of 30° with the horizontal whereas 5 kg block hangs freely. Find the acceleration of the system and the tension in the string.
Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion
Let a be the acceleration of the system and T be the Tension in the string. Equations of motions for 5 kg and 2 kg blocks are
5g – T = 5a
T – 2g sin θ – f = 2a
where f = force of limiting friction
= µR = µ mg cos θ = 0.3 x 2 g x cos 30°
Solving (1) & (2)
a = 4.87 m/s2

Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion Read More »

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 1.
Construct a quadratic equation with roots 7 and -3.
Solution:
Let the given roots be α = 7 and β = -3
Sum of the roots α + β = 7 + (-3)
α + β = 7 – 3 = 4
Product of the roots αβ = (7)(-3)
αβ = -21
The required quadratic equation is
x2 – (sum of two roots) x + Product of the roots = 0
x2 – 4x – 21 = 0

Question 2.
A quadratic polynomial has one of its zeros 1 + \(\sqrt{5}\) and it satisfies p(1) = 2. Find the quadratic polynomial.
Solution:
Given α = 1 + \(\sqrt{5}\) So, β = 1 – \(\sqrt{5}\)
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 1
The quadratic polynomial is
p(x) = x2 – (α + β)x + αβ
p(x) = k (x2 – 2x – 4)
p( 1) = k(1 – 2 – 4) = -5 k
Given p (1) = 2
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 2

The Polynomial Roots Calculator will find the roots of any polynomial with just one click.

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 3.
If α and β are the roots of the quadratic equation x2 + \(\sqrt{2}\)x + 3 = 0, form a quadratic polynomial with zeroes 1/α, 1/β.
Solution:
α and β are the roots of the equation x2 + \(\sqrt{2}\)x + 3 = 0
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 3
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 4

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 4.
If one root of k(x – 1)2 = 5x – 7 is double the other root, show that k = 2 or – 25.
Solution:
k(x – 1)2 = 5x – 7
(i.e.,) k(x2 – 2x + 1) – 5x + 7 = 0
x2 (k) + x(-2k – 5) + k + 1 = 0
kx2 – x(2k + 5) + (k + 7) = 0
Here it is given that one root is double the other.
So let the roots to α and 2α
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 5
2(4k2 + 25 + 20k) = 9k (k + 7)
2(4k2 + 25 + 20k) = 9k2 + 63k
8k2 + 50 + 40k – 9k2 – 63k = 0
-k2 – 23k + 50 = 0
k2 + 23k – 5o = 0
(k + 25)(k – 2) = 0
k = -25 or 2

Question 5.
If the difference of the roots of the equation 2x2 – (a + 1)x + a – 1 = 0 is equal to their product then prove that a = 2.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 6

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 6.
Find the condition that one of the roots of ax2 + bx + c may be
(i) negative of the other
(ii) thrice the other
(iii) reciprocal of the other.
Solution:
(i) Let the roots be α and -β
Sum of the roots = – b/a = 0 ⇒ b = 0

(ii) Let the roots be α, 3α
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 7

Question 7.
If the equations x2 – ax + b = 0 and x2 – ex + f = 0 have one root in common and if the second equation has equal roots that ae = 2(b + f).
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 8

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 8.
Discuss the nature of roots of
(i) -x2 + 3x + 1 = 0
(ii) 4x2 – x – 2 = 0
(iii) 9x2 + 5x = 0
Solution:
(i) -x2 + 3x + 1 = 0
x2 – 3x – 1 = 0 ———- (1)
Compare this equation with the equation
ax2 + bx + c = 0 ——– (2)
we have a = 1, b = -3, c = -1
Discriminant = b2 – 4ac
b2 – 4ac = (-3)2 – 4 × 1 × – 1
= 9 + 4 =13
b2 – 4ac = 13 > 0
∴ The two roots are real and distinct.

(ii) 4x2 – x – 2 = 0
4x2 – x – 2 = 0 ——(3)
Compare this equation with the equation
ax2 + bx + c = 0 (4)
we have a = 4 , b = – 1, c = – 2
Discriminant = b2 – 4ac
b2 – 4ac = (-1)2 – 4 (4) (-2)
= 1 + 32
= 33
b2 – 4ac = 33 >0
∴ The two roots are real and distinct.

(iii) 9x2 + 5x = 0
9x2 + 5x = 0 ——- (5)
Compare this equation with the equation
ax2 + bx + c = 0 ——– (6)
we have a = 9, b = 5 , c = 0
Discriminant = b2 – 4ac
b2 – 4ac = 52 – 4 × 9 × 0
b2 – 4ac = 25 > 0
∴ The two roots are real and distinct.

Question 9.
Without sketching the graphs find whether the graphs of the following functions will intersect the x- axis and if so in how many points.
(i) y = x2 + x + 2
(ii) y = x2 – 3x – 1
(iii) y = x2 + 6x + 9
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 20
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 21

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Completing the Square Calculator is a free online tool that displays the variable value for the quadratic equation using completing the square method.

Question 10.
Write f(x) = x2 + 5x + 4 in completed square form.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 22

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 Additional Questions

Question 1.
Find the values of k so that the equation x2 = 2x (1 + 3k) + 7(3 + 2k) = 0 has real and equal roots.
Solution:
The equation is x2 – x(2) (1 + 3k) + 7 (3 + 2k) = 0
The roots are real and equal
⇒ ∆ = 0 (i.e.,) b2 – 4ac = 0
Here a = 1, b = -2 (1 + 3k), c = 7(3 + 2k)
So b2 – 4ac = 0
⇒ [-2 (1 + 3k)]2 – 4(1) (7) (3 + 2k) = 0
(i.e.,) 4 (1 + 3k)2 – 28 (3 + 2k) = 0
(÷ by 4) (1 + 3k)2 – 7(3 + 2k) = 0
1 + 9k2 + 6k – 21 – 14k = 0
9k2 – 8k – 20 = 0
(k – 2)(9k + 10) = 0
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 23
To solve the quadratic inequalities ax2 + bx + c < 0 (or) ax2 + bx + c > 0

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 2.
If the sum and product of the roots of the quadratic equation ax2 – 5x + c = 0 are both equal to 10 then find the values of a and c.
Solution:
The given equation is ax2 – 5x + c = 0
Let the roots be α and β Given α + β = 10 and αβ = 10
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 24

Question 3.
If α and β are the roots of the equation 3x2 – 4x + 1 = 0, form the equation whose roots are \(\frac{\alpha^{2}}{\beta}\) and \(\frac{\beta^{2}}{\alpha}\)
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 25
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 26

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 4.
If one root of the equation 3x2 + kx – 81 = 0 is the square of the other then find k.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 27

Question 5.
If one root of the equation 2x2 – ax + 64 = 0 is twice that of the other then find the value of a.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 28

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Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

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Resolve the following rational expressions into partial fractions

Question 1.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 1
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 2

Question 2.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 3
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 4

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Question 3.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 5
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 45
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 55

Question 4.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 6
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 7

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Question 5.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 8
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 9
Equating nuemarator on bothsides we get
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 98

Question 6.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 146
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 11
Equating numerator on both sides
(x – 2)2 = A(x2 + 1) + (Bx + c)(x)
Put x = 0
1 = A
Equating co-eff of x2
1 = A + B
(i.e.,) 1 + B = 1 ⇒ B = 0
put x = 1
A(2) + B + C = 0 (i.e.,) 2A + B + C = 0
2 + 0 + C = 0 ⇒ C = -2
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 12

Question 7.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 13
Solution:
Since numerator and denominator are of same degree
we have divide the numerator by the denominator
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 14
Substituting the value in ….(1)
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 145

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Question 8.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 15
Solution:
Numerator is of greater degree than the denominator
So dividing Numerator by the denominator
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 16
⇒ 21x + 31 = A(x + 3) + B(x + 2)
Put x = -3
-63 + 31 = B(-1)
B = 32
Put x = -2
-42 + 31 = A(1) + B(0)
A = -11
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 17

Question 9.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 18
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 19
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 20

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Question 10.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 21
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 22
Equating Numerator on both sides we get
6x2 – x + 1 = A(x2 + 1) + (Bx + c)(x + 1)
6 + 1 + 1 = A(2) + 0 ⇒ 2A = 8 ⇒ A = 4
Equating co-eff of x2
6 = A + B
(i.e.,) 4 + B = 6 ⇒ B = 6 – 4 = 2
put x = 0
1 = A+ C
4 + C = 1 ⇒ C = 1 – 4 = -3
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 23

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Question 11.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 24
Solution:
Since Numerator and are of same degree divide Numerator by the denominator
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 25
equating Numerator on both sides we get
x – 5 = A(x + 3) + B(x – 1)
Put x = -3
-3 -5 = A(0) + B(-4)
-4B = -8 ⇒ B = 2
Put x = 1
1 – 5 = A(4) + B(0)
4A = -4 ⇒ A = -1
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 26

Question 12.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 27
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 28

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 Additional Questions

Question 1.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 29
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 30

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Question 2.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 31
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 32

Question 3.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 33
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 34
Equating nemerator on b/s
9 = A(x+2)2 + B(x – 1)(x + 2) + C(x – 1)
Put x = -2
9 = A(0) + B(0) + C(-3)
-3C = 9 ⇒ C = -3
Put x = 1
9 = A (1 + 2)2 + B (0) + C(0)
9A = 9
A = 1
Put x = 0
9 = 4A – 2B – C
9 = 4(1) – 2B + 3
9 – 7 = -2B
2 = -2B
B = -1
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 35

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Question 4.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 36
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 133

Question 5.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 38
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 39
0 = 0 + B(1 + 2)
3B = 0 ⇒ B = 0
Put x = -2
(-2)3 – 1 = A(-2 – 1) + B(0)
-8 – 1 = -3A
-9 = -3A
A = 9/3 ⇒ A = 3
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 40

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Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Intext Questions

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.
Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Intext Questions

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Intext Questions

Recap (Textbook Page No. 108)

Question 1.
Count the objects in the following figure and complete the table that follows :
Samacheer Kalvi 6th Maths Term 1 Chapter 5 Statistics Intext Questions Q1
From Fig 5.1 and the table, answer the following question.
(i) The total number of objects in the above picture is ______
(ii) The difference between the number of squares and the number of bats is ______
(iii) The ratio of the number of balls to the number of bats is _______
(iv) What are the objects equal in number?
(v) How many more balls are there than the number of bats?
Solution:
(i) 24
(ii) 0
(iii) \(\frac{8}{6}=\frac{4}{3}\)
(iv) Bat and Square
(v) 2 balls more
Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Intext Questions

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