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Students can Download Accountancy Chapter 10 Depreciation Accounting Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Accountancy 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 Accountancy Solutions Chapter 10 Depreciation Accounting

Samacheer Kalvi 11th Accountancy Depreciation Accounting Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

Question 1.
Under straight line method, the amount of depreciation is ……………..
(a) Incfeasing every year
(b) Decreasing every year
(c) Constant for all the years
(d) Fluctuating every year
Answer:
(c) Constant for ail the years

Question 2.
If the total charge of depreciation and maintenance cost are considered, the method that provides a uniform charge is ……………..
(a) Straight line method
(b) Diminishing balance method
(c) Annuity method
(d) Insurance policy method
Answer:
(b) Diminishing balance method

Question 3.
Under the written down value method of depreciation, the amount of depreciation is ……………..
(a) Uniform in all the years
(b) Decreasing every year
(c) Increasing every year
(d) None of the above
Answer:
(b) Decreasing every year

Question 4.
Depreciation provided on machinery is debited to ……………..
(a) Depreciation account
(b) Machinery account
(c) Trading account
(d) Provision for depreciation account
Answer:
(a) Depreciation account

Question 5.
Cash received from sale of fixed asset is credited to ……………..
(a) Profit and loss account
(b) Fixed asset account
(c) Depreciation account
(d) Bank account
Answer:
(b) Fixed asset account

Question 6.
Depreciation is provided on ……………….
(a) Fixed assets
(b) Current assets
(c) Outstanding charges
(d) All assets
Answer:
(a) Fixed assets

Question 7.
Depreciation is caused by ……………..
(a) Lapse of time
(b) Usage
(c) Obsolescence
(d) a, b and c
Answer:
(d) a, b and c

Question 8.
Depreciation is the process of ……………..
(a) Allocation of cost of the asset to the period of its useful life
(b) Valuation of assets
(c) Maintenance of an asset in a state of efficiency
(d) Adding value to the asset
Answer:
(a) Allocation of cost of the asset to the period of its useful life

Question 9.
For which of the following assets, the depletion method is adopted for writing off cost of the asset?
(a) Plant and machinery
(b) Mines and quarries
(c) Buildings
(d) Trademark
Answer:
(b) Mines and quarries

Question 10.
A depreciable asset may suffer obsolescence due to ……………..
(a) Passage of time
(b) Wear and tear
(c) Technological changes
(d) None of the above
Answer:
(c) Technological changes

Question 11.
Which method shall be efficient, if repairs and maintenance cost of an asset increases as it grows older?
(a) Straight line method
(b) Reducing balance method
(c) Sinking fund method
(d) Annuity method
Answer:
(b) Reducing balance method

Question 12.
Depreciation is to be calculated from the date when ……………..
(a) Asset is put to use
(b) Purchase order is made
(c) Asset is received at business premises
(d) Invoice of assets is received
Answer:
(a) Asset is put to use

Question 13.
If the rate of depreciation is same, then the amount of depreciation under straight line method vis – a – vis written down value method will be
(a) Equal in all years
(b) Equal in the first year but higher in subsequent years
(c) Equal in the first year but lower in subsequent years
(d) Lower in the first year but equal in subsequent years
Answer:
(b) Equal in the first year but higher in subsequent years

Question 14.
Residual value of an asset means the amount that it can fetch on sale at the of its useful life.
(a) Beginning
(b) End
(c) Middle
(d) None
Answer:
(b) End

II. Very Short Answer Questions

Question 1.
What is meant by depreciation?
Answer:
The process of allocation of the relevant cost of a fixed asset over its useful life is known as depreciation. It is an allocation of cost against the benefits derived from a fixed asset during an accounting period.

Question 2.
List out the various methods of depreciation.
Answer:

1. Straight line method or fixed instalment method or Original cost method.
2. Written down value method or Diminishing balance method or Reducing balance method.
3. Sum of years digits method.
4. Machine hour rate method.
5. Depletion method.
6. Annuity method.
7. Revaluation method.
8. Sinking fund method.
9. Insurance Policy method.

Question 3.
Give the formula to find out the amount and rate of depreciation under straight line method of depreciation.
Answer:
1. Amount of depreciation per year =
2. Rate of depreciation =

Question 4.
What is annuity method?
Answer:
Under this method, not only the original cost of the asset but also the amount of interest on the investment is taken into account while computing depreciation. The idea of considering interest is that if the investment is made in any other asset instead of the relevant fixed asset, it . would have earned a certain rate of interest. To calculate the amount of depreciation, annuity factor is used. Annuity factor can be found out from the annuity table or by using formula. Amount of depreciation is computed as follows:
Amount of depreciation = Annuity factor x original cost of the asset.

Question 5.
What is sinking fund method?
Answer:
This method is adopted especially when it is desired not merely to write off an asset but also to provide enough funds to replace an asset at the end of its working life. Under this method, the amount charged as depreciation is transferred to depreciation fund and invested outside the business. The investment is made in safe securities which offer a certain rate of interest. Interest is received annually and reinvested every year along with the amount of annual depreciation. On the expiry of the life of the asset, the investments are sold and the sale proceeds are used for replacement of the asset. This method of depreciation is suitable for assets of higher value. This method is also known as depreciation fund method.

III. Short Answer Questions

Question 1.
What are the objectives of providing depreciation?
Answer:

1. To find out the true profit or loss
2. To present the true and fair view of financial position
3. To facilitate replacement of fixed assets
4. To avail tax benefits
5. To comply with legal requirements

Question 2.
What are the causes for depreciation?
Answer:

1. Wear and tear
2. Efflux of time
3. Obsolescence
4. Inadequacy for the purpose
5. Lack of maintenance
6. Abnormal factors

Question 3.
State the advantages and limitations of straight line method of depreciation.
Answer:
Advantages:

• Simple and easy to understand
• Equality of depreciation burden
• Assets can be completely written off
• Suitable for the assets having fixed working life

Limitations:

• Ignores the actual use of the asset
• Ignores the interest factor
• Total charge on the assets will be more when the asset becomes older
• Difficulty in the determination of scrap value

Question 4.
State the advantages and limitations of written down value method of depreciation.
Answer:
Advantages:

• Equal charge against income
• Logical method

Limitations:

• Assets cannot be completely written off
• Ignores the interest factor
• Difficulty in determining the rate of depreciation
• Ignores the actual use of the asset

Question 5.
Distinguish between straight line method and written down value method of providing depreciation.
Answer:

IV. Exercises

Straight line method:

Question 1.
A firm purchased a plant for ₹ 40,000. Erection charges amounted to ₹ 2,000. Effective life of the plant is 5 years. Calculate the amount of depreciation per year under straight line method.
Answer:
Calculation of amount of depreciation
Amount of depreciation =
Original cost = Purchase of plant + Erection charges = ₹ 40,000 + ₹ 2,000 = ₹ 42,000
Estimated life = 5 years = \(\frac{₹ 42000-0}{5 \text { years }}\) = ₹ 8,400/-

Question 2.
A company purchased a building for ₹ 50,000. The useful life of the building is 10 years and the residual value is ₹ 2,000. Find out the amount and rate of depreciation under straight line method.
Answer:
(1) Calculation of amount of depreciation:
Amount of depreciation =
Original cost = ₹ 50,000
residual value = ₹ 2,000
Estimated life = 10 years
\(\frac{50,000-2,000}{10 \text { years }}\) = \(\frac { 48,000 }{ 10 }\) = ₹ 4,800/-

(2) Calculation of rate of depreciation:
rate of depreciation = = \(\frac { 48,000 }{ 50,000 }\) x 100 = 9.6%

Question 3.
Furniture was purchased for ₹ 60,000 on 1-7-2016. It is expected to last for 5 years. Estimated scrap at the end of five years is ₹ 4,000. Find out the rate of depreciation under straight line method.
Answer:
(1) Amount of depreciation =
Original cost = ₹ 60,000
Scrap value = ₹ 4,000
Estimated life = 5 years
= \(\frac{60,000-4,000}{5 \text { years }}\) = \(\frac { 56,000 }{ 5 }\) = ₹ 11,200/-

Question 4.
Calculate the rate of depreciation under straight line method from the following information: Purchased a second hand machinery on 1.1.2018 for ₹ 38,000 On 1.1.2018 spent ₹ 12,000 on its repairs
Expected useful life of machine is 4 years
Estimated residual value ₹ 6,000
Answer:
Original cost – residual value
(1) Calculation of amount of depreciation =
Original cost = Purchase of machinery + repair charges = 38,000 + 12,000 = 50,000
Residual value = 6,000
Estimated life = 4 years = \(\frac{50,000-6,000}{4 \text { years }}\) = ₹ 11,000/-
(2) Rate of depreciation =

= \(\frac { 11,000 }{ 50,000 }\) x 100 = 22 %

Question 5.
Calculate the rate of depreciation under straight line method.
Purchase price of a machine ₹ 80,000
Expenses to be capitalised ₹ 20,000
Estimated residual value ₹ 4,000
Expected useful life ₹ 4 years
Answer:
Original cost – residual value =
Original cost = Machine purchased + capitalised expenses
80,000 + 20,000 = 1,00,000
Residual value = 4,000
Estimated life = 4 years
= \(\frac{1,00,000-4,000}{4 \text { years }}\) = \(\frac { 96,000 }{ 4 }\) = ₹ 24,000/-

Question 6.
Machinery was purchased on 1^st January 2015 for ₹ 4,00,000. ₹ 15,000 was spent on its erection and ₹ 10,000 on its freight charges. Depreciation is charged at 10% per annum on straight line method. The books are closed on 31^st March each year. Calculate the amount of depreciation on machinery for the first two years.
Answer:
Calculation of depreciation:
Original cost = Machinery purchased + erection charges + freight charges
= 4,00,000 + 15,000 + 10 ,000
= 4,25,000

First year depreciation = ₹ 10,625
Second year depreciation = ₹ 42,500

Question 7.
An asset is purchased on 1.1.2016 for ₹ 25,000. Depreciation is to be provided annually according to straight line method. The useful life of the asset is 10 years and its residual value is ₹ 1,000. Accounts are closed on 31^st December every year. You are required to find out the rate of depreciation and give journal entries for first two years.
Answer:
Amount of depreciation =
Original cost = ₹ 25,000
Residual value = ₹ 1,000
Estimated life = 10 years
= \(\frac{25,000-1,000}{10 \text { years }}\) = \(\frac { 24,000 }{ 10 }\) = ₹ 24,000/-

(2) Rate of depreciation =
= \(\frac { 2,400 }{ 25,000 }\) x 100 = 9.6%
Journal Entries for first two years

Question 8.
From the following particulars, give journal entries for 2 years and prepare machinery account under straight line method of providing depreciation:
Machinery was purchased on 1.1.2016
Price of the machine ₹ 36,000
Freight charges ₹ 2,500
Installation charges ₹ 1,500
Life of the machine 5 years
Answer:
Calculation of Asset of depreciation:
Amount of depreciation =
Original cost = Price of the machine + Freight charges + Installation charges
= 36,000 + 2,500 + 1,500 = ₹ 40,000
= \(\frac { 40,000 – 0 }{ 5 years }\) = ₹ 8,000
Journal Entries for 2 years

Machinery

Question 9.
A manufacturing company purchased on 1^st April, 2010, a plant and machinery for ₹ 4,50,000 and spent ₹ 50,000 on its installation. After having used it for three years, it was sold for ₹ 3,85,000. Depreciation is to be provided every year at the rate of 15% per annum on the fixed instalment method. Accounts are closed on 31^st March every year. Calculate profit or loss on sale of machinery.
Answer:
Calculation of Profit or Loss on sale of Machinery

Note: If the selling price is more than the book value is called profit.
Selling price – Book value = Profit 3,85,000 – 2,75,000 = 1,10,000
Profit on sale of Machinery is = ₹ 1,10,000.

Question 10.
On 1^st April 2008, Sudha and Company purchased machinery for ₹ 64,000. To instal the machinery expenses incurred was ₹ 28,000. Depreciate machinery 10% p.a. under straight line method. On 30^th June, 2010 the worn out machinery was sold for ₹ 52,000. The books are closed on 31^st December every year. Show machinery account.
Answer:
Workings:

If Book value is more than the selling price it is called loss.
Book value – selling price = loss
71,300 – 52,000 = 19,300
Loss on sale of machinery is = 19,300
Machinery

Question 11.
Ragul purchased machinery on April 1, 2014 for ₹ 2,00,000. On 1^st October 2015, a new machine costing ₹ 1,20,000 was purchased. On 30^th September 2016, the machinery purchased on April 1, 2014 was sold for ₹ 1,20,000. Books of accounts are closed on 31^st March and depreciation is to be provided at 10% p.a. on straight line method. Prepare machinery account and depreciation account for the years 2014 – 15 to 2016 – 17.
Answer:
Workings

Machinery

Written down value method

Question 12.
An asset is purchased for ₹ 50,000. The rate of depreciation is 15% p.a. Calculate the annual depreciation for the first two years under diminishing balance method.
Answer:
Workings: Calculation of depreciation of Machinery

Question 13.
A boiler was purchased on 1st January 2015 from abroad for ₹ 10,000. Shipping and forwarding charges amounted to ₹ 2,000. Import duty ₹ 7,000 and expenses of installation amounted to ₹ 1,000. Calculate depreciation for the first 3 years @10% p.a. on diminishing balance method assuming that the accounts are closed 31^st December each year.
Answer:
Calculation of amount of depreciation on diminishing balance method:

2015 Depreciation ₹ 2,000
2016 Depreciation ₹ 1,800
2017 Depreciation ₹ 1,620

Question 14.
A furniture costing ₹ 5,000 was purchased on 1.1.2016, the installation charges being ₹ 1,000. The furniture is to be depreciated @ 10% p.a. on the diminishing balance method. Pass journal entries for the first two years.
Answer:

Journal Entries

Question 15.
A firm acquired a machine on 1^st April 2015 at a cost of ₹ 50,000. Its life is 6 years. The firm writes off depreciation @ 30% p.a. on the diminishing balance method. The firm closes its books on 31^st December every year. Show the machinery account and depreciation account for three years starting from 1^st April 2015.
Answer:
Workings:

Machinery

Depreciation A/c

Question 16.
A firm purchased a machine for ₹ 1,00,000 on 1-7-2015. Depreciation is written off at 20% on reducing balance method. The firm closes its books on 31^st December each year. Show the machinery account upto 31-12-2017.
Answer:
Workings:

MachineryA/c

Question 17.
On 1^st October 2014, a truck was purchased for ₹ 8,00,000 by Laxmi Transports Ltd. Depreciation was provided @ 15% p.a. under diminishing balance method. On 31^st March 2017, the above truck was sold for ₹ 5,00,000. Accounts are closed on 31^st March every year. Find out the profit or loss made on the sale of the truck.
Answer:
Calculation of Profit or loss on sale of truck:

Note: If Book value is more than the selling price it is called loss:
Book value – selling price = Loss
5,34,650 – 5,00,000 = 34,650
∴ Loss on sale of truck = ₹ 34,650

Question 18.
On 1^st January 2015, a second hand machine was purchased for ₹ 58,000 and ₹ 2,000 was spent on its repairs. On 1^st July 2017, it was sold for ₹ 28,600. Prepare the machinery account for the years 2015 to 2017 under written down value method by assuming the rate of depreciation as 10% p.a. and the accounts are closed on 31^st December every year.
Answer:
Calculation of profit or loss on sale of machinery

Note: If Book value is more than the selling price it is called loss.
Book value – selling price = loss
46,170 – 28,600 = 17,570
∴ Loss on sale of machinery = ₹ 17,570

Question 19.
Raj & Co purchased a machine on 1^st January 2014 for ₹ 90,000. On 1^st July 2014, they purchased another machine for ₹ 60,000. On 1^st January 2015, they sold the machine purchased on 1^st January 2014 for ₹ 40,000. It was decided that the machine be depreciated at 10% per annum on diminishing balance method. Accounts are closed on 31^st December every year. Show the machinery account for the years 2014 and 2015.
Answer:
Workings

Note: If Book value is more than the selling price is called loss.
Book value – selling price = loss
81,000 – 40,000 = 41,000
Machinery

Textbook Case Study Solved

Question a.
Lucky & Co’s income statement shows a loss of ₹ 3,000. The owner thinks that there is no need to provide for depreciation as the company has made a loss. He also suggests his accountant to change the method of depreciation for the next year so as to avoid the loss. But, the accountant is hesitant to make the necessary changes suggested by his owner.
Now, discuss on the following points:

Question 1.
Do you agree on the point that there is no need to charge depreciation when the company has made a loss?
Answer:
No, I don’t agree on the point that there is no need to charge depreciation when the company has made a loss. We have to charge depreciation whether profit or loss, otherwise we cannot find out the actual profit or loss.

Question 2.
Why does the accountant hesitate to make the changes suggested by his owner?
Answer:
The accountant hesitates to make the changes suggested by his owner because it will differ the profit or loss for the business. The depreciation is a necessary one, so it must be deducted every year.

Question 3.
What are the accounting principles not followed if the accountant agrees to his owner’s suggestion?
Answer:
If the accountant agrees to his owner’s suggestion, they will not follow double entry system.

Question 4.
Do you think charging depreciation could be the only reason for the company’s loss?
Answer:
No, charging depreciation could not be the only reason for the company’s loss, because the business activities are subject to change but the depreciation is compulsory when the business is running.

Samacheer Kalvi 11th Accountancy Depreciation Accounting Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer

Question 1.
Depreciation is the gradual and permanent decrease in the value of an asset from any cause ……………….
(a) Owen
(b) Wheeler
(c) Spicer and Pegler
(d) R.N. Carter
Answer:
(d) R.N. Carter

Question 2.
Certain assets whether used or not become potentially less useful with the passage of time ……………….
(a) Efflux of time
(b) Lack of maintenance
(c) Abnormal factors
(d) Wear and tear
Answer:
(a) Efflux of time

Question 3.
The normal use of a tangible asset results in physical deterioration which is called ……………….
(a) Wear and tear
(b) Abnormal factors
(c) Obsolescence
(d) Efflux of time
Answer:
(a) Wear and tear

Question 4.
Allocation of acquisition cost of intangible fixed assets such as goodwill is called ……………….
(a) Abnormal factors
(b) Wear and tear
(c) Amortization
(d) Obsolescence
Answer:
(c) Amortization

Question 5.
………………. is also known as residual value.
(a) Book value
(b) Scrap value
(c) Amortization
(d) Wear and tear
Answer:
(b) Scrap value

Question 6.
The following formula is used to complete the rate of depreciation under ……………….
(a) Written down value method
(b) Straight line method
(c) Machine hour rate method
(d) Annuity method
Answer:
(a) Written down value method

II. Very Short Answer Questions

Question 1.
State R.N. Carter’s definition of depreciation.
Answer:
According to R.N. Carter, “Depreciation is the gradual and permanent decrease in the value of an asset from any cause”.

Question 2.
What is wear and tear?
Answer:
The normal use of a tangible asset results in physical deterioration which is called wear and tear. When there is wear and tear, the value of the asset decreases proportionately.

Question 3.
What is obsolescence?
Answer:
It is a reduction in the value of assets as a result of the availability of updated alternative assets. This happens due to new inventions and innovations.

Question 4.
What is Straight line method?
Answer:
Under this method, a fixed percentage on the original cost of the asset is charged every year by way of depreciation. Hence it is called original cost method. As the amount of depreciation remains equal in all years over the useful life of an asset, it is also called as fixed instalment method.

Question 5.
What is written down value method?
Answer:
Under this method, depreciation is charged at a fixed percentage on the written down value of the asset every year. Hence, it is called written down value method.

III. Short Answer Questions

Question 1.
Write a note on sum of years of digits method.
Answer:
This method is similar to the diminishing balance method. The amount of depreciation goes on decreasing year after year in proportion to the unexpired life of the asset. This method is suitable for those assets having more profitability of obsolescence and increased repair charges as the assets grow older. Under this method, amount of depreciation per year is calculated by multiplying the cost of the asset and the number of remaining years of life and dividing it by the sum of the digits of all years of life of the asset.

Question 2.
What is machine hour rate method?
Answer:
Under this method, depreciation per machine hour is calculated. The cost of the machinery after deducting the residual value, if any, is divided by the estimated working hours of the machine to find the depreciation per hour. The actual depreciation for any given period depends upon the working hours during that year. The special feature of this method is that depreciation is found directly in proportion to the actual use of the asset. Under this method life of the asset is estimated in hours and not in years.

Question 3.
What is Depletion method?
Answer:
Depletion means exhaustion of natural resources. That is depletion means quantitative reduction in the content of assets. This is applicable to those assets that get exhausted due to extraction and exploitation. Examples: mines and oil fields, etc. Under this method, depreciation rate is calculated on the basis of the estimated quantities of the output during the whole life of the asset.

Students can Download Samacheer Kalvi 10th Maths Model Question Paper 5 English Medium Pdf, Samacheer Kalvi 10th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Model Question Paper 5 English Medium

Instructions

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III and IV are to be attempted separately.
• Question numbers 1 to 14 in Part I are Multiple Choice Quèstions of one-mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and.writing the option code and the corresponding answer.
• Question numbers 15 to 28 in Part II àre two-marks questions. These are to be answered in about one or two sentences.
• Question numbers 29 to 42 in Part III are five-marks questions. These are to be answered in about three to five short sentences.
• Question numbers 43 to 44 in Part IV are eight-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 100

PART-I

I. Choose the correct answer. Answer all the questions. [Answers are in bold] [14 × 1 = 14]

Question 1.
If n(A × B) = 6 and A = {1, 3} then n (B) is ………….. .
(1) 1
(2) 2
(3) 3
(4) 6
Answer:
(3) 3

Question 2.
Given F₁ = 1, F₂ = 3 and F_n = F_n-1+ F_n-2 then F₅ is ………….. .
(1) 3
(2) 5
(3) 8
(4) 11
Answer:
(4) 11

Question 3.
In an A.P., the first term is 1 and the common difference is 4. How many terms of the A.P must be taken for their sum to be equal to 120?
(1) 6
(2) 7
(3) 8
(4) 9
Answer:
(3) 8

Question 4.
f = {(2, a), (3, b), (4, b), (5, c)} is a ………….. .
(1) identity function
(2) one-one function
(3) many-one function
(4) constant function
Answer:
(3) many-one function

Question 5.
The number of points of intersection of quadratic polynomial x² + 4x + 4 with the x axis is ………….. .
(1) 0
(2) 1
(3) 0 (or) 1
(4) 2
Answer:
(2) 1

Question 6.
The non-diagonal elements in any unit matrix are ………….. .
(1) 0
(2) 1
(3) m
(4) n
Answer:
(1) 0

Question 7.
If A is a 2′ 3 matrix and B is a 3′ 4 matrix, how many columns does AB have?
(1) 3
(2) 4
(3) 2
(4) 5
Answer:
(2) 4

Question 8.
In figure CP and CQ are tangents to a circle with centre at O. ARB is another tangent touching the circle at R. If CP = 11 cm and BC = 7 cm then the length of BR is ………….. .

(1) 6 cm
(2) 5 cm
(3) 8 cm
(4) 4 cm
Answer:
(4) 4 cm

Question 9
The slope of the line joining (12, 3), (4, a) is \(\frac { 1 }{ 8 }\). The value of ‘a’ is ………….. .
(1) 1
(2) 4
(3) -5
(4) 2
Answer:
(4) 2

Question 10.
If x = a tan θ and y = b sec θ then ………….. .
(1) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\)
(2) \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
(3) \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
(4) \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0\)
Answer:
(1) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\)

Question 11.
A letter is chosen at random from the letter of the word “PROBABILITY”. Find the probability that it is not a vowel.
(1) \(\frac{1}{5}\)
(2) \(\frac{2}{3}\)
(3) \(\frac{1}{3}\)
(4) \(\frac{3}{5}\)
Answer:
(2) \(\frac{2}{3}\)

Question 12.
The height of a right circular cone whose radius is 5 cm and slant height is 13 cm will be ………….. .
(1) 12 cm
(2) 10 cm
(3) 13 cm
(4) 5 cm
Answer:
(1) 12 cm

Question 13.
If the mean and co-efficient of variation of a data are 4 and 87.5% then the standard deviation is ………….. .
(1) 3.5
(2) 3
(3) 4.5
(4) 2.5
Answer:
(1) 3.5

Question 14.
Variance of first 20 natural numbers is ………….. .
(1) 32.25
(2) 44.25
(3) 33.25
(4) 30
Answer:
(3) 33.25

PART-II

II. Answer any ten questions. Question No. 28 is compulsory. [10 × 2 = 20]

Question 15.
Define a function.
Answer:
A relation “f” between two non-empty sets X and Y is called a function from X to Y if for each x ∈ X there exists only one y ∈ Y such that (x, y) ∈ f

Question 16.
Compute x such that 10⁴ ≡ x (mod 19)
Answer:
10² = 100 ≡ 5 (mod 19)
10⁴ = (10²)² ≡ 5² (mod 19)
10⁴ = 25 (mod 19)
10⁴ = 6 (mod 19)
(since 25 ≡ 6 (mod 19))
Therefore, x = 6

Question 7.
Simlify \(\frac{4 x^{2} y}{2 z^{2}} \times \frac{6 x z^{3}}{20 y^{4}}\)
Answer:

Question 18.
Pari needs 4 hours to complete the work. His friend Yuvan needs 6 hours to complete the work. How long will it take to complete if they work together?
Answer:
Let the work done by Pari and Yuvan together be x
Work done by Pari = \(\frac { 1 }{ 4 }\)
Work done by Yuvan = \(\frac { 1 }{ 6 }\)
By the given condition
\(\frac{1}{4}+\frac{1}{6}=\frac{1}{x} \Rightarrow \frac{3+2}{12}=\frac{1}{x}\)
\(\frac{5}{12}=\frac{1}{x}\)
5x = 12 ⇒ x = \(\frac{12}{5}\)
x = \(2 \frac{2}{5}\) hours (or) 2 hours 24 minutes

Question 19.
Find the values of x, y and z from the following equation \(\left( \begin{matrix} 12 & 3 \\ x & \frac { 3 }{ 2 } \end{matrix} \right) =\left( \begin{matrix} y & z \\ 3 & 5 \end{matrix} \right) \)
Answer:
Since the given matrices are equal then all the corresponding elements are equal. y = 12, z = 3, x = 3
The value of x = 3, y = 12 and z = 3

Question 20.
What length of ladder is needed to reach a height of 7 ft along the wall when the base of the ladder is 4 ft from the wall? A
Answer:

Let x be the length of the ladder. BC = 4 ft, AC = 7 ft.
By Pythagoras theorem we have, AB² = AC² + BC²
x² = 7² + 4² gives x² = 49 + 16
x² = 65. Hence, x = √65
The number √65 is between 8 and 8.1.
8² = 64 < 65 < 65.61 = 8.1²
Therefore, the length of the ladder is approximately 8.1 ft.

Question 21.
Prove that \(\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\) = cosec θ + cot θ
Answer:

Question 22.
The radius of a sphere increases by 25%. Find the percentage increase in its surface area.
Answer:
Let the radius of the be “r”
Surface area of the sphere = 4πr²sq.units …… (1)
If the radius is increased by 25%
New radius = \(\frac { 25 }{ 100 }\) × r + r
= \(\frac { r }{ 4 }\) + 4
= \(\frac{r+4 r}{4}=\frac{5 r}{4}\)
Surface area of the sphere
= 4π\(\left(\frac{5 r}{4}\right)^{2}\) sq.units
= 4 × π × \(\frac{25 r^{2}}{16}\)
= \(\frac{25 \pi r^{2}}{4}\) sq.units
Difference in surface area
= \(\frac{25 \pi r^{2}}{4}-4 \pi r^{2}\)
= \(\pi r^{2}\left(\frac{25}{4}-4\right)\)
= \(\pi r^{2}\left(\frac{25-16}{4}\right)\)
= \(\pi r^{2}\left(\frac{9}{4}\right)=\frac{9 \pi r^{2}}{4}\)
Percentage of increase in surface area

Percentage of increase in surface area = 56.25%

Question 23.
The standard deviation and mean of a data are 6.5 and 12.5 respectively. Find the coefficient of variation.
Answer:
Standard deviation of a data (σ) = 6.5
Mean of the data (x̄) = 12.5
Coefficient of variation = \(\frac{\sigma}{x} \times 100 \%\) ⇒ \(\frac{6.5}{12.5} \times 100 \%\) = 52%
Coefficient of variation = 52%

Question 24.
If f{x) = 3 + x, g(x) = x – 4, then check whether fog = gof
Answer:
f(x) = 3 + x ; g(x) = x – 4
fog = f[g(x)]
= f(x – 4)
= 3 + x – 4
= x – 1
gof = g [f(x)}
= g(3 + x)
= 3 + x – 4
= x – 1
∴ fog = gof

Question 25.
An organization plans to plant saplings in 25 streets in a town in such a way that one sapling for the first street, three for the second, nine for the third and so on. How many saplings are needed to complete the work?
Answer:
Here n = 25, a = 1, r = 3
S_n = \(a \frac{\left(r^{n}-1\right)}{r-1}\)
S₂₅ = \(\frac{1\left(3^{25}-1\right)}{3-1}\)
= \(\frac{3^{25}-1}{2}\)

Question 26.
Find the 19th term of an A.P. – 11, -15, -19,……..
Answer:
First term (a) = -11
Common difference (d) = -15 – (-11)
= -15 + 11 = -4
n = 19 .
t_n = a + (n – 1) d
t_n = -11 + 18(-4)
= -11 – 72
t₁₉ = -83
∴ 19th term of an A.P. is – 83

Question 27.
Find the value of ZB AC in the given triangle.
Answer:

In right triangle ABC

θ = tan^-1 \(\left(\frac{4}{5}\right)\) = tan^-1 (0.8)
θ = 38.7° (since tan 38.7° = 0.8011)
∠BAC = 38.7°

Question 28.
The vertices of a triangle are A(-1,3), B(-1, 1) and C(5, 1). Find the length of the median through the vertex C.
Answer:

Mid point of AB = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)
= \(\left(\frac{-1+1}{2}, \frac{3-1}{2}\right)\)
= (0,1)
Length of the median CD = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
= \(\sqrt{(5-0)^{2}+(1-1)^{2}}\)
= \(\sqrt{25}\) = 5

PART – III

III. Answer any ten questions. Question No. 42 is compulsory. [10 × 5 = 50]

Question 29.
Let f be a function f : N → N be defined by f(x) = 3x + 2, x ∈ N
(i) Find the images of 1, 2,3
(ii) Find the pre – images of 29, 53
(iii) Identify the type of function.
Answer:
The function f: N → N is defined by f(x) = 3x + 2
(i) If x = 1, f(1) = 3(1) + 2 = 5
If x = 2, f(2) = 3(2) + 2 = 8
If x = 3, f(3) = 3(3) + 2=11
The images of 1, 2, 3 are 5, 8, 11 respectively.

(ii) If x is the pre-image of 29, then f(x) = 29 . Hence 3x + 2 = 29
3x = 21 ⇒ x = 9.
Similarly, if x is the pre-image of 53, then f(x) = 53.
Hence 3x + 2 = 53 3x = 51 ⇒ x = 17.
Thus the pre-images of 29 and 53 are 9 and 17 respectively.

(iii) Since different elements of N have different images in the co-domain, the function f is one – one function.
The co-domain of f is N.
But the range of f = {5, 8, 11, 14, 17,…} is a proper subset of N.
Therefore f is not an onto function. That is, f is an into function.
Thus f is one – one and into function.

Question 30.
Let f: A → B be a function defined by f(x) = \(\frac { x }{ 2 }\) – 1, where A = {2,4,6,10,12} and B = {0,1,2, 4,5,9}. Represent f by
(i) set of ordered pairs (ii) a table (iii) an arrow diagram (iv) a graph
Answer:
A= {2, 4, 6, 10, 12} and B = {0, 1, 2, 4, 5, 9}
f(x) = \(\frac { x }{ 2 }\) – 1
f(2) = \(\frac { 2 }{ 2 }\) – 1 = 1 – 1 = 0
f(4) = \(\frac { 4 }{ 2 }\) – 1 = 2 – 1 = 1
f(6) = \(\frac { 6 }{ 2 }\) – 1 = 3 – 1 = 2
f(10) = \(\frac { 10 }{ 2 }\) – 1 = 5 – 1 = 4
f(12) = \(\frac { 12 }{ 2 }\) – 1 = 6 – 1 = 5

(i) Set of ordered pairs
f = {(2, 0) (4, 1) (6, 2) (10, 4) (12, 5}

(ii) Table

(iii) Arrow diagram

(iv) Graph

Question 31.
The ratio of 6^th and 8^th term of an A.P is 7:9. Find the ratio of 9^th term to 13^th terms.
Answer:
Given t₆ : t₈ = 7 : 9 (using t_n = a + (n – 1)d
a + 5d : a + 7d =7 : 9
9(a + 5d) = 7(a + 7d)
9a + 45d = 7a + 49d
9a – 7a = 49d – 45d
2a = 4d
a = 2d
To find t₉ : t₁₃
t₉ : t₁₃ = a + 8d : a + 12d
= 2d + 8d : 2d + 12d
= 10d : 14d
= 5 : 7
∴ t₉ : t₁₃ = 5 : 7

Question 32.
The sum of first n, In and 3n terms of an A.P. are S₁ S₂ and S₃ respectively. Prove that S₃ = 3 (S₂ – S₁).
Answer:
If S₁ S₇ and S₃ are sum of first n, 2n and 3n terms of an A.P. respectively then

Question 33.
Find the values of m and n if the exprtession \(\frac{1}{x^{4}}-\frac{6}{x^{3}}+\frac{13}{x^{2}}+\frac{m}{x}+n\) is a perfect a square.
Answer:

Since it is a perfect square
\(\frac { 1 }{ 2 }\)(m + 12) = 0
m + 12 = 0
m = -12
n – 4 = 0
n = 4
∴ The value of m = -12 and n = 4

Question 34.
If α, β are the roots of the equation 2x² – x – 1 = 0 then form the equation whose roots are α²β, β²α.
Answer:
2x² – x – 1 = 0 ⇒ Here a = 2, b = -1,c = -1
α + β = \(\frac{-b}{a}=\frac{-(-1)}{2}=\frac{1}{2}\)
α + β = \(\frac{c}{a}=\frac{-1}{2}\)
Given roots are α²β, β²α
Summ of the roots α²β + β²α = αβ (α + β) = \(-\frac{1}{2}\left(\frac{1}{2}\right)=-\frac{1}{4}\)
Product of the roots (α²β) × (β²α) = α³β³ = (αβ)³ =
\(\left(-\frac{1}{2}\right)^{3}=-\frac{1}{8}\)
The required equation is x² – (Sum of the roots) x + (Product of the roots) = 0
\(x^{2}-\left(-\frac{1}{4}\right) x-\frac{1}{8}=0\) gives 8x² + 2x – 1 = 0

Question 35.
P and Q are the mid-points of the sides CA and CB respectively of a ∆ABC, right angled at C. Prove that 4(AQ² + BP²) = 5AB².
Answer:

Since, ∆AQC is a right triangle at C, AQ² = AC² + QC² ……. (1)
Also, ∆BPC is a right triangle at C, BP² = BC² + CP² ……… (2)
From (1) and (2), AQ² + BP² = AC² + QC² + BC² + CP²
4(AQ² + BP²) = 4AC² + 4QC² + 4BC² + 4CP²
= 4AC² + (2QC)² + 4BC² + (2CP)²
= 4AC² + BC² + 4BC² + AC² (Since P and Q are mid points)
= 5(AC² + BC²)
4(AQ² + BP²) = 5AB² (By Pythagoras Theorem)

Question 36.
Find the equation of a straight line passing through (1, -4) and has intercepts which are in the ratio 2:5.
Answer:
Let the x-intercept be 2a and the y intercept 5a
The equation of a line is \(\frac{x}{a}+\frac{y}{a}=1\) ⇒ \(\frac{x}{2 a}+\frac{y}{5 a}=1\)
The line passes through the point (1, -4)
\(\frac{1}{2 a}+\frac{(-4)}{5 a}=1\) ⇒ \(\frac{1}{2 a}-\frac{4}{5 a}=1\)
Multiply by 10a
(L.C.M of 2a and 5a is 10a)
5 – 8 = 10a ⇒ -3 = 10a ⇒ a = \(\frac{-3}{10}\)
The equation of the line is \(\frac{x}{2(-3 / 10)}+\frac{y}{5(-3 / 10)}=1\)
\(\frac{x}{-3 / 5}+\frac{y}{-3 / 2}=1\) ⇒ \(\frac{5 x}{-3}+\frac{2 y}{-3}=1\)
\(\frac{-5 x}{3}-\frac{2 y}{3}=1\)
Multipy by 3
– 5x – 2y = 3 ⇒ – 5x – 2y – 3 = 0
5x + 2y + 3 = 0
The equation of a line is 5x + 2y + 3 = 0

Question 37.
From the top of the tower 60m high the angles of depression of the top and bottom of a vertical lamp post are observed to be 38° and 60° respectively. Find the height of the lamp post (tan 38° = 0.7813, √3 = 1.732)
Answer:

Let the height of the lamp post be “h”
The height of the tower (BC) = 60 m
∴ EC = 60 – h
Let AB be x
In the right ∆ ABC
tan 60° = \(\frac{B C}{A B}\)
√3 = \(\frac{60}{x}\)
x = \(\frac{60}{\sqrt{3}}\) …… (1)
In the right ∆ DEC, tan 38° = \(\frac{E C}{D E}\)
0.7813 = \(\frac{60-h}{x}\)
x = \(\frac{60-h}{0.7813}\) ….. (2)
Froom (1) and (2) we get
\(\frac{60}{\sqrt{3}}=\frac{60-h}{0.7813}\)
60 × 0.7813 = 60√3 – √3h
46.88 = 60√3 – √3h
√3h = 60√3 – 46.88
= 60 × 1.732 – 46.88
= 103.92 – 46.88
1.732 h = 57.04 ⇒ h = \(\frac{57.04}{1.732}\)
h = \(\frac{5704}{1732}\) = 32.93m
∴ Height of the lamp post = 32.93 m

Question 38.
Calculate the weight of a hollow brass sphere if the diameter is 14 cm and thickness is 1 mm, and whose density is 17.3 g/cm³.
Answer:
Let r and R be the inner and outer radii of the hollow sphere.
Given that, inner diameter d = 14 cm; inner radius r = 7 cm; thickness = 1 mm = \(\frac{1}{10}\) cm
Outer radius R = 7 + \(\frac{1}{10}=\frac{71}{10}\) = 7.1 cm
Volume of hollow sphere = \(\frac{4}{3}\) π (R³ – r³) cu. cm
= \(\frac{4}{3} \times \frac{22}{7}\)(357.91 -343) = 62.48 cm³
But, weight of brass in 1 cm³ = 17.3 gm
Total weight = 17.3 × 62.48 = 1080.90 gm
Therefore, total weight is 1080.90 grams.

Question 39.
Find the Co-efficient of variation of 24, 26,33,37,29,31
Answer:
Arrange in ascending order we get 24, 26,29, 31,33,37
Assumed mean = 29


\(\frac{4.32}{30} \times 100\) = \(\frac{432}{30}=14.4\)
Coefficient of variation = 14.4%

Question 40.
Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. W hat is the probability that the sum of the two numbers appearing on the top of the dice is ………………. .
Answer:
When we roll two dice
Sample space = {(1. 1) (1,2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6. 4) (6, 5) (6, 6)}
n(S) = 36
(i) Let A be the event of getting the sum of two number is 8
A = {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)} 5
n(A) = 5
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{5}{36}\)

(ii) Let B be the event of getting the sum of the two numbers is 13.
B = { }
n(B) = 0
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{0}{36}=0\)

(iii) Let C be the event of getting the sum of the two numbers is less than 12 or equal to 12.
n(C) = 36
P(C) = \(\frac{n(\mathrm{C})}{n(\mathrm{S})}=\frac{36}{36}=1\)

Question 41.
Find two consecutive positive integers, sum of whose squares is 365.
Answer:
Let the two consecutive positive integers be “x” and x + 1
Sum of squares = 365
x² + (x+1)² = 365
2x² + 2x – 364 = 0
x² + x – 182 = 0

(x + 14) (x – 13) = 0
x + 14 = 0 or x – 13 = 0
x = -14 or x = 13 (rejecting -14. Given positive integer)
The consecutive terms are 13 and 14.

Question 42.
A cylindrical bucket of 32 cm high and with radius of base 18 cm, is filled with sand completely. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Answer:
Cylindrical bucket
Height of the bucket (h) = 32 cm
Radius of the bucket (r) = 18 cm

Conical heap
Height of the cone (H) = 24 cm
Let the radius of the conical heap be “P ”
By the given condition
Volume of the conical heap = Volume of the cylindrical bucket
\(\frac { 1 }{ 3 }\)πR²H = πr²
\(\frac { 1 }{ 3 }\) × R² × 24 = r²H
\(\frac { 1 }{ 3 }\) × R² × 24 = 18 × 18 × 32
R² = \(\frac{18 \times 18 \times 32 \times 3}{24}\)
= \(\frac{18 \times 18 \times 4 \times 3}{3}\)
= 18 × 18 × 4
= 36 cm
Slant height of the cone (l) = \(\sqrt{H^{2}+R^{2}}\)
= \(\sqrt{24^{2}+36^{2}}\)
= \(\sqrt{(12 \times 2)^{2}+(12 \times 3)^{2}}\)
= 12\(\sqrt{2^{2}+3^{2}}\)
= 12√13
Slant height of the cone = 12√13 cm
Radius of the cone = 36 cm

PART – IV

IV. Answer all the questions. [2 × 8 = 16]

Question 43.
(a) PQ is a chord of length 8 cm to a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length of the tangent TP.
Answer:

Let TR = y. Since, OT is perpendicular bisector of PQ
PR = QR = 4 cm
In ∆ORP, OP² = OR² + PR²
QR² = OP² – PR²
OR² = 5² – 4² = 25 – 16 = 9
OR = 3 cm
OT = OR + RT = 3 + y …. (1)
In ∆PRT , TP² = TR² + PR² …. (2)
and ∆OPT we have, OT² = TP² + OP²
OT² = (TR² + PR²) + OP² (substitute for TP² from (2)
(3 + y)² = y² + 4² + 5²(substitute for OT from (1))
9 + 6y + y² = y² + 16 + 5 There fore y = TR = \(\frac{16}{3}\)
6y = 41 – 9 we get y = \(\frac{16}{3}\)
From (2) TP² = TR² + PR²
TP² = \(\left(\frac{16}{3}\right)^{2}+4^{2}=\frac{256}{9}+16=\frac{400}{9} \mathrm{so}, \mathrm{TP}=\frac{20}{3} \mathrm{cm}\)

[OR]

(b) Draw a triangle ABC of base BC = 8 cm, ∠A = 60° and the bisector of ∠A meets BC at D such that BD = 6 cm.
Answer:


Step 1 : Draw a line segment BC = 8 cm.
Step 2 : At B, draw’ BE such that ∠CBE = 60° .
Step 3 : At B, draw BF such that ∠EBF = 90° .
Step 4 : Draw the perpendicular bisector to BC, w’hich intersects BF at O and BC at G.
Step 5 : With O as centre and OB as radius draw a circle.
Step 6 : From B, mark an arc of 6 cm on BC at D.
Step 7 : The perpendicular bisector intersects the circle at I. Joint ID.
Step 8 : ID produced meets the circle at A. Now join AB and AC.
Then ∆ABC is the required triangle.

Question 44.
(a) Draw the graph of y = x² + 3x – 4 and hence use it to solve x² + 3x – 4 = 0.
Answer:
Let y = x² + 3x – 4
(i) Draw the graph of y = x² + 3x – 4

(ii) Plot the points (-5, 6), (-4, 0), (-3, -4), (-2, -6), (-1, -6), (0, -4), (1, 0), (2, 6), (3, 14) on the graph using suitable scale.
(iii) Join the points by a free hand smooth curve.
The smooth curve is the graph of y = x² + 3x – 4
(iv) To solve x² + 3x – 4 = 0, subtract x² + 3x – 4 = 0 from y = x² + 3x – 4.
y = o
∴ The point of intersection with the X – axis is the solution set.
The solution set is -4 and 1.

[OR]

(b) A motor boat whose speed is 18 km/hr in still water takes 1 hour more to go to 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Answer:
Let the speed of the stream be “x” km/hr
Speed of the motor boat to go for upstream = (18 – x) km/hr
Speed of the motor boat to go for down stream = (18 + x) km/hr
Time taken to go for upstream = \(\frac{24}{18-x}\) hour
Time taken to go for down stream = \(\frac{24}{18-x}\)
By the given condition
\(\frac{24}{18-x}-\frac{24}{18+x}=1\)
\(\frac{24(18+x)-24(18-x)}{(18-x)(18+x)}=1\)
432 + 24x + 432 + 24x = (18 – x)(18 – x)
48x = 324 – x²
x² + 48x – 324 = 0
x² + 54x – 6x – 324 = 0
x(x + 54) – 6 (x + 54) = 0
(x + 54) (x – 6) = 0
x + 54 = 0 or x – 6 = 0
x = -54 or x = 6 (speed cannot be negative)
∴ Speed of the boat = 6 km/hr

Students can Download Bio Zoology Chapter 6 Evolution Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Bio Zoology 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 12th Bio Zoology Solutions Chapter 6 Evolution

Samacheer Kalvi 12th Bio Zoology Evolution Text Book Back Questions and Answers

Question 1.
The first life on Earth originated __________
(a) in air
(b) on land
(c) in water
(d) on mountain
Answer:
(c) in water

Question 2.
Who published the book “Origin of species by Natural Selection” in 1859?
(a) Charles Darwin
(b) Lamarck
(c) Weismann
(d) Hugo de Vries
Answer:
(a) Charles Darwin

Question 3.
Which of the following was the contribution of Hugo de Vries?
(a) Theory of mutation
(b) Theory of natural Selection
(c) Theory of inheritance of acquired characters
(d) Germplasm theory
Answer:
(a) Theory of mutation

Question 4.
The wings of birds and butterflies is an example of __________
(a) Adaptive radiation
(b) convergent evolution
(c) divergent evolution
(d) variation
Answer:
(b) convergent evolution

Question 5.
The phenomenon of “ Industrial Melanism” demonstrates __________
(a) Natural selection
(b) induced mutation
(c) reproductive isolation
(d) geographical isolation
Answer:
(a) Natural selection

Question 6.
Darwin’s finches are an excellent example of __________
(a) connecting links
(b) seasonal migration
(c) adaptive radiation
(d) parasitism
Answer:
(c) adaptive radiation

Question 7.
Who proposed the Germplasm theory?
(a) Darwin
(b) August Weismann
(c) Lamarck
(d) analysis of bones
Answer:
(b) August Weismann

Question 8.
The age of fossils can be determined by __________
(a) electron microscope
(b) weighing the fossils
(c) carbon dating
(d) analysis of bones
Answer:
(c) carbon dating

Question 9.
Fossils are generally found in __________
(a) igneous rocks
(b) metamorphics
(c) volcanic rocks
(d) sedimentary rocks
Answer:
(d) sedimentary rocks

Question 10.
Evolutionary history of an organism is called __________
(a) ancestry
(b) ontogeny
(c) phylogeny
(d) paleontology
Answer:
(c) phylogeny

Question 11.
The golden age of reptiles was __________
(a) Mesozoic era
(b) Cenozoic era
(c) Paleozoic era
(d) Proteroic era
Answer:
(a) Mesozoic era

Question 12.
Which period was called “Age of fishes”?
(a) Permian
(b) Triassic
(c) Devonian
(d) Ordovician
Answer:
(c) Devonian

Question 13.
Modem man belongs to which period?
(a) Quaternary
(b) Cretaceous
(c) Silurian
(d) Cambrian
Answer:
(a) Quaternary

Question 14.
The Neanderthal man had the brain capacity of __________
(a) 650 – 800cc
(b) 1200cc
(c) 900cc
(d) 1400c
Answer:
(d) 1400c

Question 15.
List out the major gases seem to fie found in primitive Earth.
Answer:
C0₂, NH₃, UV and Water vapour

Question 16.
Explain the three major categories in which fossilization occur.
Answer:
(i) Actual remains are the most common method of fossilization. When marine animals die, their hard parts such as bones and shells, etc. are covered with sediments and are protected from further deterioration. They get preserved as such as they are preserved in vast ocean; the salinity in them prevents decay. The sediments become hardened to form definite layers or strata. For example, Woolly Mammoth that lived 22 thousand years ago were preserved in the frozen coast of Siberia as such. Several human beings and animals living in die ancient city of Pompeii were preserved intact by volcanic ash which gushed out from Mount Vesuvius.

ii) Petrifaction – When animals die the original portion of their body may be replaced molecule for molecule by minerals and the original substance being lost through disintegration. This method of fossilization is called petrifaction. The principle minerals involved in this type fossilization are iron pyrites, silica, calcium carbonate and bicarbonates of calcium and magnesium. ‘

iii) Natural moulds and casts – Even after disintegration, the body of an animal might leave indelible impression on the soft mud which later becomes hardened into stones. Such impressions are called moulds. The cavities of the moulds may get filled up by hard minerals and get fossilized, which are called casts. Hardened faecal matter termed as coprolites, occur as tiny pellets. Analysis of the coprolites enables us to understand the nature of diet, the pre¬historic animals thrived.

Question 17.
Differentiate between divergent evolution and convergent evolution with one example for each.
Answer:
Divergent Evolution:

1. Divergent evolution is a result of homology.
2. Eg: The wings of bird and the forelimbs of human both are homologous structures modified according to functions. In birds, it is used for flight and in humans used for writing and other purposes.

Convergent Evolution:

1. Convergent evolution is a result of analogy,
2. E.g: Root modification in sweet potato, and stem modification in potato are analogous structures both performing same function i.e., storage,

Question 18.
How does Hardy-Weinberg’s expression (p² + 2pq + q² = 1) explain that genetic equilibrium is maintained in a population? List any four factors that can disturb the genetic equilibrium.
Answer:
The allele frequencies in a population are stable and are constant from generation to generation in the absence of gene flow, genetic drift, mutation, recombination and natural selection. If a population is in a state of Hardy Weinberg equilibrium, the frequencies of alleles and genotypes or sets of alleles in that population will remain same over generations. Evolution is a change in the allele frequencies in a population over time. Hence population in Hardy Weinberg is not evolving.

Suppose we have a large population of beetles, (infinitely large) and appear in two colours ’ dark grey (black) and light grey, and their colour is determined by ‘A’ gene. ‘AA’ and ‘Aa’ beetles are dark grey and ‘aa’ beetles are light grey. In a population let’s say that ‘ A’ allele has frequency (p) of 0.3 and ‘a’ allele has a frequency (q ) of 0.7. Then p+q= 1.

If a population is in Hardy Weinberg equilibrium the genotype frequencycan be estimated by Hardy Weinberg equation.
(p + q)² = p² + 2pq + q²
p² = frequency of AA
2pq = frequency of Aa
q² = frequency of aa
p = 0.3, q = 0.7 then,
p² = (0.3)² = 0.09 = 9 %AA
2pq = 2(0.3) (0.7) = 0.42 = 42 % Aa
q2 = (0.7)² 0.49 = 49 % aa
Hence the beetle population appears to be in Hardy- Weinberg equilibrium. When the beetles in Hardy- Weinberg equilibrium reproduce the allele and genotype frequency in the next generation would be: Let’s assume that the frequency of ‘A’ and ‘a’ allele in the pool of gametes that make the next generation would be the same, then there would be no variation in the progeny. The genotype frequencies of the parent appears in the next generation.
(i.e. 9% AA, 42% Aa and 49% aa).

If we assume that the beetles mate randomly (selection of male gamete and female gamete in the pool of gametes), the probability of getting the offspring genotype depends on the genotype of the combining parental gametes.

Question 19.
Explain how mutations, natural selection and genetic drift affect Hardy Weinberg equilibrium.
Answer:
Natural selection occurs when one allele (or combination of alleles of differences) makes an organism more or less fit to survive and reproduce in a given environment. If an allele reduces fitness, its frequencies tend to drop from one generation to the next.

The evolutionary path of a given gene (i.e) how its allele’s change in frequency in the population across generation, may result from several evolutionary mechanisms acting at once. For example, one gene’s allele frequencies might be modified by both gene flow and genetic drift, for another gene, mutation may produce a new allele, that is favoured by natural selection.

Genetic drift / Sewall Wright Effect is a mechanism of evolution in which allele frequencies of a population change over a generation due to chance (sampling error). Genetic drift occurs in all population sizes, but its effects are strong in a small population. It may result in a loss of some alleles (including beneficial ones) and fixation of other alleles. Genetic drift can have major effects, when the population is reduced in size by natural disaster due to the bottleneck effect or when a small group of population splits from the main population to form a new colony due to the founder’s effect.

Although mutation is the original source of all genetic variation, mutation rate for most organisms is low. Hence new mutations on allele frequencies from one generation to the next is usually not large.

Question 20.
How did Darwin explain the fitness of organisms?
Answer:
Organisms struggle for food, space and mate. As these become a limiting factor, competition exists among the members of the population. Darwin denoted struggle for existence in three ways

Intra-specific struggle between the same species for food, space and mate Inter-specific struggle with different species for food and space. Struggle with the environment to cope with the climatic variations, flood, earthquakes and drought, etc.

According to Darwin, nature is the most powerful selective force. He compared origin of species by natural selection to a small isolated group. Darwin believed that the struggle for existence resulted in the survival of the fittest. Such organisms become better adapted to the changed environment.

Question 21.
Mention the main objections to Darwinism.
Answer:
Some objections raised against Darwinism were

• Darwin failed to explain the mechanism of variation.
• Darwinism explains the survival of the fittest but not the arrival of the fittest.
• He focused on small fluctuating variations that are mostly non-heritable.
• He did not distinguish between somatic and germinal variations.
• He could not explain the occurrence of vestigial organs, overspecialization of some organs like large tusks in extinct mammoths and oversized antlers in the extinct Irish deer, etc.

Question 22.
Taking the example of Peppered moth, explain the action of natural selection. What do you call the above phenomenon?
Answer:
Natural selection can be explained clearly through industrial melanism. Industrial melanism is a classical case of Natural selection exhibited by the peppered moth, Bistort betularia. These were available in two colours, white and black. Before industrialization peppered moth both white and black coloured were common in England. Pre-industrialization witnessed white colpured background of the wall of the buildings hence the white coloured moths escaped from their predators. Post industrialization, the tree trunks became dark due to smoke and soot let out from the industries.

The black moths camouflaged on the dark bark of the trees and the white moths were easily identified by their predators. Hence the dark coloured moth population was selected and their number increased when compared to the white moths. Nature offered positive selection pressure to the black coloured moths. The above proof shows that in a population, organisms that can adapt will survive and produce more progenies resulting in increase in population through natural selection.

Question 23.
Darwin’s finches and Australian marsupials are suitable examples of adaptive radiation – Justify the statement.
Answer:
Darwin’s finches are the birds whose common ancestor arrived on the Galapagos about 2 million years ago. During that time, Darwin’s finches have evolved into 14 recognized species differing in body size, beak shape and feeding behavior. Changes in the size and form of the beak have enabled different species to utilize different food resources such as insects, seeds, nectar from cactus flowers and blood from iguanas, all driven by Natural selection. Genetic variation in the ALX1 gene in the DNA of Darwin finches is associated with variation in the beak shape. Mild mutation in the ALX1 gene leads to phenotypic change in the shape of the beak of the Darwin finches.

Marsupials in Australia and placental mammals in North America are two subclasses of mammals they have adapted in similar way to a particular food resource, locomotory skill or climate. They were separated from the common ancestor more than 100 million years ago and each lineage continued to evolve independently. Despite temporal and geographical separation, marsupials in Australia and placental mammals in North America have produced varieties of species living in similar habitats with similar ways of life. Their overall resemblance in shape, locomotory mode, feeding and foraging are superimposed upon different modes of reproduction. This feature reflects their distinctive evolutionary relationships.

Over 200 species of marsupials live in Australia along with many fewer species of placental mammals. The marsupials have undergone adaptive radiation to occupy the diverse habitats in Australia, just as the placental mammals have radiated across North America.

Question 24.
Who disproved Lamarck’s Theory of acquired characters? How?
Answer:
Lamarck’s “Theory of Acquired characters” was disproved by August Weismann who conducted experiments on mice for twenty generations by cutting their tails and breeding them. All mice bom were with tail. Weismann proved that change in the somatoplasm will not be transferred to the next generation but changes in the germplasm will be inherited.

Question 25.
How does the Mutation Theory of De Vries differ from Lamarck and Darwin’s view in the origin of new species?
Answer:
According to de Vries, sudden and large variations were responsible for the origin of new species, whereas Lamarck and Darwin believed in the gradual accumulation of all variations as the causative factors in the origin of new species.

Question 26.
Explain stabilizing, directional and disruptive selection with examples.
Answer:
i. Stabilising selection (centipetal selection): This type of selection operates in a stable environment as shown in fig. The organisms with average phenotypes survive whereas the extreme individuals from both ends are eliminated. There is no speciation but the phenotypic stability is maintained within the population over a generation. For example, measurements of sparrows that survived the storm clustered around the mean, and the sparrows that failed to survive the storm clustered around the extremes of the variation showing stabilizing selection.

ii. Directional Selection: The environment” which undergoes gradual change is subjected to directional selection, as shown in fig. This type of selection removes the individuals from one end towards the other end of phenotypic distribution. For example, size differences between male and female sparrows. Both male and female look alike externally but differ in body weight. Females show directional selection in relation to body weight.

iii. Disruptive selection: (centrifugal selection) When homogenous environment changes into heterogenous environment this type of selection is operational as shown in fig. The organisms of both the extreme phenotypes are selected, whereas individuals with average phenotype are eliminated. This results in splitting of the population into sub population/species. This is a rare form of selection but leads to formation of two or more different species. It is also (called adaptive radiation. (E.g:) Darwin’s finches beak size in relation to seed size inhabiting Galapagos islands. Group selection and sexual selection are other types of selection. The two major group selections are Altrusim and Kin selection.

Question 27.
Rearrange the descent in human evolution.
Answer:
Australopithecus → Homo erectus → Homo sapiens → Ramapithecus →Homo habilis
Ramapithecus → Australopithecus → Homo habilis → Homo erectus → Homo sapiens

Question 28.
Differentiate between the eating habit and brain size of Australopithecus and Ramapithecus.
Answer:

	Australopithecus	Ramapithecus
Eating Habit	Herbivores	Omnivores
Brain Size	350- 450 cc	200 – 300 cc

Question 29.
How does the Neanderthal man differ from the modern man in appearance?
Answer:
Neanderthal man differ from the modem human in having a semierect posture, flat cranium, sloping forehead, thin large orbits, heavy brow ridges, protruding jaws and no chin.

Question 30.
Mention any three similarities found common in Neanderthal man and Homo sapiens. Common characters showed by Neanderthal man and Homo sapiens are:
Answer:

1. Usage of Fire
2. Burying of dead bodies
3. Protecting themselves from predators

Question 31.
According to Darwin, organic evolution is due to
(а) Intraspecific competition
(b) Interspecific competition
(c) Competition within closely related species.
(d) Reduced feeding efficiency in one species due to the presence of interfering species.
Answer:
(d) Reduced feeding efficiency in one species due to the presence of interfering species.

Question 32.
A population will not exist in Hardy – Weinberg equilibrium if
(a) Individuals mate selectively
(b) There are no mutations
(c) There is no migration
(d) The population is large
Answer:
(a) Individuals mate selectively

Samacheer Kalvi 12th Bio Zoology Evolution Additional Questions and Answers

1 – Mark Questions

Question 1.
Identify the incorrect statement in concern with Neanderthals.
(a) Neanderthal human was found in Germany.
(b) They possessed flat cranium.
(c) They used to bury their dead.
(d) Their brain size is of 650 – 800 cc
Answer:
(d) Their brain size is of 650 – 800 cc

Question 2.
Which of the following statement does not satisfy the Hardy Weinberg’s principle?
(a) A population undergoing random mating
(b) Small-sized population
(c) Population where there is no mutation or gene flow
(d) Absence of natural selection
Answer:
(b) Small-sized population

Question 3.
Match column I with column II

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

Question 4.
Placental mammals develop during _______
(a) Eocene
(b) Oligocene
(c) Pliocene
(d) Paleocene
Answer:
(d) Paleocene

Question 5.
Identify the correct sequence from oldest to youngest
(а) Cambrian → Permian → Devonian → Silurian → Ordovician
(b) Permian → Silurian → Devonian → Ordovician → Cambrian
(c) Permian → Devonian → Silurian → Cambrian → Ordovician
(d) Cambrian → Ordovician → Silurian → Devonian → Permian
Answer:
(d) Cambrian → Ordovician → Silurian Devonian → Permian

Question 6.
Match the scientists with their terminologies used
(a) Biogenesis (i) Oparin
(b) Prebiotic soup (ii) Henry Bastin
(c) Coacervates (iii) Thomas Huxley
(d) Abiogenesis (iv) Haldane
(a) a – iii b – iv c – ii d – i
(b) a – ii b – iv c – i d – iii
(c) a – iii b – i c – iv d – ii
(d) a – i b – iv c – iii d – ii
Answer:
(b) a – ii b – iv c – i d – iii

Question 7.
Anatomical structures that have similar functions but not similar structures are called
(a) Homologous structures
(b) Vestigial structures
(c) Analogous structures
(d) Generalized structures
Answer:
(c) Analogous structures

Question 8.
Who propounded the theory of recapitulation?
(a) Ernst Von Haeckel
(b) Charles Darwin
(c) Thomas Huxley
(d) Oparin
Answer:
(c) Ernst Von Haeckel

Question 9.
Mammal in human male is
(a) Atavistic organ
(b) Rudimentary Organ
(c) Vestigial organ
(d) Homologous structure
Answer:
(c) Vestigial organ

Question 10.
Which of the following is/are not examples of analogous structure
(a) Wings of Birds and Bats
(b) Wings of Birds and Insects
(c) Thom of Bougainvillea and Tendril of cururbita
(d) Flippers of Penguins and Dolphins
(i) a, b, c
(ii) a and c
(iii) b and d
(iv) All the above
Answer:
(ii) a and c

Question 11.
identify the mismatched pairs
(a) Thom of Bougainvillea and Tenrdril of – Analogy
(b) Forelimbs of whale and cat – Analogy
(c) Octopus eye & Mammalian eye – Homology
(d) Root of sweet potato & stem of potato – Homology
Answer:
(a) Thorn of Bougainvillea & Terdril of crucurbita – Analogy

Question 12.
Witnesses for evolution are found in
(a) Rocks
(b) Ocean beds
(c) Fossils
(d) Desert
Answer:
(c) Fossils

Question 13.
Assertion (A): Oparin used the term coacervates
Reason (R): Coacervates are colloidal particles in an aqueous environment
(a) Both A and Rare incorrect
(b) Both A and R are correct
(c) Both A and R are correct. R explains A.
(d) A is correct R is incorrect
Answer:
(c) Both A and R are correct. R explains A.

Question 14.
According to the theory of spontaneous generation, life originated from
(a) Cosmic particles
(b) Non-living materials
(c) Coacervates
(d) Sea
Answer:
(b) Non-living materials

Question 15.
Assertion (A): Hardy – Weinberg principle states that the allelic frequency of a population remains constant
Reason (R) : Constancy is maintained through natural selection and mutation
(a) A is true R is false
(b) A is false R is true
(c) Both A and R are true
(d) R explains
Answer:
(a) A is true R is false

Question 16.
Calculate the allelic frequency of Aa. frequency of 0.7
(a) 0.67
(b) 0.42
(c) 0.36
Answer:
(b) 0.42

Question 17.
Match the following

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

Question 18.
Genetic drift leads to
(a) Mutation
(b) Bottle neck effect
(c) Immigration
(d) Isolation
Answer:
(b) Bottleneck effect

Question 19.
Atavism refers to
(a) Inheritance of triat by mother
(b) Inheritance of triat by father
(c) Criss-cross inheritance
(d) Inheritance of characters not shown by parents
Answer:
(d) Inheritance of characters not shown by parents

2 – Mark Questions

Question 1.
State the theory of spontaneous generation.
Answer:
According to the theory of spontaneous generation or Abiogenesis, living organisms originated from non-living materials and occurred through stepwise chemical and molecular evolution over millions of years. Thomas Huxley coined the term abiogeneis.

Question 2.
List the four eras of geological time scale.
Answer:

1. Precambrian era
2. Paleozoic era
3. Mesozoic era
4. Cenozoic era

Question 3.
Which periods of paleozoic era are referred as

1. Age of fishes
2. Invertebrates

Answer:

1. Age of fishes – Devonian period
2. Age of invertebrates – Cambrian period

Question 4.
Point out the epochs of carboniferous period.
Answer:

1. Pennsylvanian
2. Mississippian

Question 5.
Compare relative dating with absolute dating.
Answer:
Relative dating is used to determine a fossil by comparing it to similar rocks and fossils of known age. Absolute dating is used to determine the precise age of a fossil by using radiometric dating to measure the decay of isotopes

Question 6.
Wing of a cockroach and the wing of parrot. What do you infer from this statement with reference to evolution?
Answer:
Both the wings of cockroach and bird are different in structure but similar in their function. Thus, they are analogous structure that brings about convergent evolution.

Question 7.
Name the scientists who propounded the following theories.

1. Mutation theory
2. Chemical theory of evolution

Answer:

1. Mutuation theory was propounded by Hugo de Vries.
2. Chemical theory of evolution was propounded by Oparin and Haldane

Question 8.
Define fossilization and mention its types.
Answer:
Fossilization is the process by which plant and animal remains are preserved in sedimentary rocks. It is of three major types,

1. Actual remains
2. Petrifaction
3. Natural moulds and casts.

Question 9.
Name the principle minerals involved in petrifaction.
Answer:
Iron pyrites, silica, calcium carbonate and bicarbonates of calcium and magnesium.

Question 10.
What is meant by petrifaction?
Answer:
When animals die the original portion of their body may be replaced molecule for a molecule by minerals and the original substance being lost through disintegration. This method of fossilization is called petrifaction. The principal minerals involved in this type of fossilization are iron pyrites, silica, calcium carbonate and bicarbonates of calcium and magnesium.

Question 11.
Define analogous organ with an example.
Answer:
Organisms having different structural patterns but similar function are termed as analogous structures. For example, the wings of birds and insects are different structurally but perform the same function of flight that brings about convergent evolution.

Question 12.
Mention any four organs homologous to human hand.
Answer:
Flippers of whale, wings of bat, wings of bird and forelimb of horse.

Question 13.
Thorn of Bougainvillea and tendrils of Pisum sativum represent homology. How?
Answer:
The thorn of Bougainvillea and the tendrils of Curcurbita and Pisum sativum represent homology. The thorn in former is used as a defence mechanism from grazing animals and the tendrils of latter is used as a support for climbing.

Question 14.
Which type of evolution is brought out by homologous structures and analogous structures?
Answer:
Homologous structures brings about divergent evolution. Analogous structures brings about convergent evolution.

Question 15.
What are vestigial organs? Give example.
Answer:
Structures that are of no use to the possessor, and are not necessary for their existence are called vestigial organs. Vestigial organs may be considered as remnants of structures which were well developed and functional in the ancestors, but disappeared in course of evolution due to their non-utilization.
E.g: Human appendix.

Question 16.
Human appendix is a vestige. Give reason.
Answer:
Human appendix is the remnant of caecum which is functional in the digestive tract of herbivorous animals like rabbit. Cellulose digestion takes place in the caecum of these animals. Due to change in the diet containing less cellulose, caecum in human became functionless and is reduced to a vermiform appendix, which is vestigial.

Question 17.
What are connecting link? Give example.
Answer:
The organisms which possess the characters of two different groups (transitional stage) are called connecting links. Example Peripatus (connecting link between Annelida and Arthropoda) Archaeopteryx (connecting link between Reptiles and Aves).

Question 18.
Name one fossilised connecting link between reptiles and Aves also one living connecting link between Annelida and Arthropoda.
Answer:
Archaeopteryx – connecting link between Reptiles and Aves.
Peripatus – Connecting link between Annelida and Arthropoda.

Question 19.
Why it is considered as a connecting link?
Answer:
Peripatus is a worm that shown the characters of both Annelidia and Arthropoda. Hence it is a connecting link between Annelida and Arthropoda.

Question 20.
Atavistic organs – comment.
Answer:
Sudden appearance of vestigial organs in highly evolved organisms is called atavistic organs. For example, the presence of tail in human baby is an atavistic organ.

Question 21.
Define Ontogeny and Phytogeny.
Answer:

1. Ontogeny refers to the life history of an individual.
2. Phytogeny refers to the evolutionary history of a race.

Question 22.
Who proposed the theory of recapitulation? State the theory.
Answer:
Ernst Von Haeckel proposed the theory of recapitulation, which states that life history of an individual briefly repeats the evolutionary history of the race.

Question 23.
Name few Neo – Lamarckists.
Answer:
Cope, Osborn, Packard and Spencer.

Question 24.
Who proposed the theory of acquired characters? Also, mention the scientist who disproved it.
Answer:
The theory of acquired characters was proposed by Jean-Baptise de Lamarck and it was disproved by August Weismann.

Question 25.
Point out the basic principles of Darwin’s theory of evolution.
Answer:
Overproduction, the struggle for existence, Universal occurence of variation, Survival of fittest and Natural selection.

Question 26.
Name any four Neo – Darwinists.
Answer:
Gregor Mendel, August Weismann, Russel Wallace and Heinrich.

Question 27.
Enumerate the salient features of mutation theory.
Answer:

1. Mutations or discontinuous variations are transmitted to other generations.
2. In naturally breeding populations, mutations occur from time to time.
3. There are no intermediate forms, as they are fully-fledged.
4. They are strictly subjected to natural selection.

Question 28.
Who proposed the Mutation theory? Name the organism on which the experiment was carried out.
Answer:
Mutation theory was put forth by Hugo de Vries. Based on the experiments in Oenothera Lamarckian (The evening primrose plant).

Question 29.
What are the basic factors of modern synthetic theory that leads to evolution?
Answer:
Gene mutation, Chromosomal mutation, Genetic recombination, Natural selection and Reproductive isolation.

Question 30.
Name the scientists who supported modern synthetic theory.
Answer:
Sewell Wright, Dobzhansky, Huxley and Simpson.

Question 31.
Define point mutation.
Answer:
Gene mutation refers to the changes in the structure of the gene. It is also called gene/point mutation. It alters the phenotype of an organism and produces variations in their offsprings.

Question 32.
Point out the factors that alters the allelic frequency of a population.
Answer:
Natural selection, Genetic drift, Mutation and Geneflow

Question 33.
Mention any two differences between Homo habilis and Homo erectus
Answer:

1. Homo habilis: The brain capacity was between 650-800 cc. They were probably vegetarians.
2. Homo erectus: The brain capacity was around 900 cc. They probably ate meat.

Question 34.
Write a brief note on Homo sapiens with respect to evolution.
Answer:
Homo sapiens or modem humans arose in Africa some 25,000 years ago and moved to other continents and developed into distinct races. They had a brain capacity of 1300 – 1600 cc. “They started cultivating crops and domesticating animals.

Question 35.
Define evolution.
Answer:
The term evolution describes heritable changes in one or more characteristics of a population of species from one generation to the other.

3 – Mark Question

Question 36.
Write a short note on Big Bang theory.
Answer:
Big bang theory explains the origin of the universe as a singular huge explosion in physical terms. The primitive Earth had no proper atmosphere but consisted of ammonia, methane, hydrogen, and water vapour. The climate of the Earth was extremely high. UV rays from the Sun split up water molecules into hydrogen and oxygen. Gradually the temperature cooled and the water vapour condensed to form rain. Rainwater filled all the depressions to form water bodies. Ammonia and methane in the atmosphere combined with oxygen to form carbon dioxide and other gases.

Question 37.
Theory of chemical evolution states that organisms have evolved from inorganic substances. If so, what was the atmospheric condition that favoured evolution?
Answer:
The atmosphere was devoid of O₂, and with high level of CO₂, NH₃ and UV radiations.

Question 38.
Name the periods of Mesozoic era. Also mention the flora and fauna dominates during that periods.
Answer:

1. Mesozoic era is divided into three periods namely Triassic, Jurassic and Cretaceous.
2. Dominating Fauna : Reptiles and Dinosaurs Dominating
3. Flora : Conifers, Ferns and Ginkgon.

Question 39.
Which era is referred as Age of Mammals? What are the periods of that era? And also mention the fauna during the periods.
Answer:
Cenozoic era is called as Age of Mammals. Tertiary and Quaternary are the two periods of Cenozoic era. Tertiary periods marks the abundance of mammalian fauna. Quaternary period marks the beginning of human social life.

Question 40.
Write a short note on Cenozoic era.
Answer:
Cenozoic era (Age of mammals) is subdivided into two periods namely Tertiary and Quaternary. Tertiary period is characterized by abundant mammalian fauna. This period is subdivided into five epochs namely, Paleocene (placental mammals, Eocene (Monotremes except duck billed Platypus and Echidna, hoofed mammals and carnivores), Oligocene (higher placental mammals appeared), Miocene (origin of first man like apes) and Pliocene (origin of man from man like apes). Quaternary period witnesses decline of mammals and beginning of human social life.

Question 41.
Name the gaseous mixture used in Urey – Miller’s experiment. Which type of physical force is applied to generate amino acids?
Answer:
Ammonia, Methane, Hydrogen, Water vapour are the gaseous mixture allowed to circulate over electric discharge from a tungsten electrode.

Question 42.
Which is the most common methods of fossilization? Explain how it occurs.
Answer:
Actual remains – The original hard parts such as bones, teeth or shells are preserved as such in the Earth’s atmosphere. This is the most commpn method of fossilization. When marine animals die, their hard parts such as bones and shells, etc., are covered with sediments and are protected from further deterioration.

They get preserved as such as they are preserved in vast ocean; the salinity in them prevents decay. The sediments become hardened to form definite layers or strata. For example, Woolly Mammoth that lived 22 thousand years ago were preserved in the frozen coast of Siberia as such. Several human beings and animals living in the ancient city of Pompeii were preserved intact by volcanic ash which gushed out from Mount Vesuvius.

Question 43.
What are coprolites? Mention its role in phytogeny.
Answer:
Coprolites are the hardened faecal matters occurs as small pieces. Analysing the coprolites helps to understand the nature of diet of pre-historic animals.

Question 44.
What are moulds and casts?
Answer:
Even after disintegration, the body of an animal might leave indelible impression on the soft mud which later becomes hardened into stones. Such impressions are called moulds. The cavities of the moulds may get filled up by hard minerals and get fossilized, which are called casts.

Question 45.
How will you compute the age of fossil?
Answer:
The age of fossils can be determined using two methods namely, relative dating and absolute dating. Relative dating is used to determine a fossil by comparing it to similar rocks and fossils of known age. Absolute dating is used to determine the precise age of a fossil by using radiometric dating to measure the decay of isotopes.

Question 46.
“Ontogeny recapitulates phylogeny” – comment on the statement with example.
Answer:
The embryonic stages of a higher animal resemble the adult stage of its ancestors. Appearance of pharyngeal gill slits, yolk sac and the appearances of tail in human embryos are some of the examples.

Question 47.
Biogenetic law is not universal – justify.
Answer:
The biogenetic law is not universal and it is now thought that animals do not recapitulate the adult stage of any ancestors. The human embryo recapitulates the embryonic history and not the adult history of the organisms.

Question 48.
How macro molecules like DNA and RNA play their crucial role in evolutionary history?
Answer:
Molecular evolution is the process of change in the sequence composition of molecules such as DNA, RNA and proteins across generations. It uses principles of evolutionary biology and population genetics to explain patterns in the changes of molecules.

One of the most useful advancement in the development of molecular biology is proteins and other molecules that control life processes are conserved among species. A slight change that occurs over time in these conserved molecules (DNA, RNA and protein) are often called molecular clocks. Molecules that have been used to study evolution are cytochrome (respiratory pathway) and rRNA (protein synthesis).

Question 49.
Explain the principles of Lamarckian theory.
Answer:

1. The theory of use and disuse – Organs that are used often will increase in size and those that are not used will degenerate. Neck in giraffe is an example of use and absence of limbs in snakes is an example for disuse theory.
2. The theory of inheritance of acquired characters – Characters that are developed during ’ the life time of an organism are called acquired characters and these are then inherited.

Question 50.
Write a note on Mutation theory.
Answer:
Hugo de Vries put forth the Mutation theory. Mutations are sudden random changes that occur in an organism that is not heritable. De Vries carried out his experiments in the Evening Primrose plant (Oenothera Lamarckian) and observed variations in them due to mutation. According to de Vries, sudden and large variations were responsible for the origin of new species whereas Lamarck and Darwin believed in gradual accumulation of all variations as the causative factors in the origin of new species.

Question 51.
What do you mean by “adaptive radiation”? Give example.
Answer:
The evolutionary process which produces new species diverged from a single ancestral form becomes adapted to newly invaded habitats is called adaptive radiation. Adaptive radiations are best exemplified in closely related groups that have evolved in relatively short time. Darwin’s finches and Australian marsupials are best examples for adaptive radiation.

Question 52.
Darwins finches are the classical examples studied for adaptive radiation. Explain.
Answer:
Darwin’s finches are the birds whose common ancestor arrived on the Galapagos about 2 million years ago. During that time, Darwin’s finches have evolved into 14 recognized species differing in body size, beak shape and feeding behavior. Changes in the size and form of the beak have enabled different species to utilize different food resources such as insects, seeds and nectar from cactus flowers and blood from iguanas, all driven by Natural selection. Genetic variation in the ALX1 gene in the DNA of Darwin finches is associated with variation in the beak shape. Mild mutation in the ALX1 gene leads to a phenotypic change in the shape of the beak of the Darwin finches.

Question 53.
What is microevolution?
Answer:
Microevolution (evolution on a small scale) refers to the changes in allele frequencies within a population. Allele frequencies in a population may change due to four fundamental forces of evolution such as natural selection, genetic drift, mutation and gene flow.

Question 54.
Name the major types of Natural Selection.
Answer:

1. Stabilising Selection
2. Directional Selection
3. Disruptive Selection

Question 55.
What do you mean by gene flow?
Answer:
The movement of genes through gametes or movement of individuals in (immigration) and out (emigration) of a population is referred to as gene flow. Organisms and gametes that enter the population may have new alleles or may bring in existing alleles but in different proportions than those already in the population. Gene flow can be a strong agent of evolution.

Question 56.
Give an account on Genetic drift. Mention its impact on a population.
Answer:
Genetic drift is a mechanism of evolution in which allele frequencies of a population change over a generation due to chance (sampling error). Genetic drift occurs in all population sizes, but its effects are strong in a small population. It may result in a loss of some alleles (including beneficial ones) and fixation of other alleles. Genetic drift can have major effects, when the population is reduced in size by natural disaster due to bottleneck effect or when a small group of population splits from the main population to form a new colony due to the founder’s effect.

Question 57.
State Hardy – Weinberg equilibrium.
Answer:
The allele frequencies in a population are stable and are constant from generation to generation in the absence of gene flow, genetic drift, mutation, recombination and natural selection.

Question 58.
Write in brief about the characters of Australian ape-man.
Answer:
Australopithecus lived in East African grasslands about 5 mya and was called the Australian ape man. He was about 1.5 meters tall with bipedal locomotion, omnivorous, semi-erect, and lived in caves. Low forehead, brow ridges over the eyes, protruding face, lack of chin, low brain capacity of about 350 – 450 cc, human-like dentition, lumbar curve in the vertebral column were his distinguishing features.

Question 59.
Who is Cro-Magnon?
Answer:
Cro-Magnon was one of the most talked about forms of modem human found from the rocks of Cro-Magnon, France and is considered as the ancestor of modem Europeans. They were not only adapted to various environmental conditions but were also known for their cave paintings, figures on floors and walls.

5 – Mark Questions

Question 60.
Explain Oparin – Haldane hypothesis on evolution.
Answer:
According to the theory of chemical evolution, primitive organisms in the primordial environment of the Earth evolved spontaneously from inorganic substances and physical forces such as lightning, UV radiations, volcanic activities, etc. Oparin (1924) suggested that the organic compounds could have undergone a series of reactions leading to more molecules. He proposed that the molecules formed colloidal aggregates or ‘coacervates’ in an aqueous environment. The coacervates were able to absorb and assimilate organic compounds from the environment. Haldane (1929) proposed that the primordial sea served as a vast chemical laboratory powered by solar energy.

The atmosphere was oxygen-free and the combination of CO₂, NH₂ and UV radiations gave rise to organic compounds. The sea became a ‘hot’ dilute soup containing large populations of organic monomers and polymers. They envisaged that groups of monomers and polymers acquired lipid membranes and further developed into the first living cell. Haldane coined the term prebiotic soup and this became the powerful symbol of the Oparin-Haldane view on the origin of life (1924-1929). Oparin and Haldane independently suggested that if the primitive atmosphere was reducing – and if there was an appropriate supply of energy such as lightning or UV light then a wide range of organic compounds can be synthesized.

Question 61.
How Urey – Miller’s experiment supports the origin of life?
Answer:
Urey and Miller (1953), paved way for understanding the possible synthesis of organic compounds that led to the appearance of living organisms is depicted in the Figure In their experiment, a mixture of gases was allowed to circulate over electric discharge from a tungsten electrode. A small flask was kept boiling and the steam emanating from it was made to mix with the mixture of gases (ammonia, methane, and hydrogen) in the large chamber that was connected condensed to form water which ran down the ‘U’ tube.

Experiment was conducted continuously for a week and the liquid was analyzed. Glycine, alanine, beta-alanine, and aspartic acid were identified. Thus Miller’s experiments had an insight as to the possibility of abiogenetic synthesis of large amount of variety of organic compounds in nature from a mixture of sample gases in which the only source of carbon is methane. Later in similar experiments, the formation of all types of amino acids, and nitrogen bases were noticed.

Question 62.
Give a detailed account of Modern Synthetic Theory.
Answer:
Sewell Wright, Fisher, Mayer, Huxley, Dobzhansky, Simpson and Haeckel explained Natural Selection in the light of Post-Darwinian discoveries. According to this theory gene mutations, chromosomal mutations, genetic recombinations, natural selection and reproductive isolation are the five basic factors involved in the process of organic evolution.

1. Gene mutation refers to the changes in the structure of the gene. It is also called gene/ point mutation. It alters the phenotype of an organism and produces variations in their off +springs.
2. Chromosomal mutation refers to the changes in the structure of chromosomes due to deletion, addition, duplication, inversion or translocation. This too alters the phenotype of an organism and produces variations in their offspring.
3. Genetic recombination is due to crossing over of genes during meiosis. This brings about genetic variations in the individuals of the same species and leads to heritable variations.
4. Natural selection does not produce any genetic variations but once such variations occur it favours some genetic changes while rejecting others (driving force of evolution).
5. Reproductive isolation helps in preventing interbreeding between related organisms

Higher Order Thinking Skills (HO’ts) Questions

Question 1.
Name the connecting link for the following groups of organisms.

1. Annelida and Arthropoda
2. Reptiles and Aves
3. Pisces and Amphibians
4. Reptiles and Mammals

Answer:

1. Peripatus
2. Archeopteryx
3. Lungfish
4. Platypus

Question 2.
Point out any four condition under which Hardy Weinberg’s equilibrium is not attained.
Answer:

1. Selected mating
2. Flow of genes (either by immigration or emigration)
3. Occurance of mutation
4. Definite population size

Question 3.
Why are analogous structures a result of convergent evolution?
Answer:
Analogous structures are not anatomically similar though they perform same function.

Question 4.
Organs which are of no use to the organism is called as vestige. Name any four vestigal organs that can be noticed in your body.
Answer:
Wisdom teeth, Mammae in male, Body hair and Coccyx.

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Tamilnadu Samacheer Kalvi 12th Computer Applications Solutions Chapter 6 PHP Conditional Statements

Samacheer Kalvi 12th Computer Applications PHP Conditional Statements Text Book Back Questions and Answers

PART – I
I. Choose The Correct Answer

Question 1.
What will be the output of the following PHP code?
<?php
$x;
if ($x) print “hi”;
else
print “how are u”;?>
(a) how are u
(b) hi
(c) error
(d) no output
Answer:
(c) error

Question 2.
What will be the output of the following PHP code?
<?php
$x = 0;
if ($x++)
print “hi”;
else
print “how are u”;
?>
(a) hi
(b) no output
(c) error
(d) how are u
Answer:
(d) how are u

Question 3.
What will be the output of the following PHP code?
<?php
$x;
if ($x = 0)
print “hi”;
else
print “how are u”;
print “hello”
?>
(a) how are uhello
(b) hihello
(c) hi
(d) no output
Answer:
(d) no output

Question 4.
Statement which is used to make choice between two options and only option is to be performed is written as
(a) if statement
(b) if else statement
(c) then else statement
(d) else one statement
Answer:
(b) if else statement

Question 5.
What will be the output of the following PHP code?
<?php
$a =“”;
if ($a)
print “all”;
if
else
print “some”;
?>
(a) all
(b) some
(c) error
(d) no output
Answer:
(c) error

Question 6.
What will be the output of the following PHP code?
<?php
$a = “”;
if ($a)
print “all”;
if
else
print “some”;
?>
(a) all
(b) some
(c) error
(d) no output
Answer:
(c) error

Question 7.
What will be the output of the following PHP code?
<?php
$x = 10;
$y = 20;
if ($x > $y + $y != 3)
print “hi”;
else
print “how are u”;
?>
(a) how are u
(b) hi
(c) error
(d) no output
Answer:
(b) hi

Question 8.
What will be the output of the following PHP code?
<?php
$x = 10;
$y = 20;
if ($x > $y && 1||1)
print “hi” ;
else
print “how are u”;
?>
(a) how are u
(b) hi
(c) error
(d) no output
Answer:
(b) hi

Question 9.
What will be the output of the following PHP code?
<?php
if (-100)
print “hi”;
else
print “how are u”;
?>
(a) how are u
(b) hi
(c) error
(d) no output
Answer:
(b) hi

PART – II
II. Short Answer

Question 1.
Define Conditional Statements in PHP?
Answer:
Conditional statements are useful for writing decision making logics. It is most important feature of many programming languages, including PHP. They are implemented by the following types:

1. if Statement
2. if…else Statement
3. if…elseif….else Statement
4. switch Statement

Question 2.
Define if statement in PHP?
Answer:
If statement executes a statement or a group of statements if a specific condition is satisfied as per the user expectation.
Syntax: if (condition)
{
Execute statement(s) if condition is true;
}

Question 3.
What is if else statement in PHP?
Answer:
If else statement in PHP:

1. If statement executes a statement or a group of statements if a specific condition is satisfied by the user expectation.
2. When the condition gets false (fail) the else block is executed.

Syntax:
if (condition)
{
Execute statement(s) if condition is true;
}
else
{
Execute statement(s) if condition is false;
}

Question 4.
List out Conditional Statements in PHP?
Answer:
if Statement

1. if…else Statement
2. if…elseif….else Statement
3. switch Statement

Question 5.
Write Syntax of the If else statement in PHP?
Answer:
if (condition)
{
Execute statement(s) if condition is true;
}
else
{
Execute statement(s) if condition is false;
}

Question 6.
Define if…elseif….else Statement in PHP?
Answer:

1. If-elseif-else statement is a combination of if-else statement.
2. More than one statement can execute the condition based on user needs.

Question 7.
Usage of Switch Statement in PHP?
Answer:

1. The switch statement is used to perform different actions based on different conditions.
2. Switch statements work the same as if statementsbut they can check for multiple values at a time.

Question 8.
Write Syntax of Switch statement?
Answer:
switch (n) {
case label 1:
code to be executed if n=label 1; break; case label2:
code to be executed if n=label2; break; case label3:
code to be executed if n=label3; break;
default:
code to be executed if n is different from all labels;

Question 9.
Compare if and if else statement?
Answer:
If:
If statement executes a statement or a group of statements if a specific condition is satisfied as per the user expectation. When the condition fails, nothing happens.

If else:
If statement executes a statement or a group of statements if a specific condition is satisfied by the user expectation. When the condition gets false (fail) the else block is executed.

PART – III
III. Explain in Brief Answer

Question 1.
Write the features Conditional Statements in PHP?
Answer:
PHP Conditional statements:

1. Conditional statements are useful for writing decision making logics.
2. It is most important feature of many programming languages, including PHP.
3. They are implemented by the following types:
4. if Statement
5. if…else Statement
6. if…elseif….else Statement
7. switch Statement

Question 2.
Write the purpose of if elseif else statement?
Answer:

1. If statement executes a statement or a group of statements if a specific condition is satisfied by the user expectation. When the condition gets false (fail) the else block is executed.
2. If-elseif-else is a combination of if-else statement.
3. More than one statement can execute the condition based on user needs.
4. User can select from many choices.

Question 3.
Differentiate Switch and if else statement?
Answer:
Switch:

1. Switch statement uses single expression for multiple choices.
2. If all the cases are not matched default will be executed.
3. It test only for equality.
4. It evaluates only for character or integer

If-else:

1. If else statement uses multiple statement for multiple choices.
2. Nothing happens.
3. It tests for equality as well as for logical ‘ expressions.
4. any data type is allowed.

Question 4.
Write Short notes on Switch statement?
Answer:

1. The switch statement is used to perform different actions based on different conditions.
2. It tests for equality only. –
3. It uses default value, when all the case values are not matched.
4. It can have multiple ease values.

Question 5.
Differentiate if statement and if elseif else statement?
Answer:
if elseif else statement:

1. If-elseif-else statement is a combination of if-else statement. .
2. More than one statement can execute the condition based on user needs.
3. If the condition is false more alternatives are there.

PART – IV
IV. Explain in detail

Question 1.
Explain Function Conditional Statements in PHP?
Answer:
Function Conditional Statements:

1. Function conditional statement is the function specified inside the conditional statements.
2. We can’t call conditional function before its definition.

Syntax:
if(expression)
{
function function_name( )
{
block of statements;
}
}
function_name( ); // calling function.
Eg:
<? php
display( );
if(TRUE)
{
function display( )
{
echo “condition and function”;
}
}
Output: condition and function

Question 2.
Discuss in detail about Switch statement with an example?
Answer:
Switch Case:

1. The switch statement is used to perform different actions based on different conditions.
2. Switch statement tests only for equality.
3. More case values can be given.
4. When all the case values are not matched, then default will be executed.

Syntax:
switch (n) {
case label 1:
code to be executed if n=labe11;
break;
case labe12:
code to be executed if n=labe12;
break;
case labe13:
code to be executed if n=labe13; ,
break;
…….
default:
code to be executed if n is different from all labels;
}

Question 3.
Explain the process Conditional Statements in PHP?
PHP Conditional statements:
Answer:
Conditional statements are useful for writing decision making logics. It is most important feature of many programming languages, including PHP. They are implemented by the following types:

(i) if Statement:
If statement executes a statement or a group of statements if a specific condition is satisfied as per the user expectation.

(ii) if…else Statement:
If statement executes a statement or a group of statements if a specific condition is satisfied by the user expectation. When the condition gets false (fail) the else block is executed.

(iii) if…elseif….else Statement:
If-elseif-else statement is a combination of if-else statement. More than one statement can execute the condition based on user needs.

(iv) Switch Case:
The switch statement is used to perform different actions based on different conditions based on different conditions.

Question 4.
Explain concepts of if elseif else statement?
Answer:
If elseif else statement in PHP:
If-elseif-else statement is a combination of if-else statement. More than one statement can execute the condition based on user needs.
Syntax:
if (1st condition)
{
Execute statement(s) if condition is true;
}
elseif(2nd condition)
{
Execute statement(s) if 2nd condition is true;
}
else
{
Execute statement(s) if both conditions are false;
}
Example:
<?php
$Pass_Mark=35;
$first_class=60;
$Student_Mark=70;
if ($Student_Mark>= $first_class)
{
esho “The Student is eligible for the promotion with First Class”;
elseif ($Student_Mark>= $Pass_Mark)
{
echo “The Student is eligible for the promotion”;
}
else
{
echo “The Student is not eligible for the promotion”;
}?>

Question 5.
Explain if else statement in PHP?
Answer:
If else statement in PHP:
If statement executes a statement or a group of statements if a specific condition is satisfied by the user expectation. When the condition gets false (fail) the else block is executed.
Syntax:
if (condition)
{
Execute statement(s) if condition is true;
} else
{
Execute statement(s) if condition is false;
}
Example:
<?php
$Pass_Mark=35;
$Student_Mark=70;
if ($Student_Mark>= $Pass_Mark)
{
echo “The Student is eligible for the promotion”;
}
else
{
echo “The Student is not eligible for the promotion”; }
?>

Samacheer Kalvi 12th Computer Applications Solutions PHP Conditional Statements Additional Question and Answer

I. Choose the Best Answer

Question 1.
How many types of php conditional statements are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 2.
When the condition in If-else block fails then …………………………. block will be executed?
(a) if
(b) else
(c) Nested if
(d) while
Answer:
(b) else

Question 3.
The ……………………….. statement is used to perform different actions based on different conditions.
Answer:
switch

Question 4.
The …………………………… block will be executed if it is not matched with any of the case values in switch.
Answer:
default

Question 5.
The if statement contains ………………………….. expressions.
(a) arithmetic
(b) logical
(c) boolean
(d) Terriary
Answer:
(c) boolean

Question 6.
If statements have to be given in …………………………
(a) ( )
(b) <>
(c) [ ]
(d) { }
Answer:
(a) ( )

Question 7.
The block of statements have to be enclosed with
(a) ( )
(b) <>
(c) [ ]
(d) { }
Answer:
(d) { }

Question 8.
Which of the following can check for multiple values at a time?
(a) If
(b) If else
(c) Nested else
(d) Switch
Answer:
(d) Switch

II. Short Answer

Question 1.
Give the Syntax for If else if else statements in php?
Answer:
if (1st condition)
{
Execute statement(s) if condition is true;
}
elseif(2nd condition)
{
Execute statement(s) if 2nd condition is true;
}
else
{
Execute statement(s) if both conditions are false;
}

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Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 12 Final Accounts of Sole Proprietors – I

Samacheer Kalvi 11th Accountancy Final Accounts of Sole Proprietors – I Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

Question 1.
Closing stock is an item of ……………..
(a) Fixed asset
(b) Current asset
(c) Fictitious asset
(d) Intangible asset
Answer:
(b) Current asset

Question 2.
Balance sheet is ……………..
(a) An account
(b) A statement
(c) Neither a statement nor an account
(d) None of the above
Answer:
(b) A statement

Question 3.
Net profit of the business increases the ……………..
(a) Drawings
(b) Receivables
(c) Debts
(d) Capital
Answer:
(d) Capital

Question 4.
Carriage inwards will be shown ……………..
(a) In the trading account
(b) In the profit and loss account
(c) On the liabilities side
(d) On the assets side
Answer:
(a) In the trading account

Question 5.
Bank overdraft should be shown ……………..
(a) In the trading account
(b) Profit and loss account
(c) On the liabilities side
(d) On the assets side
Answer:
(c) On the liabilities side

Question 6.
Balance sheet shows the …………….. of the business.
(a) Profitability
(b) Financial position
(c) Sales
(d) Purchases
Answer:
(b) Financial position

Question 7.
Drawings appearing in the trial balance is ……………..
(a) Added to the purchases
(b) Subtracted from the purchases
(c) Added to the capital
(d) Subtracted from the capital
Answer:
(d) Subtracted from the capital

Question 8.
Salaries appearing ill the trial balance is shown on the ……………..
(a) Debit side of trading account
(b) Debit side of profit and loss account
(c) Liabilities side of the balance sheet
(d) Assets side of the balance sheet
Answer:
(b) Debit side of profit and loss account

Question 9.
Current assets does not include ……………..
(a) Cash
(b) Stock
(c) Furniture
(d) Prepaid expenses
Answer:
(c) Furniture

Question 10.
Goodwill is classified as ……………..
(a) A current asset
(b) A liquid asset
(c) A tangible asset
(d) An intangible asset
Answer:
(d) An intangible asset

II. Very Short Answer Questions

Question 1.
Write a note on trading account.
Answer:
Trading refers to buying and selling of goods with the intention of making profit. The trading account is a nominal account which shows the result of buying and selling of goods for an accounting period. Trading account is prepared to find out the difference between the revenue from sales and cost of goods sold.

Question 2.
What are wasting assets?
Answer:
These are the assets which get exhausted gradually in the process of excavation. Examples: mines and quarry.

Question 3.
What are fixed assets?
Answer:
Fixed assets are those assets which are acquired or constructed for continued use in the business and last for many years such as land and building, plant and machinery, motor vehicles, furniture, etc.

Question 4.
What is meant by purchases returns?
Answer:
Goods purchased which are returned to suppliers are termed as purchases returns or returns outward.

Question 5.
Name any two direct expenses and indirect expenses.
Answer:
Direct expenses:

• Carriage inwards or freight inwards
• Wages Indirect expenses:

Indirect expenses:

• Office and administrative expenses
• Selling and distribution expenses

Question 6.
Mention any two differences between trial balance and balance sheet.
Answer:

S.No.	Basis	Trial Balance	Balance Sheet
1.	Nature	Trial balance is a list of ledger balances on a particular date.	Balance sheet is a statement showing the position of assets and liabilities on a particular date.
2.	Purpose	Trial balance is prepared to check the arithmetical accuracy of the accounting entries made.	Balance sheet is prepared to ascertain the financial position of a business.

Question 7.
What are the objectives of preparing trading account?
Answer:

1. Provides information about gross profit or gross loss.
2. Provides an opportunity to safeguard against possible losses.

Question 8.
What is the need for preparing profit and loss account?
Answer:

1. Ascertainment of net profit or net loss
2. Comparison of profit
3. Control on expenses
4. Helpful in the preparation of balance sheet.

III. Short Answer Questions

Question 1.
What are final accounts? What are its constituents?
Answer:
Businessmen want to know the profitability and the financial position of the business. These can be ascertained by preparing the final accounts or financial statements. The final accounts or financial statements include the following:

1. Income statement or trading and profit and loss account; and
2. Position statement or Balance sheet.

Question 2.
What is meant by closing entries? Why are they passed?
Answer:
Balances of all the nominal accounts are required to be closed on the last day of the accounting year to facilitate the preparation of trading and profit and loss account. It is done by passing necessary closing entries in the journal proper. Purchases has debit balance and purchases returns has credit balance. At the end of the accounting year, the balance in purchases returns account is closed by transferring to purchase account.

Question 3.
What is meant by gross profit and net profit?
Answer:

1. If the amount of sales exceeds the cost of goods sold, the difference is gross profit.
   Sales – Cost of goods sold = Gross profit.
2. If the total of the credit side of the profit and loss account exceeds the debit side, the difference is termed as net profit.

Question 4.
“Balance sheet is not an account” – Explain.
Answer:
A balance sheet is a part of the final accounts. However, the balance sheet is a statement and not an account. It has no debit or credit sides and as such the words ‘To’ and ‘By’ are not used before the names of the accounts shown therein.

Question 5.
What are the advantages of preparing a balance sheet?
Answer:
Balance sheet discloses the financial position of a business on a particular date, it gives
the balances only for the date on which it is prepared. It shows the financial position of the business according to the going concern concept.

Question 6.
What is meant by grouping and marshalling of assets and liabilities?
Answer:
1. The term ‘grouping’ means showing the items of similar nature under a common heading. For example, the amount due from various customers will be shown under the head‘sundry debtors’.

2. ‘Marshalling’ is the arrangement of various assets and liabilities in a proper order. Marshalling can be made in one of the following two ways:

• In the order of liquidity
• In the order of permanence

IV. Exercises

Question 1.
Prepare trading account in the books of Sivashankar from the following figures:

Answer:
Trading account of Sivashankar

Question 2.
Prepare trading account in the books of Mr. Sanj ay for the year ended 31^st December 2017:

Answer:
Trading account of Mr. Sanjay for the year ended 31^st December, 2017

Question 3.
Prepare trading account in the books of Mr. Sanj ay for the year ended 31^st December 2017:

Answer:
Trading account of Saravanan for the year ended 31^st December, 2017

Question 4.
From the following details for the year ended 31^st March, 2018, prepare trading account

Answer:
Trading account for the year ended 31^st March, 2018

Question 5.
Ascertain gross profit or gross loss from the following:

Answer:
Profit and Loss Account

Question 6.
From the following balances taken from the books of Victor, prepare trading account for the year ended December 31,2017:

Answer:
Trading Account of Victor for the year ended 31^st December, 2017

Hint : Closing stock will not appear in trading account because of adjusted purchases have been given.

Question 7.
Compute cost of goods sold from the following information:

Answer:
Cost of goods sold = Opening stock + Net purchases + Direct expenses – Closing stock
= 10,000 + 80,000 + 7,000 – 15,000
= ₹ 82,000
Note: Indirect expenses do not form part of cost of goods sold.

Question 8.
Find out the amount of sales from the following information:

Answer:
Cost of goods sold = Opening stock + Net purchases + Direct expenses – Closing stock
= 30,000 + 2,00,000 + 0 – 20,000
= ₹ 2,10,000

Therefore, percentage of gross profit on cost of goods sold is
\(\frac { 30 }{ 70 }\) x 100 = 42.85% (42.857142 ……….)
Gross profit = 42.85% on 2,10,000 i.e., \(\frac { 42.85 }{ 100 }\) x 2,10,000 = ₹ 90,000
Sales = Cost of goods sold + Gross Profit
= 2,10,000 + 90,000 (Fractions to be rounded)
= ₹ 3,00,000

Question 9.
Prepare profit and loss account in the books of Kirubavathi for the year ended 31^st December, 2016 from the following information:

Answer:
Profit and loss account of Kirubavathi for the year ended 31^st Dec, 2016

Question 10.
Ascertain net profit or net loss from the following:

Answer:
Profit and loss account

Question 11.
From the following details, prepare profit and loss account.

Answer:
Profit and loss account

(Hint: Freight inwards will not appear in profit and loss account as it is a direct expense)

Question 12.
From the following information, prepare profit and loss account for the year ending 31^st December, 2016.

Answer:
Profit and loss account for the year ended 31^st December, 2016

Question 13.
From the following balances obtained from the books of Mr. Ganesh, prepare trading and profit and loss account:

Answer:
Trading and Profit & loss account of Mr. Ganesh

Question 14.
From the following balances extracted from the books of a trader, ascertain gross profit and net profit for the year ended March 31,2017:

Answer:
Trading and Profit & Loss account for the year ended 31^st March, 2017

Question 15.
From the following particulars, prepare balance sheet in the books of Bragathish as on 31^st December, 2017:

Answer:
Balance Sheet of Bragathish as on 31^st December, 2017

Question 16.
Prepare trading and profit and loss account in the books of Ramasundari for the year ended 31^st December, 2017 and balance sheet as on that date from the following information:

Answer:
Trading and Profit & Loss a/c of Ramasundari for the year ended 31 Dec, 2017 Cr.

Balance Sheet of Ramasundari as on 31^st March, 2018

Question 17.
From the Trial balance, given by Saif, prepare final accounts for the year ended 31^st March, 2018 in his books:

Answer:
Trading and Profit & Loss a/c of Saif for the year ended 31 March, 2018

Balance Sheet of Saif as on 31^st March, 2018

Question 18.
Prepare trading and profit and loss account and balance sheet in the books of Deri, a trader, from the following balances as on March 31, 2018.

Answer:
Trading and Profit & Loss a/c of Deri for the year ended 31^st March, 2018

Balance Sheet of Deri as on 31^st March, 2018

Textbook Case Study Solved

Question 1.
Mr. Abhinav started a small shop of selling dairy products. He wanted to maintain proper books of accounts. But, he had very little knowledge of accounting. He maintained only three books – purchases, sales and cash book by himself. He bought some dairy products and a refrigerator to store the milk products for which the payment was made by cheque but recorded the same in the purchases book. He also spent for the transportation charges and paid some money to the person who unloaded the stock. He recorded the same in the cash book.

He made both cash and credit sale for the next few weeks. He entered the entire sales in the sales book. In the middle of the month, he was in need of some money for his personal use. So he took some money, but did not record in the books.
Now, discuss on the following points:

Question 1.
Do you think Mr. Abhinav needs an accountant? Why do you think so?
Answer:
Yes, Mr. Abhinav needs an accountant because he records all cash and credit transactions.

Question 2.
Does he maintain enough books of accounts?
Answer:
Yes, he maintains enough books of accounts.

Question 3.
What other books do you think that he needs to maintain?
Answer:
He needs to maintain a petty cashbook.

Question 4.
What will be the impact on the profit, if he records the purchase of refrigerator in the purchases book?
Answer:
The purchase book will be overcast because this transaction will be recorded in proper journal.

Question 5.
Is it important to record the money taken for personal use? Will it affect the final accounts?
Answer:
Yes, then only the actual profit or loss can be found out in the business.

Question 6.
Identify some of the accounting principles relevant to this situation.
Answer:
Some of the accounting principles relevant to this situation are: matching principles, business entity concept, money measurement concept, dual output concept, periodicity concept and going concern concept.

Samacheer Kalvi 11th Accountancy Final Accounts of Sole Proprietors – I Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer

Question 1.
Income statement is divided into ……………….. parts.
(a) one
(b) two
(c) three
(d) four
Answer:
(b) two

Question 2.
The first part of the income statement is ………………..
(a) Final account
(b) Trading account
(c) Profit and Loss account
(d) Balance Sheet
Answer:
(b) Trading account

Question 3.
Balances of all ……………….. accounts are required to be closed on the last day of the accounting year.
(a) Nominal
(b) Personal
(c) Real
(d) Representative personal
Answer:
(a) Nominal

Question 4.
Trading account is a ……………….. account.
(a) Personal
(b) Nominal
(c) Real
(d) Representative personal
Answer:
(b) Nominal

Question 5.
Sales – Gross Profit = ………………..
(a) Sales
(b) Cost of goods sold
(c) Gross profit
(d) Gross loss
Answer:
(b) Cost of goods sold

Question 6.
……………….. account is the second part of income statement.
(a) Trading
(b) Profit and Loss
(c) Balance sheet
(d) Final
Answer:
(b) Profit and Loss

Question 7.
Which one is correctly matched?
(a) Bad debts – Indirect expense
(b) Wages – Asset
(c) Salary – Trading account
(d) Net Profit – Asset
Answer:
(a) Bad debts – Indirect expense

Question 8.
Balances of all the personal and real account are shown in ………………..
(a) Trading account
(b) Profit and loss account
(c) Income statement
(d) Balance sheet
Answer:
(d) Balance sheet

Question 9.
A balance sheet is a part of the ……………….. account.
(a) Trading
(b) Profit and Loss
(c) Income statement
(d) Final
Answer:
(d) Final

Question 10.
……………….. is a summary of the personal and real accounts.
(a) Balance sheet
(b) Final account
(c) Trading account
(d) Profit and loss account
Answer:
(a) Balance sheet

Question 11.
The balance sheet of business concern can be presented in the ……………….. forms.
(a) two
(b) three
(c) four
(d) six
Answer:
(a) two

Question 12.
Marshalling can be made in one of the ……………….. ways.
(a) three
(b) two
(c) four
(d) five
Answer:
(b) two

Question 13.
These are the assets which get exhausted gradually in the process of excavation.
(a) Wasting assets
(b) Nominal assets
(c) Liquid assets
(d) Current assets
Answer:
(a) Wasting assets

Question 14.
……………….. liabilities are not shown in the balance sheet.
(a) Contingent
(b) Current
(c) Liquid
(d) Fixed
Answer:
(a) Contingent

Students can Download Computer Applications Chapter 9 Connecting PHP and MYSQL Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Computer Applications 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 12th Computer Applications Solutions Chapter 9 Connecting PHP and MYSQL

Samacheer Kalvi 12th Computer Applications Connecting PHP and MYSQL Text Book Back Questions and Answers

PART – I
I. Choose The Correct Answer

Question 1.
Which one of the following statements instantiates the mysqli class?
(a) mysqli = new mysqli( )
(b) $mysqli = new mysqli( )
(c) $mysqli->new.mysqli( )
(d) mysqli->new.mysqli( )
Answer:
(b) $mysqli = new mysqli( )

Question 2.
which one is correct way, we can retrieve the data in the result set of MySQL using PHP?
(a) mysql_fetch_row
(b) mysql_fetch_array
(c) mysql_fetch_object
(d) All the above
Answer:
(d) All the above

Question 3.
How Can we Create a Database Using PHP and MySQL?
(a) mysqli_create_db(“Database Name”)
(b) mysqli_create_db(“Data”)
(c) create_db(“Database Name”)
(d) create_db(“Data”)
Answer:
(a) mysqli_create_db(“Database Name”)

Question 4.
Which is the correct function to execute the SQL queries in PHP?
(a) mysqli_query(“Connection Object”,“SQL Query”)
(b) query(“Connection Object”, “SQL Query”)
(c) mysql_query(“Connection Object”,“SQL Query”)
(d) mysql_query(“SQL Query”)
Answer:
(a) mysqli_query(“Connection Object”,“SQL Query”)

Question 5.
Which is the correct function Closing Connection in PHP?
(a) mysqli_close(“Connection Object”)
(b) close(“Connection Object”);
(c) mysql_close(“Connection Object”)
(d) mysqli_close(“Database Object”);
Answer:
(a) mysqli_close(“Connection Object”)

Question 6.
Which is the correct function-to establish Connection in PHP?
(a) mysqli_connect(“Server Name”,“User Name”,“Password”,“DB Name”);
(b) connect(“Server Name”,“User Name”,“Password”,“DB Name”);
(c) mysql_connect(“Server Name”,“User Name”,“Password”,“DB Name”);
(d) mysqli connect (“Database Object”);
Answer:
(a) mysqli_connect(“Server Name”,“User Name”,“Password”,“DB Name”);

Question 7.
Which is the not a correct MySQL Function in PHP?
(a) Mysqli_connect( ) Function
(b) Mysqli_close( ) Function
(c) mysqli_Select_data( ) Function
(d) mysqli_affected_rows( ) Function
Answer:
(c) mysqli_Select_data( ) Function

Question 8.
How many parameter are required for MYSQLi connect function in PHP?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 9.
How many parameter are required for MYSQLi query function in PHP?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 10.
How many parameter are required for MYSQLi Close function in PHP?
(a) 1
(b) 2
(c) 3
(d) 5
Answer:
(a) 1

Question 11.
Which version of PHP supports MySQLi functions?
(a) Version 2.0
(b) Version 3.0
(c) Version 4.0
(d) Version 5.0
Answer:
(d) Version 5.0

PART – II
II. Short Answer

Question 1.
What are the MySQLi function available PHP?
Answer:

1. Mysqli_connect( ) Function
2. Mysqli_close( ) Function
3. mysqli_select_db( ) Function
4. mysqli_affected_rows( ) Function
5. mysqli_connect_error( ) Function
6. mysqlifetchassoc( ) Function

Question 2.
What is MySQLi function?
Answer:
MySQLi functions:

1. The Mysqli functions allows you to access MySQL database servers.
2. For the MySQL functions to be available, you must compile PHP with support for the MySQLi extension.

Question 3.
What are the types MySQLi function available PHP?
Answer:
Types of MySQL Functions in PHP:

1. Database connections
2. Managing Database connections
3. Performing Queries
4. Closing connection

Question 4.
Difference between Connection and Close function?
Answer:
Connection:

1. Before accessing MySQL Database, connect to Database Server machine via PHP scripting language using Mysqli_ connect( ) Function.
2. This function requires 4 parameters.

Close function:

1. mysqli close( ) Function is used to close an existing opened database connection between PHP scripting and MySQL Database Server.
2. This function requires only one parameter.

Question 5.
Give few examples of MySQLi Queries?
Answer:
$sql=“SELECT student_name,student_age FROM student”;mysqli_query($con,$sql); sql stmt = “SELECT * FROM my contacts”;
$result = mysqli_query($connection,$sql_stmt).

Question 6.
What is Connection string?
Answer:
The variables are used to connect to the Database server. They are

1. $servername → Database Server IP address
2. $usemame → Database Server User Name
3. $password → Database Server Password
4. $DB_Name → Database Name

The mysqli connect function uses these variables and connect Database server from PHP scripting. If connection gets fail, output will be printed with MySQL error code. Otherwise connection is success.

Question 7.
What is web Database?
Answer:
Web Database:
A web database is a wide term for managing data online. A web database gives you the ability to build your own databases.
Ex. Bank, airline and rental car reservation.

Question 8.
What is mysqli_fetch_assoc( ) Function?
Answer:
mysqli_fetch_assoc( );
Fetches a result row as an associative array.
Syntax:
mysqli_fetch_assoc(result);

Question 9.
Define mysqli_connect_error( ) Function?
Answer:
mysqli connect_error( ):
It returns the error description from the last connection error.
Syntax:
mysqli_connect_ertror( ):

Question 10.
Define mysqii_affected_rows( ) Function?
Answer:
my sql i_affected_rows( ):
mysqli_affected_rows( ) returns the number of affected rows in the previous MYSQL operation.

Syntax:
mysqli_affected_rows(connection)

PART – III
III. Explain in Brief Answer

Question 1.
Write the Syntax for MySQLi Queries?
Answer:
Syntax:
mysqli_query(“Connection Object’’,“SQL Query”)

Question 2.
Write is the purpose of MySQLi function available?
Answer:
MySQL Function in PHP
In PHP Scripting language many functions are available for MySQL Database connectivity and executing SQL queries.MySQLi is extension in PHP scripting language which gives access to the MYSQL database. MySQLi extension was introduced version 5.0.0.

The MySQLi extension contains the following important functions which are related to MySQL database connectivity and management.

1. Mysqli_connect( ) Function
2. Mysqli_close( ) Function
3. mysqli_select_db( ) Function
4. mysqli_affected_rows( ) Function
5. mysqli_connect_error( ) Function
6. mysqli_fetch_assoc( ) Function

Question 3.
Differentiate mysqli_affected_rows( ) Function and mysqli_fetch_assoc( ) Function?
Answer:
Fetch assoc( )
mysqli_fetch_assoc( ):
Fetches a result row as an associative array.

Syntax:
mysqli fetch_assoc(result)

Affected rows( )
my sqli_affected_ro ws( ):
mysqli affected_rows( ) returns the number of affected rows in the previous MYSQL operation.

Syntax:
mysqli_affected_rows(connection)

Question 4.
Write MySQL Connection Syntax with example?
Answer:
Syntax:
mysqli_eonnect(“Server Name”, “User Name”, “password”, “DB Name”);
Example:
<?php
Sservername = “localhost”;
Susername = “username”;
Spassword = “password”;
$DB_name = “SchooLDB”;
// Create connection
$conn = mysqli_connect($servemame, Susername, Spassword,$DB_name);

Question 5.
Write a note PHP MySQL database connection?
Answer:
Database Connections:
Before accessing MySQL Database, connect to Database Server machine via PHP scripting language using Mysqli_connect() Function.

Syntax:
mysqli_connect(“Server Name”, “User Name”,“Password”,“DB Name”);
This function requires four parameters to connect to database server. Database Server name, Database username, password and Database Name.

Managing Database Connections:
The below code snippet describes managing database connection methods and features. c?php
$servername = “localhost”;
$username = “username”;
$password = “password”;
$DBname = “SchoolDB”;
// Create connection
$conn = mysqli_connect($servemame, Susername, Spassword,$DB_name).

PART – IV
IV. Explain in detail

Question 1.
Discuss in detail about MySQL functions with example?
Answer:
MySQL Function in PHP:
In PHP Scripting language many functions are available for MySQL Database connectivity and executing SQL queries. MySQLi is extension in PHP scripting language which gives access to the MYSQL database. MySQLi extension was introduced version 5.0.0.

The MySQLi extension contains the following important functions which are related to MySQL database connectivity and management.

1. Mysqli_connect( ) Function
2. Mysqli_close( ) Function
3. mysqli_select_db( ) Function
4. mysqli_affected_rows( ) Function
5. mysqli_connect_error( ) Function
6. mysqli_fetch_assoc( ) Function

1. Database Connections:
Before accessing MySQL Database, connect to Database Server machine via PHP scripting language using Mysqli_connect() Function.
Syntax:
mysqli_connect(“Server Name”,“User Name”,“Password”,“DB Name”);
This function requires four parameters to connect to database server. Database Server name, Database username, password and Database Name.

2. Managing Database Connections:
The below code snippet describes managing database connection methods and features.
<?php
$servemame = “localhost”;
$usemame = “username”;
$password = “password”;
$DB_name = “School_DB”;
// Create connection
$conn = mysqli_connect($servemame, Susemame, $password,$DB_name);
The mysqli connect function uses these variables and connect Database server from PHP scripting. If connection gets fail, output will be printed with MySQL error code. Otherwise connection is success.

3. Performing Queries:
The main goal of MySQL and PHP connectivity is to retrieve and manipulate the data from MySQL database server. The SQL query statements are helping with PHP MySQL extension to achieve the objective of MySQL and PHP connection, “mysqliquery” is a function, helps to execute the SQL query statements in PHP scripting language.
Syntax:
mysqli_query(“Connection Object”,’’SQL Query”)

Example:
$con=mysqli_connect(“localhost”,“my_user”,“my_password”,“Student_DB “); $sql=”SELECT student_name,student_age FROM student”;mysqli_query($con,$sql);

4. Closing Connection:
mysqli_close( ) Function is used to close an existing opened database connection between PHP scripting and MySQL Database Server.

Question 2.
Explain the Database error handling and management process in PHP?
Answer:
Managing Database Connections:
The below code snippet describes managing database connection methods and features.
<?php
$servemame = “localhost”;
$username = “username”;
$password = “password”;
$DBname = “SchoolDB”;
// Create connection
Sconn = mysqli_connect($servemame, Susername, Spassword,SDBname);
// Check connection
if (! Sconn) {
die(“Ccnnection failed: “ . mysqli_connect_error( ));
}.
echo “Connected successfully”;
?>,
In the above code snippet, three variables are used to connect to the Database server. They are

1. $servemame → Database Server IP address
2. $usemame → Database Server User Name
3. $password → Database Server Password
4. $DB_Name → Database Name

The mysqli_connect function uses these variables and connect Database server from PHP scripting. If connection gets fail, output will be printed with MySQL error code. Otherwise connection is success.

Question 3.
Explain in details types of MySQL connection method in PHP?
Answer:
Database Connections:
Before accessing MySQL Database, connect to Database Server machine via PHP scripting language using Mysqli_connect( ) Function.

Syntax:
mysqli_connect(“Server Name”,“User Name”,“Password”,“DB Name”);
This function requires four parameters to connect to database server. Database Server name, Database username, password and Database Name.

Managing Database Connections:
The below code snippet describes managing database connection methods and features.
<?php
$servemame = “localhost”;
$usemame = “username”;
$password = “password”;
SDB_name = “SchoolDB”;
// Create connection
$conn = mysqli_connect($servemame, Susemame, Spassword,$DB_name);
// Check connection
if(!$conn){
die(“Connection failed: “ . mysqli_connect_error( ));
}
echo “Connected successfully”;
?>
In the above code snippet, three variables are used to connect to the Database server. They are

1. $servemame → Database Server Server IP address
2. $username → Database Server User Name
3. $password → Database Server Password
4. $DBName → Database Name

The mysqli_connect function uses these variables and connect Database server from PHP scripting. If connection gets fail, output will be printed with MySQL error code. Otherwise connection is success.

Question 4.
Explain MySQLi Queries with examples?
Answer:
Performing Queries:
The main goal of MySQL and PHP connectivity is to retrieve and manipulate the data from MySQL database server. The SQL query statements are helping with PHP MySQL extension ^to achieve the objective of MySQL and PHP connection. “mysqli_query” is a function, helps to execute the SQL query statements in PHP scripting language.

Syntax:
mysqli_query(“Connection Object” “SQL Query”)
Example:
$con=mysqli_connect(“localhost”,“my_user”,“my_password”,“Student_DB”); $sql=“SELECT student_name,student_age FROM student”;mysqli_query($con,$sql);

Closing Connection:
mysqli_close( ) Function is used to close an existing opened database connection between. PHP scripting and MySQL Database Server.

Syntax:
mysqli_close(“Connection Object”);
<?php
$con=mysqli_connect(“localhost”,“$user”,“$password”,“SCHOOLDB”);
// ….some PHP code… mysqli_close($con);
?>

Example of PHP and MySQL Program:
<?php .
$servemame = “localhost”;
$usemame = “username”;
$password = “password”;
$dbname = “schoolDB”;
$connection = mysqli_connect(“$servemame”, “$usemame”, “$password” “$dbname”);
if (mysqli_connect_error ( ))
{
echo “Failed to connect to MySQL:”
mysqli_connect_error( );
}
sql stmt = “SELECT * FROM mycontacts”; //SQL select query
$result = mysqli_query($connection,$sql_stmt);//execute SQL statement$rows =
mysqli__num_r°ws($result);// get number of rows returned
if($rows) {
while ($row = mysqli_fetch_array($result)) {
echo ‘ID:’. $row[‘id’]. ‘<br>’;

Samacheer Kalvi 12th Computer Applications Solutions Connecting PHP and MYSQL Additional Question and Answer

I. Choose The Best Answer

Question 1.
The combination of PHP and MYSQL has become very popular …………………………. web scripting language in internet.
Answer:
Server side

Question 2.
According to recent Survey, approximately ………………………….. websites are running using PHP.
Answer:
544 million

Question 3.
Which is used to convert PHP code into C++?
(a) LPLP
(b) HPHP
(c) BPBP
(d) APAP
Answer:
(b) HPHP

Question 4.
Expand HPHP:
(a) W\gh Power High Power
(b) Heavy Processor Heavy Processor
(c) Hip Hop
(d) High Hop
Answer:
(c) Hip Hop

Question 5.
Which of the following has written alternative version of PHP?
(a) Facebook
(b) Twitter
(c) Instagram
(d) Whatsapp
Answer:
(a) Facebook

Question 6.
RAD meAnswer: ………………………..
(a) Rigid Application Design
(b) Rapid Application Development
(c) Rare App Design
(d) Raster Audio Development
Answer:
(b) Rapid Application Development

Question 7.
………………………….. is a server side scripting language designed for web development.
Answer:
php

Question 8.
Which is a request for data or ‘information’ from a database table?
(a) row
(b) record
(c) query
(d) report
Answer:
(c) query

Question 9.
Which is an open-source relational database management system?
(a) MySQL
(b) Foxpro
(c) MS-Access
(d) Excel
Answer:
(a) MySQL

Question 10.
Pick the odd one out.
(a) PRADO
(b) Cake
(c) cake PHP
(d) Symphony
Answer:
(b) Cake

II. Short Answer

Question 1.
Mention Some RDBMS Softwares?
Answer:
Relational Database Management System (RDMS) softwares are MySQL, Oracle, IBM DB2, and Microsoft SQLSERVER etc.

Question 2.
What is the main goal of providing connectivity between MySql and php?
Answer:
The main goal of MySQL and PHP connectivity is to retrieve and manipulate the data from MySQL database server.

1. The SQL query statements are helping with PHP MySQL extension to achieve the objective of MySQL and PHP connection.
2. “mysqli_query” is a function, helps to execute the SQL query statements in PHP scripting – language.

III. Explain in Brief Answer

Question 1.
Write a Snippet code to check database connection?
Answer:
// Check connection
if(!$conn){
die(“Connection failed: “ . mysqli_connect_error( ));
}
echo “Connected successfully”;
?>

Question 2.
Name the parameters of connect functions?
Answer:

1. $servemame → Database Server Server IP address
2. $username → Database Server User Name
3. $password → Database Server Password
4. $DBName → Database Name

Question 3.
Write any 3 special features of PHP?
Answer:

1. PHP can embed easily with HTML and client side scripting language
2. PHP has built-in function which is easily connect to MySQL database
3. PHP scripting language has been supported by many Software frameworks

Question 4.
Mention some web frameworks design structures to promote rapid application development (RAD)?
Answer:
Some of these include PRADO, CakePHP, Symfony, Codeigniter, Laravel, Yii Framework, Phalcon and Zend Framework, offering features similar to other web frameworks.

Students can Download Computer Applications Chapter 11 Network Examples and Protocols Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Computer Applications 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 12th Computer Applications Solutions Chapter 11 Network Examples and Protocols

Samacheer Kalvi 12th Computer Applications Network Examples and Protocols Text Book Back Questions and Answers

PART – I
I. Choose The Correct Answer

Question 1.
The ……………………… “the Net,” is a worldwide system of computer networks
(a) Internet
(b) mobile
(c) communication
(d) protocol
Answer:
(a) Internet

Question 2.
Which one of the following will be easy the way to uses Internet technology and the public telecommunication system to securely share business’s information with suppliers, vendors, partners and customers.
(a) Extranet
(b) Intranet
(c) arpanet
(d) arcnet
Answer:
(a) Extranet

Question 3.
Match the following and choose the correct Answer:wer
(i) HTTP -The core protocol of the World Wide Web.
(ii) FTP- enables a client to send and receive complete files from a server.
(iii) SMTP – Provide e-mail services.
(iv) DNS- Refer to other host computers by using names rather than numbers.
(a) (i), (ii), (iii), (iv)
(b) (ii), (iii), (iv), (i)
(c) (iii), (iv), (i), (ii)
(d) (iv), (iii), (ii), (i)
Answer:
(a) (i), (ii), (iii), (iv)

Question 4.
Communication over …………………………… is be made up of voice, data, images and text messages.
(a) Social media
(b) mobile network
(c) whatsapp
(d) software
Answer:
(b) mobile network

Question 5.
Wi-Fi stands for ……………………………
(a) Wireless Fidelity
(b) wired fidelity
(c) wired optic fibre
(d) wireless optic fibre
Answer:
(a) Wireless Fidelity

Question 6.
A TCP/IP network with access restricted to members of an organization
(a) LAN
(b) MAN
(c) WAN
(d) Intranet
Answer:
(d) Intranet

Question 7.
RFID stands for …………………………
(a) Radio Free identification
(b) real Frequency identity
(c) Radio Frequency indicators
(d) Radio Frequency Identification.
Answer:
(d) Radio Frequency Identification.

Question 8.
It guarantees the sending of data is successful and which checks error on operation at OSI layer is ………………………………
(a) Application layer
(b) Network layer
(c) Transport Layer
(d) Physical layer
Answer:
(c) Transport Layer

Question 9.
Which one of the following will secure data on trAnswer:missions?
(a) HTTPS
(b) HTTP
(c) FTP
(d) SMTP
Answer:
(a) HTTPS

Question 10.
………………………… provides e-mail service.
(a) DNS
(b) TCP
(c) FTP
(d) SMTP
Answer:
(d) SMTP

Question 11.
………………………… refer to other host computers by using names rather than numbers.
(a) DNS
(b) TCP
(c) FTP
(d) SMTP
Answer:
(a) DNS

Question 12.
TCP/IP is a combination of two protocols:
(i) TrAnswer:mission Control Protocol (TCP)
(ii) Internet Protocol (IP)
(iii) Selection Protocol (SP)
(iv) Captial Protocol (CP)
(a) (i), (ii)
(b) (i), (iii)
(c) (iii), (iv)
(d) (ii), (iii)
Answer:
(a) (i), (ii)

PART – II
II. Short Answer

Question 1.
Define Intranet?
Answer:
INTRANET:

1. It is a private network within an enterprise to share company data and computing resources between the employees.
2. It may consist of many interlinked local area networks.

Question 2.
What is the uses of mobile networks?
Answer:
The common application of mobile networks is mobile phones, tablets, etc.. In past, wireless communications largely used circuit switching to carry only voice over a network, but now currently both data and voice are being trAnswer:mitted over both circuit via switched networks and packet-switched networks.

Question 3.
List out the benefits of WiFi/
Answer:

1. It provides mobility.
2. It provides connection to Internet.
3. Flexibility of LAN.
4. Ensures connectivity.
5. It allows places that are remote to benefit from connectivity.
6. Low cost, high benefit.

Question 4.
How many types of RFID System available and what are they?
Answer:
Two types of RFID Systems:
1. Active RFID system:
the tag has its own power source. These systems used for larger distances and to track high value goods like vehicles.

2. Passive RFID system:
the tag gets power through power from a reader antenna to the tag „ antenna. They are used for shorter range trAnswer:mission.

Question 5.
Expand HTTP, HTTPS, FTP?
Answer:

1. HTTP – Hypertext Transfer Protocol
2. HTTPS – Hypertext Transfer Protocol Secure
3. FTP – File Transfer Protocol

PART – III
III. Explain in Brief Answer

Question 1.
Compare Internet, Intranet and Extranet?
Answer:

Question 2.
List out the components of a RFID enabled system?
Answer:
Main Components of a RFID System

1. A RFID tag: It has silicon microchip attached to a small antenna and mounted on a substrate.
2. A reader: It has a scanner with antennas to trAnswer:mit and receive signals, used for communication.
3. A Controller: It is the host computer with a Microprocessor which receives the reader input and process the data.

Question 3.
Write short notes on HTTP, HTTPS, FTP?
Answer:

1. HTTP – A protocol used between a web client and a web server protects non-secure data trAnswer:missions. The core protocol of the World Wide Web.
2. HTTPS – A protocol used between a web client and a web server permits secure data trAnswer:missions.
3. FTP – Used between computers for sending and receiving data. Enables a client to send and receive complete files from a server.

Question 4.
What are the layers available in TCP/IP Reference Model?
Answer:
There are four total layers of TCP/IP protocol, each of which is listed below with a brief description.

1. Network Access Layer – concerned with building packets.
2. Internet Layer – describes how packets are to be delivered.
3. Transport Layer – ensure the proper transfer mission of data.
4. Application Layer – application network processes. These processes include File Transfer Protocol (FTP), Hypertext Transfer Protocol (HTTP), and Simple Mail Transfer Protocol (SMTP).

Question 5.
Expand ARP, ICMP, SMTP and DNS?
Answer:

1. Address Resolution Protocol (ARP)
2. Internet Control Message Protocol (ICMP)
3. Transfer mission Control Protocol (TCP)
4. Simple Mail TrAnswer:fer Protocol (SMTP)
5. Domain Name System (DNS).

PART – IV
IV. Explain in detail

Question 1.
Explain about Internet, Intranet and Extranet?
Internet/Intranet/Extranet
INTERNET:
Answer:

1. The Internet, “the Net,” is a worldwide system of computer networks.
2. A network of networks where the users at any one computer can, if they have permission, get information from any other computer.
3. The Internet is a network of global connections – comprising private, public, business, academic and government networks – linked by guided, wireless and fiber-optic technologies.
4. It was perceived by the Advanced Research Projects Agency (ARPA) of the U.S. government in 1969 and was first recognized as the ARPANet.
5. The unique aim was to generate a network that would permit users of a research computer from one university to “talk to” research computers on other universities.
6. The Internet denotes to the global communication system, including infrastructure and hardware, whereas the web is one of the services interconnected over the Internet.

INTRANET:

1. It is a private network within an enterprise to share company data and computing resources between the employees.
2. It may consist of many interlinked local area networks.
3. It includes connections through one or more gateway (connects two networks using different protocols together known as protocol convertor) computers to outside Internet.

EXTRANET:
It is a private network that uses Internet technology and the public – telecommunication system to securely share business’s information with suppliers, vendors, partners, customers, or other businesses.

Question 2.
Discuss about OSI model with its layers?
Answer:
OSI Model:
Open System Interconnection (OSI) model was found in the year 1934, general framework that enables network protocols along with software and systems to be developed based on general set of guidelines. It describes the standards for the inter-computer communication.

OSI Layers:
(i) Physical Layer:
This is the 1st layer, it defines the electrical and physical specifications for devices.

(ii) Data Link Layer:
It is the 2nd layer and it guarantees that the data trAnswer:mitted are free of errors. This layer has simple protocols like “802.3 for Ethernet” and “802.11 for Wi-Fi”.

(iii) Network Layer:
It is the 3rd layer determining the path of the data packets. At this layer, routing of data packets is found using IP Addressing.

(iv) Transport Layer:
It is the 4th layer that guarantees the trAnswer:portation/sending of data is successful. It includes the error checking operation.

(v) Session Layer:
It is the 5th layer, identifies the established system session between different network entities. It controls dialogues between computers. For instance, while accessing a system remotely, session is created between your computer and the remote system.

(vi) Presentation Layer:
It is the 6th layer that does the trAnswer:lation of data to the next layer (Prepare the data to the Application Layer). Encryption and decryption protocols occur in this layer such as, Secure Socket Layer (SSL).

(vii) Application Layer:
It is the 7th layer, which acts as the user interface platform comprising of software within the system.

Question 3.
Difference between TCP/IP and OSI Reference Model?
Answer:

TCP/Ip:

1. TCP/IP stands for Transmission control protocol/Intemet protocol
2. TCP/IP has 4 layers
3. TCP/IP are set of rulers/protocols defined for communication over the network
4. TCP/IP is a client-server model
5. It follows top to bottom approach

OSI:

1. OSI means open system interconnect
2. OSI has 7 layers
3. OSI is a model based on the concept of layering
4. OSI is a conceptual model
5. It follows bottom-up approach

Question 4.
Explain about the development, merits and demerits in Mobile networks?
Merits of Mobile Networks:

1. Higher efficiency.
2. Increased ability to communicate in and out of the workspace.
3. Greater access to modem apps and services.
4. Improved networking capabilities.
5. Quality and flexibility of services.
6. Rapid developments in cloud technologies.

Demerits of Mobile Networks:

1. Cost
2. Vulnerable to security risks.
3. Additional training is needed to use new technology.
4. Cyber crime

Development:
The generations of mobile networks are as follows.

1. First Generation(lG) 1981- NMT launch
2. Second Generation(2G) 1991-GSM Launch
3. Second to Third Generation Bridge (2.5)2000 – GPRS launch
4. Third Generation! 3G) 2003- first UK 3G launch
5. Fourth Generation (4G) 2007
6. Fifth Generation (5G) 2019+

1. First Generation (1G) 1981 – NMT launch:

1. During the initial periods the mobile systems were based on analog trAnswer:mission.
2. NMT stands for Nordic Mobile Telephone communication.
3. And a very poor voice quality.

2. Second Generation(2G) 1991-GSM Launch:

1. Later the second generation of mobile systems were placed on digital trAnswer:mission with GSM.
2. GSM stands for (Global System for Mobile communication) was most popular standard which is used in second generation, using 900MHz and 1800MHz for the frequency bands.
3. The transfer mission used as TMDA stands for (Time Division Multiple Access) and CDMA One stands for (Code Division Multiple’Access) method to increase the amount of information trAnswer:ported on the network

3. Second to Third Generation Bridge (2.5)2000 – GPRS launch:

1. GPRS was introduced here GPRS stands for (General Packet Radio Service). GPRS is a data service which enables mobile devices to send and receive messages, picture messages and e-mails.
2. GSM data transfer mission rates typically reached 9.6kbit/s.

4. Third Generation( 3G) 2003- first UK 3G launch:
1. This generation of mobile systems merges different mobile technology standards, and uses higher frequency bands for transfer mission and Code Division Multiple Access to delivery data rates of up to 2Mbit/s supports multimedia services (MMS: voice, video and data).

2. Data transfer mission used a WCDMA. WCDMA stands for (Wideband Code Division Multiple Access).

3. Few 3G suppliers use ATM (Asynchronous Transfer Mode) for their ‘over the air’ network with in MPLS (Multiprotocol Label Switching) or IP for theirs backbone network.

5. Fourth Generation (4G) 2007:

1. 4G is at the research stage. 4G was based on an adhoc networking model where there was no need for a fixed infrastructure operation.

2. Adhoc networking requires global mobility features (e.g. Mobile IP) and connectivity to a global IPv6 network to support an IP address for each mobile device.

3. Logically roaming in assorted IP networks (for example: 802.11 WLAN, GPRS and UMTS) were be possible with higher data rates, from 2Mbit/s to 10-100Mbit/s, offering reduced delays and newly services.

6. Fifth Generation (5G) 2019+:

1. 5G is the stage succeeds the 4G (LTE/ WiMAx), 3G(umts) and 2G(GSM) syetems.

2. 5G targets to performance the high data rate, reduced latency, energy saving, cost reduction, higher system, capacity, and massive device connectivity.

3. The ITU IMT – 2020 provides speeds up to 20 gigabits per second it has been demonstrated with millimeter waves of 15 gigahertz and higher frequency.

Samacheer Kalvi 12th Computer Applications Solutions Network Examples and Protocols Additional Question and Answer

I. Choose the Best Answer

Question 1.
Internet Protocol delivers packets from source to destination through
(a) TCP
(b) datagram
(c) packets header
(d) http
Answer:
(c) packets header

Question 2.
IP connectionless datagram service was developed in the year
(a) 1972
(b) 1974
(c) 1976
(d) 1978
Answer:
(b) 1974

Question 3.
IP connectionless datagram service was developed by
(i) Vint cerf
(ii) Bob FrAnswer:ton
(iii) Bob Kahnin
(iv) Dan Bricklin
(a) i, ii
(b) ii, iii
(c) ii, iv
(d) i, iii
Answer:
(d) i, iii

Question 4.
TCP/IP stands for …………………………..
Answer:
Transmission Control Protocol/Internet Protocol

Question 5.
Which protocols have to do end-to-end process of secure on time and manage data or communication.
(a) Physical
(b) TCP
(c) Network
(d) ARPC
Answer:
(c) Network

Question 6.
Pick the odd one out:
(a) HTTP
(b) SFTL
(c) SSL
(d) SNMP
Answer:
(d) SNMP

Question 7.
Find the Network management protocol from the following
(a) ICMP
(b) HTTP
(c) TCP/IP
(d) SFTP
Answer:
(a) ICMP

Question 8.
Find the wrongly matched pair.
(a) Network Communication Protocol – HTTP, TCP/IP
(b) Network Security Protocol – ICMP
(c) Network Management Protocol – SNMP
Answer:
(b) Network Security Protocol – ICMP

Question 9.
ARPANET was first recognized in the year
(a) 1964
(b) 1956
(c) 1972
(d) 1969
Answer:
(d) 1969

Question 10.
The ……………………….. is one of the services interconnected over the Internet.
Answer:
web

Question 11.
Gateway connects two networks using different protocols together known as …………………………..
Answer:
Protocol Converter

Question 12.
Find the correct statement from the following.
I. Internet or things refers to the digital interconnection of everyday objects with the Internet
II. Extranet is a private network within an enterprise to share company data and computing resources between the employees.
(a) I-True, II-False
(b) I-False, II-True
(c) I, II-True
(d) I, II-False
Answer:
(a) I-True, II-False

Question 13.
Sending e-mail to a friend is an example for ………………………..
(a) Internet
(b) Intranet
(c) Extranet
(d) IP address
Answer:
(a) Internet

Question 14.
A …………………………. as it is made up of a large number of signal areas called cells.
(a) Mobile Network
(b) Cellular Network
(c) Both a & b
(d) Chatting
Answer:
(c) Both a & b

Question 15.
The NMT launch comes ………………………….. generation of mobile networks
(a) 1G
(b) 2G
(c) 3G
(d) 4G
Answer:
(a) 1G

Question 16.
Expand NMT.
(a) North Mobile Telephone
(b) Nordic Mobile Telephone
(c) Nordic Movement Telephone
(d) Nurture Mobile Telephone
Answer:
(b) Nordic Mobile Telephone

Question 17.
Match the following:
(i) 1G – (I) GSM
(ii) 2G – (II) UMTS
(iii) 3G – (III) GPRS
(iv) 2-3G Bridge – (IV) NMT
(a) IV I II III
(b) I II III IV
(c) IV III II I
(d) II I IV III
Answer:
(a) IV I II III

Question 18.
GSM means ……………………………… communication.
Answer:
Global System for Mobile

Question 19.
SIM means …………………………. technology.
Answer:
Subscriber Identity Module

Question 20.
TMDA means …………………………
Answer:
Time Division Multiple Access

Question 21.
CMDA stands for
(a) Code Divide Multiple Access
(b) Code Divide Mobile Access
(c) Code Division Multiple Access
(d) Code Division Mobile Access
Answer:
(c) Code Division Multiple Access

Question 22.
GPRS stands for ………………………
Answer:
General Packet Radio Service

Question 23.
Expand EDGE
(a) Extra Data rates for Global Evolution
(b) Enhanced Data rates for Global Evolution
(c) Entry Data rates for Global Evolution
(d) Extra Dual rate for Global Evolution
Answer:
(b) Enhanced Data rates for Global Evolution

Question 24.
UMTS stands for …………………………..
Answer:
Universal Mobile Telecommunication Systems

Question 25.
WCDMA stands for …………………………
Answer:
Wideband Code Division Multiple Access

Question 26.
ATM stands for ………………………….
Answer:
Asynchronous Transfer Mode

Question 27.
First Generation (1G) was developed in the year …………………………
(a) 1980
(b) 1983
(c) 1982
(d) 1981
Answer:
(d) 1981

Question 28.
Find the wrongly matched pair.
(a) First Generation – 1981
(b) Second Generation – 1991
(c) Third Generation – 2004
(d) Fourth Generation – 2007
Answer:
(c) Third Generation – 2004

Question 29.
Pick the odd one out.
(a) 1G
(b) 2G
(c) 2.5G
(d) 3.5G
Answer:
(d) 3.5G

Question 30.
Analog trAnswer:mission have taken place in …………………………
(a) 1G
(b) 2G
(c) 3G
(d) 4G
Answer:
(a) 1G

Question 31.
Which technology is used to authenticate a user for identification and billing purposes and to encrypt the data to prevent listen without permission.
(a) GSM
(b) NMT
(c) CDMA
(d) SIM
Answer:
(d) SIM

Question 32.
Which method increased the amount of information trAnswer:ported on the network?
(I) SIM
(II) NMT
(III) TDMA
(IV) CDMA
(a) I, II
(b) II, III
(c) III, IV
(d) IV, I
Answer:
(c) III, IV

Question 33.
General Packet Radio Service was introduced in
(a) 1G
(b) 2G
(c) 2.5G
(d) 3G
Answer:
(c) 2.5G

Question 34.
……………………………… is a data service which enables mobile devices to send and receive messages, picture messages and e-mails.
(a) GPS
(b) GPRS
(c) CDMA
(d) SIM
Answer:
(b) GPRS

Question 35.
What is the data trAnswer:mission rate of GSM?
(a) 9.6 kb/s
(b) 11.2 kb/s
(c) 12.3 kb/s
(d) 8.5 kb/s
Answer:
(a) 9.6 kb/s

Question 36.
MMS is the combination of …………………………
(I) Image, Voice, Video
(II) Voice, Video, Speed
(III) Voice, Video, Data
(IV) Speed, Data, Image
(a) III
(b) II
(c) I
(d) IV
Answer:
(a) III

Question 37.
The data trAnswer:mission used by 3G is ……………………………
(a) TDMA
(b) CDMA
(c) EDGE
(d) WCDMA
Answer:
(d) WCDMA

Question 38.
MPLS means ……………………….
Answer:
Multiprotocol Label Switching

Question 39.
Few 3G suppliers use ATM for their ……………………………. network with in MPLS or IP for their network.
Answer:
Over the Air

Question 40.
Which generation of mobile network is considered as a research stage?
(a) 1G
(b) 2G
(c) 3G
(d) 4G
Answer:
(d) 4G

Question 41.
Li-Fi is the short form of …………………………..
(a) Light Fidelity
(b) LAN Fidelity
(c) Light Fix
(d) Low Frequency
Answer:
(a) Light Fidelity

Question 42.
The term Li-Fi was first used by …………………………..
Answer:
Harald Haas

Question 43.
How many phases of 5G are there?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 44.
The two phases of 5G are ………………………….
(I) Release-14
(II) Release-15
(III) Release-16
(IV) Release-17
(a) I, II
(b) II, III
(c) I, III
(d) III, IV
Answer:
(b) II, III

Question 45.
ITU stands for ……………………………..
Answer:
International Telecommunication Union

Question 46.
Which is used for radio wares to read and capture information stored on a tag attached to an object?
(a) RF
(b) RFID
(c) Li-Fi
(d) Wi-Fi
Answer:
(b) RFID

Question 47.
How many parts are there in RFID tags are there?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 48.
RFID has a ……………………… and a ……………………….. or a ……………………………..
Answer:
reader, tag and label

Question 49.
Ffow many types of RFID tags are there?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 50.
How many components of a RFID system are there?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 51.
………………………… is a scanner with antennas to trAnswer:mit and receive signals used for communication
(a) RFID tag
(b) Reader
(c) Controller
(d) Em-ware
Answer:
(b) Reader

Question 52.
Find the statement which is true.
(а) Active RFID System – larger distances
(b) Active RFID System – shorter range transfer mission
Answer:
(а) Active RFID System – larger distances

Question 53.
Which method is used in passive RFID system?
(a) TCP
(b) UDP
(c) Induction coupling
(d) Direct coupling
Answer:
(c) Induction coupling

Question 54.
OSI model means ……………………………..
Answer:
Open System Interconnection

Question 55.
OSI model was found in the year
(a) 1927
(b) 1932
(c) 1935
(d) 1934
Answer:
(d) 1934

Question 56.
Which one describes the standards for the inter-computer communication?
(a) OSI
(b) BSI
(c) DSI
(d) LSI
Answer:
(a) OSI

Question 57.
How many OSI layers are there?
(a) 5
(b) 6
(c) 7
(d) 8
Answer:
(c) 7

Question 58.
The protocols for Ethernet in datalink layer is ………………………….
(a) 8.023
(b) 80.23
(c) 802.3
(d) 8023
Answer:
(c) 802.3

Question 59.
Which is the third layer in the OSI model?
(a) Datalink
(b) Physical
(c) port
(d) Network
Answer:
(d) Network

Question 60.
In which layer, routing of data packets is found using IP addressing?
(a) Data link
(b) Network
(c) port
(d) Physical
Answer:
(b) Network

Question 61.
SSL stands for ………………………
Answer:
Secure Socket Layer

Question 62.
Match the following
(i) Physical Layer – (I) controls dialogues between computer
(ii) Network Layer – (II) Specifications for devices
(iii) TrAnswer:port Layer – (III) data packets
(iv) Session Layer – (IV) error checking
(v) Application Layer – (V) Encryption, Decryption
(a) II, III, IV, I, V, VI
(b) I, II, III, IV, VI, VI
(c) VI, V, IV, III, II, I
(d) III, IV, V, VI, II, I
Answer:
(a) II, III, IV, I, V, VI

Question 63.
Which protocol is accountable for guaranteeing the trustworthy trAnswer:mission of data?
(a) TCP
(b) IP
(c) HTTP
(d) SMTP
Answer:
(a) TCP

Question 64.
google.com is the ………………………. for Google.
Answer:
domian name

Question 65.
How many different layers of TCP/IP are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 66.
FTP stands for …………………………
(a) File Transfer Protocol
(b) File Transmission Protocol
(c) File Transport Protocol
(d) File Type Protocol
Answer:
(a) File Transfer Protocol

Question 67.
HTTP stands for …………………………..
(a) High Text Transmission Protocol
(b) Hyper Text Transmission Protocol
(c) Hyper Text Transfer Protocol
(d) Height Typed Transfer Protocol
Answer:
(c) Hyper Text Transfer Protocol

Question 68.
SMTP stands for …………………………………
(a) Single Mail Transfer Protocol
(b) Single Message Transfer Protocol
(c) Simple Mail Text Protocol
(d) Simple Mail Transfer Protocol
Answer:
(d) Simple Mail Transfer Protocol

Question 69.
Pick the odd one out.
(a) FTP
(b) RFID
(c) HTTP
(d) SMTP
Answer:
(b) RFID

Question 70.
Which one of the following is not a network protocol.
(a) SSL
(b) ARP
(c) TCPI
(d) FTP
Answer:
(c) TCPI

Question 71.
HTTPs stands for …………………………..
Answer:
Hyper Text Transfer Protocol Secure

Question 72.
ARP stands for ………………………….
Answer:
Address Resoloution Protocol

Question 73.
Expand MAC
(a) Mega Access Code
(b) Multiple Access Code
(c) Medium Access Code
(d) Medium Access Control
Answer:
MEdium Access Control

Question 74.
Which one of the following is a hardware identification number that uniquely identifies each device on a network?
(a) MIC
(b) MAC
(c) MOC
(d) MUC
Answer:
(b) MAC

Question 75.
ICMP stands for ………………………….
Answer:
Internet Control Message Protocol

Question 76.
IGMP stands for …………………………..
Answer:
Internet Group Management Protocol

Question 77.
Which is used to send group communication messages to multiple IP addresses?
(a) ICMP
(b) IGMP
(c) IDMP
(d) IKMP
Answer:
(b) IGMP

Question 78.
The two main protocols mainly used in the trAnswer:port layers are
(I) TCP
(II) UDP
(III) UTP
(IV) ATP
(V) STP
(a) I, II
(b) III, IV
(c) II, V
(d) IV, V
Answer:
(a) I, II

Question 79.
UDP stands for …………………………….
Answer:
User Daragram Protocol

Question 80.
DNS meAnswer: ………………………..
Answer:
Domian Name System.

I. Short Answer

Question 1.
Write note on Extranet?
Answer:
EXTRANET:
It is a private network that uses Internet technology and the public telecommunication system to securely share business’s information with suppliers, vendors, partners, customers, or other businesses.

Question 2.
Mention the various generations of Mobile networks?
Answer:
The generations of mobile networks are as follows.

1. First Generation(lG) 1981- NMT launch
2. Second Generation(2G) 1991-GSM Launch
3. Second to Third Generation Bridge (2.5)2000 – GPRS launch ,
4. Third Generation( 3G) 2003- UK 3G launch
5. Fourth Generation (4G) 2007
6. Fifth Generation (5G) 2019+

Question 3.
What is meant by Li-Fi?
Answer:
Li-Fi is a wireless technology which uses light-emitting diodes (LEDs) for data transmission Li-Fi is the short form of Light Fidelity.

Question 4.
Define Wi-Fi?
Answer:
Wi-Fi stands for Wireless Fidelity. It is a wireless network technology that permits computers and alternative devices to be connected to every alternative into a local area network and to the net without wires and cables.

Question 5.
Write note on Network protocols?
Answer:
Network protocols is that the usual procedures, rules, formal standards and policies comprised of formats which allocates communication between more than one device Which connected to the network.

II. Explain in Brief Answer

Question 1.
Explain the types of Networking protocols?
Answer:
The broad types of networking protocols, including:

1. Network communication protocols is that the Basic data communication protocols which specific as HTTP and TCP/IP.
2. Network security protocols is that which implement security over network communications and include HTTP, SFTP and SSL.
3. Network management protocols will Provide network governance and maintenance and include ICMP and SNMP.

Question 2.
Define Internet of things?
Answer:
Internet of Things refers to the digital interconnection of everyday objects (home applicances, wearable devices or automobiles) with the Internet. The ‘thing’ in refers to an objects that have been assigned an IP address and have the ability to collect and transfer data over a network without manual assistance or intervention.

Question 3.
Write about Mobile Networks?
Answer:
Mobile Networks:
A mobile network or cellular network as it is made up of a large number of signal areas called cells. These cells join to form a large coverage area. Communication over mobile network is be made up of voice, data, images and text messages.

Question 4.
Differentiate Wi-Fi?
Answer:
Wi-Fi stands for Wireless Fidelity. It is a wireless network technology that permits computers and alternative devices to be connected to every alternative into a local area network and to the net without wires and cables.

Question 5.
Write note on Network protocols?
Answer:
Network protocols is that the usual procedures, rules, formal standards and policies comprised of formats which allocates communication between more than one device Which connected to the network.

III. Explain in Brief Answer

Question 1.
Explain the types of Networking protocols?
Answer:
The broad types of networking protocols, including:

1. Network communication protocols is that the Basic data communication protocols which specific as HTTP and TCP/IP.
2. Network security protocols is that which implement security over network communications and include HTTP, SFTP and SSL.
3. Network management protocols will Provide network governance and maintenance and include ICMP and SNMP.

Question 2.
Define Internet of things?
Answer:
Internet of Things refers to the digital interconnection of everyday objects (home applicances, wearable devices or automobiles) with the Internet. The ‘thing’ in IoT refers to an objects that have been assigned an IP address and have the ability to collect and trAnswer:fer data over a network without manual assistance or intervention.

Question 3.
Write about Mobile Networks?
Answer:
Mobile Networks:
A mobile network or cellular network as it is made up of a large number of signal areas called cells. These cells join to form a large coverage area. Communication over mobile network is be made up of voice, data, images and text messages.

Question 4.
Differentiate Wi-Fi and Li-Fi?
Answer:
Wi-Fi:

1. Wireless Fidelity
2. Wi-Fi uses radio frequencies for data transmission.

Li-Fi:

1. Li-Fi is a wireless technology which uses light-emitting diodes (LEDs) for data transmission.
2. Li-Fi is the short form of Light Fidelity.

Question 5.
What is RFID?
Answer:

1. RFID stands for Radio -Frequency Identification (RFID).
2. RFID used for radio waves to read and capture information stored on a tag attached to an object. Tag can be read from several feet away and does not need to be in direct-line-of-sight of the reader to be tracked.
3. RFID has been made up of two parts a reader and a tag or a label.
4. RFID tags are installed with a transmitter and receiver.

Question 6.
List out the two types of RFID tags?
Answer:
Two types of RFID tags were Active RFID and Passive RFID systems.

1. Passive RFID tag will be used the reader radio wave energy to really its stored information back to the reader.
2. Battery powered RFID tag is installed with small battery that powers the broadcast of information

Question 7.
Write note on Network Interface Layer?
Answer:
Network Interface Layer:

1. It is the bottom most level layer.
2. It is comparable to that of the Open System Interconnection Physical and Data Link layers.
3. Different TCP/IP protocols are being used at this layer, Ethernet and Token Ring for local area networks and protocols such as X.25, Frame Relay, and ATM for wide area networks.
4. It is assumed to be an unreliable layer.

Question 8.
Write note on Application Layer?
Answer:
The Application layer of the TCP/IP model is similar to the Session, Presentation, and Application layers of the OSI Reference Model. The most popular Application layer protocols are:
(i) Hypertext Transfer Protocol (HTTP):
The core protocol of the World Wide Web.

(ii) File Transfer Protocol (FTP):
enables a client to send and receive complete files from a server.

(iii) Telnet:
connect to another computer on the Internet.

(iv) Simple Mail Transfer Protocol (SMTP):
Provide e-mail services.

(v) Domain Name System (DNS):
Refer to other host computers by using names rather than numbers

IV. Explain in detail

Question 1.
Explain any 5 applications of Internet Intranet and Extranet?
Answer:

Question 2.
Explain the working of Passive and Active RFID systems with diagram?
Working of Passive RFID System
A Passive RFID system using Induction coupling method:
Answer:
The RFID tag gets power from the reader through the inductive coupling method.

A Passive RFID system using EM wave propagation method:
The reader antenna trAnswer:mits the electromagnetic waves that are received by the antenna.

Working Of Active RFID System
The reader sends signal to the tag using an antenna.

Question 3.
Discuss in detail about TCP/IP protocols?
Answer:
TCP/IP:
Transmission Control Protocol/Internet Protocol, TCP/IP is a set of protocols which governs communications among all computers on the Internet. TCP/IP protocol tells how information should be packaged, sent, and received, as well as how to get to its destination.

TCP WORKING: TCP/IP is a combination of two protocols: TrAnswer:mission Control Protocol (TCP) and Internet Protocol (IP). The Internet Protocol typically specifies he logistics of the packets that are sent out over networks; it specifies the packets which have to go, where to go and how to get there. The TrAnswer:mission Control Protocol is accountable for guaranteeing the trustworthy trAnswer:mission of data. It sees that the packets for errors and submits the requests for re-trAnswer:missions incase any of them are missing’.

Frequent TCP/IP Protocols

• HTTP – It is used between a web client and a web server and it guarantees non-secure data trAnswer:missions.
• HTTPS – It is used between a web client and a web server ensures secure data trAnswer:missions.
• FTP – It is used between computers for sending and receiving file.

Domain Names and TCP/IP Addresses
The address for any website is not as easy as to remember, domain name are used instead. For example, 216.58.216.164 is one of the IP address for Google and google.com is the domain name.

The Different Layers of TCP/IP
There are four total layers of TCP/IP protocol, each of which is listed below with a brief description.

1. Network Access Layer – concerned with building packets.
2. Internet Layer – describes how packets are to be delivered.
3. Transport Layer – ensure the proper transmission of data.
4. Application Layer – application network processes. These processes include File Transfer Protocol (FTP), Hypertext TrAnswer:fer Protocol (HTTP), and Simple Mail Transfer Protocol (SMTP).

Question 4.
Explain the important protocols present in

1. Network layer
2. Transport Layer

Answer:
1. Network Layer:
It is the layer where data is addressed, packaged, and routed among networks. The important Internet protocols that operate at the Network layer are:

2. Internet Protocol (IP):
Routable protocol which uses IP addresses to deliver packets. It is an unreliable protocol, does not guarantee delivery of information.

3. Address Resolution Protocol (ARP):
Resolves IP addresses to MAC (Medium Access Control) addresses. (A MAC address is a hardware identification number that uniquely identifies each device on a network.)i.e., to map IP network addresses to the hardware addresses.

4. Internet Control Message Protocol (ICMP):
Used by network devices to send error messages and operational information. Example: A host or router might not be reached or a requested service is not presented.

5. Internet Group Management Protocol (IGMP):
It is a communication protocol used by hosts and routers to send Multicast (group Communication) messages to multiple IP addresses at once.

6. port Layer:
The sessions are recognized and data packets are swapped between hosts in this layer. Two main protocols established at this layer are:

7. Mission Control Protocol (TCP):
Provides reliable connection oriented trAnswer:mission between two hosts. It ensures delivery of packets between the hosts.

8. User Datagram Protocol (UDP):
Provides connectionless, unreliable, one-to-one or one- to-many delivery.

Students can Download Accountancy Chapter 14 Computerised Accounting Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Accountancy 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 Accountancy Solutions Chapter 14 Computerised Accounting

Samacheer Kalvi 11th Accountancy Computerised Accounting Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

Question 1.
In accounting, computer is commonly used in the following areas:
(a) Recording of business transactions
(b) Payroll accounting
(c) Stores accounting
(d) All the above
Answer:
(d) All the above

Question 2.
Customised accounting software is suitable for ………………
(a) Small, conventional business
(b) Large, medium business
(c) Large, typical business
(d) None of the above
Answer:
(b) Large, medium business

Question 3.
Which one is not a component of computer system?
(a) Input unit
(b) Output unit
(c) Data
(d) Central Processing Unit
Answer:
(c) Data

Question 4.
An example of output device is ………………
(a) Mouse
(b) Printer
(c) Scanner
(d) Keyboard
Answer:
(b) Printer

Question 5.
One of the limitations of computerised accounting system is ………………
(a) System failure
(b) Accuracy
(c) Versatility
(d) Storage
Answer:
(a) System failure

Question 6.
Expand CAS ………………
(a) Common Application Software
(b) Computerised Accounting System
(c) Centralised Accounting System
(d) Certified Accounting System
Answer:
(b) Computerised Accounting System

Question 7.
Which one of the following is not a method of codification of accounts?
(a) Access codes
(b) Sequential codes
(c) Block codes
(d) Mnemonic codes
Answer:
(a) Access codes

Question 8.
TALLY is an example of ………………
(a) Tailor – made accounting software
(b) Ready – made accounting software
(c) In – built accounting software
(d) Customised accounting software
Answer:
(b) Ready – made accounting software

Question 9.
People who write codes and programs are called as ………………
(a) System analysts
(b) System designers
(c) System operators
(d) System programmers
Answer:
(d) System programmers

Question 10.
Accounting software is an example of ………………
(a) System software
(b) Application software
(c) Utility software
(d) Operating software
Answer:
(b) Application software

II. Very Short Answer Questions

Question 1.
What is a computer?
Answer:
A computer can be described as an electronic device designed to accept raw data as input, processes them and produces meaningful information as output. It has the ability to perform arithmetic and logical operations as per given set of instructions called program. Today, computers are used all over the world in several areas for different purposes.

Question 2.
What is CAS?
Answer:
Computerised accounting system (CAS) refers to the system of maintaining accounts using computers. It involves the processing of accounting transactions through the use of hardware and software in order to keep and produce accounting records and reports.

Question 3.
What is hardware?
Answer:
The physical components of a computer constitute its hardware. Hardware consists of input devices and output devices that make a complete computer system.

Question 4.
What is meant by software?
Answer:
A set of programs that form an interface between the hardware and the user of a computer system are referred to as software.

Question 5.
What is accounting software?
Answer:
The main function of CAS is to perform the accounting activities in an organisation and generate reports as per the requirements of the users. To obtain the desired results optimally, need based software or packages are to be installed in the organisation.

Question 6.
Name any two accounting packages.
Answer:

1. Readymade software
2. Customised software

Question 7.
Give any two examples of readymade software.
Answer:

1. Tally
2. Busy

Question 8.
What is coding?
Answer:
Code is an identification mark, generally, computerised accounting involves codification of accounts.

Question 9.
What is grouping of accounts?
Answer:
In any organisation, the main unit of classification is the major head which is further divided, into minor heads. Each minor head may have number of sub – heads. After classification of accounts into various groups.

Question 10.
What are mnemonic codes?
Answer:
A mnemonic code consists of alphabets or abbreviations as symbols to codify a piece of information.

III. Short Answer Questions

Question 1.
What are the various types of accounting software?
Answer:

1. Readymade software
2. Customised software and
3. Tailormade software

Question 2.
Mention any three limitations of computerised accounting system.
Answer:
Heavy cost of installation, Cost of training and fear of unemployment

Question 3.
State the various types of coding methods.
Answer:
Sequential codes, block codes and mnemonic codes.

Question 4.
List out the various reports generated by computerised accounting system.
Answer:

1. Liabilities and capital
2. Assets
3. Revenues and
4. Expenses.

Under Liabilities and Capital:
Capital, Non – current liabilities and current liabilities.

Under Assets:
Non – current assets and Current assets.

Question 5.
State the input and output devices of a computer system.
Answer:
Input devices: keyboard, optical scanner, mouse, joystick, touch screen and stylus.
Output devices: Monitor and printer.

Textbook Case Study Solved

Question 1.
The manager of a medium – sized business is considering the introduction of computerised accounting system. Some staff feels that it is an opportunity to learn new skill. The manager has promised free framing for their staff, So, the staff realise that their own skill can be enhanced. Also, there is a demand for highly skilled staff. But, some staff feels threatened by these changes. They feel that they may not be able to leam new skill. Moreover, some of them are nearing their retiring age. So they think that it is not needed for them. But the manager expects the cooperation from all the staff.
Now, discuss on the following points:

Question 1.
Will it be expensive for the business to introduce computerised accounting system?
Answer:
No, it will not be expensive. It may be integrated with enhanced MIS, multi – lingual and data organisation capabilities to simplify all the business processes easily and cost – effectively.

Question 2.
Will everyone get the access to use the computers? In such a case, how to protect data?
Answer:
It is an opportunity to leam new skill at free of cost. Retrieval of data is easier as the records are kept in soft copy in data base. By giving instructions, data can be retrieved quickly.

Question 3.
“People at the retirement age are not required to leam new skill” – Do you think so?
Answer:
No, the computerised accounting system is easy to leam by everyone. It is not a difficult one to retiring people also.

Question 4.
What are the factors to be considered by the managers before introducing CAS?
Answer:
The manager has to give an opportunity to leam new skill to his employees. It will take time but he has to face the employees problems:

1. Heavy cost of installation.
2. Cost of training.
3. Fear of unemployment.
4. Disruption of work and so on.

Samacheer Kalvi 11th Accountancy Computerised Accounting Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer

Question 1.
Which one is output device?
(a) Monitor
(b) Keyboard
(c) Mouse
(d) Optical scanner
Answer:
(a) Monitor

Question 2.
Components of CAS can be classified into …………….. categories.
(a) Six
(b) Seven
(c) Five
(d) Three
Answer:
(a) Six

Question 3.
Which one is operating system?
(a) File manager
(b) COBOL
(c) Windows
(d) PASCAL
Answer:
(c) Windows

Question 4.
Which one is matched correctly?
(a) Land & building – current assets
(b) Goodwill – non – current liabilities
(c) Patents – intangible assets
(d) Sales – expenses
Answer:
(c) Patents – intangible assets

Question 5.
There are …………….. methods of codification.
(a) Two
(b) Three
(c) Four
(d) Five
Answer:
(b) Three

II. Very Short Answer Questions

Question 1.
Expands of MIS and CPU.
Answer:

1. MIS – Management Information System
2. CPU – Central Processing Unit

Question 2.
Mention any two features of CAS.
Answer:

1. Simple and integrated
2. Speed

Question 3.
What is utility software?
Answer:
These are designed specifically for managing the computer device and its resources.

Question 4.
What is system operators?
Answer:
People who operate the systems and use it for different purposes. They are also called as end users.

Question 5.
What do you mean by DATA?
Answer:
The facts and figures that are fed into a computer for further processing are called data.

III. Short Answer Questions

Question 1.
What are the types of people interacting with a computer system?
Answer:

1. System analysts
2. System programmers
3. System operators

Question 2.
What is opearating system? Give two examples.
Answer:
A set of tools and programs to manage the overall working of a computer using a defined set of hardware components is called an operating system.
Example:

1. DOS
2. Windows.

Question 3.
Mention any three advantages of CAS.
Answer:

1. Faster processing
2. Accurate information and
3. Reliability

Question 4.
What are the three types of procedures in a computer system?
Answer:

1. Hardware oriented procedure
2. Software oriented procedure and
3. Internal procedure

Question 5.
Can you explain data?
Answer:
The facts and figures that are fed into a computer for further processing are called data. Data are raw input until the computer system interprets them using machine language, stores them in memory, classifies them for processing and produces results in conformance with the instructions given to it. Processed and useful data are called information which is used for decision making.

Students can Download Accountancy Chapter 13 Final Accounts of Sole Proprietors – II Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Accountancy 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 Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Samacheer Kalvi 11th Accountancy Final Accounts of Sole Proprietors – II Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

Question 1.
A prepayment of insurance premium will appear in ………………
(a) The trading account on the debit side
(b) The profit and loss account on the credit side
(c) The balance sheet on the assets side
(d) The balance sheet on the liabilities side
Answer:
(c) The balance sheet on the assets side

Question 2.
Net profit is ………………
(a) Debited to capital account
(b) Credited to capital account
(c) Debited to drawings account
(d) Credited to drawings account
Answer:
(b) Credited to capital account

Question 3.
Closing stock is valued at ………………
(a) Cost price
(b) Market price
(c) Cost price or market price whichever is higher
(d) Cost price or net realisable value whichever is lower
Answer:
(d) Cost price or net realisable value whichever is lower

Question 4.
Accrued interest on investment will be shown ………………
(a) On the credit side of profit and loss account
(b) On the assets side of balance sheet
(c) Both (a) and (b)
(d) None of these
Answer:
(c) Both (a) and (b)

Question 5.
If there is no existing provision for doubtful debts, provision created for doubtful debts is ………………
(a) Debited to bad debts account
(b) Debited to sundry debtors account
(c) Credited to bad debts account
(d) Debited to profit and loss account
Answer:
(d) Debited to profit and loss account

II. Very Short Answer Questions

Question 1.
What are adjusting entries?
Answer:
Adjustment entries are the journal entries made at the end of the accounting period to account for items which are omitted in trial balance and to make adjustments for outstanding and prepaid expenses and revenues accrued and received in advance.

Question 2.
What is outstanding expense?
Answer:
Expenses which have been incurred in the accounting period but not paid till the end of the accounting period are called outstanding expenses.

Question 3.
What is prepaid expense?
Answer:
Prepaid expenses refer to any expense or portion of expense paid in the current accounting year but the benefit or services of which will be received in the next accounting period. They are also called as unexpired expenses.

Question 4.
What are accrued incomes?
Answer:
Accrued income is an income or portion of income which has been earned during the current accounting year but not received till the end of that accounting year.

Question 5.
What is provision for discount on debtors?
Answer:
Cash discount is allowed by the suppliers to customers for prompt payment of amount due either on or before the due date. A provision created on sundry debtors for allowing such discount is called provision for discount on debtors.

III. Short Answer Questions

Question 1.
What is the need for preparing final accounts?
Answer:

1. To record omissions in trial balance such as closing stock, interest on captial, interest on drawings, etc.
2. To bring into account outstanding and prepaid expenses.
3. To bring into account income accrued and received in advance.
4. To create reserves and provisions.

Question 2.
What is meant by provision for doubtful debts? Why is it created?
Answer:
Provision for bad and doubtful debts refers to amount set aside as a charge against profit to meet any loss arising due to bad debt in future. The amount of doubtful debts is calculated on the basis of some percentage on debtors at the end of the accounting period after deducting further bad debts (if any). A provision for doubtful debts is created and is charged to profit and loss account.

Question 3.
Explain how closing stock is treated in final accounts?
Answer:
The unsold goods in the business at the end of the accounting period are termed as closing stock. As per As-2 (Revised), the stock is valued at cost price or net realisable value, whichever is lower.

Presentation in final accounts:

1. In the trading account: Shown on the credit side.
2. In the balance sheet: Shown on the assets side under current assets.

Question 4.
Give the adjusting entries for interest on capital and interest on drawings.
Answer:
Adjusting Entry: Interest on Capital

Adjusting Entry: Interest on Drawings

Question 5.
Explain the accounting treatment of bad debts, provision for doubtful debts and provision for discount on debtors.
Answer:

1. Bad Debts: When it is definitely known that amount due from a customer (debtor) to whom goods were sold on credit, cannot be realised at all, it is treated as bad debts.
2. Provision for bad and doubtful debts refers to amount set aside as a charge against profit to meet any loss arising due to bad debt in future.
3. Cash discount is allowed by the suppliers to customers for prompt payment of amount due either on or before the due date.

IV. Exercises

Question 1.
Pass adjusting entries for the following:
(a) The closing stock was valued at ₹ 5,000
(b) Outstanding salaries ₹ 150
(c) Insurance prepaid ₹ 450
(d) ₹ 20,000 was received in advance for commission.
(e) Accrued interest on investments is ₹ 1,000.
Answer:
Adjusting Entries

Question 2.
For the fol owing adjustments, pass adjusting entries:
(a) Outstanding wages ₹ 5,000.
(b) Depreciate machinery by ₹ 1,000.
(c) Interest on capital @ 5% (Capital: ₹ 20,000)
(d) Interest on drawings ₹ 50
(e) Write off bad debts ₹ 500
Answer:
Adjusting Entries

Question 3.
On preparing final accounts of Suresh, bad debt account has a balance of ₹ 800 and sundry debtors account has a balance of ₹ 16,000 of which ₹ 1,200 is to be written off as further bad debts. Pass adjusting entry for bad debts. And also show how it would appear in profit and loss account and balance sheet.
Answer:
Adjusting Entry

Profit and Loss Account

Balance Sheet

Question 4.
The trial balance on March 31, 2016 shows the following:
Sundry debtors ₹ 30,000; Bad debts ₹ 1,200
It is found that 3% of sundry debtors is doubtful of recovery and is to be provided for. Pass journal entry for the amount of provision and also show how it would appear in the profit and loss account and balance sheet.
Answer:

Profit and Loss Account for the year ended 31.03.2016

Balance Sheet as on 31.03.2016

Question 5.
The trial balance of a trader on 31^st December, 2016 shows debtors as ₹ 50,000.
Adjustments:
(a) Write off ₹ 1,000 as bad debts
(b) Provide 5% for doubtful debts
(c) Provide 2% for discount on debtors
Show how these items will appear in the profit and loss A/c and balance sheet of the trader.
Answer:
Profit and Loss Account for the year ended 31^st December, 2016

Balance Sheet as on 31^st December, 2016

Question 6.
On 1^st January, 2016, provision for doubtful debts account had a balance of ₹ 3,000. On December 31, 2016, sundry debtors amounted to ₹ 80,000. During the year, bad debts to be written off were ₹ 2,000. A provision for 5% was required for next year. Pass journal entries and show how these items would appear in the final accounts.
Answer:
Adjusting Entries

Profit and Loss Account for the year ended 31^st December, 2016

Balance Sheet as on 31.12.2016

Question 7.
The following are the extracts from the trial balance.
Sundry debtors ₹ 30,000; Bad debts ₹ 5,000 Additional information:
(a) Write off further bad debts ₹ 3,000.
(b) Create 10% provision for bad and doubtful debts.
You are required to pass necessary adjusting entries and show how these items will appear in profit and loss account and balance sheet.
Answer:
Adjusting Entries

Profit and Loss Account

Balance Sheet

Question 8.
The following are the extracts from the trial balance.

Additional information:
(a) Additional bad debts ₹ 3,000.
(b) Keep a provision for bad and doubtful debts @ 10% on sundry debtors.
You are required to pass necessary adjusting entries and show how these items will appear in profit and loss account and balance sheet.
Answer:
Adjusting Entries

Profit and Loss Account

Balance Sheet

Question 9.
The accounts of Lakshmi traders showed the following balance on 31^st March, 2016.
Sundry debtors ₹ 60,000; Bad debts ₹ 2,000
Provision for doubtful debts ₹ 4,200
At the time of preparation of final accounts on 31^st March, it was found that out of sundry debtors, ₹ 1,000 will be irrecoverable. It was decided to create a provision of 2% on debtors to meet any future possible bad debts.
Pass necessary journal entries and show how these items would appear in the final accounts.
Answer:
Adjusting Entries

Profit and Loss Account for the year ended 31.03.2016

Balance Sheet as on 31.03.2016

Question 10.

The following are the extracts from the trial balance.
Additional information:
(a) Create a provision for doubtful debts @ 10% on sundry debtors.
(b) Create a provision for discount on debtors @ 5% on sundry debtors.
You are required to pass necessary adjusting entries and show how these items will appear in the final accounts.
Answer:
Adjusting Entries

Profit and Loss Account

Balance Sheet

Question 11.
Following are the extracts from the trial balance.

Additional information:
(a) Additional bad debts 1,000
(b) Create a provision for doubtful debts @ 5% on sundry debtors.
(c) Create a provision for discount on debtors @ 2% on sundry debtors.
You are required to pass necessary journal entries and show how these items will appear in the final accounts.
Answer:
Adjusting Entries

Profit and Loss Account

Balance Sheet

Question 12.
The following are the extracts from the trial balance.

Answer:
Profit and Loss Account

Balance Sheet

Question 13.
Prepare trading account of Archana for the year ending 31^st December, 2106 from the following information.

Adjustments:
(a) Closing stock ₹ 1,00,000
(b) Wages outstanding ₹ 12,000
(c) Freight inwards paid in advance ₹ 5,000
Answer:
Trading A/c of Archana for the year ended 31.12.2016

Question 14.
Prepare profit and loss account of Manoj for the year ending on 31^st March, 2016

Adjustments:
(a) Salary outstanding ₹ 400
(b) Rent paid in advance ₹ 50
(c) Commission receivable ₹ 100
Answer:
Profit and Loss A/c of Manoj for the year ended 31.03.2016

Question 15.
From the trial balance of Sumathi and the adjustments prepare the trading and profit and loss account for the year ended 31^st March, 2016, and a balance sheet as on that date.

Adjustments
(a) Six months interest on loan is outstanding.
(b) Two months rent is due from tenant, the monthly rent being ₹ 25.
(c) Salary for the month of March 2016, ₹ 75 is unpaid.
(d) Stock in hand on March 31, 2016 was valued at ₹ 1,030.
Answer:
Trading and Profit & Loss A/c of Sumathi
for the year ended 31^st March, 2016

Balance Sheet of Sumathi as on 31.03.2016

Question 16.
The following trial balance was extracted from the books of Arun Traders as on 31^st March, 2018.

Answer:
Prepare trading and profit and loss account for the year ending 31^st March, 2018 and balance sheet as on that date after considering the following:
(a) Depreciate Plant and machinery @ 20%
(b) Wages outstanding amounts to ₹ 750.
(c) Half of repairs and maintenance paid is for the next year.
(d) Closing stock was valued at ₹ 15,000.
Answer:
Trading and Profit & Loss A/c of Arun Traders for the year ended 31.03.2018

Balance Sheet of Arun Traders as on 31.03.2018

Question 17.
The following is the trial balance of Babu as on 31^st December, 2016.

Prepare trading and profit and loss account for the year ended 31^st December, 2016 and a balance sheet as on that date after the following adjustments.
(a) Salaries outstanding ₹ 500
(b) Interest on investments receivable at 10%.
(c) Provision required for bad debts is 5%.
(d) Closing stock is valued at ₹ 9,900.
Answer:
Trading and Profit & Loss A/c for the year ended 31.12.2016

Balance Sheet as on 31.12.2016

Question 18.
From the following trial balance of Ramesh as on 31^st March, 2017, prepare the trading and profit and loss account and the balance sheet as on that date.

Answer:
Adjustments:
Closing stock was valued at ₹ 35,000
(b) Unexpired advertising ₹ 250
(c) Provision for bad and doubtful debts is to be increased to ₹ 3,000
(d) Provide 2% for discount on debtors.
Answer:
Trading and Profit & Loss A/c of Ramesh for the year ended 31.03.2017

Balance Sheet of Ramesh as on 31.03.2017

Question 19.
Following are the ledger balances of Devi as on 31^st December, 2016.

Prepare trading and profit and loss account for the year ended 31^st December, 2016 and balance sheet as on that date.
(a) Stock on 31^st December, 2016 ₹ 5,800.
(b) Write off bad debts ₹ 500.
(c) Make a provision for bad debts @ 5%.
(d) Provide for discount on debtors @ 2%.
Answer:
Trading and Profit & Loss A/c of Devi for the year ended 31.12.2016

Balance Sheet of Devi as on 31.12.2016

Question 20.
From the following trial balance of Mohan for the year ended 31^st March, 2017 and additional information, prepare trading and profit and loss account and balance sheet.

Additional information:
(a) Closing stock is valued at ₹ 15,500
(b) Write off ₹ 500 as bad debts and create a provision for bad debts @ 10% on debtors.
(c) Depreciation @ 10% required
Answer:
Trading and Profit and Loss A/c of Mohan for the year ended 31.03.2017

Balance Sheet of Mohan as on 31.03.2017

Question 21.
From the following trial balance of Subramaniam, prepare his trading and profit and loss account and balance sheet as on 31^st December, 2016.

Take into account the following adjustments:
(a) Charge interest on drawings at 8%.
(b) Outstanding salaries ₹ 3,000
(c) Closing stock was valued at ₹ 48,000
(d) Provide for 5% interest on capital.
Answer:
Trading and Profit & Loss A/c of Subramaniam for the year ended 31.12.2016

Balance Sheet of Subramaniam as on 31.12.2016

Question 22.
Prepare trading and profit and loss account and balance sheet from the following trial balance of Madan as on 31^st March, 2018.

Adjustments:
(a) The closing stock was ₹ 80,000
(b) Provide depreciation on plant and machinery @ 20%
(c) Write off ₹ 800 as further bad debts
(d) Provide the doubtful debts @ 5% on sundry debtors
Answer:
Trading and Profit & Loss A/c for the year ended 31.03.2018

Balance Sheet as on 31.03.2018

Question 23.
From the following information prepare trading and profit and loss account and balance sheet of Kumar for the year ending 31^st December, 2017.

Adjustments:
(a) The closing stock on 31st December, 2017 was valued at ₹ 3,900.
(b) Carriage inwards prepaid ₹ 250
(c) Rent received in advance ₹ 100
(d) Manager is entitled to receive commission @ 5% of net profit after providing such commission.
Answer:
Trading and Profit & Loss Account of Kumar for the year ended 31.12.2017

Balance Sheet of Kumar as on 31.12.2017

Question 24.
From the following information, prepare trading and profit and loss account and balance sheet in the books of Sangeetha for the year ending 31^st March, 2018.

Adjustments:
(a) Stock on 31st March, 2018 ₹ 14,200
(b) Income tax of Sangeetha paid ₹ 800
(c) Charge interest on drawings @ 12% p.a.
(d) Provide managerial remuneration @ 10% of net profit before charging such commission.
Answer:
Trading and Profit & Loss Account of Sangeetha for the year ended 31.03.2018

Balance Sheet of Sangeetha as on 31.03.2018

Textbook Case Study Solved

Question 1.
James is a trader who sells washing machines on credit. But, he does not remember the due date to collect the money from his debtors. Some of his customers do not pay on time. His cash inflow is becoming worse. As a result, he could not pay his telephone bill and rent at the end of the accounting period. Hence, he showed only the amount paid as expense. He has many washing machines unsold at the year end. He is worried about the performance of his business. So, he is planning to appoint a manager to take care of his business. The new manager insists James to apply the accounting principle of prudence and matching and also to allow cash discount.
Now, discuss on the following points:

Question 1.
Why does James sell on credit?
Answer:
James sells goods on credit to increase the sales volume and reduce the stock.

Question 2.
Are there any ways to encourage his debtors to make the payment on time?
Answer:
Yes, there are many ways to encourage his debtors to make the payment on time by way of cash discount and trade discount.

Question 3.
What might happen if the debtors do not pay?
Answer:
If the debtors do not pay, the bad debts will be increased in the business.

Question 4.
In what ways prudence and matching principles can be applied for the business of James?
Answer:
Prudence principle can be applied for the business here closing stock was valued on cost price or market price whichever is lower under the prudence principle. Matching principle can be applied here for revenue and expense.

Question 5.
What will be the impact on income statement and the balance sheet, if the outstanding expenses are not adjusted?
Answer:
Outstanding expenses to be added with the concerned expenditure in the income Statement and the outstanding expenses will be recorded in liabilities side.

Question 6.
On what basis the unsold washing machines should be valued?
Answer:
The unsold washing machine should be valued at cost price or market price, whichever is lower under prudence principle. Managerial commission can be given to motivate the new manager to retain him in the business of James.

Samacheer Kalvi 11th Accountancy Final Accounts of Sole Proprietors – II Additional Questions and Answers

I. Multiple Choice Questions (Other important questions)
Choose the correct answer

Question 1.
If closing stock is already adjusted, adjusted purchase account and ………………. stock will appear in trial balance.
(a) Opening
(b) Closing
(c) Average
(d) None of these
Answer:
(b) Closing

Question 2.
Outstanding expense account is a ………………. account.
(a) Nominal
(b) Real
(c) Representative personal
(d) Personal
Answer:
(c) Representative personal

Question 3.
When bad debts already appears in the trial balance, it is taken only to debit side of ………………. account.
(a) Profit and Loss
(b) Balance sheet
(c) Asset side
(d) None of these
Answer:
(a) Profit and Loss

Question 4.
Income tax paid by the business for the proprietor is treated as ……………….
(a) Expense
(b) Profit and Loss A/c
(c) Drawings
(d) None of these
Answer:
(c) Drawings

Question 5.
Commission on net profit after charging such commission:

Answer:

Students can Download Computer Applications Chapter 13 Network Cabling Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Computer Applications 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 12th Computer Applications Solutions Chapter 13 Network Cabling

Samacheer Kalvi 12th Computer Applications Network Cabling Text Book Back Questions and Answers

PART – I
I. Choose The Correct Answer

Question 1.
ARPANET stands for
(a) American Research Project Agency Network
(b) Advanced Research Project Area Network
(c) Advanced Research Project Agency Network
(d) American Research Programs And Network
Answer:
(c) Advanced Research Project Agency Network

Question 2.
WWW was invented by …………………………..
(a) Tim Berners Lee
(b) Charles Babbage
(c) Blaise Pascal
(d) John Napier
Answer:
(a) Tim Berners Lee

Question 3.
Which cable is used in cable TV to connect with setup box?
(a) UTP cable
(b) Fibre optics
(c) Coaxial cable
(d) USB cable
Answer:
(c) Coaxial cable

Question 4.
Expansion of UTP is …………………………
(a) Uninterrupted Twisted Pair
(b) Uninterrupted Twisted Protocol
(c) Unshielded Twisted Pair
(d) Universal Twisted Protocol
Answer:
(c) Unshielded Twisted Pair

Question 5.
Which medium is used in the optical fibre cables to transmit data?
(a) Microwave
(b) infra red
(c) light
(d) sound
Answer:
(c) light

Question 6.
Which of the following is a small peripheral device with a sim slot to connect the computers to Internet?
(a) USB
(b) Dongles
(c) Memory card
(d) Mobiles
Answer:
(b) Dongles

Question 7.
Which connector is used in the Ethernet cables?
(a)RJll
(b) RJ21
(c) RJ61
(d) RJ45
Answer:
(d) RJ45

Question 8.
Which of the following connector is called as champ connector?
(a) RJll
(b) RJ21
(c) RJ61
(d) RJ45
Answer:
(b) RJ21

Question 9.
How many pins are used in RJ45 cables?
(a) 8
(b) 6
(c) 50
(d) 25
Answer:
(a) 8

Question 10.
Which wiring standard is used for connecting two computers directly?
(a) straight Through wiring
(b) Cross Over wiring
(c) Rollover wiring
(d) None
Answer:
(b) Cross Over wiring

Question 11.
pick the odd one out from the following cables
(a) roll over
(b) cross over
(c) null modem
(d) straight through
Answer:
(c) null modem

Question 12.
Match the following
(i) Ethernet – Port
(ii) RJ45 connector – Ethernet
(iii) RJ45 jack – Plug
(iv) RJ45 cable – 802.3
(a) (i)-1, (ii)-2, (iii)-4, (iv)-3
(b) (i)-4, (ii)-1, (iii)-3, (iv)-2
(c) (i)-4, (ii)3, (iii)-1, (iv)-2
(d) (i)-4, (ii)-2, (iii)-1, (iv)-3
Answer:
(c) (i)-4, (ii)3, (iii)-1, (iv)-2

PART – II
II. Short Answer

Question 1.
Write a note on twisted pair cable?
Answer:

1. Twisted cable has 13 wires which are twisted to ignore electromagnetic interferefice
2. Two types of twisted pair cables are Unshielded Twisted Pair (UTP) and Shielded Twisted pair (STP).

Question 2.
What are the uses of USB cables?
Answer:
Uses of USB Cables:

1. The Universal Serial Bus are used to connect keyboard, mouse and other peripheral devices.
2. But there are some special network devices used to connect the Internet through the USB called dongles.
3. The dongle is a small peripheral device which has a compatible of mobile broadband with a sim slot in it and connects the Internet and acts as a modem to the computer.

Question 3.
Write a note on the types of RJ45 connector?
Answer:
Cat5, Cat6, Cat7, Cat3, Cat6e.

Question 4.
What is an Ethernet port?
Answer:
Ethernet port is an opening which is a part of an Ethernet card. It accepts RJ45 connector with Ethernet cable. It is also called as RJ45 jack. It is found on personal computers, laptops, routers, switches, hubs and modems.

Question 5.
What is the use of Crimping tool?
Answer:
The crimping tool is a physical tool which is used to connect the patch wire and the Ethernet connector. The tool will puncture the connector and makes the wire set in the connector.

Question 6.
What are the types of twisted pair cables?
Answer:
There are two types of twisted pair cables, Unshielded Twisted Pair (UTP) and Shielded Twisted pair (STP). The UTP is used nowadays as modem cables for Internet and they are lower in cost and installation and maintenance is easy compared to the coaxial cables.

Question 7.
What is meant by champ connector?
Answer:
The RJ-21 connector has 50 pins with 25 pins at one end and 25 pins at the other end. It is also called as champ connector or Amphenol connector. The Amphenol is a connector manufacturer. The RJ-21 interface is typically used for data communication trucking applications.

PART – III
III. Explain in Brief Answer

Question 1.
Write a note on crossover cables?
Answer:

1. If you require a cable to connect two computers or Ethernet devices directly together without a hub, then you will need to use a Crossover cable instead.
2. The easiest way to make a crossover cable is to make one end to T568A colour coding and the other end to T568B.
3. Another way to make the cable is to remember the colour coding used in this type. Here Green set of wires at one end are connected with the Orange set of wires at another end and vice versa.

Question 2.
Write a short note on RJ45 connector?
Answer:
RJ45 Connector:

1. The RJ45 connector is a small plastic cup which will be used to connect the wire inside the connector and ready to connect the Internet.
2. The RJ45 connector looks similar like a telephone jack but it looks a slightly wider. The Ethernet cables are sometime called as RJ45 cables.
3. In RJ45 the “RJ” stands for the Registered Jack and the “45” simply refers to the number of interface standard in the cable.

Question 3.
What are the differences between serial and parallel ports?
Answer:
Serial Ports:

1. Serial port transmits data one but after another.
2. Data transmisison is slow
3. 9 pin or 25 pin male connectors.
4. Eg. COM, COM2, RS232
5. It can be connected to monitor, scanners, models

Parallel Ports:

1. It transmits all 8 bits of a byte is parallel.
2. It is fast
3. 25 pin female connector
4. Eg. LPTI Port
5. It is used to connect printer, joysticks
6. external hard drives

Question 4.
What is meant by null modem cable?
Answer:
RS-232 cable is also used for interconnecting two computers without modem. So it is also a null modem cable. A cable interconnecting two devices directly is known as a null modem cable.

Question 5.
What are the components involved in Ethernet cabling?
Answer:

1. Patch Cable (Twisted pair)
2. RJ45 Connector
3. Ethernet Ports
4. Crimping Tool

Patch Cable (Twisted Pair):
These Cables are generally made up of 8 wires in different colors. Four of them are solid colours, and the others are striped. The eight colors are white green, green, white orange, blue, white blue, orange, white brown and brown.

The RJ45 connector has eight small pins inside to connect eight small wires in the patch ’cable. The eight cables has eight different colours
The Ethernet port is the jack where the Ethernet cable is to be connected. This port will be there in both the computers and the LAN port.
The crimping tool is a physical tool which is used to connect the patch wire and the Ethernet connector(RJ45).

Question 6.
What are the types of Fibre optic cables?
Answer:
There are two types of fiber optic cables available, One is single-mode (100BaseBx) another one is Multimode (100BaseSX). Single-mode cables are used for long distance transmission and at a high cost whereas the multimode cables are used for short distance transmission at a very low cost.

PART – IV
IV. Explain in detail

Question 1.
What is meant by Registered Jack? Explain briefly the types of Jacks?
Answer:
Types of JacksRegistered Jacks:
A Registered Jack commonly known as RJ is a network interface used for network cabling, wiring and jack construction. The primary function of the registered jack is to connect different data equipment and telecommunication devices.

The commonly known registered jacks are RJ-11, RJ-45, RJ-21, and RJ-28. The registered jack refers to the male physical connector (Plug), a female physical connector (Jack) and it’s wiring. We will talk some of the variety of registered jack below with some defintions.

It is the most popular modem form of registered j ack. It is found in home and office. This registered jack is mainly used in telephone and landlines. When we look the pin details of the RJ-11, there are 6 pin where the two pins give the transmission configuration, the two pins give the receiver configuration and the other two pins will be kept for reserved. The two pin will have the positive terminal and the negative terminal.

RJ-14 and RJ-61:
The RJ-14 is the same as RJ-11 which will be used for telephone lines where same it as 6 pins whereas the RJ-61 will have 8 pins. This RJ-61 will use the twisted pair cable with a modular 8 connection.

RJ-21:
The RJ-21 connector has 50 pins with 25 pins at one end and 25 pins at the other end. It is also called as champ connector or Amphenol connector. The Amphenol is a connector manufacturer. The RJ-21 interface is typically used for data communication trucking applications.

Question 2.
Explain wiring techniques used in Ethernet cabling?
Answer:
There are three types of wiring techniques to construct the Ethernet cable. It is also known as color coding techniques. They are:

1. Straight-Through Wiring
2. Cross-over Wiring
3. Roll-over Wiring Straight-Through Wiring

In general, the Ethernet cables used for Ethernet connections are “straight-through cables”. These cable wires are in the same sequence at both ends of the cable, which means that pin 1 of the plug on one end is connected to pin 1 of the plug on the other end (for both standard – T568A & T568B). the straight through wiring cables are mostly used for connecting PC / NIC card to a hub. This is a simple physical connection used in printers, computers and other network interfaces.

Cross-over Wiring:
If you require a cable to connect two computers or Ethernet devices directly together without a hub, then you will need to use a Crossover cable instead. Then the pairs(Tx and Rx lines) will be crossed which means pin 1 & 2 of the plug on one end are connected with pin 3 & 6 of the plug on other end, and vice versa (3 & 6 to pin 1 & 2).

The easiest way to make a crossover cable is to make one end to T568A colour coding and the other end to T568B. Another way to make the cable is to remember the colour coding used in this type. Here Green set of wires at one end are connected with the Orange set of wires at another end and vice versa. Specifically, connect the solid Green (G) with the solid Orange, and connect the green/white with the orange/white.

Roll-over WiringRollover cable is a type of null-modem cable that is often used to connect a device console port to make programming changes to the device. The roll over wiring have opposite pin arrangements, all the cables are rolled over to different arrangements.

In the rollover cable, The coloured wires are reversed on other end i.e. The pins on one end are connected with other end in reverse order (i.e. pin 1 to 8, 2 to 7, 3 to 6, 4 to 5, 5 to 4, 6 to 3, 7 to 2, 8 to 1). Rollover cable is also known as Yost cable or Console cable. It is typically flat (and light blue color) to distinguish it from other types of network cabling.

These all the three arrangements are used to perform an interface change. But, all the three arrangements transmits the data at the same speed only.

Question 3.
Explain about RJ45 connector?
Answer:

1. The RJ45 connector is a small plastic cup which will be used to connect the wire inside the connector and ready to connect the Internet.
2. The RJ45 connector looks similar like a telephone jack but it looks a slightly wider.
3. The Ethernet cables are sometime called as RJ45 cables.
4. In RJ45 the “RJ” stands for the Registered Jack and the “45” simply refers to the number of interface standard in the cable.
5. Each RJ45 connector has eight pins and connected to each end of the Ethernet cable.
6. Since it has 8-position, 8-contact (8P8C) modular plug, It is also known as 8P8C connector.

The two main signals of the pins: the one is the TX which is transmission of data and RX
which is Receiver of data.
“In the Figure the position of pin no. 1 describes the transmit data or bidirectional, the bidirectional means it can be sent to both connection with the TX (positive). The TX+ has positive terminal and the color is used in this position is white green.

• In the position of pin no. 2 describes the transmit data or bidirectional with the TX (negative). The TX- has negative terminal and the color is used in this position is Green.
• In the position of pin no.
• Describes the receive data or bidirectional with the RX (positive). The RX is compatible to receive the data which has positive terminal and the color is used in this position is white orange.
• In this position of pin no. 4 describes that it is not connected or bidirectional. It means there will be no transmitting or receiving will exist. But it can be used later for some other connection. The color used in this pin is blue.

• In this position of pin no. 5 describes that it is not connected or bidirectional and the color is white blue.
• In this position of pin no. 6 describes the receive data or bidirectional with the RX (Negative) which has a negative terminal and the color used in this position is orange.
• In this position of pin no. 7 describes that it is not connected or bidirectional and the color is white brown.
  In this position of pin no. 8 describes that it is not connected or bidirectional and the color is brown.

Question 4.
Explain the components used in Ethernet cabling?
Answer:
The main components are used in the Ethernet cabling are

1. Patch Cable (Twisted pair)
2. RJ45 Connector
3. Ethernet Ports
4. Crimping Tool

1. Patch Cable (Twisted Pair):
1. These Cables are generally made up of 8 wires in different colors.

2. Four of them are solid colours, and the others are striped.
3. The eight colors are white green, green, white orange, blue, white blue, orange, white brown and brown. The following figure 13.8 shows the patch cable.
Ethernet cables are normally manufactured in several industrial standards such as Cat 3, Cat 5, Cat 6, Cat 6e and cat 7. “Cat” simply stands for “Category,”. Increasing the size of the cable also lead to slower transmission speed.

4. The cables together with male connectors (RJ45) on each end are commonly referred as Ethernet cables. It is also called as RJ45 cables, since Ethernet cable uses RJ45 connectors.

2. RJ45 Connector:

• The RJ45 connector is a small plastic cup which will be used to connect the wire inside the connector and ready to connect the Internet.
• The RJ45 connector looks similar like a telephone jack but it looks a slightly wider.
• The Ethernet cables are sometime called as RJ45 cables.
• In RJ45 the “RJ” stands for the Registered Jack and the “45” simply refers to the number of interface standard in the cable.
• Each RJ45 connector has eight pins and connected to each end of the Ethernet cable.
• Since it has 8-position, 8-contact (8P8C) modular plug, It is also known as 8P8C connector. Th£se plugs (connector) are then inserted into Ethernet port of the network card.

3. Ethernet card and Port:

• Ethernet card is a Network Interface Card (NIC) that allows computers to connect and transmit data to the devices on the network. It may be an expansion card or built-in type.
• Expansion card is a separate circuit board also called as PCI Ethernet card which is inserted into PCI slot on motherboard of a computer.
• Now a days most of the computers come with built-in Ethernet cards which resides on motherboard.
• Wireless Ethernet cards are also available, which uses radio waves to transmit data.
• Ethernet port is an opening which is a part of an Ethernet card. It accepts RJ45 connector with Ethernet cable. It is also called as RJ45 jack. It is found on personal computers, laptops, routers, switches, hubs and modems.
• In these days, most of the computers and laptops have a built-in Ethernet port for connecting the device to a wired network.

4. Crimping Tool:

• Crimping is the process of joining two or more pieces of metal or wire by deforming one or both of them to hold each other.
• The crimping tool is a physical tool which is used to connect the patch wire and the Ethernet connector.
• The tool will puncture the connector and makes the wire set in the connector.

Question 5.
Explain the types of network cables?
Answer:
There are many types of cables available in the networking.

1. Coaxial Cables:
Ths cable was invented at late 1880’s, which is used to connect the television sets to home antennas. Ths cable is used to transfer the information in 10 mbps. The cable is divided into thin.net and thicknet cables. These cables have a copper wire inside and insulation is covered on the top of the copper wire to provide protection to the cable.

These cables are very difficult to install and maintain, because they are too big to carry and replace. The coaxial cable got its name by the word “coax”. Nowadays coaxial cables are also used for dish TV where the setup box and the television is connected using the coaxial cable only. Some of the cable names are Media Bridge 50-feet Coaxial cable, Amazon basics CL2-Rated Coaxial cables, etc.

2. Twisted Pair Cables:
It is type of cable with two or more insulated wires twisted together. It started with the speed of 10 mbps (10BASE-T cable is used). Then the cable is improved ” and the speed was higher and went to 100 mbps and the cable-was renamed as 100BASE-TX. Then finally the cable improved more made to 10 gbps and named as 10GBASE-T. Ths twisted cable has 8 wires which are twisted to ignore electromagnetic interference.

Also the eight wires cannot be placed in a single unit there could be a difficult in spacious, so it is twisted to make as one wire. There are two types of twisted pair cables, Unshielded Twisted Pair (UTP) and Shielded Twisted pair (STP). The UTP is used nowadays as modem cables for Internet and they are lower in cost and installation and maintenance is easy compared to the coaxial cables. STP is similar to UTP, but it is covered by an additional jackets to protect the wires from External interference.

3. Fiber Optics:
This cable is different from the other two cables. The other two cables had an insulating material at the outside and the conducting material like copper inside. But in this cable it is of strands of glass and pulse of light is used to send the information. They are mainly used in Wide Area Network (WAN). The WAN is a network that extends to very large distance to connect the computers. One example of WAN is Internet.

These cables are placed in deep underground to avoid any damage to the cables. The optic cable uses light to transmit the information from one place to another. There are two types of fiber optic cables available, One is single-mode (lOOBaseBx) another one is Multimode (lOOBaseSX). Single-mode cables are used for long distance transmission and at a high cost whereas the multimode cables are used for short distance transmission at a very low cost. The optic cables are easy to maintain and install.

4. USB Cables:
The Universal Serial Bus are used to connect keyboard, mouse and other peripheral devices. But there are some special network devices used to connect the Internet through the USB called dongles. The dongle is a small peripheral device which has a compatible of mobile broadband with a sim slot in it and connects the Internet and acts as a modem to the computer.

5. Serial and Parallel cables:
Before In the year of 1980s to 1990s the Ethernet and the USB were not developed. Then the Serial and Parallel interface cables are used to connect the Internet to the system. They were sometime used for PC to PC networking. Before the USB emerged, the system will have both serial port and parallel port.

The serial port will send 1 bit at one time whereas the parallel port will send 8 bit at one time. The parallel cables are used to connect to the printer and other disk drivers. RS232 is one type of serial cable, also known as Null modem cable.

6. Ethernet Cables:
Ethernet cable is the most common type of network cable mainly used for connecting the computers or devices at home or office. This cable connects wired devices within the local area network (LAN) for sharing the resources and accessing Internet.

Samacheer Kalvi 12th Computer Applications Network Cabling Additional Question and Answers

I. Choose the Best Answer

Question 1.
Which year was the co-axial cables invented?
(a) 1880
(b) 1890
(c) 1990
(d) 2000
Answer:
(a) 1880

Question 2.
Which cable connects television stes to home Antenna’s?
(a) fiber optics
(b) Twistedpair
(c) USB
(d) co-axial
Answer:
(d) co-axial

Question 3.
Co-axial cables transfer the information in …………………………
(a) 10 kbps
(b) 10 mbps
(c) 10 GBPS
(d) 10 TBPS
Answer:
(b) 10 mbps

Question 4.
Co-axial cables are divided into
(I) Thin net
(II) Soft net
(III) Hard net
(IV) Thick net
(a) I, II
(b) II, II
(c) I, IV
(d) II, III, IV
Answer:
(c) I, IV

Question 5.
Co-axial cables are made up of ……………………..
(a) Steel
(b) Iron
(c) Copper
(d) Aluminium
Answer:
(c) Copper

Question 6.
The co-axial cable got its name by the word ………………………
(a) copper axial
(b) combined axial
(c) coaxis
(d) coax
Answer:
(d) coax

Question 7.
…………………….. is a type of cable with two or more insulated wires twisted together.
Answer:
Twisted pair cables

Question 8.
Find the wrongly matched pair.
(a) 10 BASE-T – 10 mbps
(b) 100 BASE-X – 100 MBPS
(c) 10 GBASE-T – 10 gbps
Answer:
(b) 100 BASE-X – 100 MBPS

Question 9.
Assertion (A): 8 wires of twisted cable are twisted
Reason (R): To ignore electromagnetic interference.
(a) A is true R is the reason
(b) A, R both false
(c) A is false R is true
(d) A is true, R is not the reason
Answer:
(a) A is true R is the reason

Question 11.
STP stands for ………………………
(a) Shielded Turn paper
(b) Shielded Twisted pair
(c) Soft Turn Photo
(d) Short Time processing
Answer:
(b) Shielded Twisted pair

Question 12.
Find which is true about UTP?
(i) UTP is low in coast
(ii) easy Installation and maintainance
(iii) covered by Jackets
(a) (i), (iii)
(b) (i), (ii)
(c) (i), (ii), (iii)
(d) All are false
Answer:
(b) (i), (ii)

Question 13.
Find the wrongly matched pair.
(a) coaxial cables – TV
(b) Twisted pair cables – ATP, UTP
(c) Fiber optic cables – Single-mode, Multimode
Answer:
(b) Twisted pair cables – ATP, UTP

Question 14.
How many types of fibre optic cables are available?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 15.
The two types of fibre optic cables are ……………………… and ………………………..
Answer:
Single-mode, Multi-mode

Question 16.
USB stands for …………………………..
Answer:
Universal Serial Bus

Question 17.
Which one of the following is used for connecting keyboard, mouse and other peripheral devices?
(a) USB
(b) coaxial cables
(c) Twisted pair cables
(d) Fiber optic
Answer:
(a) USB

Question 18.
Find the correct statement.
(I) Single mode cables used for long distance
(II) Single mode cables used for short distance
(III) Multimode cables used for long distances .
(IV) Multimode cables used for short distances
(a) I, IV
(b) II, III
(c) II, IV
(d) I, III
Answer:
(a) I, IV

Question 19.
Which is the special network device used to connect the Internet through the USB?
(a) NIC
(b) Ethernet
(c) Dongles
(d) LAN
Answer:
(c) Dongles

Question 20.
What is the data transfer rate of USB 3.0?
(a) 480 Mbps
(b) 320 Mbps
(c) 4.89 Mbps
(d) 4.85Gbps
Answer:
(d) 4.85Gbps

Question 21
…………………… is a miniaturized version of the USB used for connecting mobile devices.
Answer:
Micro USB

Question 22.
The latest version of USB is
(a) USB 2.0
(b) USB 2.5
(c) USB 3.0
(d) USB 3.5
Answer:
(c) USB 3.0

Question 23.
…………………… are used for PC to PC networking.
(a) Serial interface cables
(b) Fibre optic cables
(c) Parallel Interface cables
(d) a and c
Answer:
(d) a and c

Question 24.
The Serial port will send ……………………… bit at one time.
(a) 1
(b) 2
(c) 4
(d) 8
Answer:
(a) 1

Question 25.
The parallel port will send …………………………… bit at one time.
(a) 1
(b) 2
(c) 4
(d) 8
Answer:
(d) 8

Question 26.
Which one of the following is Null modem cable?
(a) CL2
(b) RS 232
(c) RS 322
(d) RS 323
Answer:
(b) RS 232

Question 27.
Which cables are used for connecting printer and other disk drivers?
(a) Serial port
(b) Parallel port
(c) USB
(d) UTP
Answer:
(b) Parallel port

Question 28.
Two PC’s or two network devices of some type are connected bty using cables?
(a) UTP
(b) STP
(c) Fibre optics
(d) Cross over Ethernet
Answer:
(d) Cross over Ethernet

Question 29.
The speed of crossover Ethernet cable is
(a) 100 gbps
(b) 100 mbps
(c) 10 gbps
(d) 10 mbps
Answer:
(c) 10 gbps

Question 30.
A cable interconnecting two devices directly is known as a ……………………….
Answer:
null modem cable

Question 31.
Which one of the following has faster Internet speed?
(a) Wireless Network
(b) Wired Network
(c) both a and b
(d) None of these
Answer:
(b) Wired Network

Question 32.
The number of systems to be introconnected depends on
(a) switches
(b) routers
(c) both a and b
(d) None of these
Answer:
(c) both a and b

Question 33.
How many small wires are present inside the patch cables?
(a) 8
(b) 4
(c) 2
(d) 1
Answer:
(a) 8

Question 34.
…………………….. is the most popular Ethernet cable.
Answer:
RJ45

Question 35.
Pick the odd one out.
(a) Patch cable
(b) rotating tool
(c) crimping tool
(d) RJ45
Answer:
(b) rotating tool

Question 36.
The patch cable has …………………… solid color cables.
(a) 8
(b) 4
(c) 2
(d) 1
Answer:
(b) 4

Question 37.
What is the another name of patch cable?
(a) RJ45
(b) crimping
(c) Twisted pair
(d) LAN cable
Answer:
(c) Twisted pair

Question 38.
“cat” stands for
(a) animal
(b) linux commands
(c) catalogue
(d) category
Answer:
(d) category

Question 39.
RJ45 is a ………………………. connector.
Answer:
male

Question 40.
In RJ45, “RJ” stands for ……………………. and 45 indicates ………………………
Answer:
Register Jack, number of interfaces

Question 41.
The Ethernet cables are also called as ……………………. cables.
Answer:
RJ45

Question 42.
RJ45 connector is also called as connector.
(a) 48PC
(b) 8P8C
(c) P8C
(d) 88PC
Answer:
(b) 8P8C

Question 43.
How many pins are there in RJ45 connectors?
(a) 12
(b) 6
(c) 8
(d) 32
Answer:
(c) 8

Question 44.
How many wiring schemes are available in RJ45?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 45.
Identify the wiring schemes available in RJ45.
(i) T-568A
(ii) T-568B
(iii) T-568C
(iv) T-568D
(a) (i), (ii)
(b) (ii), (iii)
(c) (iii), (iv)
(d) (i), (iv)
Answer:
(a) (i), (ii)

Question 46.
Identify the color which is not present in T-568A?
(a) green
(b) blue
(c) orange
(d) red
Answer:
(d) red

Question 47.
Match the following colour codings.
T- 568A:

1. green (Tx-1)
2. blue
3. orange (Rx-1)
4. brown

T-568B

1. orange (Tx-1)
2. blue
3. green (Rx-1)
4. brown

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

Question 48.
(Rx-) in T-568A is ……………………..
Answer:
receiver negative

Question 49.
…………………….. is Transmission negative.
Answer:
(Tx -)

Question 50.
NIC stands for ……………………….
Answer:
Network Interface card

Question 51.
Wireless Ethernet cards uses ……………………… waves to transmit data.
(a) radio
(b) micro
(c) UV
(d) Infrared
Answer:
(a) radio

Question 52.
……………………… is the process of joining two or more pieces of metal or wire by deforming one or both of them to hold each other
Answer:
crimping

Question 53.
The ……………………. tool is used to connect the patch wire and the Ethernet connector.
Answer:
crimping

Question 54.
Who invented Ethernet?
(a) Bob Frankston
(b) Bob Metcalfe
(c) Dan Bricklin
(d) Robert
Answer:
(b) Bob Metcalfe

Question 55.
Ethernet was invented in the year
(a) 1972
(b) 1973
(c) 1976
(d) 1978
Answer:
(b) 1973

Question 56.
PARC stands for ………………………
Answer:
Palo Alto Research Center

Question 57.
The IEEE standard was first published in
(a) 1983
(b) 1984
(c) 1985
(d) 1986
Answer:
(c) 1985

Question 58.
………………………. CSMA/CD means
Answer:
Carrrier Sense Multiple Access with Collision Detection

Question 59.
RJ stands for ………………………..
Answer:
Registered Jacks

Question 60.
Pick the odd one out.
(a) RJ-11
(b) RJ-45
(c) RJ-27
(d) RJ-28
Answer:
(c) RJ-27

Question 61.
The most popular modem form of registered Jack is
(a) RJ-11
(b) RJ-45
(c) RJ-28
(d) RJ-45
Answer:
(a) RJ-11

Question 62.
Find the wrongly matched pair.
(a) RJ-11 – 6 pins
(b) RJ-14 – 18 pins
(c) RJ-61 – 8 pins
(d) RJ-21-50 pins
Answer:
(b) RJ-14 – 18 pins

Question 63.
RJ-21 is otherwise called as ………………………
(i) champ connector
(ii) Amphenol connector
(a) (i), (ii)
(b) (ii), (iii)
(c) (i), (ii), (iii)
(d) (i), (iii)
Answer:
(a) (i), (ii)

Question 64.
Which interface is typically used for data communication trucking applications?
(a) RJ-11
(b) RJ-21
(c) RJ-28
(d) RJ-45
Answer:
(b) RJ-21

Question 65.
How many types of wiring techniques to construct the ethernet cable?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 66.
Match the following
(i) Straight-through wiring – 1. 1 to 1, 2 to 2, 3 to 3
(ii) Cross-over wiring – 2. 1 to 3, 3 to 1, 6 to 2
(iii) Roll-over wiring – 3. 1 to 8, 2 to 7, 3 to 6
(a) (i)-1 (ii)-2 (iii)-3
(b) (i)-3 (ii)-2 (iii)-1
(c) (i)-2 (ii)-1 (iii)-3
(d) (i)-1 (ii)-3 (iii)-2
Answer:
(a) (i)-1 (ii)-2 (iii)-3

Question 67.
Roll over cable is also known as
(a) yost cable
(b) console cable
(c) a and b
(d) none of these
Answer:
(c) a and b

Question 68.
In connectors, TX/RX denotes …………………………
Answer:
Transmit/Receive

II. Short Answer

Question 1.
Write note on coaxial cables?
Answer:
Coaxial Cables:
This cable was invented at late 1880’s, which is used to connect the television sets to home antennas. This cable is used to transfer the information in 10 mbps.

Question 2.
What is meant by Twisted pair cables?
Answer:
Twisted Pair Cables: It is type of cable with two or more insulated wires twisted together. It started with the speed of 10 mbps (1 OBASE-T cable is used).

Question 3.
Mention the different types of cables used to connect computer on Network?
Answer:
Computers can be connected on the network with the help of wired media (Unshielded Twisted pair, shielded Twisted pair, Co-axial cables and Optical fibre) or wireless media (Infra Red, Bluetooth, WiFi)

Question 4.
Define crimping?
Answer:
Crimping is the process of joining two or more pieces of metal or wire by deforming one or both of them to hold each other.

Question 5.
Give the Pin details of RJ-11?
Answer:
Pin details of the RJ-11, there are 6 pin where the two pins give the transmission configuration, the two pins give the receiver configuration and the other two pins will be kept for reserved. The two pin will have the positive terminal and the negative terminal.

Question 6.
Give some examples for coaxial cables?
Answer:
Some of the cable names are Media Bridge 50-feet Coaxial cable, Amazon basics CL2-Rated Coaxial cables.

III. Explain in Brief Answer

Question 1.
Compare UTP and STP?
Answer:
UTP:
The UTP is used nowadays as modem cables for Internet and they are lower in cost and installation and maintenance is easy compared to the coaxial cables.

STP:
STP is similar to UTP, but it is covered by an additional jackets to protect the wires from External interference.

Question 2.
Write note on Fibre optics?
Answer:
IT is of strands of glass and pulse of light is used to send the information. They are mainly used in Wide Area Network (WAN).

Question 3.
What are the uses of serial and parallel cables?
Answer:
Serial and Parallel interface cables are used to connect the Internet to the system. They were sometime used for PC to PC networking. Before the USB emerged, the system will have both serial port and parallel port.

Question 4.
Write about crossover Ethernet cable?
Answer:
The Crossover Ethernet cable is an example of the Null modem Cables. This cable is used to join two PCs or two network devices of the same type. This cable is the sophisticated Ethernet cable used to connect the Internet to the system. This cable works at a speed of 10 gbps and more. The Ethernet crossover cable is identical on both the ends. Nowadays Routers are being connected through the crossover cables to provide wireless network from the local network.

Question 5.
Differentiate Wireless and Wired Networks?
Answer:
Wireless:
Wireless networks enable more devices including mobiles sharing the resources and Internet connections remotely. But Compared to wireless networks, wired networks maintain a faster Internet speed and more secure.

Wired Networks:
Wired networks for larger area are more expensive. Wired networks are still used widely in the offices where need increased speed and secure connections.

Question 6.
What is NIC?
Answer:
Ethernet card is a Network Interface Card (NIC) that allows computers to connect and transmit data to the devices on the network. It may be an expansion card or built-in type.

Question 7.
List the alternative Names given to the following devices, a) Ethernet Technology b) RJ45 connector 4) RJ45 socket d) RJ45 cable?
Answer:
Devices/Technology:

1. Ethernet Technology
2. RJ45 Connector (male)
3. RJ45 socket (female)
4. RJ45 Cable

Alternative Names:

1. RJ45, 802.3 (according to IEEE)
2. RJ45 plug, Ethernet connector, 8P8C connector
3. RJ45 jack, Ethernet Port
4. Ethernet cable

Question 8.
What is RJ?
Answer:
Registered Jacks: A Registered Jack commonly known as RJ is a network interface used for network cabling, wiring and jack construction. The primary function of the registered jack is to connect different data equipment and telecommunication devices. The commonly known registered jacks are RJ-11, RJ-45, RJ-21, andRJ-28.

Question 9.
Write note on Roll-over wiring?
Answer:
Rollover cable is a type of null-modem cable that is often used to connect a device console port to make programming changes to the device. The roll over wiring have opposite pin arrangements, all the cables are rolled over to different arrangements. In the rollover cable, The coloured wires are reversed on other end i.e. The pins on one end are connected with other end in reverse order (i.e. pin 1 to 8, 2 to 7, 3 to 6, 4 to 5, 5 to 4, 6 to 3, 7 to 2, 8 to 1).

Question 10.
How will you determine the type of Ethernet cable?
Answer:
Straight-through:
The coloured wires are in the same sequence at both ends of the cable. Cross-over: The first coloured wire at one end of the cable is the third coloured wire at the other end of the cable.

Question 11.
Write note on straight-through wiring?
Answer:
In general, the Ethernet cables used for Ethernet connections are “straight-through cables”. These cable wires are in the same sequence at both ends of the cable, which means that pin 1 of the plug on one end is connected to pin 1 of the plug on the other end (for both standard – T568A & T568B). the straight through wiring cables are mostly used for connecting PC / NIC card to a hub.

IV. Explain in detail

Question 1.
Explain the Crimping process to make Ethernet cables?
Answer:
Crimping process for making Ethernet cables

1. Cut the cable with desired length
2. Strip the insulation sheath about 1 inch from both end of the cable and expose the Twisted pair wires
3. After stripping the wire, untwist the smaller wires and arrange them into the proper wiring scheme, T568B preferred generally.
4. Bring the wires tighter together and cut them down so that they all have the same length ( Vi inch).
5. Insert the all 8 coloured wires into the eight grooves in the connector. The wires should be, inserted until the plastic sheath is also inside the connector.
6. Use the crimping tool to lock the RJ45 connector on the cable. It should be strong enough to handle manual traction. Now it is ready for data transmission.
7. Use a cable tester to verify the proper connectivity of the cable, if need.