Class 11

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 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

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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.

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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

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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) \(u \sqrt{\frac{2 h}{g}}\)

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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°.

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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.

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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

SamacheerKalvi.Guru

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 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

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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.

SamacheerKalvi.Guru

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).

SamacheerKalvi.Guru

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 \(\frac{d \vec{p}}{d t}\) = \(\overline{\mathrm{F}}\)

SamacheerKalvi.Guru

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.

SamacheerKalvi.Guru

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).

SamacheerKalvi.Guru

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.

SamacheerKalvi.Guru

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

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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

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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

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 Read More »

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

Get the free “Partial Fraction Calculator” widget for your website, blog, WordPress, Blogger, or iGoogle.

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 11th Physics Solutions Chapter 5 Motion of System of Particles and Rigid Bodies

Students can Download Physics Chapter 5 Motion of System of Particles and Rigid Bodies 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 5 Motion of System of Particles and Rigid Bodies

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Textual Questions Solved

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Multiple Choice Questions

Question 1.
The center of mass of a system of particles does not depend upon,
(a) position of particles
(b) relative distance between particles
(c) masses of particles
(d) force acting on particle
Answer:
(d) force acting on particle

Question 2.
A couple produces, [AIPMT 1997, AIEEE 2004]
(a) pure rotation
(b) pure translation
(c) rotation and translation
(d) no motion [AIPMT 1997]
Answer:
(a) pure rotation

Question 3.
A particle is moving with a constant velocity along a line parallel to positive X – axis. The magnitude of its angular momentum with respect to the origin is –
(a) zero
(b) increasing with x
(c) decreasing with x
(d) remaining constant [IIT 2002]
Answer:
(d) remaining constant

Question 4.
A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force 30 N?
(a) 0.25 rad s-2
(b) 25 rad s-2
(c) 5 m s-2
(d) 25 m s-2
[NEET 2017]
Answer:
(b) 25 rad s-2

Question 5.
A closed cylindrical container is partially filled with water. As the container rotates in a horizontal plane about a perpendicular bisector, its moment of inertia,
(a) increases
(b) decreases
(c) remains constant
(d) depends on direction of rotation. [IIT 1998]
Answer:
(a) increases

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Use this kinetic energy calculator to help you find out the energy of an object in motion. This KE calculator makes use of the kinetic energy formula

Question 6.
A rigid body rotates with an angular momentum L. If its kinetic energy is halved, the angular momentum becomes,
(a) L
(b) L / 2
(c) 2 L
(d) L / 2 [AFMC 1998, AIPMT 2015]
Answer:
(d) L / 2

Question 7.
A particle undergoes uniform circular motion. The angular momentum of the particle remain conserved about –
(a) the center point of the circle.
(b) the point on the circumference of the circle
(c) any point inside the circle.
(d) any point outside the circle. [IIT 2003]
Answer:
(a) the center point of the circle.

Question 8.
When a mass is rotating in a plane about a fixed point, its angular momentum is directed along –
(a) a line perpendicular to the plane of rotation
(b) the line making an angle of 45° to the plane of rotation
(c) the radius
(d) tangent to the path [AIPMT 2012]
Answer:
(a) a line perpendicular to the plane of rotation

Question 9.
Two discs of same moment of inertia rotating about their regular axis passing through center and perpendicular to the plane of disc with angular velocities ω1 and ω1. They are brought in to contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is-
(a) \(\frac {1}{4}\) I(ω1 – ω22
(b) I(ω1 – ω22
(c) \(\frac {1}{8}\) I(ω1 – ω22
(d) \(\frac {1}{2}\) I(ω1 – ω22
Answer:
(a) \(\frac {1}{4}\) I(ω1 – ω22

Question 10.
A disc of moment of inertia Ia is rotating in a horizontal plane about its symmetry axis with a constant angular speed to. Another disc initially at rest of moment of inertia Ib is dropped coaxially on to the rotating disc. Then, both the discs rotate with same constant angular speed. The loss of kinetic energy due to friction in this process is-
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Answer:
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

SamacheerKalvi.Guru

Question 11.
The ratio of the acceleration for a solid sphere (mass m and radius R) rolling down an incline of angle 0 without slipping and slipping down the incline without rolling is –
(a) 5 : 7
(b) 2 : 3
(c) 2 : 5
(d) 7 : 5
[AIPMT 2014]
Answer:
(a) 5 : 7

Question 12.
From a disc of radius R a mass M, a circular hole of diameter R, whose rim passes through the center is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis passing through it?
(a) 15MR2/32
(b) 13MR2/32
(c) 11MR2/32
(d) 9MR2/32 [NEET 2016]
Answer:
(b) 13MR2/32

Question 13.
The speed of a solid sphere after rolling down from rest without sliding on an inclined plane of vertical height h is,
(a) \(\sqrt{\frac{4}{3} g h}\)
(b) \(\sqrt{\frac{10}{7} g h}\)
(c) \(\sqrt{2gh}\)
(d) \(\sqrt{\frac{1}{2} g h}\)
Answer:
(a) \(\sqrt{\frac{4}{3} g h}\)

Question 14.
The speed of the center of a wheel rolling on a horizontal surface is vQ. A point on the rim in level with the center will be moving at a speed of speed of,
(a) zero
(b) v0
(c) \(\sqrt{2}\)v0
(d) 2 v0
[PMT 1992, PMT 2003, IIT 2004]
Answer:
(c) \(\sqrt{2}\)v0

SamacheerKalvi.Guru

Question 15.
A round object of mass m and radius r rolls down without slipping along an inclined plane. The fractional force,
(a) dissipates kinetic energy as heat.
(b) decreases the rotational motion.
(c) decreases the rotational and transnational motion ,
(d) converts transnational energy into rotational energy [PMT 2005]
Answer:
(d) converts transnational energy into rotational energy

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Short Answer Questions

Question 1.
Define center of mass.
Answer:
The center of mass of a body is defined as a point where the entire mass of the body appears to be concentrated.

Question 2.
Find out the center of mass for the given geometrical structures.
(a) Equilateral triangle
(b) Cylinder
(c) Square
Answer:
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
(a) For equilateral triangle, center of mass lies at its centro-id.
(b) For cylinder, center of mass lies at its geometrical center.
(c) For square, center of mass lies at the point where the diagonals meet.

Question 3.
Define torque and mention its unit.
Answer:
Torque is defined as the moment of the external applied force about a point or axis of rotation. The expression for torque is,
\(\vec{\tau}\) = \(\vec{r}\) x \(\vec{F}\)

Question 4.
What are the conditions in which force cannot produce torque?
Answer:
The forces intersect (or) passing through the axis of rotation cannot produce torque as the perpendicular distance between the forces is 0 i.e. r = 0.
∴ \(\vec{\tau}\) = \(\vec{r}\) x \(\vec{F}\) = 0

Question 5.
Give any two examples of torque in day – to – day life.
Answer:

  • The opening and closing of a door about the hinges.
  • Turning of a nut using a wrench.

Question 6.
What is the relation between torque and angular momentum?
Answer:
We have the expression for magnitude of angular momentum of a rigid body as, L = I ω. The expression for magnitude of torque on a rigid body is, τ = I α.
We can further write the expression for torque as,
τ = I\(\frac {dω}{dt}\) (∴ α = \(\frac {dω}{dt}\))
Where, ω is angular velocity and α is angular acceleration. We can also write equation,
τ = \(\frac {d(Iω)}{dt}\)
τ = \(\frac {dL}{dt}\)

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Question 7.
What is equilibrium?
Answer:
A rigid body is said to be in mechanical equilibrium where both its linear momentum and angular momentum remain constant.

Question 8.
How do you distinguish between stable and unstable equilibrium?
Answer:
Stable Kquilibrium:

  • The body tries to come back to equilibrium if slightly disturbed and released.
  • The center of mass of the body shifts slightly higher if disturbed from equilibrium.
  • Potential energy of the body is minimum and it increases if disturbed.

Unstable Equilibrium:

  • The body cannot come back to equilibrium if slightly disturbed and released.
  • The center of mass of the body shifts slightly lower if disturbed from equilibrium.
  • Potential energy of the body is not minimum and it decreases if disturbed.

Question 9.
Define couple.
Answer:
A pair of forces which are equal in magnitude but opposite in direction and separated by a perpendicular distance so that their lines of action do not coincide that causes a turning effect is called a couple.

Question 10.
State principle of moments.
Answer:
Principle of moment states that when an object is in equilibrium the sum of the anticlockwise moments about a point is equal to the sum of the clockwise moments.

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Question 11.
Define center of gravity.
Answer:
The center of gravity of a body is the point at which the entire weight of the body acts, irrespective of the position and orientation of the body.

Question 12.
Mention any two physical significance of moment of inertia
Answer:
Moment of inertia for point mass,
I = \(m_{i} r_{i}^{2}\)
Moment of inertia for bulk object,
I = ∑\(m_{i} r_{i}^{2}\)

Question 13.
What is radius of gyration?
Answer:
The radius of gyration of an object is the perpendicular distance from the axis of rotation to an equivalent point mass, which would have the same mass as well as the same moment of inertia of the object.

Question 14.
State conservation of angular momentum.
Answer:
The law of conservation of angular momentum states that when no external torque acts on the body the net angular momentum of a rotating rigid body remains constant.

Question 15.
What are the rotational equivalents for the physical quantities, (i) mass and (ii) force?
Answer:
The rotational equivalents for (i) mass and (ii) force are moment of inertia and torque respectively.

SamacheerKalvi.Guru

Question 16.
What is the condition for pure rolling?
Answer:
In pure rolling, there is no relative motion of the point of contact with the surface when the rolling object speeds up or shows down. It must accelerate or decelerate respectively.

Question 17.
What is the difference between sliding and slipping?
Sliding:

  • Velocity of center of mass is greater than Rω i.e. VCM > Rω.
  • Velocity of transnational motion is greater than velocity of rotational motion.
  • Resultant velocity acts in the forward direction.

Slipping:

  • Velocity of center of mass is lesser than Rω. i.e. VCM < Rω
  • Velocity of translation motion is lesser than velocity of rotational motion.
  • Resultant velocity acts in the backward direction.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Long Answer Questions

Question 1.
Explain the types of equilibrium with suitable examples.
Answer:

  • Transnational motion – A book resting on a table.
  • Rotational equilibrium – A body moves in a circular path with constant velocity.
  • Static equilibrium – A wall – hanging, hanging on the wall.
  • Dynamic equilibrium – A ball decends down in a fluid with its terminal velocity.
  • Stable equilibrium – A table on the floor
  • Unstable equilibrium – A pencil standing on its tip.
  • Neutral equilibrium – A dice rolling on a game board.

Question 2.
Explain the method to find the center of gravity of a irregularly shaped lamina.
Answer:
There is also another way to determine the center of gravity of an irregular lamina. If we suspend the lamina from different points like P, Q, R as shown in figure, the vertical lines I PP’, QQ’, RR’ all pass through the center of gravity. Here, reaction force acting at the point of suspension and the gravitational force acting at the center of gravity cancel each other and the torques caused by them also cancel each other.
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Determination of center of gravity of plane lamina by suspending

Question 3.
Explain why a cyclist bends while negotiating a curve road? Arrive at the expression for angle of bending for a given velocity.
Answer:
Let us consider a cyclist negotiating a circular level road (not banked) of radius r with a speed v. The cycle and the cyclist are considered as one system with mass m. The center gravity of the system is C and it goes in a circle of radius r with center at O. Let us choose the line OC as X – axis and the vertical line through O as Z – axis as shown in Figure.

Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

The system as a frame is rotating about Z – axis. The system is at rest in this rotating frame. To solve problems in rotating frame of reference, we have to apply a centrifugal force (pseudo force) on the system which will be \(\frac{m v^{2}}{r}\) This force will act through the center of gravity. The forces acting on the system are,

  • gravitational force (mg)
  • normal force (N)
  • frictional force (f)
  • centrifugal force (\(\frac{m v^{2}}{r}\)).

As the system is in equilibrium in the rotational frame of reference, the net external force and net external torque must be zero. Let us consider all torques about the point A in Figure.
For rotational equilibrium,
τnet = 0
The torque due to the gravitational force about point A is (mg AB) which causes a clockwise turn that is taken as negative. The torque due to the centripetal force is I BC which causes an (\(\frac{m v^{2}}{r}\) BC) Which causes an anticlockwise turn that is taken as positive.
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
While negotiating a circular level road of radius r at velocity v, a cyclist has to bend by an angle 0 from vertical given by the above expression to stay in equilibrium (i.e. to avoid a fall).

SamacheerKalvi.Guru

Question 4.
Derive the expression for moment of inertia of a rod about its center and perpendicular to the rod.
Answer:
Let us consider a uniform rod of mass (M) and length (l) as shown in figure. Let us find an expression for moment of inertia of this rod about an axis that passes through the center of mass and perpendicular to the rod. First an origin is to be fixed for the coordinate system so that it coincides with the center of mass, which is also the geometric center of the rod. The rod is now along the x axis. We take an infinitesimally small mass (dm) at a distance (x) from the origin. The moment of inertia (dI) of this mass (dm) about the axis is, dI = (dm) x2
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

As the mass is uniformly distributed, the mass per unit length (λ) of the rod is, λ = \(\frac {M}{l}\)
The (dm) mass of the infinitesimally small length as, dm = λ dx = \(\frac {M}{l}\) dx
The moment of inertia (I) of the entire rod can be found by integrating dl,
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
As the mass is distributed on either side of the origin, the limits for integration are taken from to – l/2 to l/2.
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 5.
Derive the expression for moment of inertia of a uniform ring about an axis passing through the center and perpendicular to the plane.
Answer:
Let us consider a uniform ring of mass M and radius R. To find the moment of inertia of the ring about an axis passing through its center and perpendicular to the plane, let us take an infinitesimally small mass (dm) of length (dx) of the ring. This (dm) is located at a distance R, which is the radius of the ring from the axis as shown in figure.
The moment of inertia (dl) of this small mass (dm) is,
dI = (dm)R2
The length of the ring is its circumference (2πR). As the mass is uniformly distributed, the mass per unit length (λ) is,
λ = \(\frac {mass}{lengh}\) = \(\frac {M}{2πR}\)
The mass (dm) of the infinitesimally small length is,
dm = λ dx = \(\frac {M}{2πR}\) dx
Now, the moment of inertia (I) of the entire ring is,
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies To cover the entire length of the ring, the limits of integration are taken from 0 to 2πR.
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 6.
Derive the expression for moment of inertia of a uniform disc about an axis passing through the center and perpendicular to the plane.
Answer:
Consider a disc of mass M and radius R. This disc is made up of many infinitesimally small rings as shown in figure. Consider one such ring of mass (dm) and thickness (dr) and radius (r). The moment of inertia (dl) of this small ring is,
dI = (dm)R2
As the mass is uniformly distributed, the mass per unit area (σ) is σ = \(\frac {mass}{area}\) = \(\frac{M}{\pi R^{2}}\)
The mass of the infinitesimally small ring is,
dm = σ 2πr dr = \(\frac{\mathrm{M}}{\pi \mathrm{R}^{2}}\) 2πr dr
where, the term (2πr dr) is the area of this elemental ring (2πr is the length and dr is the thickness), dm = \(\frac{2 \mathrm{M}}{\mathrm{R}^{2}}\) r dr
dI = \(\frac{2 \mathrm{M}}{\mathrm{R}^{2}}\) r3 dr
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies The moment of inertia (I) of the entire disc is,
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 7.
Discuss conservation of angular momentum with example.
Answer:
When no external torque acts on the body, the net angular momentum of a rotating rigid body remains constant. This is known as law of conservation of angular momentum.
τ = \(\frac {dL}{dt}\)
If τ = 0 then, L = constant.
As the angular momentum is L = Iω, the conservation of angular momentum could further be written for initial and final situations as,
Iiωi = Iiωi (or) Iω = constant
The above equations say that if I increases ω will decrease and vice – versa to keep the angular momentum constant.
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
There are several situations where the principle of conservation of angular momentum is applicable. One striking example is an ice dancer as shown in Figure A. The dancer spins slowly when the hands are stretched out and spins faster when the hands are brought close to the body.

Stretching of hands away from body increases moment of inertia, thus the angular velocity decreases resulting in slower spin. When the hands are brought close to the body, the moment of inertia decreases, and thus the angular velocity increases resulting in faster spin. A diver while in air as in Figure B curls the body close to decrease the moment of inertia, which in turn helps to increase the number of somersaults in air.
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 8.
State and prove parallel axis theorem.
Answer:
Parallel axis theorem:
Parallel axis theorem states that the moment of inertia of a body about any axis is equal to the sum of its moment of inertia about a parallel axis through its center of mass and the product of the mass of the body and the square of the perpendicular distance between the two axes.

If IC is the moment of inertia of the body of mass M about an axis passing through the center of mass, then the moment of inertia I about a parallel axis at a distance d from it is – given by the relation,
I = IC + M d2
Let us consider a rigid body as shown in figure. Its moment of inertia about an axis AB passing through the center of mass is IC. DE is another axis parallel to AB at a perpendicular distance d from AB. The moment of inertia of the body about DE is I. We attempt to get an expression for I in terms of IC. For this, let us consider a point mass m on the body at position x from its center of mass.

Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
The moment of inertia of the point mass about the axis DE is, m (x + d)2. The moment of inertia I of the whole body about DE is the summation of the above expression.
I = ∑ m (x + d)2
This equation could further be written as,
I = ∑ m(x2 + d2 + 2xd)
1= ∑ (mx2 + md2 + 2 dmx)
l = ∑ mx2 + md2 + 2d ∑ mx
Here, ∑ mx2 is the moment of inertia of the body about the center of mass. Hence,IC = ∑ mx2
The term, ∑ mx = 0 because, x can take positive and negative values with respect to the axis AB. The summation (∑ mx) will be zero.
Thus, I = IC + ∑ m d2 = IC + (∑m) d2
Here, ∑ m is the entire mass M of the object (∑ m = M).
I = IC + Md2

SamacheerKalvi.Guru

Question 9.
State and prove perpendicular axis theorem.
Answer:
Perpendicular axis theorem:
This perpendicular axis theorem holds good only for plane laminar objects. The theorem states that the moment of inertia of a plane laminar body about an axis perpendicular to its plane is equal to the sum of moments of inertia about two perpendicular axes lying in the plane of the body such that all the three axes are mutually perpendicular and have a common point.

Let the X and Y – axes lie in the plane and Z – axis perpendicular to the plane of the laminar object. If the moments of inertia of the body about X and Y-axes are IX and IY respectively – and IZ is the moment of inertia about Z-axis, then the perpendicular axis theorem could be expressed as,
IZ = IX + IY

To prove this theorem, let us consider a plane laminar object of negligible thickness on which lies the origin (O). The X and Y – axes lie on the plane and Z – axis is perpendicular to it as shown in figure. The lamina is considered to be made up of a large number of particles of mass m. Let us choose one such particle at a point P which has coordinates (x, y) at a distance r from O.
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
The moment of inertia of the particle about Z – axis is, mr2.
The summation of the above expression gives the moment of inertia of the entire lamina about Z – axis as, IZ = ∑ mr2
Here, r2 = x2 + y2
Then, IZ = ∑ m (x2 + y2)
IZ = ∑ m x2 + ∑ m y2
In the above expression, the term ∑ m x2 is the moment of inertia of the body about the Y-axis and similarly the term ∑ m y2is the moment of inertia about X- axis. Thus,
IX = ∑ m y2 and IY = ∑ m x2
Substituting in the equation for Iz gives,
IZ = IX + IY
Thus, the perpendicular axis theorem is proved.

Question 10.
Discuss rolling on inclined plane and arrive at the expression for the acceleration.
Answer:
Let us assume a round object of mass m and radius R is rolling down an inclined plane without slipping as shown in figure. There are two forces acting on the object along the inclined plane. One is the component of gravitational force (mg sin θ) and the other is the static frictional force (f). The other component of gravitation force (mg cos θ) is cancelled by the normal force (N) exerted by the plane. As the motion is happening along the incline, we shall write the equation for motion from the free body diagram (FBP) of the object.

Samacheer Kalvi 11th Physics Solution C
For transnational motion, mg sin θ is the supporting force and f is the opposing force, mg sin θ f = ma
For rotational motion, let us take the torque with respect to the center of the object. Then mg sin 0 cannot cause torque as it passes through it but the frictional force f can set torque of Rf = Iα
By using the relation, a = rα, and moment of inertia I = mK2 we get,
Rf = mK2 \(\frac {a}{R}\); f = ma \(\left(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)\)
Now equation becomes,
mg sin θ – ma \(\left(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)\) = ma
mg sin θ = ma + ma \(\left(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)\)
a \(\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)\) = g sin θ
After rewriting it for acceleration, we get,
a = \(\frac{g \sin \theta}{\left(1+\frac{K^{2}}{R^{2}}\right)}\)
We can also find the expression for final velocity of the rolling object by using third equation of motion for the inclined plane.
v2 = u2 + 2as. If the body starts rolling from rest, u = 0. When h is the vertical height of the incline, the length of the incline s is, s = \(\frac {h}{sin θ}\)
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies By taking square root,
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
The time taken for rolling down the incline could also be written from first equation of motion as, v = u + at. For the object which starts rolling from rest, u = 0. Then,
t = \(\frac {v}{a}\)
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
The equation suggests that for a given incline, the object with the least value of radius of gyration K will reach the bottom of the incline first.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Conceptual Questions

Question 1.
When a tree is cut, the cut is made on the side facing the direction in which the tree is required to fall. Why?
Answer:
A cut on the tree is made on the side facing the direction in which the tree is required to fall because that side will no longer be supported by the normal force from the bottom, therefore the gravitational force tries to rotate it. So the torque given by the gravity to the tree makes the tree fall on the side as anticipated.

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Question 2.
Why does a porter bend forward while carrying a sack of rice on his back?
Answer:
When a porter carries a sack of rice, the line of action of his center of gravity will go away from the body. It affects the balance, to avoid this he bends. By which center of gravity will realign within the body again. So balance is maintained.

Question 3.
Why is it much easier to balance a meter scale on your finger tip than balancing on a match stick?
Answer:
A meter scale is larger then a match stick. So the center of gravity for meter scale is higher than a matchstick when we keep it vertically. It is easier to balance the object whose center of gravity is higher than the object whose centro of gravity is lower. So, it is hard to balance a match stick than a meter scale.

Question 4.
Two identical water bottles one empty and the other filled with water are allowed to roll down an inclined plane. Which one of them reaches the bottom first? Explain your answer.
Answer:
Mass of the empty water bottle mostly concentrated on its surface. So moment of inertia of empty water bottle is more than the bottle filled with water. As we know, moment of inertia is inversely proportional to angular velocity. Therefore, the bottle filled with water whirls with greater speed and reaches the ground first.

Question 5.
Write the relation between angular momentum and rotational kinetic energy. Draw a graph for the same. For two objects of same angular momentum, compare the moment of inertia using the graph.
Answer:
Let a rigid body of moment of inertia I rotate with angular velocity ω.
The angular momentum of a rigid body is, L = Iω
The rotational kinetic energy of the rigid body is, KE = \(\frac { 1 }{ 2 }\) Iω2.
By multiplying the numerator and denominator of the above equation with I, we get a relation between L and KE as,
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
It resembles with y = Kx2. If angular momentum is same for two objects, kinetic energy is inversely proportional to moment of inertia.
Moment of inertia of the object whose kinetic energy is lesser will have greater magnitude.

Question 6.
Three identical solid spheres move down through three inclined planes A, B and C all same dimensions. A is without friction, B is undergoing pure rolling and C is rolling with slipping. Compare the kinetic energies EA, EB and EC at the bottom.
Answer:
Even though, the three identical solid spheres of same dimensions move down through three different inclined plane, according to the law of conservation of energy, the potential energy possessed by these three solid spheres will be converted into kinetic energies. So the kinetic energies EA, EB and EC are equal at the bottom.

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Question 7.
Give an example to show that the following statement is false. Any two forces acting on a body can be combined into single force that would have same effect.
Answer:
A single force i.e. resultant of two forces acting on a body depends upon the angle between them also. The simple example for this is if two forces 5 N and 5 N acting on the object in the opposite direction, the single resultant force acting on the body is zero. But, if two forces acting on the object along the same direction, then the resultant i.e. the single force is 5 + 5 = 10 N. Hence the given statement “any two forces acting on a body can be combined into single force that would leave same effect” is wrong.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Numerical Problems

Question 1.
A uniform disc of mass 100 g has a diameter of 10 cm. Calculate the total energy of the disc when rolling along a horizontal table with a velocity of 20 cm s-2.
Answer:
Given,
Mass of the disc = 100 g = 100 x 10-3 kg = \(\frac { 1 }{ 10 }\)kg
Velocity of disc = 20 cm s-1 = 20 x 10-2 ms-1 = 0.2 ms-1
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Question 2.
A particle of mass 5 units is moving with a uniform speed of v = \(3 \sqrt{2}\) units in the XOY plane along the line y = x + 4. Find the magnitude of angular momentum.
Answer:
Given,
Mass = 5 units
Speed = v = \(3 \sqrt{2}\) units
Y = X + 4
Angular momentum = L = m(\(\bar{r} \times \bar{v}\))
= m(x\(\hat{i}\) +y\(\hat{j}\))x(v\(\hat{i}\) + v\(\hat{j}\)) = m[xv\(\hat{k}\)-vy\(\hat{k}\)] = m[xv\(\hat{k}\)– v(x + 4)\(\hat{k}\)]
L = -mv\(\hat{k}\) = -4 x 5 x \(3 \sqrt{2}\)\(\hat{k}\) = – 60\(\sqrt{2}\)\(\hat{k}\)
L = 60\(\sqrt{2}\) units.

Question 3.
A fly wheel rotates with a uniform angular acceleration. If its angular velocity increases from 20π rad/s to 40π rad/s in 10 seconds, find the number of rotations in that period.
Answer:
Given,
Initial angular velocity ω0 = 20 π rad/s
Final angular velocity ω = 40 π rad/s
Time t = 10 s
Solution:
Angular acceleration α = \(\frac{\omega-\omega_{0}}{t}\) = \(\frac {40π – 20π }{ 10 }\)
α = 2π rad/s2
According to equation of motion for rotational motion
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
The number of rotations = n = \(\frac {θ}{ 2π }\)
n = \(\frac {300π}{ 2π }\) = 150 rotations.

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Question 4.
A uniform rod of mass m and length / makes a constant angle 0 with an axis of rotation which passes through one end of the rod. Find the moment of inertia about this gravity is.
Answer:
Moment of inertia of the rod about the axis which is passing through its center of gravity is
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Moment of inertia of a uniform rod of mass m and length l about one axis which passes through one end of the rod
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 5.
Two particles P and Q of mass 1 kg and 3 kg respectively start moving towards each other from rest under mutual attraction. What is the velocity of their center of mass?
Answer:
Given,
Mass of particle P = 1 kg Mass of particle Q = 3 kg
Solution:
Particles P and Q forms a system. Here no external force is acting on the system,

Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
We know that M = \(\frac {d}{dt}\) (VCM ) = f
It means that, C.M. of an isolated system remains at rest when no external force is acting and internal forces do not change its center of mass.

Question 6.
Find the moment of inertia of a hydrogen molecule about an axis passing through its center of mass and perpendicular to the inter-atomic axis.
Given: mass of hydrogen atom 1.7 x 1027kg and inter atomic distance is equal to 4 x 10-10m.
Answer:
Given,
Inter-atomic distance : 4 x 10-10 m
Mass of H2 atom : 1.7 x 10-27 kg
Moment of inertia of H2 =
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 7.
On the edge of a wall, we build a brick tower that only holds because of the bricks’ own weight. Our goal is to build a stable tower whose overhang d is greater than the length l of a single brick. What is the minimum number of bricks you need?
(Hint: Find the center of mass for each brick and add.)
Answer:
Given:
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Length of the brick = l
Length of the overhang = d
The mono of bricks can be decided only by using the concept of position of center of gravity. The first brick is in contact with the ground and it will not fall over.
Let one end of brick 2 is coinciding with the center of brick 1 i.e. x = 0.
∴ The position of n brick is
xn = (n – 1) \(\frac {L}{4}\)
The center of gravity is in the midway between the center of brick 2 and the center of brick n.
position of G =
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
brick tower will fall when G >\(\frac {L}{4}\) it shows that n > 4.

Question 8.
The 747 boing plane is landing at a speed of 70 m s-1. Before touching the ground, the wheels are not rotating. How long a skid mark do the wing wheels leave (assume their mass is 100 kg which is distributed uniformly, radius is 0.7 m, and the coefficient of friction with the ground is 0.5)?
Answer:
The types of the plane will leave a skid mark if the speed of the types in contact with ground is lesser than the velocity of the plane. The condition for this is –
v > ω
(When the type attained an angular velocity of V/R)
The types will stop the skidding and starts the rolling.
The forces acting on the wheel after the plane touches down are,
N – P Normal force W – weight
The wheel is not accelerating means
N = ω
The torque about the center of the wheel is
τ = RF = µωR
The angular acceleration is
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
According to equation of motion, time taken to stop the skidding by the wheel is,
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies Q8
The six mark will have a length of
l = vt = 70 x 0.03 = 2.1 m
Note:
The 747 is resting on the runway, supported by 16 wheels under the wing, and 2 under the nose total length is 68.63 m. The normal force experienced by plane through its 16 wheels is ω = 232 KN.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Additional Questions Solved

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Multiple Choice Questions

Question 1.
The changes produced by the deforming forces in a rigid body are –
(a) very large
(b) infinity
(c) negligibly small
(d) small
Answer:
(c) negligibly small

Question 2.
When a rigid body moves all particles that constitute the body follows-
(a) same path
(b) different paths
(c) either same or different path
(d) circular path
Answer:
(b) different path

Question 3.
For bodies of regular shape and uniform mass distribution, the center of mass is at –
(a) the comers
(b) inside the objects
(c) the point where the diagonals meet
(d) the geometric center
Answer:
(d) the geometric center

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Question 4.
For square and rectangular objects center of mass lies at –
(a) the point where the diagonals meet
(b) at the comers
(c) on the center surface
(d) any point
Answer:
(a) the point where the diagonals meet

Question 5.
Center of mass may lie –
(a) within the body
(b) outside the body
(c) both (a) and (b)
(d) only at the center
Answer:
(c) both (a) and (b)

Question 6.
The dimension of point mass is –
(a) positive
(b) negative
(c) zero
(d) infinity
Answer:
(c) zero

Question 7.
The motion of center of mass of a system of two particles is unaffected by their internal forces –
(a) irrespective of the actual directions of the internal forces
(b) only if they are along the line joining the particles
(c) only if acts perpendicular to each other
(d) only if acting opposite
Answer:
(a) irrespective of the actual directions of the internal forces

Question 8.
A circular plate of diameter 10 cm is kept in contact with a square plate of side 10 cm. The density of the material and the thickness are same everywhere. The center of mass of the system will be
(a) inside the circular plate
(b) inside the square plate
(c) At the point of contact
(d) outside the system
Answer:
(6) inside the square plate

Question 9.
The center of mass of a system of particles does not depend on
(a) masses of particles
(b) position of the particles
(c) distribution of masses
(d) forces acting on the particles
Answer:
(d) forces acting on the particles

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Question 10.
The center of mass of a solid cone along the line from the center of the base to the vertex is at –
(a) \(\frac { 1 }{ 2 }\) th of its height
(b) \(\frac { 1 }{ 3 }\) of its height
(c) \(\frac { 1 }{ 4 }\) th of its height
(d) \(\frac { 1 }{ 5 }\) th of its height
Answer:
(d) \(\frac { 1 }{ 5 }\) th of its height

Question 11.
All the particles of a body are situated at a distance of X from origin. The distance of the center of mass from the origin is –
(a) ≥ r
(b) ≤ r
(c) = r
(d) > r

Question
A free falling body breaks into three parts of unequal masses. The center of mass of the three parts taken together shifts horizontally towards –
(a) heavier piece
(b) lighter piece
(c) does not shift horizontally
(d) depends on vertical velocity
Answer:
(c) does not shift horizontally

Question 13.
The distance between the center of carbon and oxygen atoms in the gas molecule is 1.13 A. The center of mass of the molecule relative to oxygen atom is –
(a) 0.602 Å
(b) 0.527 Å
(c) 1.13 Å
(d) 0.565 Å
Answer:
(b) 0.527 Å
Given,
Inter atomic distance = 1.13 Å
Mass of carbon atom = 14
Mass of oxygen atom = 16
Let C.M. of molecule lies at a distance of X from oxygen atom-
i.e. m1r1 = m2r2
16 X = 14(1.13 – X)
30 X = 15.82
X = 0.527 Å

Question 14.
The unit of position vector of center of mass is-
(a) kg
(b) kg m2
(c) m
(d) m2
Answer:
(c) m

Question 15.
The sum of moments of masses of all the particles in a system about the center of mass is-
(a) minimum
(b) maximum
(c) zero
(d) infinity
Answer:
(c) zero

Question 16.
The motion of center of mass depends on-
(a) external forces acting on it
(b) internal forces acting within it
(c) both (a) and (b)
(d) neither (a) nor (b)
Answer:
(a) external forces acting on it

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Question 17.
Two particles P and Q move towards with each other from rest with the velocities of 10 ms-1 and 20 ms-1 under the mutual force of attraction. The velocity of center of mass is-
(a) 15 ms-1
(b) 20 ms-1
(c) 30 ms-1
(d) zero
Answer:
(d) zero

Question 18.
The reduced mass of the system of two particles of masses 2 m and 4 m will be –
(a) 2 m
(b) \(\frac {2 }{ 3 }\)y m
(c) \(\frac {3}{ 2 }\)y m
(d) \(\frac { 4 }{ 3 }\)m
Answer:
(d) \(\frac { 4 }{ 3 }\)m

Question 19.
The motion of the center of mass of a system consists of many particles describes its –
(a) rotational motion
(b) vibratory motion
(c) oscillatory motion
(d) translator y motion
Answer:
(c) oscillatory motion

Question 20.
The position of center of mass can be written in the vector form as –
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Answer:
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 21.
The positions of two masses m1 and m2 are x1 and x2. The position of center of mass is –
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Answer:
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 22.
In a two particle system, one particle lies at origin another one lies at a distance of X. Then the position of center of mass of these particles of equal mass is –
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Answer:
(a) \(\frac {X}{2}\)

Question 23.
Principle of moments is –
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Answer:
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 24.
Infinitesimal quantity means –
(a) collective particles
(b) extremely small
(c) nothing
(d) extremely larger
Answer:
(b) extremely small

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Question 25.
In the absence of external forces the center of mass will be in a state of –
(a) rest
(b) uniform motion
(c) may be at rest or in uniform motion
(d) vibration
Answer:
(c) may be at rest or in uniform motion

Question 26.
The activity of the force to produce rotational motion in a body is called as –
(a) angular momentum
(b) torque
(c) spinning
(d) drive force
Answer:
(b) torque

Question 27.
The moment of the external applied force about a point or axis of rotation is known as –
(a) angular momentum
(b) torque
(c) spinning
(d) drive force
Answer:
(b) torque

Question 28.
Torque is given as –
(a) \(\vec{r}\) . \(\vec{F}\)
(b) \(\vec{r}\) x \(\vec{F}\)
(c) \(\vec{F}\) x \(\vec{r}\)
(d) r F cos θ
Answer:
(b) \(\vec{r}\) x \(\vec{F}\)

Question 29.
The magnitude of torque is –
(a) rF sin θ
(b) rF cos θ
(c) rF tan θ
(d) rF
Answer:
(a) rF sin θ

Question 30.
The direction of torque ácts –
(a) along \(\vec{F}\)
(b) along \(\vec{r}\) & \(\vec{F}\)
(c) Perpendicular to \(\vec{r}\)
(d) Perpendicular to both \(\vec{r}\) & \(\vec{F}\)
Answer:
(d) Perpendicular to both \(\vec{r}\) & \(\vec{F}\)

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Question 31.
The unit of torque is –
(a) is
(b) Nm-2
(c) Nm
(d) Js-1
Answer:
(c) Nm

Question 32.
The direction of torque is found using –
(a) left hand rule
(b) right hand rule
(c) palm rule
(d) screw rule
Answer:
(b) right hand rule

Question 33.
if the direction of torque is out of the paper then the rotation produced by the torque is –
(a) clockwise
(b) anticlockwise
(c) straight line
(d) random direction
Answer:
(a) clockwise

Question 34.
If the direction of the torque is inward the paper then the rotation is –
(a) clockwise
(b) anticlockwise
(c) straight line
(d) random direction
Answer:
(a) clockwise

Question 35.
if \(\vec{r}\) and \(\vec{F}\) are parallel or anti parallel, then the torque is –
(a) zero
(b) minimum
(c) maximum
(d) infinity
Answer:
(a) zero

Question 36.
The maximum possible value of torque is –
(a) zero
(b) infinity
(c) \(\vec{r}\) + \(\vec{F}\)
(d) rF
Answer:
(d) rF

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Question 37.
The relation between torque and angular acceleration is –
(a) \(\vec{τ}\) = \(\frac{1}{\alpha}\)
(b) \(\vec{α}\) = \(\frac{\vec{\tau}}{\mathrm{I}}\)
(c) \(\vec{α}\) = I \(\vec{τ}\)
(d) \(\vec{τ}\) = \(\frac{\vec{\alpha}}{\mathrm{I}}\)
Answer:
(b) \(\vec{α}\) = \(\frac{\vec{\tau}}{\mathrm{I}}\)

Question 38.
Angular momentum is –
(a) \(\vec{P}\) x \(\vec{r}\)
(b) \(\vec{r}\) x \(\vec{P}\)
(c) \(\overrightarrow{\frac{r}{\vec{p}}}\)
(d) \(\vec{r}\) . \(\vec{P}\)
Answer:
(b) \(\vec{r}\) x \(\vec{P}\)

Question 39.
The magnitude of angular momentum is given by –
(a) rp
(b) rp sin θ
(c) rp cos θ
(d) rp tan θ
Answer:
(b) rp sin θ

Question 40.
Angular momentum is associated with –
(a) rotational motion
(b) linear motion
(c) both (a) and (b)
(d) circular motion only
Answer:
(c) both (a) and (b)

Question 41.
Angular momentum acts perpendicular to –
(a) \(\vec{r}\)
(b) \(\vec{P}\)
(c) both \(\vec{r}\) and \(\vec{P}\)
(d) plane of the paper
Answer:
(c) both \(\vec{r}\) and \(\vec{P}\)

Question 42.
Angular momentum is given by –
(a) \(\frac {I}{ω}\)
(b) τω
(c) Iω
(d) \(\frac {ωI}{2}\)
Answer:
(c) Iω

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Question 43.
The rate of change of angular momentum is –
(a) Torque
(b) angular velocity
(c) centripetal force
(d) centrifugal force
Answer:
(a) Torque

Question 44.
The forces acting on a body when it is at rest –
(a) is gravitational force
(b) Normal force
(c) both gravitational as well as normal force
(d) No force is acting
Answer:
(c) both gravitational as well as normal force

Question 45.
The net force acting on a body when it is at rest is –
(a) gravitational force
(b) Normal force
(c) Sum of gravitational and normal force
(d) zero
Answer:
(d) zero

Question 46.
If net force acting on a body is zero, then the body is in –
(a) transnational equilibrium
(b) rotational equilibrium
(c) both (a) and (b)
(d) none
Answer:
(a) transnational equilibrium

Question 47.
If the net torque acting on the body is zero, then the body is in –
(a) transnational equilibrium
(b) rotational equilibrium
(c) mechanical equilibrium
(d) none
Answer:
(b) rotational equilibrium

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Question 48.
when the net force and net torque acts on the body is zero then the body is in –
(a) transnational equilibrium
(b) rotational equilibrium
(c) mechanical equilibrium
(d) none
Answer:
(d) none

Question 49.
When the net force and net torque acts on the body is zero then the body is in –
(a) static equilibrium
(b) Dynamic equilibrium
(c) both (a) and (b)
(d) transnational equilibrium
Answer:
(c) both (a) and (b)

Question 50.
When two equal and opposite forces acting on the body at two different points, it may give –
(a) net force
(b) torque
(c) stable equilibrium
(d) none
Answer:
(b) torque

Question 51.
The torque in rotational motion is analogous to in transnational motion –
(a) linear momentum
(b) mass
(c) couple
(d) force
Answer:
(d) force

Question 52.
Which of the following example does not constitute a couple?
(a) steering a car
(b) turning a pen cap
(c) ball rolls on the floor
(d) closing the door
Answer:
(c) ball rolls on the floor

Questioner 53.
If the linear momentum and angular momentum are zero, then the object is said to be in –
(a) stable equilibrium
(b) unstable equilibrium
(c) neutral equilibrium
(d) all the above
Answer:
(d) all the above

Question 54.
When the body is disturbed, the potential energy remains same, then the body is in –
(a) stable equilibrium
(b) unstable equilibrium
(c) neutral equilibrium
(d) all the above
Answer:
(c) neutral equilibrium

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Question 55
The point where the entire weight of the body acts is called as –
(a) center of mass
(b) center of gravity
(c) both (a) and (b)
(d) pivot
Answer:
(b) center of gravity

Question 56.
The forces acting on a cyclist negotiating a circular Level road is /are –
(a) gravitational force
(b) centrifugal force
(c) frictional force
(d) all the above
Answer:
(d) all the above

Question 57.
While negotiating a circular level road a cyclist has to bend by an angle θ from vertical to stay in an equilibrium is-
(a) \(\tan \theta=\frac{r g}{r^{2}}\)
(b) θ = \(\tan ^{-1}\left(\frac{v^{2}}{r g}\right)\)
(c) θ = \(\sin ^{-1}\left(\frac{r g}{r^{2}}\right)\)
(d) zero
Answer:
(b) θ = \(\tan ^{-1}\left(\frac{v^{2}}{r g}\right)\)

Question 58.
Moment of inertia for point masses –
(a) m2r
(b) rw2
(c) mr2
(d) zero
Answer:
(c) mr2

Question 59.
Moment of inertia for bulk object –
(a) rm2
(b) rw2
(c) \(m_{i} r_{i}^{2}\)
(d) \(\Sigma m_{i} r_{i}^{2}\)
Answer:
(d) \(\Sigma m_{i} r_{i}^{2}\)

Question 60.
For rotational motion, moment of inertia is a measure of –
(a) transnational inertia
(b) mass
(c) rotational inertia
(d) invariable quantity
Answer:
(c) rotational inertia

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Question 61.
Unit of moment of inertia –
(a) kgm
(b) mkg-2
(c) kgm2
(d) kgm-1
Answer:
(c) kgm2

Question 62.
Dimensional formula for moment of inertia is –
(a) [ML-2]
(b) [M2L-1]
(c) [M-2]
(d) [ML2]
Answer:
(d) [ML2]

Question 63.
Moment of inertia of a body is a –
(a) variable quantity
(b) invariable quantity
(c) constant quantity
(d) measure of torque
Answer:
(a) variable quantity

Question 64.
Moment of inertia of a thin uniform rod about an axis passing through the center of mass and perpendicular to the length is –
(a) \(\frac { 1 }{ 3 }\)Ml2
(b) \(\frac { 1 }{ 12 }\)Ml2
(c) \(\frac { 1 }{ 2 }\)M(l2 + b2 )
(d) Ml2
Answer:
(b) \(\frac { 1 }{ 12 }\)Ml2

Question 65.
Moment of inertia ofa thin uniform rod about an axis passing through one end and perpendicular to the length is-
(a) \(\frac { 1 }{ 3 }\)Ml2
(b) \(\frac { 1 }{ 12 }\)Ml2
(c) \(\frac { 1 }{ 2 }\)M(l2 + b2 )
(d) Ml2
Answer:
(a) \(\frac { 1 }{ 3 }\)Ml2

Question 66.
Moment of inertia of a thin uniform rectangular sheet about an axis passing through the center of mass and perpendicular to the plane of the sheet is-
(a) \(\frac { 1 }{ 3 }\)Ml2
(b) \(\frac { 1 }{ 12 }\)Ml2
(c) \(\frac { 1 }{ 2 }\)M(l2 + b2 )
(d) Ml2
Answer:
(c) \(\frac { 1 }{ 2 }\)M(l2 + b2 )

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Question 67.
Moment of inertia of a thin uniform ring about an axis passing through the center of gravity and perpendicular to the plane is –
(a) MR2
(b) 2 MR2
(c) \(\frac { 1 }{ 2 }\)MR2
(d) \(\frac { 3 }{ 2 }\)MR2
Answer:
(a) MR2

Question 68.
Moment of inertia of a thin uniform ring about an axis passing through the center and lying on the plane (along diameter) is –
(a) MR2
(b) 2 MR2
(c) \(\frac { 1 }{ 2 }\) MR2
(d) \(\frac { 2 }{ 3 }\)MR2
Answer:
(c) \(\frac { 1 }{ 2 }\) MR2

Question 69.
Moment of inertia of a thin uniform disc about an axis passing through the center and perpendicular to the plane is –
(a) MR2
(b) 2 MR2
(c) \(\frac { 1 }{ 2 }\) MR2
(d) \(\frac { 2 }{ 3 }\)MR2
Answer:
(c) \(\frac { 1 }{ 2 }\) MR2

Question 70.
Moment of inertia of a thin uniform disc about an axis passing through the center lying on the plane (along diameter is)
(a) MR2
(b) \(\frac { 1 }{ 2 }\) MR2
(c) \(\frac { 3 }{ 2 }\) MR2
(d) \(\frac { 1 }{ 4 }\) MR2
Answer:
(d) \(\frac { 1 }{ 4 }\) MR2

Question 71.
Moment of inertia of a thin uniform hollow cylinder about an axis of the cylinder is –
(a) MR2
(b) \(\frac { 1 }{ 2 }\) MR2
(c) \(\frac { 3 }{ 2 }\) MR2
(d) \(\frac { 1 }{ 4 }\) MR2
Answer:
(a) MR2

Question 72.
Moment of inertia of a thin uniform hollow cylinder about an axis of the cylinder is –
(a) MR2
(b) M\(\left(\frac{\mathrm{R}^{2}}{2}+\frac{l^{2}}{12}\right)\)
(c) \(\frac { 1 }{ 2 }\) MR2
(d) M\(\left(\frac{\mathrm{R}^{2}}{4}+\frac{l^{2}}{12}\right)\)
Answer:
(b) M\(\left(\frac{\mathrm{R}^{2}}{2}+\frac{l^{2}}{12}\right)\)

Question 73.
Moment of inertia of a uniform solid cylinder about an axis passing through the center and along the axis of the cylinder is –
(a) MR2
(b) M\(\left(\frac{\mathrm{R}^{2}}{2}+\frac{l^{2}}{12}\right)\)
(c) \(\frac { 1 }{ 2 }\) MR2
(d) M\(\left(\frac{\mathrm{R}^{2}}{4}+\frac{l^{2}}{12}\right)\)
Answer:
(c) \(\frac { 1 }{ 2 }\) MR2

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Question 74.
Moment of inertia of a uniform solid cylinder about as axis passing perpendicular to the length and passing through the center is –
(a) MR2
(b) M\(\left(\frac{\mathrm{R}^{2}}{2}+\frac{l^{2}}{12}\right)\)
(c) \(\frac { 1 }{ 2 }\) MR2
(d) M\(\left(\frac{\mathrm{R}^{2}}{4}+\frac{l^{2}}{12}\right)\)
Answer:
(d) M\(\left(\frac{\mathrm{R}^{2}}{4}+\frac{l^{2}}{12}\right)\)

Question 75.
Moment of inertia of a thin hollow sphere about an axis passing through the center along its diameter is
(a) \(\frac { 2 }{ 3 }\)MR2
(b) \(\frac { 5 }{ 3 }\)MR2
(c) \(\frac { 7 }{ 5 }\)MR2
(d) \(\frac { 2 }{ 5 }\)MR2
Answer:
(a) \(\frac { 2 }{ 3 }\)MR2

Question 76.
Moment of inertia of a thin hollow sphere about an axis passing through the edge along its tangent is –
(a) \(\frac { 2 }{ 3 }\)MR2
(b) \(\frac { 5 }{ 3 }\)MR2
(c) \(\frac { 7 }{ 5 }\)MR2
(d) \(\frac { 2 }{ 5 }\)MR2
Answer:
(b) \(\frac { 5 }{ 3 }\)MR2

Question 77.
torment of inertia of a uniform solid sphere about an axis passing through the center along its diameter is –
(a) \(\frac { 2 }{ 3 }\)MR2
(b) \(\frac { 5 }{ 3 }\)MR2
(c) \(\frac { 7 }{ 5 }\)MR2
(d) \(\frac { 2 }{ 5 }\)MR2
Answer:
(d) \(\frac { 2 }{ 5 }\)MR2

Question 78.
Moment of inertia of a uniform solid sphere about an axis passing through the edge along its tangent is –
(a) \(\frac { 2 }{ 3 }\)MR2
(b) \(\frac { 5 }{ 3 }\)MR2
(c) \(\frac { 7 }{ 5 }\)MR2
(d) \(\frac { 2 }{ 5 }\)MR2
Answer:
(c) \(\frac { 7 }{ 5 }\)MR2

Question 79.
The ratio of K2/R2 of a thin uniform ring about an axis passing through the center and perpendicular to the plane is-
(a) 1
(b) 2
(c) \(\frac { 7 }{ 5 }\)
(d) \(\frac { 3 }{ 2 }\)
Answer:
(a) 1

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Question 80.
The ratio of K2/ R2 of a thin uniform disc about an axis passing through the center and perpendicular to the plane is –
(a) 1
(b) 2
(c) \(\frac {1}{ 2 }\)
(d) \(\frac { 3 }{ 2 }\)
Answer:
(c) \(\frac {1}{ 2 }\)

Question 81.
When no external torque acts on the body, the net angular momentum of a rotating body.
(a) increases
(b) decreases
(c) increases or decreases
(d) remains constant
Answer:
(d) remains constant

Question 82.
Moment of inertia of a body is proportional to –
(a) ω
(b) \(\frac { 1 }{ ω }\)
(c) ω2
(d) \(\frac{1}{\omega^{2}}\)
Answer:
(b) \(\frac { 1 }{ ω }\)

Question 83.
When the hands are brought closer to the body, the angular velocity of the ice dancer –
(a) decreases
(b) increases
(c) constant
(d) may decrease or increase
Answer:
(b) increases

Question 84.
When the hands are stretched out from the body, the moment of inertia of the ice dancer –
(a) decreases
(b) increases
(c) constant
(d) may decrease or increase
Answer:
(b) increases

Question 85.
The work done by the torque is –
(a) F. ds
(b) F. dθ
(c) τ dθ
(d) r.dθ
Answer:
(c) τ dθ

Question 86.
Rotational Kinetic energy of a body is –
(a) \(\frac { 1 }{ 2 }\)mr
(b) \(\frac { 1 }{ 2 }\) Iω2
(c) \(\frac { 1 }{ 2 }\)Iv2
(d) \(\frac { 1 }{ 2 }\)mω2
Answer:
(b) \(\frac { 1 }{ 2 }\) Iω2

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Question 87.
Rotational kinetic energy is given by –
(a) \(\frac { 1 }{ 2 }\)mr
(b) \(\frac { 1 }{ 2 }\)Iv2
(c) \(\frac{\mathrm{L}^{2}}{2 \mathrm{I}}\)
(d) \(\frac{2 \mathrm{I}}{\mathrm{L}^{2}}\)
Answer:
(c) \(\frac{\mathrm{L}^{2}}{2 \mathrm{I}}\)

Question 88.
If E is a rotational kinetic energy then angular momentum is-
(a) \(\sqrt{2 \mathrm{IE}}\)
(b) \(\frac{\mathrm{E}^{2}}{2 \mathrm{I}}\)
(c) \(\frac{2 \mathrm{I}}{\mathrm{E}^{2}}\)
(d) \(\frac{E}{I^{2} \omega^{2}}\)
Answer:
(a) \(\sqrt{2 \mathrm{IE}}\)

Question 89.
The product of torque acting on a body and angular velocity is –
(a) Energy
(b) power
(c) work done
(d) kinetic energy
Answer:
(b) power

Question 90.
The work done per unit time in rotational motion is given by –
(a) \(\vec{F}\) .v
(b) \(\frac {dθ}{dt}\)
(c) τ ω
(d) I ω
Answer:
(c) τ ω

Question 91.
While rolling, the path of center of mass of an object is –
(a) straight line
(b) parabola
(c) hyperbola
(d) circle
Answer:
(a) straight line

Question 92.
In pure rolling, the velocity of the point of the rolling object which comes in contact with the surface is –
(a) maximum
(b) minimum
(c) zero
(d) 2 VCM
Answer:
(c) zero

Question 93.
In pure rolling velocity of center of mass is equal to –
(a) zero
(b) Rω
(c) \(\frac { ω }{ R }\)
(d) \(\frac { R }{ ω }\)
Answer:
(b) Rω

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Question 94.
In pure rolling, rotational velocity of points at its edges is equal to-
(a) Rω
(b) velocity of center of mass
(c) transnational velocity
(d) all the above
Answer:
(a) Rω

Question 95.
Sliding of the object occurs when –
(a) Vtrans < Vrot
(b) Vtrans = Vrot
(c) Vtrans > Vrot
(d) Vtrans = 0
Answer:
(c) Vtrans > Vrot

Question 96.
Sliding of the object occurs while –
(a) Vtrans = Vrot
(b) VCM = Rω
(c) VCM < Rω
(d) VCM > Rω
Answer:
(d) VCM > Rω

Question 97.
Slipping of the object occurs when –
(a) Vtrans < Vrot
(b) Vtrans = Vrot
(c) Vtrans > Vrot
(d) Vtrans = 0
Answer:
(a) Vtrans < Vrot

Question 98.
Slipping of the object occurs when –
(a) Vtrans = Vrot
(b) VCM = Rω
(c) VCM < Rω
(d) VCM > Rω
Answer:
(c) VCM < Rω

Question 99.
In sliding, the resultant velocity of a point of contact acts along –
(a) forward direction
(b) backward direction
(c) either (a) or (b)
(d) tangential direction
Answer:
(a) forward direction

Question 100.
In slipping, the resultant velocity of a point of contact acts along –
(a) forward direction
(b) backward direction
(c) either (a) or (b)
(d) tangential direction
Answer:
(b) backward direction

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Question 101.
When a solid sphere is undergoing pure rolling, the ratio of transnational kinetic energy to rotational kinetic – energy is –
(a) 2 : 5
(b) 5 : 2
(c) 1 : 5
(d) 5 : 1
Answer:
(b) 5 : 2

Question 102.
Time taken by the rolling object in inclined plane to reach its bottom is –
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Answer:
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 103.
The velocity of the rolling object on inclined plane at the bottom of inclined plane is –
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Answer:
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 104.
Moment of inertia of an annular disc about an axis passing through the centre and perpendicular to the plane of disc is –
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Answer:
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 105.
Moment of inertia of a cube about an axis passing through the center of mass and perpendicular to face is –
(a) \(\frac{\mathrm{Ma}^{2}}{6}\)
(b) \(\frac {1}{3}\) Ma2
(c) \(\frac {Ma}{6}\)
(d) \(\frac{\mathrm{Ma}^{2}}{12}\)
Answer:
(a) \(\frac{\mathrm{Ma}^{2}}{6}\)

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Question 106.
Moment of inertia of a rectangular plane sheet about an axis passing through center of mass and perpendicular to side b in its plane is –
(a) \(\frac{\mathrm{Ml}^{2}}{12}\)
(b) \(\frac{\mathrm{Ma}^{2}}{12}\)
(c) \(\frac{\mathrm{Mb}^{2}}{12}\)
(d) \(\frac{\mathrm{Ml}^{2}}{6}\)
Answer:
(c) \(\frac{\mathrm{Mb}^{2}}{12}\)

Question 107.
Rotational kinetic energy can be calculated by using –
(a) \(\frac{1}{2}\) I ω2
(b) \(\frac{\mathrm{L}^{2}}{2I}\)
(c) \(\frac{1}{2}\) Lω
(d) all the above
Answer:
(b) \(\frac{\mathrm{L}^{2}}{2I}\) )

Question 108.
The radius of gyration of a solid sphere of radius r about a certain axis is r. The distance of that axis from the center of the sphere is –
(a) \(\frac{2}{5}\)r
(b) \(\sqrt{\frac{2}{5}}\)r
(c) \(\sqrt{0.6r}\)
(d) \(\sqrt{\frac{5}{3}}\)
Answer:
(c) \(\sqrt{0.6r}\)
From parallel axis theorem
I = IG + Md2
mr2 = \(\frac{2}{5}\) mr2 + md2
d = \(\sqrt{\frac{3}{5}}\)r = \(\sqrt{0.6r}\)

Question 109.
A wheel is rotating with angular velocity 2 rad/s. It is subjected to a uniform angular acceleration 2 rad/s2 then the angular velocity after 10 s is
(a) 12 rad/s
(b) 20 rad/s
(c) 22 rad/s
(d) 120 rad/s
Answer:
(c) 22 rad/s
ω = ω0 + αt
Here ω0 = 2 rad/s,
α = 2 rad/s2
ω = 10 s
ω = 2 + 2 x 10 = 22 rad/s

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Question 110.
Two rotating bodies A and B of masses m and 2m with moments of inertia IA and IB (Ib > IA) have equal kinetic energy of rotation. If LA and LB be their angular momenta respectively,
then,
(a) LB > LA
(b) LA > LB
(c) LA = \(\frac{L_{B}}{2}\)
(d) LA = 2LB
Answer:
(a) LB > LA

Question 111.
Three identical particles lie in x, y plane. The (x, y) coordinates of their positions are (3, 2), (1, 1), (5, 3) respectively. The (x, y) coordinates of the center of mass are –
(a) (a, b)
(b) (1, 2)
(c) (3, 2)
(d) (2, 1)
Answer:
(c) The X and Y coordinates of the center of mass are
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 112.
A solid cylinder of mass 3 kg and radius 10 cm is rotating about its axis with a frequency of 20/π. The rotational kinetic energy of the cylinder
(a) 10 π J
(b) 12 J
(c) \(\frac{6 \times 10^{2}}{\pi}\) J
(d) 3 J
Answer:
(b) 12 J
Given,
M = 3 kg
R = 0.1 m
v = 20 / π
Angular frequency ω = 2πv = \(\frac{2π x 20}{π}\) = 40 rad/s-1
Moment of inertia of the cylinder about its axis = I = \(\frac{1}{2}\) mR2 = \(\frac{1}{2}\) x 3 x (0.1)2 = 0.015 kg m2
K.E. = \(\frac{1}{2}\) Iω2 = \(\frac{1}{2}\) x 0.015 x (40)2 = 12 J

Question 113.
A circular disc is rolling down in an inclined plane without slipping. The percentage of rotational energy in its total energy is
(a) 66.61%
(b) 33.33%
(c) 22.22%
(d) 50%
Answer:
(b) 33.33%
Rotational K.E. = \(\frac{1}{2}\)Iω2 \(\frac{1}{2}\)(\(\frac{1}{2}\)MR22 = \(\frac{1}{4}\) MR2ω2
Transnational K.E. = \(\frac{1}{2}\)MV2 = \(\frac{1}{2}\)M(Rω)2 = \(\frac{1}{2}\) MR2ω2
Total kinetic energy = Erot + Etrans = \(\frac{1}{4}\) MR2ω2\(\frac{1}{2}\)M(Rω)2 = \(\frac{3}{4}\) MR2ω2
% of Erot = \(\frac{E_{\text {rot }}}{E_{\text {Tot }}}\) x 100% = 33.33%

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Question 114.
A sphere rolls down in an inclined plane without slipping. The percentage of transnational energy in its total energy is
(a) 29.6%
(b) 33.4%
(c) 71.4%
(d) 50%
Answer:
(c) 71.4%
Rotational K.e. Erot =
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 115.
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the center of mass is –
(a) 30 m/s
(b) 20 m/s
(c) 10 m/s
(d) 5 m/s
Answer:
(c) According to law of conservation of linear momentum
MV = (M + M) VCM
VCM = \(\frac{MV}{M + M}\) = \(\frac{10 × 10}{10 + 4}\) = 10 ms-1

Question 116.
A mass is whirled in a circular path with constant angular velocity and its angular momentum is L. If the string is now halved keeping the angular velocity the same, the angular momentum is –
(a) \(\frac{L}{4}\)
(b) \(\frac{L}{2}\)
(c) L
(d) 2L
Answer:
(a) \(\frac{L}{4}\)
We know that
angular momentum L = Mr2
Here, m and co are constants L α r2
If r becomes \(\frac{r}{2}\) angular momentum becomes \(\frac{1}{4}\) th of its initial value.

Question 117.
The moment of inertia of a thin uniform ring of mass 1 kg and radius 20 cm rotating about the axis passing through the center and perpendicular to the plane of the ring is –
(a) 4 x 10-2 kg m2
(b) 1 x 10-2 kg m2
(c) 20 x 10-2 kg m2
(d) 10 x 10-2 kg m2
Answer:
(b) Moment of inertia I = MR2 = 1 x (10 x 10-2)2 = 1 x 10-2 kg m2.

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Question 118.
A solid sphere is rolling down in the inclined plane, from rest without slipping. The angle of inclination with horizontal is 30°. The linear acceleration of the sphere is –
(a) 28 ms-2
(b) 3.9 ms-2
(c) \(\frac{25}{7}\)ms-2
(d) \(\frac{1}{20}\)ms-2
Answer:
(c) \(\frac{25}{7}\)ms-2
We know that,a =
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 119.
An electron is revolving in an orbit of radius 2 A with a speed of 4 x 105 m /s. The angular momentum of the electron is [Me = 9 x 10-31 kg]
(a) 2 x 10-35 kg m2 s-1
(b) 72 x 10-36 kg m2 s-1
(c) 7.2 x 10-34 kg m2 s-1
(d) 0.72 x 10-37 kg m2 s-1
Answer:
(b) Angular momentum L = mV x r = 9 x 10-31 x 4 x 105 x 2 x 10-10 = 72 x 10-36kg m2 s-1

Question 120.
A raw egg and hard boiled egg are made to spin on a table with the same angular speed about the same axis. The ratio of the time taken by the eggs to stop is –
(a) =1
(b) < 1
(c) > 1
(d) none of these
Answer:
(d) When a raw egg spins, the fluid inside comes towards its side.
∴ “1” will increase in – turn it decreases ω. Therefore it takes lesser time than boiled egg.
∴\(\frac {time fìr raw egg}{time for boiled egg}\) < 1

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Short Answer Questions (1 Mark)

Question 1.
What is a rigid body?
Answer:
A rigid body is the one which maintains its definite and fixed shape even when an external force acts on it.

Question 2.
When an object will have procession? Give one example.
Answer:
The torque about the axis will rotate the object about it and the torque perpendicular to the axis will turn the axis of rotation when both exist simultaneously on a rigid body the body will have a procession.
Example:
The spinning top when it is about to come to rest.

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Question 3.
Define angular momentum. Give an expression for it.
Answer:
The angular momentum of a point mass is defined as the moment of its linear momentum.
\(\vec{L}\) = \(\vec{r}\) x \(\vec{p}\) or L = rp sin θ

Question 4.
When an angular momentum of the object will be zero?
Answer:
If the straight path of the particle passes through the origin, then the angular momentum is zero, which is also a constant.

Question 5.
When an object be in-mechanical equilibrium?
Answer:
A rigid body is said to be in mechanical equilibrium when both its linear momentum and angular momentum remain constant.

Question 6.
Derive an expression for the power delivered by torque.
Answer:
Power delivered is the work done per unit time. IF we differentiate the expression for work done with respect to time, we get the instantaneous
power (P).
p = \(\frac{dw}{dt}\) = τ \(\frac{dθ}{dt}\)
p = τ dω

Question 7.
A boy sits near the edge of revolving circular disc

  1. What will be the change in the motion of a disc?
  2. If the boy starts moving from edge to the center of the disc, what will happen?

Answer:

  1. As we know L = Iω = constant if the boy sits on the edge of revolving disc, its I will be increased in turn it reduces angular velocity.
  2. If the boy starts moving towards the center of the disc, its I will decrease in turn that increases its angular velocity.

Question 8.
Are moment of inertia and radius of gyration of a body constant quantities?
Answer:
No, moment of inertia and radius of gyration depends on axis of rotation and also on the distribution of mass of the body about its axis.

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Question 9.
A cat is able to land on its feet after a fall. Which principle of physics is being used? Explain.
Answer:
A cat is able to land on its feet after a fall. This is based on law of conservation of angular ~ momentum. When the cat is about to fall, it curls its body to decrease the moment of inertia and increase its angular velocity. When it lands it stretches out its limbs. By which it increases its moment of inertia and inturn it decreases its angular velocity. Hence, the cat lands safety.

Question 10.
About which axis a uniform cube will have minimum moment of inertia ?
Answer:
It will be about an axis passing through the center of the cube and connecting the opposite comers.

Question 11.
State the principle of moments of rotational equilibrium.
Answer:
∑ =\(\bar{\tau}\) = 0

Question 12.
Write down the moment of inertia of a disc of radius R and mass m about an axis in its plane at a distance R / 2 from its center.
Answer:
\(\frac { 1 }{ 2 }\) MR2

Question 13.
Can the couple acting on a rigid body produce translator motion ?
Answer:
No. It can produce only rotatory motion.

Question 14.
Which component of linear momentum does not contribute to angular momentum?
Answer:
Radial Component.

Question 15.
A system is in stable equilibrium. What can we say about its potential energy ?
Answer:
PE. is minimum.

Question 16.
Is radius of gyration a constant quantity ?
Answer:
No, it changes with the position of axis of rotation.

Question 17.
Two solid spheres of the same mass are made of metals of different densities. Which of them has a large moment of inertia about the diameter?
Answer:
Sphere of smaller density will have larger moment of inertia.

Question 18.
The moment of inertia of two rotating bodies A and B are IA and IB (IA > IB) and their angular momenta are equal. Which one has a greater kinetic energy ?
Answer:
K = \(\frac{\mathrm{L}^{2}}{2 \mathrm{I}}\) ⇒ KA > KA

Question 19.
A particle moves on a circular path with decreasing speed. What happens to its angular momentum?
Answer:
As \(\vec{L}\) = \(\vec{r}\) x m\(\vec{v}\) i.e., \(\vec{L}\) magnitude decreases but direction remains constant.

Question 20.
What is the value of instantaneous speed of the point of contact during pure rolling ?
Answer:
Zero.

Question 21.
Which physical quantity is conserved when a planet revolves around the sun ?
Answer:
Angular momentum of planet.

Question 22.
What is the value of torque on the planet due to the gravitational force of sun ?
Answer:
Zero.

Question 23.
If no external torque acts on a body, will its angular velocity be constant ?
Answer:
No.

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Question 24.
Why there are two propellers in a helicopter ?
Answer:
Due to conservation of angular momentum.

Question 25.
A child sits stationary at one end of a long trolley moving uniformly with speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, then what is the effect of the speed of the centre of mass of the (trolley + child) system ?
Answer:
No change in speed of system as no external force is working.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Short Answer Questions (2 Marks)

Question 26.
State the factors on which the moment of inertia of a body depends.
Answer:

  • Mass of body
  • Size and shape of body
  • Mass distribution w.r.t. axis of rotation
  • Position and orientation of rotational axis

Question 27.
On what factors does radius of gyration of body depend?
Answer:
Mass distribution.

Question 28.
Why the speed of whirl wind in a Tornado is alarmingly high?
Answer:
In this, air from nearly regions get concentrated in a small space, so I decreases considerably. Since Iω = constant so ω increases so high.

Question 29.
Can a body be in equilibrium while in motion? If yes, give an example.
Answer:
Yes, if body has no linear and angular acceleration then a body in uniform straight line of motion will be in equilibrium.

Question 30.
There is a stick half of which is wooden and half is of steel, (i) it is pivoted at the wooden end and a force is applied at the steel end at right angle to its length (ii) it is pivoted at the steel end and the same force is applied at the wooden end. In which case is the angular acceleration more and why?
Answer:
I (first case) > 1 (Second case)
∴ τ r = l α
⇒ α (first case) < α (second case)

SamacheerKalvi.Guru

Question 31.
If earth contracts to half of its present radius what would be the length of the day at equator?
Answer:
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 32.
An internal force cannot change the state of motion of center of mass of a body. Flow does the internal force of the brakes bring a vehicle to rest?
Answer:
In this case the force which bring the vehicle to rest is friction, and it is an external force.

Question 33.
When does a rigid body said to be in equilibrium? State the necessary condition for a body to be in equilibrium.
Answer:
For translation equilibrium
∑ Fext  = 0
For rotational equilibrium
∑ \(\overline{\mathrm{τ}}\)ext  = 0

Question 34.
How will you distinguish between a hard boiled egg and a raw egg by spinning it on a table top’
Answer:
For same external torque, angular acceleration of raw egg will be small than that of Hard boiled egg.

Question 35.
Equal torques are applied on a cylinder and a sphere. Both have same mass and radius. Cylinder rotates about its axis and sphere rotates about one of its diameter. Which will acquire greater speed and why?
Answer:
τ = I α α = \(\frac { τ }{ I }\)
α in cylinder, αC = \(\frac{\tau}{I_{C}}\)
α in sphere, αS = \(\frac{\tau}{I_{S}}\)
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 36.
In which condition a body lying in gravitational field is in stable equilibrium?
Answer:
When vertical line through center of gravity passes through the base of the body.

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Question 37.
Give the physical significance of moment of inertia. Explain the need of fly wheel in Engine.
Answer:
It plays the same role in rotatory motion as the mass does in translator y motion.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Short Answer Questions (3 Marks)

Question 38.
Three mass point m1, m2, m3 are located at the vertices of equilateral A of side ‘a’. What is the moment of inertia of system about an axis along the altitude of A passing through mi?
Answer:
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 39.
A disc rotating about its axis with angular speed ω0 is placed lightly (without any linear push) on a perfectly friction less table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in figure. Will the disc roll?
Answer:

Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
For A VA = R ω0 in forward direction
For B = VB = R ω0 in backward direction R
For C, VC = \(\frac {R}{2}\) ω0 in forward direction disc will not roll.

Question 40.
Find the torque of a force 7\(\hat{i}\) – 3\(\hat{j}\) – 5\(\hat{k}\) about the origin which acts on a particle whose position vector is \(\hat{j}\) +\(\hat{j}\) – \(\hat{j}\)
Answer:
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Numericals

Question 41.
Three masses 3 kg, 4 kg and 5 kg are located at the comers of an equilateral triangle of side 1 m. Locate the center of mass of the system.
Answer:
(x,y) = (0.54 m, 0.36 m)

Question 42.
Two particles mass 100 g and 300 g at a given time have velocities 10\(\hat{j}\) – 7\(\hat{j}\) – 3\(\hat{j}\) and 7\(\hat{i}\) – 9 \(\hat{j}\) + 6\(\hat{k}\) ms-1 respectively. Determine velocity of center of mass.
Answer:
Velocity of center of mass = \(\frac{31 \hat{i}-34 \hat{j}+15 \hat{k}}{2}\) ms-1

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Question 43.
From a uniform disc of radius R, a circular disc of radius R / 2 is cut out. The center of the hole is at R / 2 from the center of original disc. Locate the center of gravity of the resultant flat body.
Answer:
Center of mass of resulting portion lies at R/6 from the center of the original disc in a direction opposite to the center of the cut out portion.

Question 44.
The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds,

  1. What is its angular acceleration (assume the acceleration to be uniform)
  2. How many revolutions does the wheel make during this time ?

Answer:
a = 4π rad s-2
n = 576

Question 45.
A meter stick is balanced on a knife edge at its center. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm, what is the mass of the meter stick?
Answer:
m = 66.0 gm.

Question 46.
A solid sphere is rolling op a friction less plane surface about its axis of symmetry. Find ratio of its rotational energy to its total energy.
Answer:
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 47.
Calculate the ratio of radii of gyration of a circular ring and a disc of the same radius with respect to the axis passing through their centers and perpendicular to their planes.
Answer:
2 : 1

Question 48.
Two discs of moments of inertia I1 and I2 about their respective axes (normal to the disc and passing through the center), and rotating with angular speed col and ω2 are brought into contact face to face with their axes of rotation coincident,

  1. What is the angular speed of the two – disc system ?
  2. Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy ? Take ω1 ≠ ω2.

Answer:

  1. Let co be the angular speed of the two-disc system. Then by conservation of angular momentum
    Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
  2. Initial K.E. of the two discs.
    Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Hence there is a loss of rotational K.E. which appears as heat.
When the two discs are brought together, work is done against friction between the two discs.

Question 49.
In the HCL molecule, the separating between the nuclei of the two atoms is about 1.27 Å (1Å = 10-10m). Find the approximate location of the CM of the molecule, given that the chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in all its nucleus.
Answer:
As shown in Fig. suppose the H nucleus is located at the origin. Then,
x1 = 0, x2 = 1.27 Å, m1 = 1, m2 = 35.5
The position of the CM of HCl molecule is
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Thus the CM of HCl is located on the line joining H and Cl nuclei at a distance of 1.235 Å from the H nucleus.

Question 50.
A child stands at the center of turn table with his two arms out stretched. The turn table is set rotating with an angular speed of 40 rpm. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/3 times the initial value?

  1. Assume that the turn table rotates without friction
  2. Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation.

How do you account for this increase in kinetic energy ?
Answer:
Here ω = 40 rpm, I2 = \(\frac { 1 }{ 2 }\) I1
By the principle of conservation of angular momentum,
I1ω1 = I2ω2 or I1 x 4o = \(\frac {2}{5}\) I1 ω1 or ω2 = 100 rpm
(ii) Initial kinetic energy of rotation –
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
new kinetic energy of rotation –
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Thus the child’s new kinetic energy of rotation is 2.5 times its initial kinetic energy of rotation. This increase in kinetic energy is due to the internal energy of the child which he uses in folding his hands back from the out stretched position.

SamacheerKalvi.Guru

Question 51.
To maintain a rotor at a uniform angular speed of 200 rad s-1 an engine needs to transmit a torque of 180 N m. What is the power required by the engine? Assume that the engine is 100% efficient.
Here ω = 200 rad s-1, τ = 180 N m
Power, P = τω = 180 x 200 = 36,000 W = 36 kW.

Question 52.
A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its center of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front and back wheel.
Answer:
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
For transnational equilibrium of car
NF + NB = W = 1800 x 9.8 = 17640 N
For rotational equilibrium of car
1.05 NF = 0.75 NB
1.05 NF = 0.75(17640 – NF )
1.8 NF = 13230
NF = 13230 / 1.8 = 7350 N
NB = 17640 – 7350 = 10290 N
Force on each front wheel = \(\frac {7350}{ 2 }\) = 3675 N
Force on each back wheel = \(\frac {10290}{ 2 }\) = 5145 N

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Long Answer Questions (5 Marks)

Question 1.
Derive an expression for center of mass for distributed point masses.
Answer:
A point mass is a hypothetical point particle which has nonzero mass and no size or shape. To find the center of mass for a collection of n point masses, say,m1 , m2, m3 ….. mwe have to first choose an origin and an appropriate coordinate system as shown in Figure. Let, x1, x2, x3 …….. xn be the X – coordinates of the positions of these point masses in the X direction from the origin.
The equation for the X coordinate of the center of mass is,
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
where, ∑ mi ¡s the total mass M of all the particles. ( ∑ mi = M).Hence,
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Similarly, we can also find y and z coordinates of the center of mass for these distributed point masses as indicated in figure.
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
Hence, the position of center of mass of these point masses in a Cartesian coordinate system is (xCM, yCM zCM). in general, the position of center of mass can be written in a vector form as,
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
where, is the position vector of the center of mass and \(\vec{r}_{i}\) = xi \(\hat{j}\) + yi\(\hat{j}\) + zi\(\hat{k}\) is the position vector of the distributed point mass; where, \(\hat{i}\), \(\hat{j}\), and \(\hat{j}\) are the unit vectors along X, Y and Z-axis respectively.

Question 2.
Discuss the center of mass of two point masses with pictorial representation.
Answer:
With the equations for center of mass, let us find the center of mass of two point masses m1 and m2, which are at positions x1 and x2 respectively on the X – axis. For this case, we can express the position of center of mass in the following three ways based on the choice of the coordinate system.

SamacheerKalvi.Guru

(1) When the masses are on positive X-axis:
The origin is taken arbitrarily so that the masses m1 and m2 are at positions x1 and x2 on the positive X-axis as shown in figure (a). The center of mass will also be on the positive X- axis at xCM as given by the equation,
\(x_{\mathrm{CM}}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}\)

(2) When the origin coincides with any one of the masses:
The calculation could be minimized if the origin of the coordinate system is made to coincide with any one of the masses as shown in figure (b). When the origin coincides with the point
mass m1, its position x1 is zero, (i.e. x1 = 0). Then,
\(x_{\mathrm{CM}}=\frac{m_{1}(0)+m_{2} x_{2}}{m_{1}+m_{2}}\)
The equation further simplifies as,
xCM = \(\frac{m_{2} x_{2}}{m_{1}+m_{2}}\)

(3) When the origin coincides with the center of mass itself:
If the origin of the coordinate system is made to coincide with the center of mass, then, xCM = O and the mass rn1 is found to be on the negative X- axis as shown in figure (c). Hence, its position x1 is negative, (i.e. – x1).
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
The equation given above is known as principle of moments.
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 3.
Derive an expression for kinetic energy in rotation and establish the relation between rotational kinetic energy and angular momentum.
Answer:
Let us consider a rigid body rotating with angular velocity ω about an axis as shown in figure. Every particle of the body will have the same angular velocity ω and different tangential velocities v based on its positions from the axis of rotation. Let us choose a particle of mass mi situated at distance ri from the axis of rotation. It has a tangential velocity vi given by the relation, vi = ri ω. The kinetic energy KEi. of the particle is,
KEi = \(\frac{1}{2} m_{i} v_{i}^{2}\)
Writing the expression with the angular velocity,
KE = \(\frac{1}{2}\) mi(riω)2 = \(\frac{1}{2} m_{i} r_{i}^{2}\)ω2

For the kinetic energy of the whole body, which is made up of large number of such particles, the equation is written with summation as,
KE = \(\frac{1}{2}\left(\sum m_{i} r_{i}^{2}\right)\)ω2
where, the term ∑ mirir is the moment of inertia I of the whole body. ∑ mirir
Hence, the expression for KE of the rigid body in rotational motion is –
KE = \(\frac{1}{2}\) Iω2
This is analogous to the expression for kinetic energy in transnational motion.
KE = \(\frac{1}{2}\) Mv2

Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Relation between rotational kinetic energy and angular momentum
Let a rigid body of moment of inertia I rotate with angular velocity ω.
The angular momentum of a rigid body is, L = Iω
The rotational kinetic energy of the rigid body is, KE = \(\frac{1}{2}\) Iω2
By multiplying the numerator and denominator of the above equation with I, we get a relation between L and KE as,
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Question 4.
Discuss how the rolling is the combination of transnational and rotational and also be possibilities of velocity of different points in pure rolling.
Answer:
The rolling motion is the most commonly observed motion in daily life. The motion of wheel is an example of rolling motion. Round objects like ring, disc, sphere etc. are most suitable for rolling. Let us study the rolling of a disc on a horizontal surface. Consider a point P on the edge of the disc. While rolling, the point undergoes transnational motion along with its center of mass and rotational motion with respect to its center of mass.

Combination of Translation and Rotation:
We will now see how these transnational and rotational motions arc related in rolling. If the radius of the rolling object is R, in one full rotation, the center of mass is displaced by 2πR (its circumference). One would agree that not only the center of mass. but all the points Of l the disc are displaced by the same 2πR after one full rotation. The only difference is that the center of mass takes a straight path; but, all the other points undergo a path which has a combination of the transnational and rotational motion. Especially the point on the edge undergoes a path of a cyclonic as shown in the figure.
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

As the center of mass takes only a straight line path. its velocity vCM is only transnational velocity vTRANS (vCM = vTRANS). All the other points have two velocities. One is the transnational velocity vTRANS (which is also the velocity of center of mass) and the other is the rotational velocity vROT (vROT = rω). Here, r ¡s the distance of the point from the center of mass and o is the angular velocity. The rotational velocity vROT is perpendicular to the instantaneous position vector from the center of mass as shown in figure (a). The resultant of these two velocities is v. This resultant velocity y is perpendicular to the position vector from the point of contact of the rolling object with the surface on which it is rolling as shown in figure (b).
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

We shall now give importance to the point of contact. In pure rolling, the point of the rolling object which comes in contact with the surface is at momentary rest. This is the case with every point that is on the edge of the rolling object. As the rolling proceeds, all’the points on the edge, one by one come in contact with the surface; remain at momentary rest at the time of contact and then take the path of the cycloid as already mentioned.
Hence, we can consider the pure rolling in two different ways.
(i) The combination of transnational motion and rotational motion about the center of mass.
(or)
(ii) The momentary rotational motion about the point of contact.
As the point of contact is at momentary rest in pure rolling, its resultant velocity v is zero (v = o). For example, in figure, at the point of contact, vTRANS is forward (to right) and vROT is backwards (to the left).
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

That implies that, vTRANS and vROT are equal in magnitude and opposite in direction (v = vTRANS – vROT = 0). Hence, we conclude that in pure rolling, for all the points on the edge, the magnitudes of vTRANS and vROT are equal (vTRANS = vROT) As vTRANS = vCM and vROT = Rω, in pure rolling we have,
vCM = Rω

We should remember the special feature of the above equation. In rotational motion, as per the relation v = rω, the center point will not have any velocity as r is zero. But in rolling motion, it suggests that the center point has a velocity vCM given by above equation vCM – Rω. For the topmost point, the two velocities vTRANS and vROT are equal in magnitude and in the same direction (to the right). Thus, the resultant velocity v is the sum of these two velocities, v = vTRANS + vROT In other form, v = 2 vCM as shown in figure below.

SamacheerKalvi.Guru

Question 5.
Derive an expression for kinetic energy in pure rolling.
Answer:
As pure is the combination of transnational and rotational motion, we can write the total kinetic energy (KE) as the sum of kinetic energy due to transnational motion (KETRANS) and kinetic energy due to rotational motion (KEROT).
KE = KETRANS + KEROT ………(i)
If the mass of the rolling object is M, the velocity of center of mass is vCM, its moment of inertia about center of mass is ICM and angular velocity is ω, then
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies
With center of mass as reference:
The moment of inertia (ICM) of a rolling object about the center of mass is, ICM = MK2 and vCM = Rω. Here, K is radius of gyration.
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

With point of contact as reference:
We can also arrive at the same expression by taking the momentary rotation happening with respect to the point of contact (another approach to rolling). If we take the point of contact as o, then,
KE = \(\frac {1}{2}\) I0ω2

Here, I0 is the moment of inertia of the object about the point of contact. By parallel axis
theorem, I0 = ICM + MK2 Further we can write, I0 MK2 + MR2. With vCM = Rω or ω = \(\frac{v_{\mathrm{CM}}}{\mathrm{R}}\)
Samacheer Kalvi 11th Physics Solution Chapter 5 Motion of System of Particles and Rigid Bodies

As the two equations (v) and (vi) are the same, it ¡s once again confirmed that the pure tolling problems could be solved by considering the motion as any one of the following two cases.
(i) The combination of transnational motion and rotational motion about the center of mass.
(or)
(ii) The momentary rotational motion about the point of contact.

Question 6.

  1. Can a body in translator y motion have angular momentum? Explain.
  2. Why is it more difficult to revolve a stone by tying it to a longer string than by tying it to a shorter string?

Answer:
(1) Yes, a body in translatory motion shall have angular momentum unless fixed point about which angular momentum is taken lies on the line of motion of body
\(|\overrightarrow{\mathrm{L}}|\) = rp sin θ
= 0 only when θ = O° or 180°

(2) MI of stone I = ml2 (l – length of string) l is large, a is very small
τ = Iα
α = \(\frac {τ}{I}\) = \(\frac{\tau}{m l^{2}}\)
if l is large a is very small.
∴ more difficult to revolve.

SamacheerKalvi.Guru

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Samacheer Kalvi 11th Bio Botany Solutions Chapter 7 Cell Cycle

Students can Download Bio Botany Chapter 7 Cell Cycle 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 7 Cell Cycle

Samacheer Kalvi 11th Bio Botany Cell Cycle Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the correct answer

Question 1.
The correct sequence in cell cycle is …………… .
(a) S-M-G1-G2
(b) S-G1-G2-M
(c) G1-S-G2-M
(d) M-G-G2-S
Answer:
(c) G1-S-G2-M

Question 2.
If mitotic division is restricted in G1 phase of the cell cycle then the condition is known as …………… .
(a) S Phase
(b) G2 Phase
(c) M Phase
(d) G0 Phase
Answer:
(d) G0 Phase

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 3.
Anaphase promoting complex APC is a protein degradation machinery necessary for proper mitosis of animal cells. If APC is defective in human cell, which of the following is expected to occur?
(a) Chromosomes will be fragmented
(b) Chromosomes will not condense
(c) Chromosomes will not segregate
(d) Recombination of chromosomes will occur
Answer:
(b) Chromosomes will not condense

Question 4.
In S phase of the cell cycle …………… .
(a) Amount of DNA doubles in each cell
(b) Amount of DNA remains same in each cell
(c) Chromosome number is increased
(d) Amount of DNA is reduced to half in each cell
Answer:
(a) Amount of DNA doubles in each cell

Question 5.
Centromere is required for …………… .
(a) Transcription
(b) Crossing over
(c) Cytoplasmic cleavage
(d) Movement of chromosome towards pole
Answer:
(d) Movement of chromosome towards pole

Question 6.
Synapsis occur between …………… .
(a) mRNA and ribosomes
(b) Spindle fibres and centromeres
(c) Two homologous chromosomes
(d) A male and a female gamete
Answer:
(c) Two homologous chromosomes

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 7.
In meiosis crossing over is initiated at …………… .
(a) Diplotene
(b) Pachytene
(c) Leptotene
(d) Zygotene
Answer:
(b) Pachytene

Question 8.
Colchicine prevents the mitosis of the cells at which of the following stage …………… .
(a) Anaphase
(b) Metaphase
(c) Prophase
(d) Interphase
Answer:
(b) Metaphase

Question 9.
The paring of homologous chromosomes on meiosis is known as …………… .
(a) Bivalent
(b) Synapsis
(c) Disjunction
(d) Synergids
Answer:
(b) Synapsis

Question 10.
Anastral mitosis is the characteristic feature of …………… .
(a) Lower animals
(b) Higher animals
(c) Higher plants
(d) All living organisms
Answer:
(c) Higher plants

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 11.
Write any three significance of mitosis.
Answer:
Exact copy of the parent cell is produced by mitosis (genetically identical).

  1. Genetic stability – daughter cells are genetically identical to parent cells.
  2. Repair of tissues – damaged cells must be replaced by identical new cells by mitosis.
  3. Regeneration – Arms of star fish.

Question 12.
Differentiate between Mitosis and Meiosis.
Answer:
Difference Between Mitosis and Meiosis:

Difference Between Mitosis and Meiosis

Mitosis

Meiosis

1. One division 1. Two divisions
2. Number of chromosomes remains the same 2. Number of chromosomes is halved
3. Homologous chromosomes line up separately on the metaphase plate 3. Homologous chromosomes line up in pairs at the metaphase plate
4. Homologous chromosome do not pair up 4. Homologous chromosome pairup to form bivalent
5. Chiasmata do not form and crossing over never occurs 5. Chiasmata form and crossingover occurs
6. Daughter cells are genetically identical 6. Daughter cells are genetically different from the parent cells
7. Two daughter cells are formed 7. Four daughter cells are formed

Question 13.
Given an account of G0 phase.
Answer:
Some cells exit G1 and enters a quiescent stage called G0, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G0 phase. In G0 cells cease growth with reduced rate of RNA and protein synthesis. The G0 phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G0 .Many cells in animals remains in G0 unless called onto proliferate by appropriate growth factors or other extracellular signals. G0 cells are not dormant.

Question 14.
Differentiate Cytokinesis in plant cells and animal cells.
Answer:
1. Cytokinesis in Plant Cells:
Division of the cytoplasm often starts during telophase. In plants, cytokinesis cell plate grows from centre towards lateral walls centrifugal manner of cell plate formation. Phragmoplast contains microtubules, actin filaments and vesicles from golgi apparatus and ER. The golgi vesicles contains carbohydrates such as pectin, hemicellulose which move along the microtubule of the pharagmoplast to the equator fuse, forming a new plasma membrane and the materials which are placed their becomes new cell wall.

The first stage of cell wall construction is a line dividing the newly forming cells called a cell plate. The cell plate eventually stretches right across the cell forming the middle lamella. Cellulose builds up on each side of the middle lamella to form the cell walls of two new plant cells.

2. Cytokinesis in Animal Cells:
It is a contractile process. The contractile mechanism contained in contractile ring located inside the plasma membrane. The ring consists of a bundle of microfilaments assembled from actin and myosin. This fibril helps for the generation of a contractile force. This force draws the contractile ring inward forming a cleavage furrow in the cell surface dividing the cell into two.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 15.
Write about Pachytene and Diplotene of Prophase I.
Answer:
1. Pachytene: At this stage bivalent chromosomes are clearly visible as tetrads. Bivalent of meiosis I consists of 4 chromatids and 2 centromeres. Synapsis is completed and recombination nodules appear at a site where crossing over takes place between non – sister chromatids of homologous chromosome. Recombination of homologous chromosomes is completed by the end of the stage but the chromosomes are linked at the sites of crossing over. This is mediated by the enzyme recombinase.

2. Diplotene: Synaptonemal complex disassembled and dissolves. The homologous chromosomes remain attached at one or more points where crossing over has taken place. These points of attachment where ‘X’ shaped structures occur at the sites of crossing over is called Chiasmata. Chiasmata are chromatin structures at sites where recombination has been taken place. They are specialised chromosomal structures that hold the homologous chromosomes together.

Sister chromatids remain closely associated whereas the homologous chromosomes tend to separate from each other but are held together by chiasmata. This substage may last for days or years depending on the sex and organism. The chromosomes are very actively transcribed in females as the egg stores up materials for use during embryonic development. In animals, the chromosomes have prominent loops called lampbrush chromosome.

Samacheer Kalvi 11th Bio Botany Cell Cycle Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer:
Question 1.
Most of the neurons in the brain are in …………… stage.
(a) G1
(b) S
(c) G2
(d) G0
Answer:
(d) G0

Question 2.
Short, constricted region in the chromosome is …………… .
(a) Kinetochore
(b) Centromere
(c) Satellite
(d) Telomere
Answer:
(b) Centromere

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 3.
Robert Brown discovered the nucleus in the cells of …………… roots.
(a) Mirabilas
(b) Orchid
(c) Moringa
(d) Oryza
Answer:
(b) Orchid

Question 4.
Scientist who described chromosomes for the first time is …………… .
(a) Robert Brown
(b) Anton van Leeuwenhoek
(c) Boveri
(d) Anton Schneider
Answer:
(d) Anton Schneider

Question 5.
Number of chromosomes in onion cell is …………… .
(a) 8
(b) 16
(c) 32
(d) 64
Answer:
(a) 16

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 6.
Longest part of the cell cycle is …………… .
(a) Prophase
(b) G1 Phase
(c) Interphase
(d) Sphase
Answer:
(c) Interphase

Question 7.
Eukaryotic cells divides every …………… .
(a) 12
(b) 24
(c) 1
(d) 6
Answer:
(b) 24

Question 8.
Cell cycle was discovered by …………… .
(a) Singer & Nicolson
(b) Prevost & Dumans
(c) Schleider & Schwann
(d) Boveri
Answer:
(b) Prevost & Dumans

Question 9.
G0 stage is called as …………… stage.
(a) Quiescent
(b) Metabolically active
(c) Synthesis of DNA
(d) Replication
Answer:
(a) Quiescent

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 10.
…………… protein acts as major check point in phase.
(a) Porins
(b) Kinases
(c) Cyclins
(d) Ligases
Answer:
(c) Cyclins

Question 11.
Replication of DNA occurs at …………… phase.
(a) G0
(b) G1
(c) S
(d) G2
Answer:
(c) S

Question 12.
Condensation of interphase chromosomes into mitotic forms is done by …………… proteins.
(a) MPF
(b) APF
(c) AMF
(d) MAF
Answer:
(a) MPF

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 13.
Which of the following is also called as direct division?
(a) Amitosis
(b) Meiosis
(c) Mitosis
(d) Reduction division
Answer:
(a) Amitosis

Question 14.
Cells of mammalian cartilage undergoes …………… .
(a) Amitosis
(b) Meiosis
(c) Mitosis
(d) Equational division
Answer:
(a) Amitosis

Question 15.
Yeast cells undergo …………… .
(a) Open mitosis
(b) Closed mitosis
(c) Amitosis
(d) Meiosis
Answer:
(b) Closed mitosis

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 16.
…………… is the longest phase in mitosis.
(a) Anaphase
(b) Telophase
(c) Prophase
(d) Interphase
Answer:
(c) Prophase

Question 17.
The DNA protein complex present in the centromere is …………… .
(a) Cyclin
(b) Kinesis
(c) MPF
(d) Kinetochore
Answer:
(d) Kinetochore

Question 18.
…………… protein induces the break down of cohesion proteins leading to chromatid separation during mitosis.
(a) APC
(b) MPF
(c) Cyclin
(d) Kinetochore
Answer:
(a) APC

Question 19.
Regeneration of arms of star fish is due to …………… .
(a) Meiosis
(b) Amitosis
(c) Mitosis
(d) Budding
Answer:
(c) Mitosis

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 20
…………… is called as reduction division.
(a) Meiosis
(b) Mitosis
(c) Amitosis
(d) Budding
Answer:
(a) Meiosis

Question 21.
Bivalents occur at …………… stage.
(a) Zygotene
(b) Leptotene
(c) Pachytene
(d) Diplotene
Answer:
(a) Zygotene

Question 22.
Recombination of chromosomes occur at …………… .
(a) Zygotene
(b) Leptotene
(c) Pachytene
(d) Diplotene
Answer:
(c) Pachytene

Question 23.
Terminalisation of chiasmata occurs at …………… .
(a) Zygotene
(b) Leptotene
(c) Diakinesis
(d) Pachytene
Answer:
(c) Diakinesis

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 24.
Number of daughter cells formed at the end of Meiosis I is …………… .
(a) 2
(b) 4
(c) 1
(d) 0
Answer:
(a) 2

Question 25.
…………… division leads to genetic variability.
(a) Mitotic
(b) Amitotic
(c) Meiotic
(d) Equational
Answer:
(c) Meiotic

Question 26.
Crossing over occurs at …………… stage.
(a) Leptotene
(b) Zygotene
(c) Pachytene
(d) Diplotene
Answer:
(c) Pachytene

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 27.
Which of the following is not a mitogen?
(a) Giberellin
(b) Ethylene
(c) Kinetin
(d) Colchicine
Answer:
(d) Colchicine

Question 28.
In plants mitosis occurs at …………… cells.
(a) Sclerenchyma
(b) Meristem
(c) Xylem
(d) Parenchyma
Answer:
(b) Meristem

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 29.
Which of the following alone is formed in the division of plant cells?
(a) Aster
(b) Centrioles
(c) Spindle
(d) Microtubules
Answer:
(c) Spindle

Question 30.
Amphiastral type cell division is seen in …………… cells.
(a) Fungal
(b) Algal
(c) Plant cells
(d) Animal
Answer:
(d) Animal

II. Very Short Answer Type Questions (2 Marks)

Question 1.
Name the two types of nuclear division.
Answer:
The two types of nuclear division:

  1. Mitosis and
  2. Meiosis.

Question 2.
Define Cell Cycle.
Answer:
A series of events leading to the formation of new cell is known as cell cycle.

Question 3.
Who discovered the Cell Cycle?
Answer:
Prevost & Dumans in 1824.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 4.
Draw a tabular column showing the duration of various phase in the cell cycle of human cell.
Answer:
A tabular column showing the duration of various phase in the cell cycle of human cell:

Cell cycle of a proliferating human cell

Phase

Time Duration (in hrs)

1. G2 1. 11
2. S 2. 8
3. G2 3. 4
4. M 4. 1

Question 5.
Define C – Value.
Answer:
C – Value is the amount in picograms of DNA contained within a haploid nucleus.

Question 6.
Which is the longest phase of cell cycle? What happens during that phase?
Answer:
Interphase is the longest phase. Cells are metabolically active and involved in protein synthesis and growth.

Question 7.
Name the phases which comprises the Interphase.
Answer:
The phases which comprises the Interphase:

  1. G1 Phase
  2. S Phase and
  3. G2 Phase.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 8.
Name the proteins involved in the activation of genes & their proteins to perform cell division.
Answer:
Kinases & Cyclins.

Question 9.
What do you mean by G0 stage?
Answer:
G0 stage is called as quiescent stage, where the cells remain metabolically active without proliferation.

Question 10.
What is the role of MPF in Cell cycle?
Answer:
Maturation Promoting Factor (MPF) brings about condensation of interphase chromosomes into the mitotic form.

Question 11.
Distinguish between Karyokinesis & Cytokinesis.
Answer:
Between Karyokinesis & Cytokinesis:

  • Karyokinesis: Karyokinesis refers to the nuclear division.
  • Cytokinesis: Cytokinesis refers to the cytoplasmic division.

Question 12.
Point out any two cell – types which remain G0 phase.
Answer:
Mature neurons and Skeletal muscle cells.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 13.
Why amitosis is called as incipient cell division?
Answer:
Amitosis is also called incipient cell division. Since there is no spindle formation and chromatin material does not condense.

Question 14.
List out the disadvantages of Amitosis.
Answer:
The disadvantages of Amitosis:

  • Causes unequal distribution of chromosomes.
  • Can lead to abnormalities in metabolism and reproduction.

Question 15.
Mitosis also called as equational division – Justify.
Answer:
At the end of mitosis the number of chromosomes in the parent and the daughter (Progeny) cells remain the same so it is also called as equational division.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 16.
Enumerate the stages of mitosis.
Answer:
Mitosis is divided into four stages prophase, metaphase, anaphase and telophase.

Question 17.
Define an aster.
Answer:
In animal cell the centrioles extend a radial array of microtubules towards the plasma membrane when they reach the poles of the cell. This arrangement of microtubules is called an aster. Plant cells do not form asters.

Question 18.
What is metaphase plate?
Answer:
The alignment of chromosome into compact group at the equator of the cell is known as metaphase plate.

Question 19.
What is Kinetochore?
Answer:
Kinetochore is a DNA – Protein complex present in the centromere DNA, where the microtubules are attached. It is a trilaminar disc like plate.

Question 20.
How will you calculate the length of the S period.
Answer:
Length of the S period = Fraction of cells in DNA replication × generation time.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 21.
Which type of cell division occurs in reproductive cells? What will be the result?
Answer:
Meiosis takes place in the reproductive organs. It results in the formation of gametes with half the normal chromosome number.

Question 22.
Define Synapsis.
Answer:
In Zygotene, pairing of homologous chromosomes takes place and it is known as synapsis.

Question 23.
What do you understand by independent assortment?
Answer:
The random distribution of homologous chromosomes in a cell in Metaphase I is called independent assortment.

Question 24.
Define Mitogen. Give an example.
Answer:
The factors which promote cell cycle proliferation is called mitogen.
Example: gibberellin. These increase mitotic rate.

Question 25.
What are mitotic poisons.
Answer:
Certain chemical components act as inhibitors of the mitotic cell division and they are called mitotic poisons.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 26.
Distinguish between Anastral & Amphiastral.
Answer:
Between Anastral & Amphiastral:
Anastral:

  1. This is present only in plant cells.
  2. No asters or centrioles are formed only spindle fibres are formed during cell division.

Amphiastral:

  1. This is found in animal cells.
  2. Aster and centrioles are formed at each pole of the spindle during cell division.

Question 27.
Draw a simple diagram to show the Amitosis.
Answer:
The Amitosis:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 7 Cell Cycle 1

III. Short Answer Type Questions (3 Marks)

Question 1.
What is the role of nucleus in the cell?
Answer:
The role of nucleus in the cell:

  • Control activities of the cell.
  • Genetic information copied from cell to cell while the cell divides.
  • Hereditary characters are passed onto new individuals when gametic cells fuse together in sexual reproduction.

Question 2.
What are restriction points? Mention its role in Cell cycle.
Answer:
The checkpoint called the restriction point at the end of G1 it determines a cells fate whether it will continue in the cell cycle and divide or enter a stage called G0 as a quiescent stage and probably as specified cell or die.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 3.
Point out the reasons responsible for the arresting of the cell in G1 phase?
Answer:
Cells are arrested in G1 due to:

  • Nutrient deprivation
  • Lack of growth factors or density dependant inhibition
  • Undergo metabolic changes and enter into G0 state.

Question 4.
Write a note on G0 phase.
Answer:
Some cells exit G1 and enters a quiescent stage called G0, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G0 phase. In G0 cells cease growth with reduced rate of RN A and protein synthesis. The G0 phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G0. Many cells in animals remains in G0 unless called onto proliferate by appropriate growth factors or other extracellular signals. G0 cells are not dormant.

Question 5.
List out the events taking place in S – Phase.
Answer:
S Phase – Synthesis phase – cells with intermediate amounts of DNA Growth of the cell continues as replication of DNA occur, protein molecules called histones are synthesised and attach to the DNA. The centrioles duplicate in the cytoplasm. DNA content increases from 2C to 4C.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 6.
Distinguish between Karyokinesis & Cytokinesis.
Answer:
Karyokinesis:

  1. Involves division of nucleus.
  2. Nucleus develops a constriction at the center and becomes dumbellshaped.
  3. Constriction deepens and divides the nucleus into two.

Cytokinesis:

  1. Involves division of cytoplasm.
  2. Plasma membrane develops a constriction along nuclear constriction.
  3. It deepens centripetally and finally divides the cell into two cells.

Question 7.
Explain the differences between closed and open mitosis.
Answer:
Between closed and open mitosis:

  1. In closed mitosis, the nuclear envelope remains intact and chromosomes migrate to opposite poles of a spindle within the nucleus. Example: Yeast and slime molds.
  2. In open mitosis, the nuclear envelope breaks down and then reforms around the 2 sets of separated chromosome. Example: Most plants and animals cells.

Question 8.
What happens to plant cells at the end of Telophase in Mitosis?
Answer:
In plants, phragmoplast are formed between the daughter cells. Cell plate is formed between the two daughter cells, reconstruction of cell wall takes place. Finally the cells are separated by the distribution of organelles, macromolecules into two newly formed daughter cells.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 9.
Bring out the significance of Meiosis.
Answer:
The significance of Meiosis:

  • Meiosis maintains a definite constant number of chromosomes in organisms.
  • Crossing over takes place and exchange of genetic material leads to variations among species. These variations are the raw materials to evolution. Meiosis leads to genetic variability by partitioning different combinations of genes into gametes through independent assortment.
  • Adaptation of organisms to various environmental stress.

Question 10.
Differentiate between the mitosis of Plant Cell & Animal Cell.
Answer:
Plants:

  1. Centrioles are absent
  2. Asters are not formed
  3. Cell division involves formation of a cell plate
  4. Occurs mainly at meristem.

Animals:

  1. Centrioles are present
  2. Asters are formed
  3. Cell division involves furrowing and cleavage of cytoplasm
  4. Occurs in tissues throughout the body.

Question 11.
Explain briefly about Endomitosis.
Answer:
The replication of chromosomes in the absence of nuclear division and cytoplasmic division resulting in numerous copies within each cell is called endomitosis. Chromonema do not separate to form chromosomes, but remain closely associated with each other. Nuclear membrane does not rupture. So no spindle formation. It occurs notably in the salivary glands of Drosophila and other flies. Cells in these tissues contain giant chromosomes (polyteny), each consisting of over thousands of intimately associated, or synapsed, chromatids. Example: Polytene chromosome.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 12.
How G0 cells help in Closing Technology?
Answer:
Since the DNA of cells in G0, do not replicate. The researcher are able to fuse the donor cells from a sheep’s mammary glands into G0, state by culturing in the nutrient free state. The G0, donor nucleus synchronised with cytoplasm of the recipient egg, which developed into the clone Dolly.

IV. Long Answer Type Questions (5 Marks)

Question 1.
Draw and label the various stages of Prophase I.
Answer:
Label the various stages of Prophase I:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 7 Cell Cycle 2

Question 2.
Explain in detail about the various stages of Prophase I.
Answer:
The various stages of Prophase I:
1. Prophase I – Prophase I is of longer duration and it is divided into 5 substages – Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis.

2. Leptotene – Chromosomes are visible under light microscope. Condensation of chromosomes takes place. Paired sister chromatids begin to condense.

3. Zygotene – Pairing of homologous chromosomes takes place and it is known as synapsis. Chromosome synapsis is made by the formation of synaptonemal complex. The complex formed by the homologous chromosomes are called as bivalent (tetrads).

4. Pachytene – At this stage bivalent chromosomes are clearly visible as tetrads. Bivalent of meiosis I consists of 4 chromatids and 2 centromeres. Synapsis is completed and recombination nodules appear at a site where crossing over takes place between non – sister chromatids of homologous chromosome. Recombination of homologous chromosomes is completed by the end of the stage but the chromosomes are linked at the sites of crossing over. This is mediated by the enzyme recombinase.

5. Diplotene – Synaptonemal complex disassembled and dissolves. The homologous chromosomes remain attached at one or more points where crossing over has taken place. These points of attachment where ‘X’ shaped structures occur at the sites of crossing over is called.

6. Chiasmata: Chiasmata are chromatin structures at sites where recombination has been taken place. They are specialised chromosomal structures that hold the homologous chromosomes together. Sister chromatids remain closely associated whereas the homologous chromosomes tend to separate from each other but are held together by chiasmata. This substage may last for days or years depending on the sex and organism. The chromosomes are very actively transcribed in females as the egg stores up materials for use during embryonic development. In animals, the chromosomes have prominent loops called lampbrush chromosome.

7. Diakinesis – Terminalisation of chiasmata. Spindle fibres assemble. Nuclear envelope breaks down. Homologous chromosomes become short and condensed. Nucleolus disappears.

Question 3.
Describe the process of Cytokinesis in Plant cell & Animal Cell.
Answer:
1. Cytokinesis in Plant Cell: Division of the cytoplasm often starts during telophase. In plants, cytokinesis cell plate grows from centre towards lateral walls – centrifugal manner of cell plate formation. Phragmoplast contains microtubules, actin filaments and vesicles from golgi apparatus and ER. The golgi vesicles contains carbohydrates such as pectin, hemicellulose which move along the microtubule of the pharagmoplast to the equator fuse, forming a new plasma membrane and the materials which are placed there becomes new cell wall.

The first stage of cell wall construction is a line dividing the newly forming cells called a cell plate. The cell plate eventually stretches right across the cell forming the middle lamella. Cellulose builds up on each side of the middle lamella to form the cell walls of two new plant cells.

2. Cytokinesis in Animal Cells:
It is a contractile process. The contractile mechanism contained in contractile ring located inside the plasma membrane. The ring consists of a bundle of microfilaments assembled from actin and myosin. This fibril helps for the generation of a contractile force. This force draws the contractile ring inward forming a cleavage furrow in the cell surface dividing the cell into two.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 4.
What are the significances of Mitosis.
Answer:
Exact copy of the parent cell is produced by mitosis (genetically identical).

  1. Genetic stability – Daughter cells are genetically identical to parent cells.
  2. Growth – As multicellular organisms grow, the number of cells making up their tissue increases. The new cells must be identical to the existing ones.
  3. Repair of tissues – Damaged cells must be replaced by identical new cells by mitosis.
  4. Asexual reproduction – Asexual reproduction results in offspring that are identical to the parent. Example Yeast and Amoeba.
  5. In flowering plants, structure such as bulbs, corms, tubers, rhizomes and runners are produced by mitotic division. When they separate from the parent, they form a new individual. The production of large numbers of offsprings in a short period of time, is possible only by mitosis. In genetic engineering and biotechnology, tissues are grown by mitosis (i.e. in tissue culture).
  6. Regeneration – Arms of star fish

Question 5.
Explain the various phases in Cell Cycle.
Answer:
The different phases of cell cycle are as follows:
1. Interphase: Longest part of the cell cycle, but it is of extremely variable length. At first glance the nucleus appears to be resting but this is not the case at all. The chromosomes previously visible as thread like structure, have dispersed. Now they are actively involved in protein synthesis, at least for most of the interphase. C – Value is the amount in picograms of DNA contained within a haploid nucleus.

2. G1 Phase: The first gap phase – 2C amount of DNA in cells of G1 The cells become metabolically active and grows by producing proteins, lipids, carbohydrates and cell organelles including mitochondria and endoplasmic reticulum. Many checkpoints control the cell cycle. The checkpoint called the restriction point at the end of G1 it determines a cells fate whether it will continue in the cell cycle and divide or enter a stage called G0 as a quiescent stage and probably as specified cell or die. Cells are arrested in G1 due to:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 7 Cell Cycle 3
3. Nutrient deprivation: Lack of growth factors or density dependant inhibition. Undergo metabolic changes and enter into G0 state. Biochemicals inside cells activates the cell division. The proteins called kinases and cyclins activate genes and their proteins to perform cell division. Cyclins act as major checkpoint which operates in G1 to determine whether or not a cell divides.

4. G0 Phase: Some cells exit G1 and enters a quiescent stage called G0, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G0 phase. In G0 cells cease growth with reduced rate of RNA and protein synthesis. The G0 phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G0. Many cells in animals remains in G0 unless called on to proliferate by appropriate growth factors or other extracellular signals. G0 cells are not dormant.

5. S phase – Synthesis phase – cells with intermediate amounts of DNA. Growth of the cell continues as replication of DNA occur, protein molecules called histones are synthesised and attached to the DNA. The centrioles duplicate in the cytoplasm. DNA content increases from 2C to 4C.

6. G2 – The second Gap phase – 4C amount of DNA in cells of G2 and mitosis. Cell growth continues by protein and cell organelle synthesis, mitochondria and chloroplasts divide. DNA content remains as 4C. Tubulin is synthesised and microtubules are formed. Microtubles organise to form spindle fibre. The spindle begins to form and nuclear division follows.

One of the proteins synthesized only in the G2 period is known as Maturation Promoting Factor (MPF). It brings about condensation of interphase chromosomes into the mitotic form. DNA damage checkpoints operates in G1 S and G2 phases of the cell cycle.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 6.
List out the important features of Chromosomes.
Answer:
The four important features of the chromosome are:
1. The shape of the chromosome is specific: The long, thin, lengthy structured chromosome contains a short, constricted region called centromere. A centromere may occur any where along the chromosome, but it is always in the same position on any given chromosome. The number of chromosomes per species is fixed: For example the mouse has 40 chromosomes, the onion has 16 and humans have 46.

2. Chromosomes occur in pairs: The chromosomes of a cell occur in pairs, called homologous pairs. One of each pair come originally from each parent. Example, human has 46 chromosomes, 23 coming originally from each parent in the process of sexual reproduction. Chromosomes are copied: Between nuclear divisions, whilst the chromosomes are uncoiled and cannot be seen, each chromosome is copied. The two identical structures formed are called chromatids.

V. Higher Order Thinking Skills (HOTs)

Question 1.
Given that the average duplication time of E.coli is 20 minutes. How much time will two E.coli cells takes to become 32 cells?
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 7 Cell Cycle 4
One cells takes 80 minutes to form 16 cells. If 2 cells undergoes division simultaneously, it take 160 minutes (2 hours 40 minutes) to form 32 cells.

Question 2.
Complete the cell cycle by filling the gaps with respective phases.
Answer:
X= S phase or Synthesis phase
Y= M phase or Mitosis phase
Z= G0 phase
Samacheer Kalvi 11th Bio Botany Solutions Chapter 7 Cell Cycle 5

Question 3.
Telophase is reverse of prophase – Comment
Answer:
Events in Prophase:

  1. Nuclear membrane disappears
  2. Nucleolus disappear
  3. Spindle fibre begins to form
  4. Chromosomes threads condeme to form chromosomes

Events in Telophase:

  1. Nuclear membrane reappears
  2. Nucleolus reappears
  3. Spindle fibre disappears
  4. Chromosomes decondeme to form chromosomes

Question 4.
Name the pathological condition when uncontrolled cell division occurs.
Answer:
Uncontrolled cell division & abnormal growth of cells leads to the pathological condition called tumor or cancer.

Question 5.
Microspores are produced in the multiples of four, why?
Answer:
Microspores are haploid spores produced from diploid microspores mother cells. Each microspores mother cell (2n) undergoes meiosis producing four Microspores (n). Because a complete meiotic division yields 4 cells. Thus microspores are produced in multiples of four.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 6.
Between Prokaryotes & Eukaryotes, which cell has a shorter cell division time.
Answer:
Prokaryotes like bacteria undergo simple form of cell division called binary fission which will get completed with in a hour, whereas Eukaryotic cell division (mitosis) takes nearly 24 hours to get completed. Hence Prokaryotes have shorter cell division time.

Question 7.
Though Prokaryotic cell division differs from Eukaryotic cell division, both show certain common aspects during cell division. Explain.
Answer:
Whether a cell is Prokaryote or Eukaryote, while undergoing division, the following events must occur in common.

  1. Replication of DNA.
  2. Cytokinesis at the end of cell division.

Question 8.
An anther has 1204 pollen grains. How many Pollen mother cells must have been there to produce them?Explain.
Answer:
301 – Pollen mother cells: 301 Pollen mother cells undergo meiosis producing 1204 pollen grains. Because at the end of meiosis, each pollen mother cells produces 4 pollen grains.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 9.
A cell has 32 chromosomes. It undergoes mitosis. What will be the chromosome number during metaphase?
Answer:
During S phase of interphase, the genetic material of the cell is duplicated. So during metaphase, the chromosome number(chromatid number) will be doubled thus 64 chromosomes (chromatids) will be present.

Question 10.
Why sibilings show disimilarities?
Answer:
Though born to same parents, siblings show dissimilarities and variation due to the crossing over and recombination of chromosomes during meiosis.

Question 11.
Ramu’s met with an accident while riding cycle and got wounded in his leg. After few days, the wound was healed and the skin becomes normal. How?
Answer:
Ramu’s wound was healed because of the mitotic division. As a result of mitosis, new cells are produced and damaged tissues were repaired resulting the damaged skin to become normal.

Question 12.
A flower of tomato plant following the process of sexual reproduction produces 240 viable seeds. What is the minimum number of microspore mother cells involved in this process?
Answer:
60 microspore mother cells are involved in providing 240 pollen grains. Because each microspore mother cell undergoes meiosis producing four pollen grains (i.e. 60 × 4 = 240). Each pollen grain produces two male gametes of which one undergoes true fertilization of ovule producing seeds. Other male gamete participate in double fertilization.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 7 Cell Cycle Read More »

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5

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 2.5

Question 1.
Solve 2x2 + x – 15 ≤ 0.
Solution:
To find the solution of the inequality
ax2 + bx + c ≥ 0 or ax2 + bx +c ≤ 0 (for a > 0)
First we have to solve the quadratic equation ax2 + bx + c = 0
Let the roots be a and P (where a < P)
So for the inequality ax2 + bx + c ≥ 0 the roots lie outside α and β
(i.e.,) x ≤ α and x ≥ β
So for the inequality ax2 + bx + c ≤ 0. The roots lie between α and β
(i.e.,) x > α and x < β (i.e.) a ≤ x ≤ β
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 1

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5

The inequality solver will then show you the steps to help you learn how to solve it on your own.

Question 2.
Solve -x2 + 3x – 2 ≥ 0
Solution:
-x2 + 3x – 2 ≥ 0 ⇒ x2 – 3x + 2 ≤ 0
(x – 1) (x – 2) ≤ 0
[(x – 1) (x – 2) = 0
⇒ x = 1 or 2.
Here α = 1 and β = 2. Note that α < β]
So for the inequality (x – 1) (x – 2) ≤ 2
x lies between 1 and 2
(i.e.) x ≥ 1 and x ≤ 2 or x ∈ [1, 2] or 1 ≤ x ≤ 2

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 Additional Questions

Question 1.
Solve for x.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 5
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 6
Select the intervals in which (3x +1) (3x – 2) is positive
(3x + 1) > 0 and (3x – 2) > 0 or
3x +1 < 0 and 3x – 2 < 0
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 7

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5

Question 2.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 25
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 26

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5

Question 3.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 8
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 9

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