Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Accountancy Guide Pdf Chapter 5 Admission of a Partner Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 5 Admission of a Partner

12th Accountancy Guide Admission of a Partner Text Book Back Questions and Answers

I. Multiple Choice Questions

Choose the correct answer

Question 1.
Revaluation A/c is a
(a) Real A/c
(b) Nominal A/c
(c) Personal A/c
(d) Impersonal A/c
Answer:
(b) Nominal A/c

Question 2.
On revluation, the increase in the value of assets leads to
(a) Gain
(b) Loss
(c) Expense
(d) None of these
Answer:
(a) Gain

Question 3.
The profit or loss on revaluation of assets and liabilities is transferred to the capital account of
(a) The old partners
(b) The new partner
(c) All the partners
(d) The Sacrificing partners
Answer:
(a) The old partners

Question 4.
If the old profit sharing ratio is more than the new profit sharing ratio of a partner, the difference is called
(a) Capital ratio
(b) Sacrificing ratio
(a) all the partners
(b) the old partners
(c) the new partner
(d) the sacrificing partners
Answer:
(b) Sacrificing ratio

Question 6.
Which of the following statements is not true in relation to the admission of a partner
(a) Generally mutual rights of the partners change
(b) The profits and losses of the previous years are distributed to the old partners
(c) The firm is reconstituted under a new agreement
(d) The existing agreement does not come to an end
Answer:
(d) The existing agreement does not come to an end

Question 7.
Match List-I with List-II and select the correct answer using the codes given below:

Answer:
b

Question 8.
Select the odd one out
(a) Revaluation profit
(b) Accumulated loss
(c) Goodwill brought by a new partner
(d) Investment fluctuation fund
Answer:
(c) Goodwill brought by a new partner

Question 9.
James and Kamalesh are partners sharing profits and losses in the ratio of 2:1. They admit Yogesh into partnership. The new profit sharing ratio between Balaji, Kamalesh, and Yogesh is agreed to 3:1:1. Find the sacrificing ratio.
(a) 1:3
(b) 3:1
(c) 5:3
(d) 3:5
Hint:

Answer:
(c) 5:3

Question 10.
Balaji and Kamalesh are partners sharing profits and losses in the ratio of 2:1. They admit Yogesh into partnership. The new profit sharing ratio between Balaji, Kamalesh and Yogesh is agreed to 3:1:1. Find the sacrificing ratio between Balaji and Kamalesh.
(a) 1:3
(b) 3:1
(c) 2:1
(d) 1:2
Hint:
Sacrifice ratio = old ratio – new ratio
Old partner’s = Balaji, Kamalesh

Answer:
(d) 1:2

II. Very Short Answer Questions

Question 1.
What is meant by the revaluation of assets and liabilities?
Answer:
When a partner is admitted into the partnership the assets and liabilities are revealed as the current value may differ from the book value. Determination of current values of assets and liabilities is called revaluation of assets and liabilities.

Question 2.
How are accumulated profits and losses distributed among the partner at the time of admission of a new partner?
Answer:
The following and the adjustment are necessary at the time of admission of a partner.

•  Distribution of accumulated profits, reserves, and losses
•  Revaluation of assets and liabilities
•  Determination of new profit – sharing ratio and sacrificing ratio
•  Adjustment for goodwill
•  Adjustment of capital on the basis of the new profit sharing ratio (if no agreed).

Question 3.
What is sacrificing ratio?
Answer:
The sacrificing ratio is the proportion of the profit which is sacrificed or foregone by the old partners in favour of the new partner. The purpose of finding the sacrificing ratio is to share the goodwill brought in by the new partner.
Sacrifice Ratio = Old share – New share

Question 4.
Give the journal entry for writing off existing goodwill at the time of admission of a new partner.
Answer:
Old partners capital /current A/c (in old ratio) Dr. To G/W A/c

Question 5.
State whether the following will be debited or credited in the revaluation account.
Answer:

1. Depreciation on assets – Debited
2. Unrecorded liability – Debited
3. Provision for outstanding expenses – Debited
4. Appreciation of assets – Credited

III Short Answer Questions

Question 1.
What are the adjustments required at the time of admission of a partner?
Answer:
The following adjustments are necessary at the time of admission of a partner.

1. Distribution of accumulated profits, reserves, and losses
2. Revaluation of assets and liabilities
3. Determination of new profit sharing ratio and sacrificing ratio
4. Adjustment for goodwill
5. Adjustment of capital on the basis of new profit sharing ratio

Question 6.
What are the journal entries to be passed on revaluation of ssets and liabilities?
Answer:
Following are the journal entries to be passed to record the revaluation of assets and liabilities:

Question 7.
Write a short note on the accounting treatment of goodwill.
Answer:
Accounting treatment for goodwill on the admission of a partner is discussed below:

Question 1.
When a new partner brings cash towards goodwill
Answer:
When the new brings cash towards goodwill in addition to the amount of capital, it is distributed to the existing partners in the sacrificing ratio.
If the new partner does not bring goodwill in cash or in kind, his share of goodwill must be adjusted through the capital accounts of the partners.

Sometimes the new partner may bring only a part of the goodwill in cash or assets. In such a case, for the cash or the assets brought, the respective account is debited and for the amount not brought in cash or kind, the new partner’s capital account is debited.

If goodwill already appears in the books of accounts, at the time of admission if the partners decide, it can be written off by transferring it to the existing partner s capital account / current account in the old profit sharing ratio.

IV Exercises

Distribution of accumulated profits, reserves, and losses

Question 1.
Arul and Anitha are partners sharing profits and losses in the ratio of 4:3. On 31.03.2018, Ajay was admitted as partner. On the date of admissions, the book of the firm showed a general reserve of ₹ 42,000. Pass the journal entry to distribute the general reserve.
Solution:

(General reserve transferred to old partners capital A/c in the old profit sharing ratio 4:3.)
Answer :
Arul: ₹ 24,000 (Cr.); Anitha: ₹ 18,000 (Cr.)

Question 2.
Anjali and Nithya are partners of firms sharing profits and losses in the ratio of 5:3. They admit Pramila on 01.01.2018. On that date, their balance sheet showed an accumulated loss of ₹ 40,000 on the asset side of the balance sheet. Give the journal entry to transfer the accumulated loss on admission.
Solution:

(Accumulated loss transferred to old partner s capital A/cs in the old ratio.)
Answer:
Profit and loss: ₹ 25,000 (Dr.); Nithya: ₹ 15,000(Dr.)

Question 3.
Oviya and Kavya are partners in firm sharing profits and losses in the ratio of 5:3. They admit Agalya into the partnership. Their balance sheet as of 31st March 2019 is as follows:

The pass journal entry to transfer the accumulated profit and reserve on admission.
Solution:

Accumulated profits and reserve transferred to old partners capital A/c in the old ratio
Answer :
Oviya : ₹ 37,500; Kavya : ₹ 22,500

Question 4.
Hari, Madhavan, and Kesavan are partners, sharing profit and losses in the ratio of 5:32. As from 1st April 2017, Vanmathi is admitted into the partnership and the new profit sharing ratio is decided as 4:3:2:1. The folio 5 adjustments are to be made.
(a) Increase the value of premises by ₹ 60,000.
(b) Depreciate stock by ₹ 5,000, furniture by ₹ 2, 000 and machinery by ₹ 2,500.
(c) Provide for an outstanding liability of ₹ 500.
Pass journal entries and prepare revaluation account.
Solution:

Question 5.
Seenu and Siva are partners sharing profits and losses in the ratio of 5:3. In view of kowsalya admission, they decided.
(i) To increase the value of the building by ₹ 40,000.
(ii) To bring into record investment at ₹ 10,000, Which have not so far been brought into
(iii) To decrease the value of machinery by ₹ 14,000 and furniture by ₹ 12,000.
(iv) To write off sundry creditors by ₹ 16,000.
Pass journal entries and prepare revaluation account
Solution :

Answer:
Revolution Profit in Rs. 40.000

Question 6.
Sai and Shankar are partners, sharing profits and losses in the ratio of 5:3.The firm’s balance sheet as on 31st December, 2017, was as follows:


On 31st December 2017, Shanmugam was admitted into the partnership for 1/4 share of profit with ₹ 12,000 as capital subject to the following adjustments.
(a) Furniture is to be revalued at ₹ 5,000 and the building is to be revalued at ₹ 50,000.
(b) Provision for doubtful debts is to be increased to ₹ 5,500
(c) An unrecorded investment of ₹ 6,000 is to be brought into account.
(d) An unrecorded liability of ₹ 2,500 has to be recorded now.
Pass journal entries and prepare Revaluation Account and capital account of partners after admission.
Solution:


Answer:
Revaluation Profit: ₹16,000; Capital accounts; Sai: ₹ 58,000 (Cr.), Shankar: ₹ 46,000 (Cr.); Shanmugam: ₹ 12,000 (Cr.))

Question 7.
Amal and Vimal are partners in firm sharing profits and losses in the ratio of 7:5. Their valance sheet as of 31st March 2019, is as follows:

Nirmal is admitted as a new partner on 01.04.2018 by introducing a capital of 30,000 for 1/3 share in the future profit subject to the following adjustments.
(a) Stock to be depreciated by ₹ 5,000;
(b) Provision for doubtful debts to be created for ₹ 3,000
(c) Land to be appreciated by ₹ 20,000.
Prepare revaluation account and capital account of partners after admission.
Solution:

Answer:
Revaluation Profit: ₹ 12,000; Capital accounts: Amal ₹ 91,000 (Cr.), Vimal ₹ 65,000 (Cr.), Nirmal ₹ 30,000 (Cr.))

Question 8.
Praveena and Dhanya are partners sharing profits in the ration of 7:3 they admit Malini into the firm. The new ratio among Praveena, Dhanya, and Malini are 5:2:3. Calculate the sacrificing ratio.
Solution:
Sacrificing ratio = Old share – New share
OR = 7:3 NR = 5:2:3
Praveena
SR = OR – NR
\(\frac{7}{10}-\frac{5}{10}=\frac{2}{10}\)
Dhanya
= \(\frac{3}{10}-\frac{2}{10}=\frac{1}{10}\)
SR = OR – NR
SR = 2:1
Answer :
Sacrificing ratio = 2:1

Question 9.
Ananth and Suman are partners sharing profits and losses in the ratio of 3:2. They admit Saran for 1/5 share, which he acquires entirely from Ananth. Find out the new profit sharing ratio and sacrificing ratio.
Solution :

Answer:
New profit sharing ratio: 2:2:1 Sacrificing ratio 1:0

Question 10.
Raja and Ravi are partners, sharing profit in the ratio of 3:2. They admit Ram for 1/4 share of the Profit, he takes 1/20 share from Raja and 4/20 from Ravi. Calculate the new profit sharing ratio and sacrificing ratio.
Solution:


Answer :
New Profit sharing ratio: 11:4:5; Sacrificing ratio 1: 4

Question 11.
Vimala and Kamala are partners, sharing profits in the ratio of 4:3. Vinitha enters into the partnership and she acquires 1/14 from Vimala and 1/14 from Kamala. find out the new profit sharing ratio and sacrificing ratio.
Solution:

Answer :
New Profit sharing ratio 7 : 5 : 2; sacrificing ratio 1 : 1

Question 12
Govinda and Gopal are partners is a firm sharing profit^ in the ratio of 5 : 4. They admit Rahim as a partner. Govind surrenders 2/9 of his share in favour of Rahim. Gopal surrenders 1 /9 of his share in favour of R thim. Calculate the new profit sharing ratio and sacrificing ratio.
Solution:


Answer:
New profit sharing ratio 35:32:14; Sacrificing ratio 5:2

Question 13.
Prema and Chandra share profits in the ratio of 5:3. Hema has admitted a partner. Prema surrendered 1 /8 of her share and Chandra surrendered 1 /8 of her share in favour of Hema. Calculate the new profit sharing ratio and sacrificing ratio.
Solution:
Prema : Chandra → 5:3 (OR)
Sacrificing Ratio

Answer:
New profit sharing ratio 35:21:8; Sacrificing ratio 5:3

Question 14.
Karthik and Kannan are equal partners. They admit Kailash with 1/4 share of the profit. Kailash acquired his share from old partners in the ratio of 7:3 Calculate the new profit sharing ratio and sacrificing ratio.
Solution :
Sacrificing Ratio → Share of New Parter 1/4 → Sacrificed → 7 : 3
karthicks SR

Answer:
New profit sharing ratio 13:17:10; Sacrificing ratio 7:3

Question 15.
Selvam and Senthil are partners sharing profit in the ratio of 2:3. Siva is admitted into the firm with 1/5 share of profit. Siva acquires equally from Selvam and Senthil. Calculate the new profit sharing ratio and sacrificing ratio.
Solution:

Answer:
New profit sharing ratio 3:5:2; Sacrificing ratio 1:1

Question 16.
Mala and Anitha are partners, sharing profits and losses in the ratio of 3:2. Mercy is admitted into the partnership with 1/5 share in the profits. Calculate new profit sharing ratio and sacrificing ratio.
Solution:
Since share sacrificed proportion and new profit sharing ratio are not given, it is assumed that the existing partners sacrifice in their old profit sharing ratio that is 3:2
Sacrificing Ratio

Answer:
New profit sharing ratio 12:8:5; Sacrificing ratio 3:2

Question 17.
Ambika, Dharani and Padma are partners in a firm sharing profile in the ratio of 5:3:2. they admit Ramya for 25% profit. Calculate the new profit sharing ratio and sacrificing ratio.
Solution:

Answer:
New profit sharing ratio 15:9:6:10; Sacrificing ratio 5:3:2

Adjustment for goodwill

Question 18.
Aparna and Priya are partners who share profits and losses in the ratio of 3:2. Brindha joins the firm for 1/5 share of profits and brings in cash for her share of the goodwill of Rs. 10,000. Pass necessary journal entry for adjusting goodwill on the assumption that the fluctuating capital method is followed and the partners withdraw the entire amount of their share of goodwill.
Solution:


Answer:
Share of goodwill: Aparna: ₹ 6,000; Priya ₹ 4,000

Question 19.
Deepak, Senthil, and Santhosh are partners sharing profits and losses equally. They admit Jerald into a partnership for 1/3 share in future profits. The goodwill of the firm is valued at ₹ 45,000 and Jerald brought cash for his share of goodwill. The existing partners withdraw half of the amount of their share of goodwill. Pass necessary journal entries for adjusting goodwill on the assumption that the fluctuating capital method is followed.
Solution:
Jerald’s share of G/w = 45,000 x \(\frac{1}{3}\) = Rs. 15,000

Answer:
Share of goodwill: Deepak: ₹ 5,000; Senthil: ₹ 5,000; Santhosh: ₹ 5,000

Question 20.
Malathi and Shobana are partners sharing profits and losses in the ratio of 5:4. They admit Jayasri into a partnership for 1/3 share of profit. Jayasri pays cash ₹ 6,000 towards her share of goodwill. The new ratio is 3:2:1. Pass necessary journal entry for adjusting goodwill on the assumption that the fixed capital method is followed.
Solution:
= OR – NR
OR = 5:4
New Ratio = 3:2:1

Answer:
Share of goodwill: Malathi’s Current account: ₹ 2,000; Shobana’s Current account: ₹ 4,000;

Question 21.
Anu and Arul were partners in a firm sharing profits and losses in the ratio of 4:1. They have decided to admit Mano into the firm for 2/5 share of profits. The goodwill of the firm on the date f admission was valued at ₹ 25,000. Mano is not able to bring in cash for his share of goodwill. Pass necessary journal entry for goodwill on the assumption that the fluctuating Capital method is followed.
Solution:
Mano’s Share of G/w = 25,000 x 2/5 = Rs. 10,000

Answer:
Share of goodwill: Anu: ₹ 8,000; Arul: ₹ 2,000;

Question 22.
Varun and Barath are sharing profits and losses 5:4. They admit Dhamu into partnership. The new profit sharing ratio is agreed at 1:1:1. Dhamu’s share of goodwill is valued at ₹ 15,000 of which he pays ₹ 10,000 in cash. Pass necessary journal entries for adjustment of goodwill on the assumption that the fluctuating capital method is . followed. ‘
Answer:

Answer:
Share of goodwill: Varun : ₹ 10,000; Barath : ₹ 5,000;

Question 23.
Sam and Jose are partners in the firm sharing profits and losses in the ratio of 3:2. On 1st April 2018, they admitted Joel as a partner. On the date of Joel’s admission, goodwill appeared in the books of the firm at ₹ 30,000. By assuming the fluctuating capital method, pass the necessary journal entry if the partners decide
(a) Write off the entire amount of existing goodwill
(b) write off ₹ 20,000 of the existing goodwill.
Solution :
To write off the entire amount of goodwill

Answer:
Share of goodwill: (a) Sam: ₹ 18,000 (Dr); Jose : ₹ 12,000 (Dr) (b) Sam: ₹ 12,000 (Dr); Jose: ₹ 8,000 (Dr.))

Comprehensive problems:

Question 24.
Rajan and Selva are partners sharing profits and losses in the ratio of 3:1. Their balance sheet as on 31st March 2017 is as under.

On 01.04.2017, they admit Ganesan as a new partner on the following arrangements.
(i) Ganesan brings ₹ 10,000 as capital for 1/5 share of profit.
(ii) Stock and furniture are to be reduced by 10%, a reserve of 5% on debtors for doubtful debts is to be created.
(iii) Appreciate buildings by 20%
Prepare revaluation account, partner’s capital account, and the balance sheet of the firm after admission.
Solution :


Answer:
Revaluation profit:: ₹ 2,100; Capital accounts : Rajan : 27075; Selva; ₹ 15,025; Ganesan : ₹ 10,000 Balance sheet total: ₹ 89,600

Question 25.
Sundar and Suresh are partners sharing profit in the ratio of 3 : 2. Their balance sheet as on 1st January 2017 was as follows:

They decided to admit Sugumar into a partnership for 1/4 share in the profits on the following terms:
(a) Sugumar has to bring in ₹ 30,000 as capital. His share of goodwill is valued at Rs. 5,000. He could not bring cash towards goodwill.
(b) That the stock is valued at ₹ 20,000.
(c) That the furniture is depreciated by ₹ 2,000.
(d) That the value of building be depreciated by 20%
Prepare necessary ledger accounts and the balance sheet after admission.
Solution:



Answer:
Answer:
Revaluation loss : ₹ 15,000; Capital accounts: Sundar : ₹ 39,000; Suresh; ₹ 26,000; Sugumar : ₹ 25,000 Balance sheet total ₹ 1,40,000

Question 26.
The following is the balance sheet of James and Justina as on 1.1.2017. They share the profits and losses equally.

On the above date, Balan is admitted as a partner with a 1/5 share in future profits. Following are the terms for his admission:
(i) Balan brings ₹ 25,000 as capital.
(ii) His share of goodwill is ₹ 10,000 and he brings cash for it.
(iii) The assets are to be valued as under:
Building ₹ 80,000; Debtors ₹ 18,000; Stock ₹ 33,000
Prepare necessary ledger accounts and the balance sheet after admission.
Solution:


Answer:
Revaluation profit:: ₹ 11,000; Capital accounts : James : ₹ 58,000; Justina; ₹ 68,000; Balan : ₹ 25,000 Balance sheet total: ₹ 1,86,000

Question 27.
Anbu and Shankar are partners in a business sharing profits and losses in the ratio of 7:5. The balance sheet of the partners on 31.03.2018 is as follows:

Rajesh is admitted for 1/5 share on the following terms:
(i) Goodwill of the firm is valued at ₹ 80,000 and Rajesh brought cash ₹ 6,000 for his share of goodwill.
(ii) Rajesh is brought ₹ 1,50,000 as his capital.
(iii) Motor car is valued at ₹ 2,00,000; stock at ₹ 3,80,000 and debtors at ₹ 3,50,000.
(iv) Anticipated claim on workmen compensation fund is ^ 10,000
(v) Unrecorded investment of ₹ 5,000 has to be brought into account.
Prepare revaluation account, capital account, and balance sheet after Rajesh’s admission.
Solution:


Answer:
Revaluation profit:: ₹ 15,000; Capital accounts: Anbu: ₹ 5,11,417; Shankar; ₹ 3,79,583; Rajesh : ₹ 1,50,000 Balance sheet total: ₹ 11,71,000

12th Accountancy Guide Admission of a Partner Additional Important Questions and Answers

Question 1.
When A and B sharing profit and losses in the ratio of 3:2. They admit C as a partner giving him 1 /3 share of profits. This will be given by A 8t B
(a) Equally
(b) in the ration of their
(c) in the ratio of profits
Answer:
(c) in the ratio of profits

Question 2.
In order to maintain fair dealings at the time of admission, it is necessary to revalue assets 8t liabilities of the firm to their
(a) Cost Price
(b) Cost price less depreciation
(c) True value.
Answer:
(c) True value.

Question 3.
If the new share of the incoming partner is given without mentioning the details of the sacrifice made by the old partners then the presumption is that old partners sacrifice in the
(a) Old profit sharing ration
(b) Gaining ratio
(c) Capital ratio.
Answer:
(a) Old profit sharing ration

Question 4.
On admission of a new partner, the balance of general Reserve A/c should be transferred to the capital account of
(a) all partners in their new profit sharing ratio
(b) Old partners in their new old profit sharing ratio
(c) Old partners in their new profit sharing ratio.
Answer:
(b) Old partners in their new old profit sharing ratio

Question 5.
The old partners share all the accumulated profit & reserves in their
(a) new profit sharing ratio
(b) Old profit sharing ratio
(c) Capital ratio
Answer:
(b) Old profit sharing ratio

Question 6.
Hie reconstitution of the partnership requires a revision of the existing partners
(a) Profit sharing ratio
(b) Capital ratio
(c) Sacrificing ratio|
Answer:
(a) Profit sharing ratio

Question 7.
……………… ratio is computed at the time by the admission of a partner
(a) gaining ratio
(b) Capitalization
(c) Sacrificing ratio
Answer:
(c) Sacrificing ratio

Question 8.
When unrecorded liability is brought in to books of accounts, it results in
(a) profit
(b) loss
(c) Neither profit nor loss
Answer:
(b) loss

IV Additional Problems:

Question 1.
Anandan and Balaraman partners in a firm with a capital of Rs. 70,000 and Rs. 50,000 respectively. They decided to admit Chandran into the firm with a capital of Rs. 40,000. Give journal entry for Capital brought in by Chandran.
Solution :

Question 2.
Rathai and Kothai are partners sharing profits in the ratio of 3:2. They admit Kanmani for 1/5th share of future profits which she acquires 3/20th from Rathai and 1/20th from Kothai. Calculate new Pro iring ratio and. sacrifice s ratio of old partners.
Solution :

Question 3.
P and Q are partners sharing profits in the ratio of 3:2. They admit R for 1/5th Share which acquires equally from P and Q. Calculate new profit sharing ratio and sacrificing ratio of old partners.
Solution:

Question 4.
Sankar and Saleem are partner in a firm sharing profits and losses in the ration of 3:2 as on 31st March 2005. Their Balarr sheet was as under.

On 1st April 2005, they admit Solomon into partnership on the following condition: Solomon has brought Rs. 1,00,000 as capital.
The value of land and building was to be increased by Rs. 20,000.
Stock and furniture were to be depreciated by Rs. 10,000 and Rs. 5,000 respectively. Rs. 15,000 to be written off from sundry creditors as it is no longer liable.
Provision for doubtful debts is to be increased by Rs. 1,000.
Give journal entries, prepare Revaluation Account and the Balance Sheet.
Solution:

Question 5.
Amar and Akbar are partners in a firm sharing profits and losses in the ratio of 2:1  as 31st March 2005. Their Balan Sh» was as under.

On 1st April 2005, they admit Antony into partnership on the following conditions:
Antony has brought in the capital of Rs. 1,50,000 for 1/5th share of the future profits. Stock and machinery were to be depreciated by Rs. 6,000 and Rs. 15,000 respectively.
Investments Rs. 15,000 not recorded in the books brought into accounts.
Provision for doubtful debts is to be created at 5% on debtors.
A liability of Rs. 4,000 for outstanding repairs has been omitted to be recorded in the books. Give journal entries, prepare Revaluation Account, Capital Account, Bank Account, and the Balance Sheet.
Solution:

Question 6.
Sumathi and Sundari are partners of a firm sharing profit and loss in the ratio of 4 :3. Their Balance Sheet shows Rs. 14,000 as Profit and Loss A/c on the liabilities side. Pass entry.
Solution:

Question 7.
Mahalakshmi and Dhanalakshmi are partners sharing profit and loss in the ratio of 3:2. They admit Deepalakshmi on 1st January 2005. On that date, their Balance Sheet showed an amount of Rs. 25,000 as Profit and Loss A/c in the Asset side. Pass entry.
Solution:

Question 8.
Damodaran and Jagadeesan are partners sharing profits in the ratio of 3:2. They decided to admit Vijayan for 1/5th share of future profit. Goodwill of the firm is to be valued at Rs. 50,000.
Give Journal entries, if
There is no goodwill in the books of the firm.
The goodwill appears at Rs. 30,000
The goodwill appears at Rs. 60,000.
Solution:

Question 9.
Sankari and Sudha are partners sharing profit and loss in the ratio of 3:2. Their Balance Sheet as on 31st March 2005 is as under.

They decided to admit Santhi into the partnership with effect from 1st April 2005 on the following terms:
Santhi to bring in Rs. 60,000 as Capital for 1/3rd share of profits.
Goodwill was valued at Rs. 45,000
The land was valued at Rs. 1,50,000
The stock was to be written down by Rs. 8,000
The provision for doubtful debts was to be increased to Rs. 3,000
Creditors include Rs. 5,000 no longer payable and this sum was to be written off.
Investment of Rs. 10,000 be brought into books.
Prepare Revaluation A/c, Capital A/c, and Balance Sheet of the new firm.
Solution :

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.7 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.7

Question 1.
Evaluate the following
(i) \(\int_{0}^{∞}\) x⁵ e^-3x dx
Solution:

(ii) \(\int_{0}^{π/2}\) \(\frac{e^{-tanx}}{cos^6 x}\) dx
Solution:

Question 2.
\(\int_{0}^{∞}\) sin^αx² x³ dx = 32, α > 0, find α
Solution:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Physics Guide Pdf Chapter 1 Electrostatics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

12th Physics Guide Electrostatics Book Back Questions and Answers

Part – I:
I. Multiple choice questions:

Question 1.
Two identical point charges of magnitude – q are fixed as shown in the figure below. A third charge +q is placed midway between the two charges at the point P. Suppose this charge +q is displaced a small distance from the point P in the directions indicated by the arrows, in which direction(s) will +q be stable with respect to the displacement?

(a) A₁ and A₂
(b) B₁ and B₂
(c) both directions
(d) No stable
Answer:
(b) B₁ and B₂
Solution:
The potential due to an electric dipole along the equatorial line is zero.

Question 2.
Which charge configuration produces a uniform electric field?
(a) point charge
(b) uniformly charged infinite line
(c) uniformly charged infinite plane
(d) uniformly charged spherical shell
Answer:
c) uniformly charged infinite plane
Solution:

Uniform field lines are represented by equidistant parallel lines.

Question 3.
What is the ratio of the \(\left|\frac{\mathrm{q}_{1}}{\mathrm{q}_{2}}\right|\) charges for the following electric field line pattern?

(a) \(\frac{1}{5}\)
(b) \(\frac{25}{11}\)
(c) 5
(d) \(\frac{11}{25}\)
Answer:
(d) \(\frac{11}{25}\)
Solution:
Here q₁ is a negative charge and q₂ is a positive charge
Hint:
Count the number of lines on each charge.

Question 4.
An electric dipole is placed at an alignment angle of 30° with an electric field of 2 × 10⁵ NC^-1. It experiences a torque equal to 8 Nm. The charge on the dipole if the dipole length is lcm is
(a) 4 mC
(b) 8 mC
(c) 5 mC
(d) 7 mC
Answer:
(b) 8 mC
Solution:
T = PE sin θ
T = (q × 21)E sin30°
q = (q × 10^-2) × q = 8 × 10^-2c

Question 5.
Four Gaussian surfaces are given below with charges inside each Gaussian surface. Rank the electric flux through each Gaussian surface in increasing order:

(a) D < C < B < A
(b) A < B = C < D
(c) C < A – B < D
(d) D > C > B > A
Answer:
(a) D < C < B < A
Hint :
flux depends on charge

Question 6.
The total electric flux for the following closed surface which is kept inside water:

(a) \(\frac{80 \mathrm{q}}{\varepsilon_{\mathrm{o}}}\)
(b) \(\frac{\mathrm{q}}{40 \varepsilon_{0}}\)
(c) \(\frac{\mathrm{q}}{80 \varepsilon_{\mathrm{o}}}\)
(d) \(\frac{\mathrm{q}}{160 \varepsilon_{0}}\)
Answer:
(b) \(\frac{\mathrm{q}}{40 \varepsilon_{0}}\)
Solution:

Question 7.
Two identical conducting balls having positive charges q₁ and q₂ are separated by a center to center distance r. If they are made to touch each other and then separated to the same distance, the force between them will be
(a) less than before
(b) same as before
(c) more than before
(d) zero
Answer:
(c) more than before
Solution:

Question 8.
Rank the electrostatic potential energies for the given system of charges in increasing order:

(a) 1 = 4 < 2 < 3
(b) 2 = 4 < 3 < 1
(c) 2 = 3 < 1 < 4
(d) 3 < 1 < 2 < 4
Answer:
(a) 1 = 4 < 2 < 3
Solution:

Question 9.
An electric field \(\overrightarrow{\mathrm{E}}\)= 10 × Î exists in a certain region of space. Then the potential difference V = V_o – V_A, where V_o is the potential at the origin and V_A is the potential at x = 2 m is:
(a) 10 V
(b) -20 V
(c) +20 V
(d) -10 V
Answer:
(c) +20 V
Solution:
E = \(-\frac{d v}{d x}\)
dv = E.dx
= l0x
= 10 × 2
dv = 20 V

Question 10.
A thin conducting spherical shell of radius R has a charge Q which is uniformly distributed on its surface. The correct plot for electrostatic potential due to this spherical shell is
(a)

(b)

(c)

(d)
Answer:
(b)
Solution:
In a spherical shell, the electric field inside is zero. But the electric potential is constant.
\(\mathrm{V}=\frac{\mathrm{q}}{4 \pi \varepsilon_{\mathrm{o}} \mathrm{r}}\) as distance increases its potential decrease non-linearly.

Question 11.
Two points A and B are maintained at a potential of 7 V and -4 V respectively. The work done in moving 50 electrons from A to B is
(a) 8.80 × 10^-17 J
(b) -8.80 × 10^-17 J
(c) 4.40 × 10^-17 J
(d) 5.80 × 10^-17 J
Answer:
(a) 8.80 × 10^-17 J
Solution:
W_A→B = (V_A – V_B)q
=(7+4)ne
= 11 × 50 × 1.6 × 10^-19
=8.8 × 10^-17

Question 12.
If the voltage applied on a capacitor is increased from V to 2V, choose the correct conclusion.
(a) Q remains the same, C is doubled
(b) Q is doubled, C doubled
(c) C remains the same, Q doubled
(d) Both Q and C remain the same
Answer:
(c) C remains the same, Q doubled
Solution:

Question 13.
A parallel plate capacitor stores a charge Q at a voltage V. Suppose the area of the parallel plate capacitor and the distance between the plates are each doubled then which is the quantity that will change?
(a) Capacitance
(b) Charge
(c) Voltage
(d) Energy density
Answer:
(d) Energy density

Question 14.
Three capacitors are connected in a triangle as shown in the figure. The equivalent capacitance between points A and C is

(a) 1 μF
(b) 2 μF
(c) 3 μF
(d) \(\frac{1}{4}\)μF
Answer:
(b) 2 μF
Solution:

Question 15.
Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 × 10^-2 C and 5 × 10^-2 C respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is
(a) 3 × 10^-2 C
(b) 4 × 10^-2 C
(c) 1 × 10^-2 C
(d) 2 × 10^-2 C
Answer:
(a) 3 × 10^-2 C
Solution:
Total charge Q = q₁ + q₂ = 4 × 10^-2 C
charge on bigger sphere,
q₂ = Q\(\left(\frac{r_{2}}{r_{1}+r_{2}}\right)\)
= 4 × 10^-2 × \(\frac{3}{4}\)
q₂ = 3 × 10^-2 C

II. Short Answer Questions:

Question 1.
What is meant by the quantization of charges?
Answer:
The charge q on any object is equal to an integral multiple of this fundamental unit of charge e.
q = ne
Here n is an integer (0, ±1, ±2, ±3, ±4).
This is called quantisation of electric charge.

Question 2.
Write down Coulomb’s law in vector form and mention what each term represents.
Answer:
The force on the point charge q₂ exerted by another point charge q₁

where r̂₂ is the unit vector directed from charge q₁ to charge q₂, and ‘K’ is the proportionality constant K = \(\frac{1}{4 \pi \varepsilon_{0}}\); ε₀ is the permittivity of free space.

Question 3.
What are the differences between the Coulomb force and the gravitational force?
Answer:

• The gravitational force between two masses is always attractive but Coulomb’s force between two charges can be attractive or repulsive, depending on the nature of charges.
• The value of the gravitational constant G = 6.626 x 10^-11 N m² kg^-2. The value of the constant k in Coulomb law is k = 9 x 10⁹ N m² C².
• The gravitational force between two masses is independent of the medium. The electrostatic force between the two charges depends on the nature of the medium in which the two charges are kept at rest.
• The gravitational force between two point masses is the same whether two masses are at rest or in motion. If the charges are in motion, yet another force (Lorentz force) comes into play in addition to Coulomb force.

Question 4.
Write a short note on the superposition principle.
Answer:

1. The superposition principle explains the interaction of multiple charges.
2. The total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges.
3. The force on q1 exerted by the charge q<₂ is \(\overrightarrow{\mathrm{F}}_{12}=\mathrm{K} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{21}^{2}} \hat{\mathrm{r}}_{21}\)
4. The force on q1 exerted by the charge q<₃ is \(\overrightarrow{\mathrm{F}}_{13}=\mathrm{K} \frac{\mathrm{q}_{1} \mathrm{q}_{3}}{\mathrm{r}_{31}^{2}} \hat{\mathrm{r}}_{31}\)
5. \(\overrightarrow{\mathrm{F}}_{1}^{\text {tot }}=\mathrm{K}\left\{\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{21}^{2}} \hat{\mathrm{r}}_{21}+\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{\mathrm{r}_{31}^{2}} \hat{\mathrm{r}}_{31}+\ldots \frac{\mathrm{q}_{1} \mathrm{q}_{\mathrm{n}}}{\mathrm{r}_{\mathrm{n}}^{2}} \hat{\mathrm{r}}_{\mathrm{n}_{i}}\right\}\)
   

Question 5.
Define ‘electric field’.
Answer:
The electric field at the point P at a distance r from the point charge q is the force experienced by a unit charge and is given by
\(\vec { E } \) = \(\frac { \vec { F } }{ { q }_{ 0 } } \)
The electric field is a vector quantity and its SI unit is Newton per Coulomb (NC^-1).

Question 6.
What is mean by ‘electric field lines?
Answer:
Electric lines of force is an imaginary straight or curved path in which unit positive charge tends to move in the presence of an electric field.

Question 7.
The electric field lines never intersect. Justify.
Answer:
As a consequence, if some charge is placed at the intersection point, then it has to move in two different directions at the same time, which is physically impossible. Hence, electric field lines do not intersect.

Question 8.
Define ‘electric dipole’. Give the expression for the magnitude of its electric dipole moment and the direction.
Answer:

1. Two equal and opposite charges separated by a very small distance constitute an electric dipole.
   Ex:- Carbondioxide, Water.
2. The magnitude of the dipole moment is given by the product of the magnitude of the one of the charges and the distance between them.
   \(|\overrightarrow{\mathrm{P}}|=2 \mathrm{qa}\)
   \(\overrightarrow{\mathrm{P}}=\mathrm{qr}_{+}+(-\mathrm{q}) \overrightarrow{\mathrm{r}}_{-}\)
   where r̂₊ is the position of vector of +1 from the origin and r̂_– is the position vector of -q from the origin
   \(\overrightarrow{\mathrm{P}}=\mathrm{qai}-\mathrm{qa}(-\hat{i})=2 \mathrm{qai}\)
3. It is a vector quantity having direction along the dipole axis -q to +q.
4. Unit: coulomb metre (Cm).

Question 9.
Write the general definition of electro dipole moment for a collection of point charge.
Answer:
The electric dipole moment vector lies along the line joining two charges and is directed from -q to + q. The SI unit of dipole moment is coulomb meter (Cm).
\(\vec { P } \) = qa\(\hat{i} \) -qa(\(\hat{-i} \)) = 2 qa\(\hat{i} \)

Question 10.
Define ‘electrostatic potential’.
Answer:
The electric potential at a point P is equal to the work done by an external force to bring a unit positive charge with constant velocity from infinity to the point P in the region of the external electric field \(\overrightarrow{\mathrm{E}}\).

Question 11.
What is an equipotential surface?
Answer:
An equipotential surface is a surface on which all the points are at the same potential.

Question 12.
What are the properties of an equipotential surface?
Answer:

1. The work done to move a charge ‘q’ between two points lie on equipotential surface is zero.
2. The electric field is normal to an equipotential surface.

Question 13.
Give the relation between electric field and electric potential.
Answer:
Consider a positive charge q kept fixed at the origin. To move a unit positive charge by a small distance dx in the electric field E, the work done is given by dW = -E dx. The minus sign implies that work is done against the electric field. This work done is equal to an electric potential difference. Therefore,
dW = dV.
(or) dV = -Edx
Hence E = \(\frac { dV }{ dx }\)
The electric field is the negative gradient of the electric potential.

Question 14.
Define ‘electrostatic potential energy’.
Answer:
It is defined as the amount of work done in assembling the charges at their locations by bringing them in from infinity.

Question 15.
Define ‘electric flux’.
Answer:
The number of electric field lines crossing a given area kept normal to the electric field lines is called electric flux. Its unit is N m² C^-1. Electric flux is a scalar quantity.

Question 16.
What is meant by electrostatic energy density?
Answer:

1. The energy stored per unit volume of space is defined electrostatic energy density.
   \(\mathrm{U}_{\mathrm{E}}=\frac{\mathrm{U}}{\text { Volume }}=\frac{1}{2} \varepsilon_{\mathrm{o}} \mathrm{E}^{2}\)
2. U – electrostatic potential energy
3. E – electric field
4. £0 – permittivity of free space

Question 17.
Write a short note on ‘electrostatic shielding’.
Answer:
Consider a cavity inside the conductor. Whatever the charges at the surfaces and whatever the electrical disturbances outside, the electric field inside the cavity is zero. A sensitive electrical instrument which is to be protected from external electrical disturbance is kept inside this cavity. This is called electrostatic shielding.

Question 18.
What is Polarisation?
Answer:
Polarisation is defined as the total dipole moment per unit volume of the dielectric.
\(\begin{array}{l}
\vec{p}=\chi_{\mathrm{e}} \overrightarrow{\mathrm{E}}_{\text {ext }} \\
\chi_{\mathrm{e}}=\text { electric susceptibility }
\end{array}\)

Question 19.
What is dielectric strength?
Answer:
The maximum electric field the dielectric can withstand before it breakdowns is called dielectric strength.

Question 20.
Define ‘capacitance’. Give its unit.
Answer:

• It is defined as the ratio of the magnitude of the charge on either of the conductor plates to the potential difference existing between the conductors.
• Unit: farad (or) C/V

Question 21.
What is corona discharge?
Answer:
The electric field near the edge is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge. This reduces the total charge of the conductor near the sharp edge. This is called action at points or corona discharge.

III. Long Answer Questions:

Question 1.
Discuss the basic properties of electric charges.
Answer:
Basic properties of charges
(i) Electric charge:
Most objects in the universe are made up of atoms, which in turn are made up of protons, neutrons and electrons. These particles have mass, an inherent property of particles. Similarly, the electric charge is another intrinsic and fundamental property of particles. The SI unit of charge is the coulomb.

(ii) Conservation of charges:
Benjamin Franklin argued that when one object is rubbed with another object, charges get transferred from one to the other. Before rubbing, both objects are electrically neutral and rubbing simply transfers the charges from one object to the other. (For example, when a glass rod is rubbed against silk cloth, some negative charges are transferred from glass to silk. As a result, the glass rod is positively charged and silk cloth becomes negatively charged).

From these observations, he concluded that charges are neither created or nor destroyed but can only be transferred from one object to other. This is called conservation of total charges and is one of the fundamental conservation laws in physics. It is stated more generally in the following way. The total electric charge in the universe is constant and charge can neither be created nor be destroyed. In any physical process, the net change in charge will always be zero.

(iii) Quantisation of charges:
The charge q on any object is equal to an integral multiple of this fundamental unit of charge e.
q = ne
Here n is any integer (0, ±1, ±2, ±3, ± ….). This is called quantisation of electric charge. Robert Millikan in his famous experiment found that the value of e = 1.6 x 10^-19C. The charge of an electron is -1.6 x 10^-19 C and the charge of the proton is +1.6 x 10^-19C. When a glass rod is rubbed with silk cloth, the number of charges transferred is usually very large, typically of the order of 10¹⁰. So the charge quantisation is not appreciable at the macroscopic level. Hence the charges are treated to be continuous (not discrete). But at the microscopic level, quantisation of charge plays a vital role.

Question 2.
Explain in detail Coulomb’s law and its various aspects.
Answer:
1. Coulomb’s law states that “force of attraction or repulsion between two point charges is directly proportional to the product of the charges and inversely proportional to the square of distance between them.
\(\overrightarrow{\mathrm{F}}=\mathrm{k} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}}_{12}\)

2. The force on the charge q₂ exerted by the charge q&lt₁ always lies along the line joining the two charges.
r̂&lt₁2 is a unit vector pointing from q₁ to q₂.

3. In SI units, \(\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\)= 9 x l0⁹Nm²C^-2
where £0 = 8.85 × 10^-12 C² N^-1m^-2 .

4. The magnitude of the electrostatic force between two charges each of one coulomb and separated by a distance of lm is
|F| = 9 x l0⁹N
This is a huge quantity, almost equivalent to weight of one million tons.

5. Coulombs law in vacuum takes the form
\(\overrightarrow{\mathrm{F}}_{21}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}}_{12}\)
In a Medium,
\(\overrightarrow{\mathrm{F}}_{21}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}}_{12}\)
£ > £0, the force between two point charges in a medium other than vacuum is always less than that in vacuum.

6. Relative permitivitv £r = \(\frac{\varepsilon}{\varepsilon_{o}}\)
for vacuum and air £r = 1
other media, £r > 1

Question 3.
Define ’electric field’ and discuss its various aspects.
Answer:
The electric field is defined as the force experienced by unit positive charge kept at that point.
\(\overrightarrow{\mathrm{E}}=\frac{\overrightarrow{\mathrm{F}}}{\mathrm{q}_{\mathrm{o}}}\)

Important aspects of the Electric field:

1. If the charge q is positive then the electric field points away from the source charge and if q is negative, the electric field points towards the source charge q.
2. If the electric field at a point P is E, then the force experienced by the test charge q0 placed at the point P is
   \(\overrightarrow{\mathrm{F}}=\mathrm{q}_{\mathrm{o}} \overrightarrow{\mathrm{E}}\)
3. Electric field is independent of the test charge q₀ and depends only on the source charge q.
   
4. Since the electric field is a vector quantity, at every point in space, this field has unique direction.
5. The test charge is made sufficiently small such that it will not modify the electric field of the source charge.
6. For continuous and finite size charge distributions, integration techniques must be used.
7. There are two kinds of electric field: uniform or constant electric field and non-uniform electric field.
8. Uniform electric field will have the same direction and constant magnitude at all points in space. Non-uniform electric field will have different directions or different magnitudes or both at different points in space.

Question 4.
Calculate the electric field due to a dipole on its axial line and the equatorial plane.
Answer:
Case (i):

Electric field due to an electric dipole at points on the axial line.

1. Consider an electric dipole placed on the x-axis A point C is located at a distance of r from the midpoint of the dipole along the axial line.
   

2. The electric field at a point C due to +q is
\(\overrightarrow{\mathrm{E}}_{+}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{(\mathrm{r}-\mathrm{a})^{2}}\) (along BC)

3. Since the electric dipole moment vector p is from -q to +q and is directed along BC, the above equation is rewritten as
\(\overrightarrow{\mathrm{E}}_{+}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{(\mathrm{r}-\mathrm{a})^{2}} \hat{\mathrm{p}}\) ………..(1)

4. The electric field at a point C due to -q is
\(\overrightarrow{\mathrm{E}}_{-}=-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{(\mathrm{r}+\mathrm{a})^{2}} \hat{\mathrm{p}}\) ……………(2)

5. Since +q is located closer to the point C than -q, \(\overrightarrow{\mathrm{E}}+\) is stronger than \(\overrightarrow{\mathrm{E}}_{-}\).
Therefore, the length of the \(\overrightarrow{\mathrm{E}}+\) vector is drawn larger than that of \(\overrightarrow{\mathrm{E}}_{-}\) vector.
The total electric field at point C is calculated using the superposition principle of the electric field.

6. The direction of is shown in Figure

Case (ii)
1. Electric field due to an electric dipole at a point on the equatorial plane

2. Consider a point C at a distance r from the midpoint 0 of the dipole on the equatorial plane.

3. Since point C is equidistant from +q and -q and are the same.

4. The direction of \(\overrightarrow{\mathrm{E}}+\) is along BC and the direction of \(\overrightarrow{\mathrm{E}}_{-}\) is along CA.

\(\overrightarrow{\mathrm{E}}+\) and \(\overrightarrow{\mathrm{E}}_{-}\) are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it.

5. The perpendicular components \(|\overrightarrow{\mathrm{E}}|\) sinθ and \(\left|\overrightarrow{\mathrm{E}}_{-}\right|\) sinθ are oppositely directed and cancel each other.
\(\mathrm{E}_{\text {tot }}=-\left|\mathrm{E}_{+}\right| \cos \theta \hat{\mathrm{p}}-\left|\overrightarrow{\mathrm{E}}_{-}\right| \cos \theta \hat{\mathrm{p}} \cdots \ldots .(6)\)

6. The magnitudes \(\overrightarrow{\mathrm{E}}+\) and \(\overrightarrow{\mathrm{E}}_{-}\) are the same and are given by

7. By substituting equation (7) into equation (6) we get
\(\overrightarrow{\mathrm{E}}_{\mathrm{tot}}=-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{2 \mathrm{q} \cos \theta}{\left(\mathrm{r}^{2}+\mathrm{a}^{2}\right)} \hat{\mathrm{p}}\)
since \(\cos \theta=\frac{a}{\sqrt{r^{2}+a^{2}}}\)

8. At very large distances (r»a), the equation (8) becomes
\(\overrightarrow{\mathrm{E}}_{\mathrm{tot}}=-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\overrightarrow{\mathrm{p}}}{\mathrm{r}^{3}}(\mathrm{r}>>\mathrm{a})\)

Question 5.
Derive an expression for the torque experienced by a dipole due to a uniform electric field.
Answer:
1. Consider an electric dipole of dipole moment \(\overrightarrow{\mathrm{p}}\) placed in a uniform electric field . The charge+q will experience a force q\(\overrightarrow{\mathrm{E}}\) in the direction of the field and charge-q\(\overrightarrow{\mathrm{E}}\) will experience a force -qE in a direction opposite to the field. Since the external field \(\overrightarrow{\mathrm{E}}\) is uniform, the total force acting on the dipole is zero.

2. These two forces acting at different points will constitute a couple and the dipole experience a torque. This torque tends to rotate the dipole.

3. The total torque on the dipole about the point 0, is given by
\(\vec{\tau}=\overrightarrow{\mathrm{0A}} \times(-\mathrm{q} \overrightarrow{\mathrm{E}})+\overrightarrow{\mathrm{0B}} \times \mathrm{q} \overrightarrow{\mathrm{E}}\)
Torque on dipole.

4. Total torque is perpendicular to the plane of the paper and is directed into it. The magnitude of the total torque
\(\vec{\tau}=|\overrightarrow{\mathrm{0A}}|(-\mathrm{q} \overrightarrow{\mathrm{E}})|\sin \theta+| \overrightarrow{\mathrm{0B}}|\mathrm{q} \overrightarrow{\mathrm{E}}| \sin \theta\)

\(\tau\) = qE.2a sin θ

5. Where θ is the angle made by \(\overrightarrow{\mathrm{p}}\) with \(\overrightarrow{\mathrm{E}}\). Since p = 2aq, the torque is written in terms of the vector product as
\(\vec{\tau}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{E}}\)

6. The magnitude of thectrque is \(\tau=p E \sin \theta\) and is maximum when θ = 90^o.
This torque tends to rotate the dipole and align it with the electric field \(\overrightarrow{\mathrm{E}}\). once \(\overrightarrow{\mathrm{p}}\) is aligned with, the total torque on the dipole becomes zero.

7. If the electric field is nonuniform, then the force experienced by +q is different from that experienced by -q. In addition to the torque, there will be a net force acting on the dipole.

Question 6.
Derive an expression for electrostatic potential due to a point charge.
Answer:
1. Consider a positive charge ‘q’ kept fixed at the origin. Let P be a point at distance r from the charge ‘q’.

2. The electric potential at the point P is
\(\mathrm{V}=\int_{\infty}^{\mathrm{r}}(-\overrightarrow{\mathrm{E}}) \cdot \mathrm{d} \overrightarrow{\mathrm{r}}=-\int_{\infty}^{\mathrm{r}} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dr}}\)

3. Electric field due to positive point charge is
\(\begin{array}{l}
\overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}} \\
\mathrm{V}=\frac{-1}{4 \pi \varepsilon_{\mathrm{o}}} \int_{\infty}^{\mathrm{r}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}} \cdot \mathrm{d} \overrightarrow{\mathrm{r}}
\end{array}\)

4. The infinitesimal displacement vector,
d\(\overrightarrow{\mathrm{r}}\) = drr̂ and using r̂.r̂ = 1, we have
\(\mathrm{V}=\frac{-1}{4 \pi \varepsilon_{\mathrm{o}}} \int_{\infty}^{\mathrm{r}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}} \cdot \mathrm{dr} \hat{\mathrm{r}}=-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \int_{\infty}^{\mathrm{r}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \mathrm{dr}\)

5. After the integration,
\(\mathrm{V}=-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \mathrm{q}\left\{-\frac{1}{\mathrm{r}}\right\}_{\infty}^{\mathrm{r}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{\mathrm{r}}\)

6. Hence the electric potential due to a point charge q at a distance r is
\(\mathrm{V}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{\mathrm{r}}\)

Question 7.
Derive an expression for electrostatic potential due to an electric dipole.
Answer:
Consider two equal and opposite charges separated by a small distance 2a. The point P is located at a distance r from the midpoint of the dipole. Let θ be the angle between the line OP and dipole axis AB.

Potential due to electric dipole:
1. Let r₁ be the distance of point P from +q and r₂ be the distance of point P from -q.

2. Potential at P due to charge,
\(+q=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r_{1}}\)
Potential at P due to charge
\(-q=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r_{2}}\)
Total potential at the point P,
\(\mathrm{V}=\frac{\mathrm{q}}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{r}_{2}}\right)\) (1)

3. By the cosine law for triangle BOP,
r₁² =r² + a² – 2ra cos θ

4. Similarly applying the cosine law for triangle A0P,
r₂² = r² + a²  – 2racos(180-θ)
since cos(180-θ) = – cos θ
r₂² = r² + a² +2racos θ

Using Binomial theorem, we get

\(\mathrm{V}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\overrightarrow{\mathrm{p}} \cdot \hat{\mathrm{r}}}{\mathrm{r}^{2}}\)
Special Cases:

Question 8.
0btain an expression for potential energy due to a collection of three-point charges which are separated by finite
distances.
Answer:

1. The electric potential at a point at a distance r from point charge q&lt₁ is given by
   \(\mathrm{V}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}_{1}}{\mathrm{r}}\)
2. The potential V is the work done to bring a unit positive charge front infinity to the point. If the charge q, is brought from infinity to that point at a distance r from q₁.
3. The work done is stored as the electrostatic potential energy W = q₂V
   \(\mathrm{U}=\mathrm{q}_{2} \mathrm{~V}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \mathrm{q}_{1} \mathrm{q}_{2}\) ………….(1)
4. The electrostatic potential energy depends only on the distance between the two point charges.

Three charges are arranged in the following configuration.

1. Bringing a charge q₁ from infinity to point A requires no work because there are no other charges already present in the vicinity of charge q₁.

2. To bring the second charge q₂ to point B, work must be done against the electric field at B created by the charge q₁. So the work done on the charge q₂ is Wq₂V_1B Here V_1B is electrostatic potential due to the charge q₁ at point B.
\(\mathrm{U}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{12}}\) (2)

3. Similarly to bring the charge q₂ to point C, work has to be done against the total electric field due to both the charges q₁ and q₂. So the work done to bring the charge q₃ is the electrostatic potential due to charge q₁ at point C and V₂ is the electrostatic potential due to charge q₂ at point C. The electrostatic potential is
\(\mathrm{U}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{\mathrm{r}_{13}}+\frac{\mathrm{q}_{2} \mathrm{q}_{3}}{\mathrm{r}_{23}}\right)\) ………….(3)

4. Adding equations (2) and (3), the total electrostatic potential energy for the system of three charges q₁, q₂ and q₃ is
\(\mathrm{U}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{12}}+\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{\mathrm{r}_{13}}+\frac{\mathrm{q}_{2} \mathrm{q}_{3}}{\mathrm{r}_{23}}\right)\) ……………. (4)

Question 9.
Derive an expression for the electrostatic potential energy of the_dipole in a uniform electric field.
Answer:
1. Consider a dipole placed in the uniform electric field \(\overrightarrow{\mathrm{E}}\). A dipole experiences a torque when kept in an uniform electric field \(\overrightarrow{\mathrm{E}}\). This torque rotates the dipole to align it with the direction of the electric field.

2. To rotate the dipole an equal and opposite external torque must be applied on the dipole.

3. The work done by the external torque to rotate the dipole from angle θ<super>1 to θ at constant angular velocity is

4. The work done is equal to the potentiaL energy difference between the angular positions θ’ and θ.
U(θ) – U(θ’)= ΔU = – pE cos θ+pE cos θ’

5. If the initial angle is θ’ = 90° and is taken as reference point, then U(θ’)= pE cos 90°=0.

6. The potential energy stored in the system of dipole kept in the uniform electric field is given by
U= – pE cos θ = – \(\overrightarrow{\mathrm{p}}\).\(\overrightarrow{\mathrm{E}}\) ……….(3)

7. In addition to p and E, the potential energy also depends on the orientation 0 of the electric dipole with respect to the external electric field.

8. The potential energy is maximum when the dipole is aligned anti-parallel (θ = π) to the external electric field and minimum when the dipole is aligned parallel (θ = 0)

Question 10.
Obtain Gauss law from Coulomb’s law.
Answer:
1. Consider a positive point charge Q is surrounded by an imaginary sphere of radius r.

2. The total electric flux for the entire area is given by,
\(\Phi_{\mathrm{E}}=\oint \mathrm{E} \cdot \mathrm{d} \overrightarrow{\mathrm{A}}=\oint \mathrm{EdA} \cos \theta \ldots \ldots\) (1)

3. The electric field of the point charge is directed radially outward at all points on the surface of the sphere. Therefore the direction of the area element d\(\overrightarrow{\mathrm{A}}\) is along the electric field \(\overrightarrow{\mathrm{E}}\) and θ = 0°.
\(\Phi_{\mathrm{E}}=\oint \mathrm{E.dA}\) (since cos 0° = 1)
E is a uniform on the surface of the sphere

4. This equation is Gauss Law.

Question 11.
Obtain the expression for electric field due to an infinitely long charged wire.
Answer:

Electric field due to infinite long charged wire:

1. Consider an infinitely long straight wire having uniform linear charge density λ.

2. Let P be a point located at a perpendicular distance r for the wire.

3. The charged wire possesses a cylindrical symmetry.

4. Let us choose a cylindrical Gaussian surface of radius r and length L,

5. The total electric flux in this closed surface is calculated as follows.

for the curved surface. \(\overrightarrow{\mathrm{E}}\) is parallel to \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{E}}\) .dA = EdA.
For the top and bottom surfaces,
\(\overrightarrow{\mathrm{E}}\) is perpendicular to \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{E}}\)
\(\begin{aligned}
&\Phi_{\mathrm{E}}=\int_{\text {Curved }} \mathrm{EdA}=\frac{\mathrm{Q}_{\text {encl }}}{\varepsilon_{0}} \ldots \ldots . \text { (2) }\\
&\text { surface }
\end{aligned}\)

Since the electric field is constant, E is taken out of the integration and
Q_encl is given by Q_encl=λL.
E \(\int_{\begin{array}{l}
\text { Curved } \\
\text { surface }
\end{array}} \mathrm{d} \mathrm{A}=\frac{\lambda \mathrm{L}}{\varepsilon_{\mathrm{o}}}\) ……..(3)
Here,
Φ_E=∫_{curved surface} dA= total area of the curved surface = 2πrL
E.2πrL=\(=\frac{\lambda \mathrm{L}}{\varepsilon_{\mathrm{o}}}\)

6. The electric field due to the infinite charged wired depends on \(\frac{1}{\mathrm{r}}\)rather than \(\frac{1}{r^{2}}\) for a point charge.

7. Equation (5) indicates that the electric field is always along the perpendicular direction (r̂) to wire. If λ > 0 then \(\overrightarrow{\mathrm{E}}\) points perpendicular outward (r̂) from the wire and if λ < 0, then points perpendicular inward (-r̂).

Question 12.
Obtain the expression for electric field due to a charged infinite plane sheet.
Answer:

1. Consider an infinite plane sheet of charges with uniform surface charge density σ.
2. Let P be a point at a distance of r from the sheet.

Electric field due to changed infinite Planar sheet:

1. A cylindrical shaped Gaussian surface of length 2r and area A of the flat surfaces are chosen such that the infinite plane sheet passes perpendicularly through the middle part of the Gaussian surface.
2. Applying Gauss law for this cylindrical surface.
   
3. n̂ is the outward unit vector normal to the plane. The electric field depends on the surface charge density and is independent of the distance r.
4. If σ > 0, the electric field at any point P is outward perpendicular to the plane.
5. If σ < 0, the electric field points inward perpendicularly to the plane.

Question 13.
Obtain the expression for electric field due to a uniformly charged spherical shell.
Answer:
Consider a uniformly charged spherical shell of radius R and total charge Q.
Case (a) At a point outside the shell (r > R)
1. Let us choose a point P outside the shell at a distance r from the center.

2. The charge is uniformly distributed on the surface of the sphere.

3. The electric field must point radially outward if Q > 0 and point radially inward if Q < 0.

4. A spherical Gaussian surface of radius r is chosen and the total charge enclosed by this Gaussian surface is Q.

5. By Gauss law.
\(\oint_{\text {Gaussian }} \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{A}}=\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}\) ………..(1)

6. The magnitude of is also the same at all points due to the spherical symmetry of the charge distribution.
\(\mathrm{E} \oint_{\text {Gaussian }} \mathrm{d} \mathrm{A}=\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}\) ………….(2)

7. The total area of the Gaussian surface,
∫dA = 4πr²
E.4πr²=\(\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}\)
For points outside the sphere, a large spherical Gaussian surface is drawn concentric with the sphere.

8. For points outside the sphere, a spherical gaussian surface smaller than the sphere is drawn

Case (c):
The electric field due to a charged spherical cell

• The electric field is radially outward if Q > 0 and radially inward if Q < 0.

Case (b):
At a point on the surface of the spherical shell (r=R)

• The electrical field at points on the spherical shell (r=R) is given by
  
  Electric fields versus distance Gauss spherical shell of radius R

Case (c):
At point inside the spherical shell (r < R)

1. Consider a point P inside the shell at a distance r from the center.
2. A Gaussian sphere of radius r is constructed.
3. Applying Gauss law,
   \(\begin{aligned}
   &\oint_{\text {Gaussian }} \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{A}}=\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}\\
   &\mathrm{E} .4 \pi \mathrm{r}^{2}=\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}
   \end{aligned}\) …………..(5)
4. Since Gaussian surface encloses no charge, Q = 0.
   E = 0 ……………(6)
   Electric field versus distance for a spherical shell of radius R
5. The electric field due to the uniformly charged spherical shell is zero at all points inside the shell.
6. A graph can be plotted between the electric field and radial distance.

Question 14.
Discuss the various properties of conductors in electrostatic equilibrium,
i) The electric field is zero everywhere, inside the conductor. This is true regardless of whether the conductor is solid or hollow.
Answer:

1. When we apply a uniform electric field, the free electrons accelerated.
2. It causes the left side to be negatively charged and the right side to be positively charged.
3. Due to this realignment of free electrons, there will be an internal electric field created inside the conductor.
4. It increases until it nullifies the external electric field and is said to be in electrostatic equilibrium in the order of
   10^-16 s.
   

ii) There is no net charge inside the conductors. The charges must reside only on the surface of the conductors.

1. Consider an arbitrarily shaped conductor and a Gaussian surface is drawn inside the conductor.
2. It is very close to the surface of the conductor.
3. Since the electric field is zero inside the conductor, the net electric flux is zero.
4. So, there is no net charge inside the conductor.
5. No net charge inside the conductor

iii) The electric field outside the conductor is perpendicular to the surface of the conductor and has a magnitude of σ/ε. Where σ is the surface charge density at that point.

1. If the electric field has components parallel to the surface of the conductor, then free electrons of the surface of the conductor ‘would experience acceleration.
   
   (a) Electric field is along the surface
   (b) Electric field is perpendicular to the surface of the conductor
2. At electrostatic equilibrium, the electric field must be perpendicular to the surface of the conductor.
3. Consider a small cylindrical gaussian surface. 0ne half of this cylinder is embedded inside the conductor.
   
4. Since the electric field is normal to the surface of the conductor, the curved part of the cylinder has zero electric flux.
5. Inside the conductor, the electric field is zero.
6. The bottom flat part of the Gaussian surface has no electric flux.
7. The top flat part of the Gaussian surface contributes to the electric flux.
8. The electric field is parallel to the area vector and the total charge inside the surface is σA.
9. Applying Gauss law, EA = \(\frac{\sigma \mathrm{A}}{\varepsilon_{\mathrm{o}}}\)
10. In Vector form, \(\overrightarrow{\mathrm{E}}\)=\(\frac{\sigma}{\varepsilon_{\mathrm{o}}} \hat{\mathrm{n}}\)

iv) The electrostatic potential has the same value on the surface and inside of the conductor.

1. This is possible only if the electrostatic potential is constant at all points on the surface and there is no potential difference between any two points on the surface.
2. Since the electric field is zero inside the conductor, the potential is the same as the surface of the conductor.
3. At electrostatic equilibrium, the conductor is always at equipotential.

Question 15.
Explain the process of electrostatic induction.
Answer:
Definition: Charging without actual contact is called electrostatic induction.

Consider an uncharged (neutral) conducting sphere at rest on an insulating stand. A negatively charged rod is brought near the conductor without touching. (Fig. a)

1. The negative charge of the rod repels the electrons in the conductor to the opposite side. As a result, positive charges are induced near the region of the charged rod while negative charges on the farther side.
2. Before introducing the charged rod, the free electrons were distributed uniformly on the surface of the conductor and the net charge is zero. 0nce the charged rod is brought near the conductor, the distribution is no longer uniform with more electrons located on the farther side of the rod and positive charges are located closer to the rod. But the total charge is zero.

Now the conducting sphere is connected to the ground through a conducting wire, This is called grounding. Since the ground can always receive any amount of electrons, grounding removes the electron from the conducting sphere.
(Fig. b)

3. When the grounding wire is removed from the conductor, the positive charges remain near the charged rod.
(Fig. c)

4. Now the charged rod is taken away from the conductor. As soon as the charged rod is removed, the positive charge gets distributed uniformly on the surface of the conductor. By this process, the neutral conducting sphere becomes positively charged. (Fig. d)

Question 16.
Explain the dielectrics in detail and how an electric field is induced inside a dielectric.
Answer:

1. When an external electric field is applied to a conductor, the charges are aligned in such a way that an internal electric field is created which cancels the external electric field.
2. But in the case of a dielectric, which has no free electrons, the external electric field only realigns the charges so that an internal electric field is produced.
3. The magnitude of the internal electric field is smaller than that of the external electric field.
4. Therefore the net electric field inside the dielectric is not zero but is parallel to an external electric field with a magnitude less than that of the external electric field.
5. For example, let us consider a rectangular dielectric slab placed between two oppositely charged plates (capacitor).
   a)

1. The uniform electric field between the plates acts as an external electric field ext which polarizes the dielectric placed between plates.
2. The positive charges are induced on one side surface and negative charges are induced on the other side of the surface.
3. But inside the dielectric, the net charge is zero even in a small volume.
4. So the dielectric in the external field is equivalent to two oppositely charged sheets with the surface charge densities
   +σ_b and -σ_b.
5. These charges are called bound charges. They are not free to move like free electrons in conductors.

Question 17.
0btain the expression for capacitance for a parallel plate capacitor.
Answer:

1. Consider a capacitor with two parallel plates each of cross-sectional area A and separated by a distance d.
2. The electric field between two infinite parallel plates is uniform and is given by
   \(\mathrm{E}=\frac{\sigma}{\varepsilon_{\mathrm{o}}}\)
   \(\sigma=\frac{Q}{A}\)
   σ = surface charge density on the plates
   
   
3. Capacitance is directly proportional to the area of cross-section and is inversely proportional to the distance between the plates.
4. If the area of cross-section of the capacitor plates is increased, more charges can be distributed for the same potential difference. As a result, the capacitance is increased.
5. If the distance between the two plates is reduced, the potential difference between the plate decreases with the E constant.
6. The voltage difference between the terminals of the battery increases, leads to an additional flow of charge to the plates from the battery, till the voltage on the capacitor equals the battery’s terminal voltage.
7. If the distance is increased, the capacitor voltage increases and becomes greater than the battery voltage.
8. The charges flow from capacitor plates to the battery till both voltages become equal.

Question 18.
0btain the expression for energy stored in the parallel plate capacitor.
Answer:

1. Capacitor not only stores the charge but also stores energy.
2. When a battery is connected to the capacitor, electrons of total charge -Q are transferred from one plate to the other plate.
3. To transfer the charge, work is done by the battery. This work done is stored as electrostatic potential energy in the capacitor.
4. To transfer an infinitesimal charge dQ for a potential difference V, the work done is given by
   dW = V dQ ………….(1)
   
5. This stored energy is thus directly proportional to the capacitance of the capacitor and the square of the voltage between the plates of the capacitor.
   \(\mathrm{U}_{\mathrm{E}}=\frac{1}{2}\left(\frac{\varepsilon_{\mathrm{o}} \mathrm{A}}{\mathrm{d}}\right)(\mathrm{Ed})^{2}=\frac{1}{2} \varepsilon_{\mathrm{o}}(\mathrm{Ad}) \mathrm{E}^{2}\) …………(4) where Ad = volume of the space between the capacitor plates. The energy stored per unit volume of space is defined \(\mathrm{U}_{\mathrm{E}}=\frac{1}{2} \varepsilon_{\mathrm{o}} \mathrm{E}^{2}\) ……….(5)
6. Energy is stored in the electric field existing between the plates of the capacitor. 0nce the capacitor is allowed to discharge, the energy is retrieved.
7. The energy density depends only on the electric field and not on the size of the plates of the capacitor.

Question 19.
Explain in detail the effect of a dielectric placed in a parallel plate capacitor.
Answer:

1. Consider a capacitor with two parallel plates each of cross-sectional area A and are separated by distance d.
2. The capacitor is charged by a battery of voltage V₀ and the charge stored is Q₀
3. The capacitance of the capacitor without the dielectric is
   \(\mathrm{C}_{\mathrm{0}}=\frac{\mathrm{Q}_{\mathrm{0}}}{\mathrm{V}}\)

1. The modified electric field is,
   \(\mathrm{E}=\frac{\mathrm{E}_{\mathrm{o}}}{\varepsilon_{\mathrm{r}}} \ldots \ldots \ldots .(1)\)
2. Here E₀ is the electric field inside the capacitors when there is no dielectric and
   Er is the dielectric constant. Since εr >1, the electric E < E₀.
3. As a result, the electrostatic potential difference between the plates (V = Ed) is also reduced. But at the same time, the charge Q₀ will remain constant once the battery is disconnected.
4. Hence the new potential difference is,
   \(\mathrm{V}=\mathrm{Ed}=\frac{\mathrm{E}_{0}}{\varepsilon_{\mathrm{r}}} \mathrm{d}=\frac{\mathrm{V}_{\mathrm{o}}}{\varepsilon_{\mathrm{r}}}\) …………(2)
5. Thus new capacitance in the presence of a dielectric is,
   \(\mathrm{C}=\frac{\mathrm{Q}_{\mathrm{o}}}{\mathrm{V}}=\varepsilon_{\mathrm{r}} \frac{\mathrm{Q}_{\mathrm{o}}}{\mathrm{V}_{\mathrm{o}}}=\varepsilon_{\mathrm{r}} \mathrm{C}_{\mathrm{o}}\) ……………..(3)
   Since ε_r > 1, we have C > C₀ . Thus insertion of the dielectric constant ε_r increases the capacitance.
   \(\mathrm{C}=\frac{\varepsilon_{\mathrm{r}} \varepsilon_{\mathrm{o}} \mathrm{A}}{\mathrm{d}}=\frac{\varepsilon \mathrm{A}}{\mathrm{d}}\) ……………(4)
   where ε = ε_r ε₀ is the permititivity of the dielectric medium.
6. The energy stored in the capacitor before the insertion of a dielectric is given by
   \(\mathrm{U}_{\mathrm{o}}=\frac{1}{2} \frac{\mathrm{Q}_{0}^{2}}{\mathrm{C}_{0}}\) ………(5)
7. After the dielectric is inserted, Q₀ remains constant but the capacitance is increased.
8. The stored energy is decreased.
9. \(\mathrm{U}_{\mathrm{o}}=\frac{1}{2} \frac{\mathrm{Q}_{0}^{2}}{\mathrm{C}_{0}}=\frac{1}{2} \frac{\mathrm{Q}_{0}^{2}}{\varepsilon_{\mathrm{r}} \mathrm{C}_{0}}=\frac{\mathrm{U}_{\mathrm{o}}}{\varepsilon_{\mathrm{r}}}\) ………………..(6)
   Since ε_r > 1, we get U > U₀.

ii) When the battery remains connected to the capacitor:

1. The battery of voltage V₀ remains connected to the capacitor when the dielectric is inserted into the capacitor.
2. The potential difference V₀ across the plates remains constant.
3. The charge stored in the capacitor is increased by a factor ε_r.
   
4. The energy stored in the capacitor before the insertion of a dielectric is given by \(\mathrm{U}_{\mathrm{o}}=\frac{1}{2} \mathrm{C}_{\mathrm{o}} \mathrm{V}_{\mathrm{o}}^{2}\) ………..(10)
5. After the dielectric is inserted, the capacitance is increased and the stored energy is also increased
6. \(\mathrm{U}=\frac{1}{2} \mathrm{CV}_{\mathrm{o}}^{2}=\frac{1}{2} \varepsilon_{\mathrm{r}} \mathrm{C}_{\mathrm{o}} \mathrm{V}_{\mathrm{o}}^{2}=\varepsilon_{\mathrm{r}} \mathrm{U}_{\mathrm{o}}\) …………(11)
   Since ε_r>1 we have U > U₀.
7. Since the voltage between capacitor V, is constant, the electric field between the plates also remains constant.
8. The energy density is given by
   \(\mathrm{U}=\frac{1}{2} \varepsilon \mathrm{E}_{0}^{2}\) (12)
   where ε is the permittivity of the given dielectric material.

Question 20.
Derive the expression for resultant capacitance, when capacitors are connected in series and in parallel.
Answer:
i) Capacitor in series

1. Consider three capacitors of capacitance C₁, C_2, and C₃
2. connected in series with a battery of voltage V.
   
3. By these processes, each capacitor stores the same amount of charge Q. The capacitances of the capacitors are in general different so that the voltage across each capacitor is also different and are denoted as V₁, V₂ and V₃ respectively.
4. The total voltage across each capacitor must be equal to the voltage of the battery.
   V = V₁ + V₂ + V₃
5. Since, Q = CV, we have \(\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}_{1}}+\frac{\mathrm{Q}}{\mathrm{C}_{2}}+\frac{\mathrm{Q}}{\mathrm{C}_{3}}\)
   \(\mathrm{V}=\mathrm{Q}\left(\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}\right)\)
   \(\frac{\mathrm{Q}}{\mathrm{C}_{\mathrm{s}}}=\mathrm{Q}\left(\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}\right), \frac{1}{\mathrm{C}_{\mathrm{s}}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}\)
6. The inverse of the equivalent capacitance C_s of three capacitors connected in series is equal to the sum of the inverses of each capacitance.
7. The equivalent capacitance C_s is always less than the smallest individual capacitance in the series.

ii) Capacitance in parallel:

1. Consider three capacitors of capacitance C₁, C_2, and C₃ connected in parallel with a battery of voltage V.
2. The potential difference across each capacitor is the same.
   Total charge Q = Q₁ + Q₂ + Q₃
   Q₁, Q2, Q3 – Charge stored in C₁, C₂, C₃
   Now, since Q = CV, we have Q = C₁V + C₂V + C₃V
   
3. If these three capacitors are considered to form a single capacitance Cp which stores the total charge Q.
   Q = C_pV.
   C_pV= C₁V + C₂V + C₃V
   C_p = C₁ + C₂ + C₃
4. The equivalent capacitance of capacitors connected in parallel is equal to the sum of the individual capacitances.
5. The equivalent capacitance C_p in a parallel connection is always greater than the largest individual capacitance.

Question 21.
Explain in detail how charges are distributed in a conductor, and the principle behind the lightning conductor.
Answer:
1. Consider two conducting spheres A and B of radii r₁ and r₂ respectively connected to each other by a thin conducting wire.

2. If a charge Q is introduced into any one of the spheres, this charge Q is redistributed into both the spheres such that the electrostatic potential is the same in both the spheres.

3. They are now uniformly charged and attain electrostatic equilibrium.

4. Let q₁ be the charge residing on the surface of sphere A and is the charge residing on the surface of sphere B such that, Q = q₁ + q₂.

5. The charges are distributed only on the surface and there is no net charge inside the conductor.

6. The electrostatic potential at the surface of the sphere A is given by
\(V_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1}}{r_{1}}\)

7. The electrostatic potential at the surface of the sphere B is given by 1 q₂
\(V_{B}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{2}}{r_{2}}\)
V_A = V_B

8. The charge density on the surface of sphere A is σ₁, and the charge density on the surface of sphere B is σ₂. This implies that q₁ = 4πr₁² σ₁ and

q₂ = 4πr₂² σ₂ ………..(3)
σ₁r₁ = σ₂r₂ ………..(4)
σr = constant ………..(5)

9. The surface charge density σ is inversely proportional to the radius of the sphere.
Corona discharge:

Consider a charged conductor of irregular shape.

10. The smaller the radius of curvature, the larger is the charge density. The end of the conductor which has larger curvature (smaller radius) has a large charge accumulation.

11. As a result, the electric field near this edge is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge.

12. This reduces the total charge of the conductor near the sharp edge. This is called action at points or corona discharge.

Lightning conductor:
Uses to protect tall buildings from lightning strikes.

Principle:
Action at points or corona discharge.

Construction:

1. This device consists of a long thick copper rod passing from the top of the building to the ground.
2. The upper end of the rod has a sharp spike or a sharp needle.
3. The lower end of the rod is connected to the copper plate which is buried deep into the ground.

Working:

1. When a negatively charged cloud is passing above the building, it induces a positive charge on the spike.
2. Since the induced charge density on a thin sharp spike is large, it results in a corona discharge.
3. This positive charge ionizes the surrounding air which in turn neutralizes the negative charge in the cloud.
4. The negative charge pushed to the spikes passes through the copper rod and is safely diverted to the earth.
5. The lightning arrester does not stop the lightning; rather it diverts the lightning to the ground safely.

Question 22.
Explain in detail the construction and working of a Van-de-Graaff generator.
Answer:
Principle:
Electrostatic induction and action at points.

Construction:

1. Hollow metallic sphere A is mounted on an insulating pillar.
2. Pulley B is mounted at the center of the sphere.
3. Pulley C is mounted near the bottom.
4. A silk belt moves over the pulleys.
5. Pulley C is driven continuously by an electric motor.
6. D and E are the comb-shaped conductors mounted near the pulleys.
   

Working:

1. 1. 10⁴ V is given to comb D.
   2. Near the comb, D air gets ionized due to the action of points.
   3. Negative charges move towards the needle and
   4. Positive charges stick to the belt, moves up, and reach the comb E. Due to the electrostatic induction.
   5. Comb E gets a negative charge and the sphere gets a positive charge.
   6. Comb E ionize the air.
   7. Hence descending belt will be. left uncharged.
   8. The machine transfers the positive charge to the sphere.
   9. Leakage of charges can be reduced by enclosing it in a gas-filled steel chamber at very high pressure.
   10. It produces a large potential 10⁷v.

Uses:
To accelerate positive ions (protons, deuteron) for the purpose of nuclear disintegration.

IV.Exercises:

Question 1.
When two objects are rubbed with each other, approximately a charge of 50 nC can be produced in each object. Calculate the number of electrons that must be transferred to produce this charge.
Answer:

Question 2.
The total number of electrons in the human body is typically in the order of 10²⁸. Suppose, due to some reason, you and your friend lost 1% of this number of electrons. Calculate the electrostatic force between you and your friend separated at a distance of lm. Compare this with your weight. Assume the mass of each person is 60kg and use point charge approximation.
Answer:

Question 3.
Five identical charges Q are placed equidistant on a semicircle as shown in the figure. Another point charge q is kept at the center of the circle of radius R. Calculate the electrostatic force experienced by the charge q.

Answer:


\(\mathrm{F}_{\text {resultent }}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{Qq}}{\mathrm{R}^{2}}(1+\sqrt{2}) \hat{\mathrm{i}}\)

Question 4.
Suppose a charge +q on Earth’s surface and another +q charge is placed on the surface of the Moon.
(a) Calculate the value of q required to balance the gravitational attraction between Earth and the Moon
(b) Suppose the distance between the Moon and Earth is halved, would the charge q change?
(Take m_E = 5.9 × 10²⁴ kg, M_m = 7.9 × 10²² (kg)
Answer:

Question 5.
Draw the free body diagram for the following charges as shown in figure (a), (b) and (c).
a)
Answer:

b)
Answer:

c)
Answer:

Question 6.
Consider an electron travelling with a speed y0 and entering into a uniform electric field which is perpendicular to as shown in the Figure. Ignoring gravity, obtain the electron’s acceleration, velocity, and position as functions of time.

Answer:

Question 7.
A closed triangular box is kept in an electric field of magnitude E = 2 × 10^-3 NC^-1 as shown in the figure.

Calculate the electric flux through the
(a) vertical rectangular surface
Answer:

(b) slanted surface
Answer:

c) entire surface.
Answer:

Question 8.
The electrostatic potential is given as a function of x in figure (a) and (b). Calculate the corresponding electric fields in regions A, B. and D. Plot the electric field as a function of x for figure (b)

Answer:
Figure a:
\(\overrightarrow{\mathrm{E}}=-\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}} \hat{\mathrm{i}}\)
From 0 to 0.2 m,

Figure b:

Question 9.
A spark plug in a bike or a car is used to ignite the air-fuel mixture in the engine. It consists of two electrodes separated by a gap of around 0.6 mm gap as shown in the figure.

To create the spark, an electric field of magnitude 3 × 10⁶ Vm^-1 is required.
Question (a)
What potential difference must be applied to produce the spark?
Answer:
V = E × d
= 3 × 10⁶ × 0.6 × 10^-3
= 1800 V

Question (b)
If the gap is increased, does the potential difference increase, decrease, or remains the same?
Answer:
If the distance between the plate increased, then its capacitance will decrease which gives rise to an increase in potential

Question (c)
Find the potential difference if the gap is 1 mm.
Answer:
V = E × d
= 3 × 10⁶ × 1 × 10^-3
= 3000 V

Question 10.
A point charge of +10 μC is placed at a distance of 20 cm from another identical point charge of +10 μC. A point charge of -2 μC is moved from point a to b as shown in the figure. Calculate the change in potential energy of the system? Interpret your result.

Answer:

Question 11.
Calculate the resultant capacitances for each of the following combinations of capacitors.

Question a.

Answer:

Question b.

Answer:

Question c.

Answer:

Question d.

Answer:

Question e.

Answer:

Question 12.
An electron and a proton are allowed to fall through the separation between the plates of a parallel plate capacitor of voltage 5 V and separation distance h = 1 mm as shown in the figure. (Take m_p =1.6 × 10^-27 kg, m_e =9.1 × 10^-31 kg and g= 10ms^-2)

Question (a)
Calculate the time of flight for both electron and proton
Answer:
E = \(\begin{array}{l}
\mathrm{V} \\
\hline \mathrm{d}
\end{array}[latex]
= [latex]\frac{5}{10^{-3}}\)
= 5 × 10³ Vm^-1

Question (b)
Suppose if a neutron is allowed to fall, what is the time of flight?
Answer:

Question (c)
Among the three, which one will reach the bottom first?
Answer:
Hence electron will reach the bottom first.

Question 13.
During a thunderstorm, the movement of water molecules within the clouds creates friction, partially causing the bottom part of the clouds to become negatively charged. This implies that the bottom of the cloud and the ground act as a parallel plate capacitor. If the electric field between the cloud and ground exceeds the dielectric breakdown of the air (3 × 10⁶ Vm^-1), lightning will occur.

Question (a)
If the bottom part of the cloud is 1000m above the ground, determine the electric potential difference that exists between the cloud and ground.
Answer:
\(\mathrm{E}=\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}}\)
V = E.x
= 3 ×10⁶ ×10³
= 3 × 10⁹ V

Question (b)
In a typical lightning phenomenon, around 25 C of electrons are transferred from cloud to ground. How much electrostatic potential energy is transferred to the ground?
Answer:

Question 14.
For the given capacitor configuration
a) Find the charges on each capacitor
b) potential difference across them
c) energy stored in each capacitor.

Answer:

Question 15.
Capacitors P and Q have identical cross-sectional areas A and separation d. The space between the capacitors is filled with a dielectric of dielectric constant Er as shown in the figure. Calculate the capacitance of capacitors P and Q.

Answer:
1. The arrangement can be supposed to be a parallel combination of two capacitors each with plate area A/2 and separation. d. Total capacitance C_p = C_air + C_dielectric

2. The arrangement can be supposed to be a series combination of two capacitors, each with plate area A and separation d/2.

Part – II:

12th Physics Guide Electrostatics Additional Important Questions and Answers

I. Matching Type Questions:

Question 1.
Match Column I and Column II.

Answer:
(A) → (2)
(B) → (1)
(C) → (4)
(D) → (3)

Question 2.
Match Column I and Column II.

Answer:
(A) → (2)
(B) → (3)
(C) → (1)
(D) → (4)

Question 3.
Match the entries of Column I and Column II.

Answer:
(A) → (2)
(B) → (4)
(C) → (3)
(D) → (1)

Question 4.
When a dielectric slab is inserted between the plates of one of the two identical capacitors shown in the figure then match the following:

Answer:
(A) → (1)
(B) → (1)
(C) → (2)
(D) → (2)

II. Assertion – Reason Type Questions:

a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion.
c) Assertion is correct, reason is incorrect
d) Assertion is incorrect, reason is correct.

Question 1.
Assertion:
Four-point charges q_1, q₂, q_3, and q₄ are as shown in figure. The flux over the shown Gaussian surface depends only on charges q₁ and q₂.
Reason:
Electric field at all points on Gaussian surface depends only on charges q₁ and q₂,

Answer:
(d) Assertion is incorrect, reason is correct.

Question 2.
Assertion:
On disturbing an electric dipole in stable equilibrium in an electric field, it returns back to its stable equilibrium orientation.
Reason :
A restoring torque acts on the dipole on being disturbed from its stable equilibrium.
Answer:
(a) Assertion is correct, reason is correct; reason is a correct explanation for the assertion.

Question 3.
Assertion:
Work done in moving a charge between any two points in an electric field is independent of the path followed by the charge, between these points.
Reason :
Electrostatic force is a non-conservative force.
Answer:
(c) Assertion is correct, reason is incorrect

Question 4.
Assertion:
Dielectric polarisation means the formation of positive and negative charges inside the dielectric.
Reason:
Free electrons are formed in this process.
Answer:
(c) Assertion is correct, reason is incorrect

III. Statement Type Questions:

Question 1.
Which of the following about potential difference between any two points is true?
I. It depends only on the initial and final position.
II. It is the work done per unit positive charge in moving from one point to other.
III. It is more for a positive charge of two units as compared to a positive charge of one
Answer:
I and II

Question 2.
Select the correct statements from the following.
I. The electric field due to a charge outside the Gaussian surface contributes zero net flux through the surface.
II. Total flux linked with a closed body, not enclosing any charge will be zero.
III. Total electric flux, if a dipole is enclosed by a surface is zero.
Answer:
I, II, and III

Question 3.
An electric dipole of moment \(\overrightarrow{\mathrm{P}}\) is placed in a uniform electric field \(\overrightarrow{\mathrm{E}}\). Then
I. the torque on the dipole is \(\overrightarrow{\mathrm{P}}\) x \(\overrightarrow{\mathrm{E}}\).
II. the potential energy of the system is \(\overrightarrow{\mathrm{P}}\) x \(\overrightarrow{\mathrm{E}}\).
III. the resultant force on the dipole is zero.
Which of the above statements is/are correct.
Answer:
I and III

Question 4.
Select the incorrect statements from the following.
I. Polar molecules have permanent electric dipole moment.
II. CO₂ molecule is a polar molecule
III. H₂O is a non-polar molecules.
Answer:
II and III

IV. Choose the incorrect statement:

Question 1.
Which of the following statements is incorrect?
I. The charge q on a body is always given by q=ne, where n is any integer positive or negative.
II. By convention, the charge on an electron is taken to be negative charge.
III. The fact that electric charge is always an integral multiple of e is termed as quantisation of charge.
IV. The quantisation of charge was experimently demonstrated by Newton in 1912.
(a) only I
(b) only II
(c) only lV
(d) only III
Answer:
(c) only IV

Question 2.
The energy stored in a parallel plate capacitor is given by \(\mathrm{V}_{\mathrm{E}}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}\). Now which of the following statements is not true?
I. The work done in charging a capacitor is stored in the form of electrostatic potential energy \(\mathrm{V}_{\mathrm{E}}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}\)
II. The net charge on the capacitor is Q.
III. The magnitude of the net charge on one plate of a capacitor is Q.
(a) I only
(b) II only
(c) I and II
d) I, II and III
Answer:
(b) II only

Question 3.
Which of the following is not true ?
I. For a point charge, electrostatic potential varies as 1/r .
II. For a dipole, the potential depends on the magnitude of position vector, and dipole moment vector.
III. The electric potential varies as is at large distance.
IV. For a point charge, the electrostatic field varies as \( \frac{1}{\mathrm{r}^{2}}\).
(a) I only
(b) II only
(c) III only
(d) I, II and III
Answer:
(c) III only

V. Diagram – Type Question:

Question 1.
The figure shows a charge +q at point P held in equilibrium in air with the help of four + q charges situated at the vertices of a square. The net electrostatic force on q is given by

(a) Gauss’s law
(b) Coulomb’s law
(c) Principle of superposition
(d) net electric flux out the position of +q.
Answer:
(c) Principle of superposition
Solution:
The weight mg of the charge hole in air is in equilibrium with the net electrostatic force exerted by the four charges situated at the corners. The net electrostatic force is given by the charges at the corners. This is the principle of superposition.

Question 2.
Which of the following graphs shows the correct variation of force when the distance r between two charges varies?


Answer:

Solution:
From Coulombs’ law \(\mathrm{F}=\frac{\mathrm{Kq}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}\) i.e., \(\mathrm{F} \propto \frac{1}{\mathrm{r}^{2}}\) which is correctly shown by graph (d).

Question 3.
Figure shows some of the electric field lines corresponding to an electric field. The figure suggests that

(a) E_A > E_B > E_C
(b)E_A = E_B = E_C
(c) E_A > E_C > E_B
(d) E_A > E_C > E_B
Answer:
(c) ) E_A > E_C > E_B

VI. Choose the odd man out:

Question 1.
(a) lightning arrestor
(b) Van-de-Graf generator
(c) Photocopying
(d) AC generator
Answer:
(d) AC generator

Question 2.
(a) Gold
(b) Silver
(c) electric potential
(c) Aluminium
(d) ebonite
Answer:
(d) ebonite

Question 3.
(a) Farad
(b) \(\frac{J}{C} m^{-1}\)
(c) Vm^-1
(d) NC^-1
Answer:
(a) Farad

Question 4.
(a) electric field
(b) electric force
(c) electric potential
(d) electric dipole moment
Answer:
(c) electric potential

VII. Choose The Incorrect Pair:
Question 1.

Answer:
C) Electric flux – Vector product

Question 2.

Answer:
A) Coulomb’s law – Force is directly proportional to square of distance

VIII. Choose the correct pair:

Question 1.

Answer:
D) Electric potential due to a uniform charged solid non-conducting sphere – Varies inversely of radius

Question 2.

Answer:
A) Inside a conductor placed in an external electric field – Electric field = 0

IX. Fill In the blanks:

Question 1.
The charge acquired by 5 ×  10¹⁰ electrons _______.
Answer:
q = ne
q = 5 × 10¹⁰ × l.6 × 10^-19 = 8 × 10^-9 C

Question 2.
The unit of permittivity is…………………
Answer:
C²N^-1m^-2

Question 3.
At sharp points, the charge density is ________.
Answer:
maximum

Question 4.
The charges in an electrostatic field are analogous to ………………… in a gravitational field.
Answer:
mass

X. Choose the best answer:

Question 1.
If electric field in a region is radially outward with magnitude E = Ar, the charge contained in a sphere of radius r centred at the origin is
(a) \(\frac{1}{4 \pi \varepsilon_{o}} A r^{3}\)
(b) \(4 \pi \varepsilon_{\mathrm{o}} \mathrm{Ar}^{3}\)
(c) \(\frac{1}{4 \pi \varepsilon_{0}} \frac{A}{r^{3}}\)
(d) \(\frac{4 \pi \varepsilon_{0}}{\mathrm{r}^{3}}\)
Answer:
b) \(4 \pi \varepsilon_{\mathrm{o}} \mathrm{Ar}^{3}\)
Solution:

Question 2.
The electric field intensity just sufficient to balance the earth’s gravitational attraction on an electron will be: (given
(a) -5.6 × 10^-11N/C
(b) -4.8 × 10^-15N/C
(c) -1.6 × 10^-19N/C
(d) -3.2 × 10^-19N/C
Answer:
(a) -5.6 × 10^-11N/C
Solution:

Question 3.
The insulation property of air breaks down when the electric field is 3 × 10⁶ vm^-1 The maximum charge that can be given to a sphere of diameter 5m is approximately
(a) 2 × 10^-2 C
(b) 2 × 10^-3 C
(c) 2 × 10^-4 C
(d) 2 × 10^-5 C
Answer:
(b) 2 × 10^-3 C
Solution:

Question 4.
ABC is an equilateral triangle. Charges +q are placed at each corner as shown in fig. The electric intensity at centre 0 will be

(a) \(\frac{1}{4 \pi \epsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}}\)
(b) \(\frac{1}{4 \pi \epsilon_{0}} \frac{q}{r^{2}}\)
(c) \(\frac{1}{4 \pi \epsilon_{o}} \frac{3 q}{r^{2}}\)
(d) zero
Answer:
(d) zero
Solution:
(Unit positive charge at 0 will be repelled equally by three charges at the three corners of the triangle. By symmetry, the resultant at 0 would be zero.)

Question 5.
A hollow insulated conduction sphere is given a positive charge of 10μC. What will be the electric field at the centre of the sphere if its radius is 2m?
(a) Zero
(b) 5μCm^-2
(c) 20μCm^-2
(d) 8μCm^-2
Answer:
(a) zero
Solution:
(Charge resides on the outer surface of a conducting hollow sphere of radius R. We consider a spherical surface of radius r< R.)

Question 6.
Five balls marked 1, 2, 3, 4, and 5 are suspended by separate threads. The pairs (1, 2) (2, 4) and (4, 1) show mutual attraction and the pairs (2, 3) and (4, 5) show repulsion. The nature of ball marked as 1 is
a) positive
b) negative
c) neutral
d) can’t determine
Answer:
c) neutral

Question 7.
The resultant capacitance of four plates, each is having an area A, arranged, as shown above, will be (plate separation is d)

(a) \(\frac{A \varepsilon_{0}}{d}\)
(b) \(\frac{\mathrm{A} \varepsilon_{\mathrm{o}}}{2 \mathrm{~d}}\)
(c) \(\frac{2 \mathrm{~A} \varepsilon_{\mathrm{o}}}{\mathrm{d}}\)
(d) \(\frac{3 A \varepsilon_{o}}{d}\)
Answer:
(c) \(\frac{2 \mathrm{~A} \varepsilon_{\mathrm{o}}}{\mathrm{d}}\)

Question 8.
At infinity, the electrostatic potential is
(a) Infinity
(b) maximum
(c) minimum
(d) zero
Answer:
(d) zero

Question 9.
The electric field at a point on the equatorial line of a dipole and direction of the dipole moment

(a) will be parallel
(b) will be in the opposite direction
(c) will be perpendicular
(d) are not related
Answer:
(b) will be in the opposite direction
Solution:
(The direction of the electric field at equatorial point A or B will be in opposite direction, like that of direction of dipole moment.)

Question 10.
Debye is the unit of
(a) electric flux
(b) electric dipole moment
(c) electric potential
(d) electric field intensity
Answer:
(b) electric dipole moment

Question 11.
In the given diagram a point charge +q is placed at the origin 0. Work done in taking another point charge -Q from point A to point B is:

(a) \(\frac{q Q}{4 \pi \epsilon_{0} a^{2}}\left(\frac{a}{\sqrt{2}}\right)\)
(b) zero
(c) \(\left[\frac{-\mathrm{q} \mathrm{Q}}{4 \pi \epsilon_{0}} \frac{1}{\mathrm{a}^{2}}\right] \sqrt{2} \mathrm{a}\)
(d) \(\left[\frac{\mathrm{qQ}}{4 \pi \epsilon_{0}} \frac{1}{\mathrm{a}^{2}}\right] \sqrt{2} \mathrm{a}\)
Answer:
(b) zero

Question 12.
The total electric flux emanating from a closed surface enclosing an α -particle is (e-electronic charge)
(a) \(\frac{2 e}{\varepsilon_{0}}\)
(b) \(\frac{e}{\varepsilon_{0}}\)
(c) \(\mathrm{e} \varepsilon_{\mathrm{o}}\)
(d) \(\frac{\varepsilon_{0} \mathrm{e}}{4}\)
Answer:
(a) \(\frac{2 e}{\varepsilon_{0}}\)
Solution:
(According to Gauss’s law total electric flux \(\frac{1}{\varepsilon_{0}}\) through a closed surface is time the total charge inside that surface.
Electric flux, Φ_E = \(\frac{\mathrm{q}}{\varepsilon_{\mathrm{o}}}\)
Charge on a-particle = 2e)
Φ_E = \(\frac{2 e}{\varepsilon_{o}}\)

Question 13.
A coil of the area of cross-section 0.5 m² with 10 turns is in a plane that is parallel to a uniform electric field of 100 N/C. The flux through the plane is?
(a) 100 V.m
(b) 500 V.m
(c) 20 V.m
(d) zero
Answer:
(b) 500 V.m

Question 14.
0n moving a charge of Q coulomb by χ cm, WJ of work is done, then the potential difference between the point is
(a) \(\frac{\mathrm{W}}{\mathrm{Q}} \mathrm{v}\)
(b) QWV
(c) \(\frac{Q}{W} V\)
(d) \(\frac{Q^{2}}{W} V\)
Answer:
(a) \(\frac{\mathrm{W}}{\mathrm{Q}} \mathrm{v}\)
Solution:
Potential difference between two points in an electric field is.
V_A – V_B = \(\frac{\mathrm{W}}{\mathrm{q}_{\mathrm{o}}}\)

Question 15.
The positive terminal of 12 V battery is connected to the ground. Then the negative terminal will be
(a) – 6 V
(b) +12 V
(c) zero
(d)- 12V
Answer:
(d) – 12V
Solution:
When negative terminal is grounded, the positive terminal of battery is at +12 V. When positive terminal is grounded, the negative terminal will be at -12 V.

Question 16.
The maximum electric field that a dielectric medium can withstand without break-down is called is
(a) permittivity
(b) dielectric constant
(c) electric susceptibility
(d) dielectric strength
Answer:
(d) dielectric strength

Question 17.
The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E is
(a) \(\varepsilon_{0} \mathrm{E}^{2} \mathrm{Ad}\)
(b) \(\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2} \mathrm{Ad}\)
(c) \(\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2} / \mathrm{Ad}\)
(d) \(\varepsilon_{0} \mathrm{E}^{2} / \mathrm{Ad}\)
Answer:
(a) \(\varepsilon_{0} \mathrm{E}^{2} \mathrm{Ad}\)
Solution:
(Energy required to charge the capacitor is W = u = QV)
∵ E = V/d

Question 18.
Energy is stored in a capacitor in the form of
(a) electrostatic energy
(b) magnetic energy
(c) light energy
(d) heat energy
Answer:
(a) electrostatic energy

Question 19.
Dimension and unit of Electric flux is
(a) ML²T^-3 A^-2, Nm² C^-1
(b) ML³ T^-3 A^-1, Nm² C^-1
(c) ML² T^-1 A^-2, Nm² C^-1
(d) ML^-4 T^-3 A^-2, Nm² C^-1
Answer:
b) ML³ T^-3 A^-1, Nm² C^-1

Question 20.
An air-core capacitor is charged by a battery. After disconnecting it from the battery, a dielectric slab is fully inserted in between its plates. Now, which of the following quantities remains constant?
(a) Energy
(b) voltage
(c) Electric field
(d) Charge
Answer:
(d) Charge

Question 21.
In a charged capacitor, the energy is stored in
(a) the negative charges
(b) the positive charges
(c) the field between the plates
(d) both (a) and (b)
Answer:
(c) the field between the plates

Question 22.
The potential gradient at which the dielectric of a condenser just gets punctured is called
(a) dielectric constant
(b) dielectric strength
(c) dielectric resistance
(d) dielectric number
Answer:
(b) dielectric strength

Question 23.
The unit of permittivity is
(a) C²N^-1m^-2
(b) Nm²C^-2
(c) Hm^-1
(d) NC^-2m^-2
Answer:
(a) C²N^-1

Question 24.
In the case of a Van de Graaff generator, the breakdown field of air is
(a) 2 × 10⁸ Vm^-1
(b) 3 × 10⁶ Vm^-1
(c) 2 × 10⁸ Vm^-1
(d) 2 × 10⁴ Vm^-1
Answer:
(b) 3 × 10⁶ Vm^-1

Question 25.
Van de Graaff generator is used to
(a) store electrical energy
(b) build up the high voltage of a few million volts
(c) decelerate charged particle-like electrons
(d) both (a) and (b)
Answer:
(b) build up the high voltage of a few million volts

XI. Two mark questions:

Question 1.
What is meant by triboelectric charging?
Answer:
Charging the objects through rubbing is called triboelectric charging.

Question 2.
When does an object is said to be electrically neutral?
Answer:
If the net charge is zero in an object, It is said to be electrically neutral.

Question 3.
State Gauss’s Law?
Answer:
Definition:
Gauss’s law states that if a charge Q is enclosed by an arbitrary closed surface, then the total electric flux OE through the closed surface is Φ_E = \(\oint { \vec { E } } \) .d\(\vec { A } \) = \(\frac {{ q }_{encl}}{{ ε }_{0}}\)

Question 4.
State coulomb’s law.
Answer:
Coulomb’s law states that the force of attraction or repulsion between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Question 5.
What is meant by dielectric?
Answer:
A dielectric is a non-conducting material and has no free electrons. The electrons in a dielectric are bound within the atoms. Ebonite, glass, and mica are some examples of dielectrics.

Question 6.
What is meant by the superposition principle?
Answer:
The total force on a given charge is the vector sum of the forces exerted on it due to all other charges. The force on q, due to charges q < ₂, q < ₃ ………… q_n

Question 7.
What are polar molecules? Give examples.
Answer:
In polar molecules, the centers of the positive and negative charges are separated even in the absence of an external electric field. They have a permanent dipole moment.
The net dipole moment is zero in the absence of an external electric field. Examples of polar molecules are H₂O, N₂O, HCl, NH₃.

Question 8.
Explain the working of a microwave oven?
Answer:

1. It works on the principle of torque acting on an electric dipole.
2. The food we consume has water molecules which are permanent electric dipoles.
3. Oven produces microwaves that are oscillating electromagnetic fields and produce a torque on the water molecules.
   
4. Due to this torque on each water molecule, the molecules rotate very fast and produce thermal energy.
5. Thus, the heat generated is used to cook the food.

Question 9.
What are permittivity and relative permittivity? How are they related?
Answer:

• Permittivity of a medium = permittivity of free space × relative permittivity, ε = ε₀ ε_r
  Its unit is C² N^-1 m^-2
• Relative permitivity = \(\frac{\text { permitivity of a medium }}{\text { permitivity of vacuum }}=\varepsilon_{\mathrm{r}}=\frac{\varepsilon}{\varepsilon_{\mathrm{o}}}\)
• It has no unit.

Question 8.
What is a capacitor?
Answer:
The capacitor is a device used to store electric charge and electrical energy. Capacitors are widely used in many electronic circuits and have applications in many areas of science and technology.

Question 11.
State Gauss law?
Answer:
Gauss law states that the total flux of the electric field E over any closed surface is equal to l/£0 times of the net charge enclosed by the surface,
\(\phi_{\mathrm{E}}=\frac{\mathrm{q}}{\varepsilon_{\mathrm{o}}}\)

Question 12.
What is meant by electrostatic shielding?
Answer:

• It is the process of isolating a certain region of space from external field.
• It is based on the fact that electric field inside a conductor is zero.

Question 13.
What is meant by electrostatic Induction?
Answer:

• The phenomenon of producing induced charges without any contact with another charge is known as electrostatic induction.
• Electrostatic induction is used in electrostatic machines like Van de Graaff generators and capacitors.

Question 14.
Define farad.
Answer:
A conductor has a capacitance of one farad; if a charge of 1 coulomb given to it, rises its potential by 1 volt.

Question 15.
What are dielectrics? Give examples.
Answer:
A dielectric is an insulating material in which all the electrons are tightly bound to the nucleus of the atom. There are no free electrons to carry current.
E.g:- Ebonite, mica, and oil.

Question 16.
Can two equipotential surfaces intersect? Give reason.
Answer:
Since the electric field is normal to the equipotential surface and also the potential difference between any two points on the surface is nullified, the intersection is not possible.

Question 17.
Distinguish between polar and non-polar molecule?
Answer:

Question 18.
What is meant by dielectric breakdown?
Answer:
When the external electric field applied to a dielectric is very large, it tears the atoms apart so that the bound charges become free charges. Then the dielectric starts to conduct electricity. This is called dielectric breakdown.

Question 19.
The electric field outside a conductor is perpendicular to its surface. Justify.
Answer:

• The electric field has components parallel to the surface of the conductor.
• The free electrons in the conductor would experience acceleration.
• The conductor is not in equilibrium. At electrostatic equilibrium, the electric field must be perpendicular to its surface.

Question 20.
What is called ‘fringing field’ is a capacitor? when does it ignore?
Answer:
For finite-sized plates, the electric field is not strictly uniform between the plates. At both edges, the electric field is bent outward. Under the condition (d²<< A), this effect can be ignored.

Question 21.
How does a capacitor is used in a Computer keyboard?
Answer:
When the key is pressed, the separation between the plates decreases leading to an increase in the capacitance. This is turn triggers the electronic circuits in the computer to identify which key is pressed.

Question 22.
What will happen to the charge, voltage, electric field, the capacitance of a dielectric placed capacitor when it is connected with a battery and then disconnected?
Answer:

Question 23.
Calculate the number of electrons in one Coulomb of negative charge.
Solution:
According to the quantisation of charge q = ne
Here q = 1C. So the number of electrons in 1 coulomb of charge is
\(\mathrm{N}=\frac{q}{e}=\frac{1 \mathrm{C}}{1.6 \mathrm{X} 10^{-19}}=6.25 \times 10^{18} \text { electrons }\)

Question 24.
Define electric potential energy of two point charges?
Answer:
It is equal to work done to assemble the charges or work done in bringing each charge or work done in bringing a charge from an infinite distance.

Question 25.
Why is it safer to be inside a car than standing under a tree during lightning?
Answer:

• The metal body of the car provides electrostatic shielding.
• The electric field inside the car is zero.
• During lightning, the electric discharge passes through the body of the car. So during lightning, it is safer to sit inside a car than on open ground or under a tree.

Question 26.
Define the electric potential energy of an electric dipole placed in an electric field.
Answer:
Electric potential energy of an electric dipole in an electrostatic field is the work done in rotating the dipole to the desired position in the field.

Question 27.
What happens if a polar molecule is placed in an electric field?
Answer:

1. When a polar molecule is placed in an electric field, the dipoles orient themselves in the direction of the electric field. Hence a net dipole moment is produced.
2. The alignment of dipole moments in the direction of the applied electric field is called polarisation of electric polarisation.
3. The magnitude of the induced dipole moment p is directly proportional to the electric field E. The dipole moment, p = αE when α is called molecular polarisability.

Question 28.
Define volt.
Answer:

1. The unit of potential difference is volt.
2. The potential difference between the two points is I volt if 1 joule of work is done in moving 1 coulomb of charge from one point to another against the electric force.

Question 29.
What does an electric dipole experience when kept in a uniform electric field and non-uniform electric field?
Answer:
Uniform electric field:  When a dipole is kept in a uniform electric field at an angle θ, the net force is zero. It experiences a torque \(\vec{\tau}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{E}} \quad \tau=p \mathrm{E} \sin \theta\)

Non-uniform electric field: If the dipole is placed in a non-uniform electric field at an angle θ in addition to torque, it also experiences a force.

Question 30.
What is the Effective Capacitance of capacitors connected in series?
Answer:
When a number of capacitors are connected in series, the reciprocal of the effective capacitance is equal to the sum of reciprocal of the capacitances of the individual capacitors.

Question 31.
What is the effective capacitance of capacitors connected in parallel?
Answer:
The effective capacitance of the capacitors connected in parallel is equal to the sum of the capacitances of the individual capacitors.
C_p = C₁ + C₂ + ……….. + C_n

Question 32.
What are conductors and insulators? Give examples.
Answer:

1. Bodies which allow the charges to pass through are called conductors.
   e.g. Metals, human body, Earth, etc.
2. Bodies that do not allow the charge to pass through are called insulators.
   (e.g) Glass, mica, ebonite, plastic, etc.

Question 33.
What is meant by point charge?
Answer:
Any charge which occupies a space with dimensions less than its distance away from an observation point considered a point charge.

Question 34.
What is a capacitor?
Answer:
A capacitor is a device used to store electric charge and electrical energy.

Question 35.
Why does a balloon after rubbing stick to a wall?
Answer:
It is due to the polarisation of the wall due to the electric field due to a balloon.

Question 36.
How does the lightning conductor prevent the lightning stroke from the damage of the building?
Answer:

1. When a negatively charged cloud passes over the building, a positive charge will be induced on the pointed conductor.
2. The positively charged sharp points will ionize the air in the vicinity.
3. This will partly neutralise the negative charge of the cloud, thereby lowering the potential of the cloud.
4. The negative charges that are attracted to the conductor travel down to the earth. Thereby preventing the lightning stroke from the damage of the building.

Question 37.
Define the physical quantity whose unit is Vm, and state whether it is scalar or Vector.
Answer:

1. Electric flux has the unit Vm.
2. The number of electric field lines passing through a given area.
3. q = EA cosθ
4. It is a scalar quantity.
5. The other unit is Nm²c^-1.

XII. Three Mark Questions:

Question 1.
Give some important inference, over the expression of the electric field due to an electric dipole on its axial line and equational line.
Answer:
Important inferences
i) The magnitude of the electric field at points on the dipole axis is twice the magnitude of the electric field at points on the equatorial plane. The direction of the electric field at points on the dipole axis is directed along the direction of dipole moment vector \(\overrightarrow{\mathrm{p}}\) but at points on the equatorial plane it is directed opposite to the dipole moment vector, that is along \(\overrightarrow{\mathrm{p}}{-}\).

ii) At very large distances, the electric field due to a dipole varies as \(\frac{1}{\mathrm{r}^{3}}\) electric field due to a dipole at very large distances goes to zero faster than the electric field due to a point charge.

iii) The distance 2a approaches zero and q approaches infinity such that the product 2aq = p is finite, then the dipole is called a point dipole.

Question 2.
Mention some important points on an expression of electric potential due to point charge.
Answer:
Important points :
i) If the source charge q is positive, V > 0. If q is negative, then V is negative and equal to
\(\mathrm{V}=-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{\mathrm{r}}\)
ii) The potential due to positive charge decreases as the distance increases, but for a negative charge the potential increases as the distance is increased. At infinity (r = 0) electrostatic potential is zero (V=0).
iii) The electric potential at a point P due to a collection of charges q₁, q₂, q₃,……..q_n is equal to sum of the electric potentials due to individual charges.

Question 3.
Mention some important points over the derivation of electric potential due to an electric dipole.
Answer:
Important points:
i) The potential due to an electric dipole falls as \(\frac{1}{r^{2}}\) and the potential due to a single point
charge falls as \(\frac{1}{\mathrm{r}}\). Thus the potential due to the dipole falls faster than that due to a monopole. As the distance increases from electric dipole, the effects of positive and negative charges nullify each other.

ii) The potential due to a point charge is spherically symmetric since it depends only on the distance r. But the potential due to a dipole is not spherically symmetric because the potential depends on the angle between \(\overrightarrow{\mathrm{p}}\) and position vector \(\overrightarrow{\mathrm{r}}\) of the point.

iii) However the dipole potential is axially symmetric. If the position vector \(\overrightarrow{\mathrm{r}}\) is rotated about \(\overrightarrow{\mathrm{p}}\) by keeping θ fixed, then all points on the cone at the same distance r will have the same potential

Question 4.
Small mercury drops of the same size are charged to the same potential V. If n such drops coalesce to form a single large drop, then calculate its potential.
Solution:
Let r be the radius of a small drop and R that of the large drop. Then, since the volume remains conserved,
\(\frac { 1 }{ 2 }\) πR² = \(\frac { 4 }{ 3 }\) πR³n
⇒ R³ = r³n
R = r³(n)^1/3
Further, since the total charge remains conserved, we have, using Q = CV
C_large V = n C_small v
Where V is the potential of the large drop.
4πε₀ RV = n (4πε₀r)v
V = \(\frac { nrv }{ R }\) = \(\frac { nrv }{{ r(n) }^{1/3}}\)
V = vn^2/3

Question 5.
Derive an expression for electric flux in a non-uniform electric field and an arbitrarily shaped area.
Answer:
1. Suppose the electric field is not uniform and the area is not flat, then the entire area is divided into n small area segments such that each area element is almost flat and the electric field through each area element is considered to be uniform.

2. The electric flux for the entire area A is approximately written as
\(\begin{array}{l}
\Delta \overrightarrow{\mathrm{A}}_{1}, \Delta \overrightarrow{\mathrm{A}}_{2}, \Delta \overrightarrow{\mathrm{A}}_{3} \ldots \ldots \Delta \overrightarrow{\mathrm{A}}_{\mathrm{n}} \\
\Phi_{\mathrm{E}}=\overrightarrow{\mathrm{E}}_{1} \cdot \Delta \overrightarrow{\mathrm{A}_{1}}+\overrightarrow{\mathrm{E}}_{2} \cdot \Delta \overrightarrow{\mathrm{A}}_{2}+\overrightarrow{\mathrm{E}}_{3} \cdot \Delta \overrightarrow{\mathrm{A}}_{3} \ldots . \overrightarrow{\mathrm{E}}_{\mathrm{n}} \cdot \Delta \overrightarrow{\mathrm{A}_{\mathrm{n}}} \\
=\sum_{\mathrm{i}=1}^{\mathrm{n}} \overrightarrow{\mathrm{E}}_{\mathrm{i}} \cdot \Delta \overrightarrow{\mathrm{A}_{\mathrm{i}}}
\end{array}\)

3. By taking the limit \(\Delta \overrightarrow{\mathrm{A}}_{i} \rightarrow 0\) (for all i) the summation in equation becomes integration. The total electric flux for the entire area is given by \(\Phi_{\mathrm{E}}=\int \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{A}}\)

Electric flux for non-uniform electric Field:

1. The electric flux for a given surface depends on both the electric field pattern on the surface area and orientation of the surface with respect to the electric field.

Question 6.
The electric field is zero everywhere inside the conductor. This is true regardless of whether the conductor is solid or hollow. Justify.
Answer:

• This is an experimental fact. Suppose the electric field is not zero inside the metal, then there will be a force on the mobile charge carriers due to this electric field.
•  As a result, there will be a net motion of the mobile charges, which contradicts the conductors being in electrostatic equilibrium. Thus the electric field is zero everywhere inside the conductor.
• We can also understand this fact by applying an external uniform electric field on the conductor.

Question 7.
The electrostatic potential has the same value on the surface and inside the conductor. Justify.
Answer:

• We know that the conductor has no parallel electric component on the surface without doing any work.
• This is possible only if the electrostatic potential is constant at all points on the surface and there is no potential difference between any two points on the surface.
• Since the electric field inside the conductor is zero inside the conductor, the potential is same as the surface of the conductor.
• Thus at electrostatic equilibrium, the conductor is always at equipotential.

Question 8.
Explain Faraday Cage’s experiment? Faraday cage is an instrument to demonstrate the effect of electrostatic shielding.
Answer:

• It is made up of metal bars configured in the instrument.
•  If an artificial lightning bolt is created outside the person inside is not affected.
• This is because the metal bar provides electrostatic shielding.
• The Electric field inside becomes zero.
• The charges flow through the metal bar to the ground with no effect on the person inside.

Question 9.
What do you understand from the expression of capacitance in a parallel plate capacitor?
Answer:
Capacitance is directly proportional to the area of cross-section and is inversely proportional to the distance between the plates.
1. If the area of cross-section of the capacitor plates is increased, more charges can be distributed for the same potential difference. As a result, the capacitance is increased.

2. If the distance d between the two plates is reduced, the potential difference between the plates (V = Ed) decreases with E constant. As a result, the voltage difference between the terminals of the battery increases which in turn leads to an additional flow of charge to the plates from the battery, till the voltage on the capacitor equals to the battery’s terminal voltage. Suppose the distance is increased, the capacitor voltage increases and becomes greater than the battery voltage. Then, the charges flow from capacitor plates to the battery till both voltages become equal.

Question 10.
Mention the applications of capacitors?
Answer:

1. The flash which comes from the digital camera when we take photographs is due to the energy released from the capacitor, called a flash capacitor.
2. During cardiac arrest, a device called heart defibrillator is used to give a sudden surge of a large amount of electrical energy to the patient’s chest to retrieve the normal heart function.
3. Capacitors are used in the ignition system of automobile engines to eliminate sparking.
4. Capacitors are used to reduce power fluctuations in power supplies and to increase the efficiency of power transmission.

XIII – Five mark questions:

Question 1.
Explain the historical background of electric charges?
Answer:

• Two millenniums ago, Greeks noticed that amber after rubbing with animal fur attracted small pieces of leaves and dust.
• The amber possessing this property is said to be charged.
• A glass rod rubbed with a silk cloth, attracts a piece of paper. So glass rod also becomes ‘charged’ when rubbed with a suitable material.
• Consider a charged rubbed rod hanging from a thread.
• Suppose another charged rubber rod is brought near the first rubber rod; the rods repel each other.
• Now if we bring a charged glass rod close to the charged rubber rod, they attract each other.
• At the same time, if a charged glass rod is brought near another charged glass rod both the rods repel each other.

Inferences:
i) The charging of the rubber rod and that of the glass rod are different from one another.
ii) The charged rubber rod repels another charged rubber rod, which implies that Tike charges repel each other’.
iii) The charged amber rod attracts the charged glass rod, implying that the charge in the glass rod is not the same kind of charge present in the rubber. Thus unlike charges attract each other.
Therefore, two kinds of charges exist in the universe. Benjamin Franklin called one type of charge as positive (+) and another type of charge as negative (-). Based on Franklin’s convention, rubber and amber rods are negatively charged while the glass rod is positively charged. If the net charge is zero in the object, it is said to be electrically neutral.

iv) The atom is electrically neutral and is made up of negatively charged electrons, positively charged protons, and neutrons which have zero charges. When an object is rubbed with another object, some amount of charge is transferred from one object to another due to the friction between them, and the object is then said to be ‘electrically charged’. Charging the object through rubbing is called ‘triboelectric charging’.

Question 2.
Derive an expression for electric field due to a system of point charges.
Answer:
Suppose a number of point charges are distributed in space, to find the electric field at some point P due to this collection of point charges, superposition principle is used.

The electric field due to a collection of point charges at an arbitrary point is simply equal to the vector sum of the electric fields created by the individual point charges. This is called the superposition of electric fields.

Question 3.
Discuss the electric flux of a uniform electric field?
Answer:
Electric flux for uniform Electric field:

1. Consider a uniform electric field in a region of space. Let us choose an area A normal to the electric field lines.
Φ_E = EA ……….(1)

2. Suppose the same area A is kept parallel to the uniform electric field, then no electric field lines pass through area A. The electric flux for this case is zero.
Φ_E > E = 0 ………..(2)

3. If the area is inclined at an angle e with the field, then the component of the electric field perpendicular to the area alone contributes to the electric flux.

4. The electric field component parallel to the surface area will not contribute to the electric flux. For this case, the electric flux  Φ_E = (E cosθ) A …………..(3)

5. Further, e is also the angle between the electric field and the direction normal to the area, Hence in general, for a uniform electric field, the electric flux is defined as
Φ_E = \(\overrightarrow{\mathrm{E}}\) . \(\overrightarrow{\mathrm{A}}\) = EA cosθ ………………(4)

6. \(\overrightarrow{\mathrm{A}}\) is the area vector \(\overrightarrow{\mathrm{A}}\) = A n̂. Its magnitude is simply the area A and the direction are along the unit vector n̂ perpendicular to the area.

7. Using this definition for flux,
Φ_E = \(\overrightarrow{\mathrm{E}}\) . \(\overrightarrow{\mathrm{A}}\)
equations (2) and (3) can be obtained as special cases.
In Figure (a), θ=0° Φ_E = \(\overrightarrow{\mathrm{E}}\) . \(\overrightarrow{\mathrm{A}}\) = EA
In Figure (b), θ = 90° Φ_E = \(\overrightarrow{\mathrm{E}}\) . \(\overrightarrow{\mathrm{A}}\) = 0
Here \(\overrightarrow{\mathrm{A}}\) = An̂

Question 4.
Discuss some salient points about Gauss law?
Answer:
1. The total electric flux through the closed surface depends on the charges enclosed by the surface and the charges present outside the surface will not contribute to the flux and the shape of the closed surface which can be chosen arbitrarily.

2. The total electric flux is independent of the location of the charges inside the closed surface.

3. Chosen imaginary surface is called a Gaussian surface. The shape of the Gaussian surface to be chosen depends on the type of charge configuration and the kind of symmetry existing in that charge configuration.

4. The electric field E is due to charges present inside and outside the Gaussian surface but the charge Qencl denotes the charges which lie only inside the Gaussian surface.

5. The Gaussian surface cannot pass through any discrete charge but it can pass through continuous charge distributions.

6. Gauss law is another form of Coulomb’s law and it is also applicable to the charges in motion. Because of this reason, Gauss law is treated as much more general law than Coulomb’s law.

Question 5.
Derive an expression for electric field due to two parallel charged infinite sheet.
Answer:

1. Consider two infinitely large charged plane sheets with equal and opposite charge densities +σ and -σ which are placed parallel to each other.
2. The electric field between the plate and outside the plates is found using Gauss law.
3. The magnitude of the electric field due to an infinite charged plane sheet is \(\frac{\sigma}{2 \varepsilon}\) and it points perpendicularly outward if σ>0 and points inward if σ<0.
   
4. At the points P₁ and P₂, the electric field due to both plates are equal in magnitude and opposite in direction. As a result, electric field at a point outside the plate is zero. But inside the plate, electric fields are in the same direction
5. The direction of the electric field inside the plates is directed from positively charged plate to the negatively charged plate and is uniform everywhere inside the plate.
   \(\mathrm{E}_{\text {inside }}=\frac{\sigma}{2 \varepsilon_{0}}+\frac{\sigma}{2 \varepsilon_{0}}=\frac{\sigma}{\varepsilon_{0}}\)

Question 6.
Explain the principle of a capacitor?
Answer:

1. A capacitor is a device used to store electric charge and electric energy.
2. It consists of two conducting objects separated by some distance.
3. A simple capacitor consists of two parallel metal plates separated by a small distance.
4. When a capacitor is connected to a battery of potential difference V, the electrons are transferred from the battery to one plate and the other plate to the battery so that one plate becomes negatively charged with a charge of – Q and the other plate becomes positively charged with + Q.
5. The potential difference between the plates is equivalent to the battery’s terminal voltage. If the battery voltage is increased, the amount of charges stored in the plates also increase.
6. In general, the charge stored in the capacitor is proportional to the potential difference between the plates.
   Q α V, So that Q = CV, where C is the proportionality constant called capacitance.
   
7. The capacitance of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between conductors.
8. Unit:- Farad or Coulomb per volt. Farad is a very large unit of capacitance.
9. In Practice, we use micro & picofarad.
10. Total charge stored in the capacitor is zero when we say the capacitor stores charges, it means that amount of charge that can be stores charge, it means that amount of charge that can be stored in any one of the plates.
11. Nowadays it is available in different shapes and types. It is used in various kinds of electronic circuits.

Question 7.
Write the properties of lines of force:
Answer:

1. It starts from positive charge and ends at negative charge .
2. For a positive charge, the electric field lines point radially outward and for a negative charge, electric field lines point radially inward.
3. Tangent drawn at any point gives the direction of the electric field at the point.
4. The electric field lines are denser in a region where the electric field has a larger magnitude and less dense in a region of smaller magnitude.
5. They do not intersect
6. Field lines emanating from the positive charge are proportional to the magnitude of the charge.

XlV. Additional Problems (Two Marks):

Question 1.
An ele1tric dipole of charges 2 × l0^-10C separated by a distance 5 mm, is placed at an angle of 60 to a uniform field of 10 Vm^-1. Find the
i) magnitude and direction of the force acting on each charge and
ii) torque exerted by the field.
Data:
+q = 2 × 10^-10 C
-q = – 2 × 10^-10 C
θ = 60°
E = 10Vm^-1
2d = 5mm = 5 × 10^-3 m
i) Magnitude and directions of the force acts on each charge =?
ii) Torque exerted by the field =?
Solution:

1. Positive charge experience a force in the direction of the field.
2. Negative charge experiences a force opposite to the direction of the field.
   i) To find the force on each charge:
3. Force acts on positive charge: F = qE
   F = 2 × 10^-10 × 10
   F = 2 × l0^-9 × N in the direction of E
   Force acts on negative charge:
   F = qE = 2 × 10^-10 × 10
   F= 2 × l0^-9 N in the direction of E

ii) To find Torque exerted bj the field:

Question 2.
An electric dipole of charges 2 × 10^-6 C, – 2 × 10^-6 C is separated by a distance of 1 cm. Calculate the electric field due to dipole at a points
i) axial line 1 m from its centre and
ii) equatorial line 1 m from its centre.
Answer:


i) Electric field at a point on axial line = 360NC^-1
ii) Electric field at a point on equatorial line = 180NC^-1

Question 3.
Two charges +q and -3q are separated by a distance of lm. At what point in between the charges on its axis is the potential zero?
Answer:
Data :
q₁ = +q; q₂ = -3q; r = 1m
Solution:
Let the potential be zero at point 0 at a d . 1 from +q charge.


The potential is zero at a distance of 0.25m from the charge +q.

Question 4.
A parallel plate capacitor is maintained at some potential difference. A 3 mm thick slab is introduced between the plates. To maintain the plates at the same potential difference, the distance between the plates is increased by 2.4mm. Find the dielectric constant of the slab.
Data:
t = 3 mm; d’- d = 2.4 mm
V is same;
ε_r = ?
Solution:
Capacitance of air-filled parallel plate capacitor:

Since the potential are made the same capacitance in the above two cases is also the same.

The dielectric constant of the slab ε_r = 5

Question 5.
Three capacitors each of capacitance 9 pF are connected in series.
i) What is the total capacitance of the combination?
ii) What is the potential difference across each capacitor if the combination is connected to 120 V supply?
Answer:
Data:
C₁ = C₂ = C₃ = 9pF
V = 120V
C_s = ?
V₁, V₂, V₃ =?
To find effective capacitance:
Three capacitors are in series

∵ \(\frac{1}{\mathrm{C}_{\mathrm{S}}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}\)
C_s = 3pF
C_s = 3 × 10^-12 F

V₁, V₂, V₃ are the potential differences across the three capacitors as shown in the figure.

i) C_s = 3 × 10^-12 F
ii) V₁, V₂, V₃ = 40 V

XV. Five Mark Problems:

Question 1.
Four-point charges +q, +q, and -q is to be arranged respectively at the four corners of a square PQRS of side r. Find the work needed to assemble this arrangement.
Solution:

Here PQ = QS = SR= RP = r
\(\mathrm{PS}^{\prime}=\mathrm{RQ}=\mathrm{r} \sqrt{2}\)
W = Potential Energy of the system of four charges.
\(=\frac{(-q)(+q)}{4 \pi \varepsilon_{o} r}+\frac{(+q)(+q)}{4 \pi \varepsilon_{o} r}+\frac{(+q)(-q)}{4 \pi \varepsilon_{o} r}+\frac{(-q)(-q)}{4 \pi \varepsilon_{o} r}\)

Question 2.
Two small charged spheres repel each other with a force of 2 × 10^-3 N. The charge on one sphere is twice that of the other. When one of the charges is moved 10 cm away from the other, the force is 5 × 10^-4 N. Calculate the charges and the initial distance between them.
Answer:
Data:
F₁= 2 × 10^-3N when distance = r
F₂ = 5 × 10^-4N
when distance (i.e.) r’= (r+ 0. 1)rn
r = r + 10cm
q₁ = ?, q₂ = ?, r =?
Solution:
According to Coulomb’s law
\(\mathrm{F}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \cdot \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{2}}\)
Let q₁ = q and q₂ = 2q and distance between the charges be r experience a repulsive force F₁.


We know,
q₁ = q = 33.3 × 10^-9 C
q₂ = 2q = 2 × 33.3 × 10^-9 C
q₂ =66.6 × 10^-9 C
Initial distance r = 0.1m
Charges:
q₁ = 33.3 × 10^-9 C
q₂ = 66.6 × 10^-9 C

Question 3.
Two capacitors of unknown capacitance are connected in series and parallel. If net capacitances in two combinations are 6µF and 25µF respectively. Find their capacitances.
Answer:
Given:
Cs = 6µF
Cp = 25µF
\(\frac{1}{C_{S}}=\frac{1}{C_{1}}+\frac{1}{C_{2}} \quad \cdots \cdots\)

Question 4.
Three charges + 1 µC, + 3 µC, and – 5 µC are kept at the vertices of an equilateral triangle of sides 60 cm. Find the electrostatic potential energy of the system of charges.
Answer:


The electrostatic potential energy of the system of charges = – 0.255 J

Question 5.
Calculate the force between electron and proton in a Hydrogen atom.
(e = 1.6 × 10^-9 and r₀ = 0.53Å)
Solution:
The electron and proton attract each other. The force between these two particles is given by

Question 6.
Four point charges are placed at the four corners of a square in two ways (a) and (b) as shown in figure. Will the (i) electric potential and (ii) electric field, at the centre of the square be the same or different in the two configurations, and why?

Answer:
a) Electric potential is a scalar quantity. 0pposite charges cancel each other So, in both the configurations electric potential will be zero and the same.
b) Electric field is a vector quantity. It depends on positions of charge.
In the first diagram Electricfield,
\(E=2 \sqrt{2} \frac{\mathrm{kq}}{(\mathrm{r} / \sqrt{2})^{2}}\)
In the second diagram, E= 0
Because opposite charges are placed at the corner.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Accountancy Guide Pdf Chapter 4 Goodwill in Partnership Accounts Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 4 Goodwill in Partnership Accounts

12th Accountancy Guide Goodwill in Partnership Accounts Text Book Back Questions and Answers

I Multiple Choice Questions

Choose the correct answer

Question 1.
Which of the following statement is true?
(a) Goodwill is an intangible asset
(b) Goodwill is a current asset
(c) Goodwill is a fictitious asset
(d) Goodwill cannot be acquired
Answer:
(a) Goodwill is an intangible asset

Question 2.
Super profit is the difference between
(a) Capital employed and average profit
(b) Assets and liabilities
(c) Average profit and normal profit
(d) Current year’s profit and average profit
Answer:
(c) Average profit and normal profit

Question 3.
The average rate of return of similar concerns is considered as
(a) Average profit
(b) Normal rate of return
(c) Expected rate of return
(d) None of these
Answer:
(b) Normal rate of return

Question 4.
Which of the following is true?
(a) Super profit = Total profit / number of years
(b) Super profit = Weighted profit / number of years .
(c) Super profit = Average profit – Normal profit
(d) Super profit = Average profit × Years of purchase
Answer:
(c) Super profit = Average profit – Normal profit

Question 5.
Identify the incorrecte pair
(a) Good will under average profit method – Average profit × Number of years of purchase
(b) Goodwill under Super profit method – Super profit × Number of years of purchase
(c) Goodwill under Annuity method – Average Profit × Present value annuity factor
(d) Goodwill under Weighted average profit method – Weighted average profit x Number of years of purchase
Answer:
(c) Goodwill under Annuity method – Average Profit x Present value annuity factor

Question 6.
When the average profit is ₹ 25,000 and the normal profit ₹ 15,000 super profit is
(a) ₹ 25,000
(b) ₹ 5,000
(c) ₹ 10,000
(d) ₹ 15,000

Hints: Find out Super Profit
Super Profit = Average Profit – Normal Profit
= 25,000- 15,000
= ₹ 10,000
Answer:
(c) ₹ 10,000

Question 7.
Book profit of 2017 is ₹ 35,000; non -recurring income included in the profit is ₹ 1,000 and abnormal loss charged in the year 2017 was ₹ 2,000, then the adjusted profit is
(a) ₹ 36,000
(b) ₹ 35,000
(c) ₹ 38,000
(d) ₹ 34,000
Hints:

Answer:
(a) ₹ 36,000

Question 8.
The total capitalised value of a business is ₹ 1,00,000; asset are ₹ 1,50,000 and liabilities ₹ 80,000. the value of goodwill as per the capitalisation method will be
(a) ₹ 40,000
(b) ₹ 70,000
(c) ₹ 1,00,000
(d) ₹ 30,000
Hints:
Goodwill = Total capitalized value of the business – Actual capital employed
i) Capital employed = Fixed Assets + Current assets – Current Liabilities
= 1,50,000-80,000
= ₹ 70,000
ii) Capitalized Value = ₹ 1,00,000
= 1,00,000 – 70,000
= ₹ 30,000
Answer:
(d) ₹ 30,000

II. Very Short Answer Questions

Question 1.
What is goodwill?
Answer:
Goodwill is the good name or reputation of the business which brings benefit to the business. It enables the business to earn more profit. It is the present value of a firm’s future excess earnings.

Question 2.
What has acquired goodwill?
Answer:
Goodwill acquired by making payment in cash or kind is called acquired or purchased goodwill. When a firm purchases an existing business, the price paid of the purchase of such a business may exceed the net assets (Assets – Liabilities) of the business acquired. The excess of purchase consideration over the value of net assets acquired is treated as acquired goodwill.

Question 3.
What is super profit?
Answer:
Super profit is the excess of average profit over the normal profit of a business.
Super Profit = Average Profit – Normal Profit

Question 4.
What is normal rate of return?
Answer
:It is the rate at which profit is earned by similar business entities in the industry under normal circumstances.

Question 5.
State any two circumstances under which goodwill of a partnership firm is valued
Answer:

1. When there is a change in the profit-sharing ratio
2. When a new partner is admitted into a firm
3. When an existing partner retires from the firm or when a partner dies.
4. When a partnership firm is dissolved.

III Short Answer Questions

Question 1.
State any six-factor determining goodwill.
Answer:

1. Profitability of the firm.
2. Favourable location of the business enterprise.
3. Good quality of goods or services offered.
4. Tenure of the business enterprise.
5. Efficiency of management.
6. Degree of competition.
7. Other factors.

Question 2.
How is goodwill calculated under the super profits method?
Answer:
Under these methods, super profit is the base for calculation of the value of goodwill. Super profit is the excess of average profit over the normal profit of a business.
Super profit = Average profit – Normal profit

Average profit is calculated by dividing the total of adjusted actual profits of certain number of year by the total number of such years. Normal profit is the profit earned by the similar business firms under normal conditions.
Normal profit = Capital employed × Normal rate of return
Capital employed = Fixed assets + Currents assets – Current liabilities.

Question 3.
How is the value of goodwill calculated under the capitalisation method?
Answer:
Under these methods, super profit is the base for calculation of the value of goodwill. Super profit is the excess of average profit over the normal profit of a business.
Goodwill = Total capitalised value of the business – Actual capital employed
The total capitalised value of the business is calculation by capitalising the average profit on the basis of the normal rate of return.
Capitalised value of the business = \(\frac{\text { Average profit }}{\text { Normal rate of return }} \times 100\)
Actual capital employed = Fixed assets (excluding goodwill) + Current assets – Current liabilities

Question 4.
Computer average profit from the following information.
2016: ₹ 8,000; 2017: ₹ 10,000; 2018: ₹ 9,000
Solution:

Question 5.
Calculate the value of goodwill at 2 Years purchase of average profit when average profit is? 15,000.
Solution:
Goodwill = Average profit x No. of years of purchase
Goodwill = 15,000 x 2 Years
= Rs. 30,000

IV Exercises

Simple average profit method

Question 1.
The following are the profits of a firm in the last five years:
2014 : ₹ 10,000; 2015 : ₹ 11,000;
2016 :  ₹ 12,000; 2017 : ₹ 13,000; and 2018 : ₹ 14,000
Calculate the value of goodwill at 2 years purchase of average profit of five years.
Solution:
G/W = Average profit x No.of years of purchase

Answer :
Average profit: ₹ 12,000, Goodwill: ₹ 24,000

Question 2.
From, following information, calculate the value of goodwillOn the basis of 3 years purchase of average profits of last four years.

Solution:
G/W = Average profit x No.of years of purchase

Gs/W = 4,000 × 3 years = Rs. 12,000
Answer:
Average profit: ₹ 4,000; Goodwill: ₹ 12,000

Question 3.
From the following information relating to a partnership firm, find out the value of its goodwill based on 3 years purchas of average profits of the last 4 years.
(a) Profit of the years 2015, 2016, 2017 and 2018 are ₹ 10,000, ₹ 12,500, ₹ 12,000 and ₹ 11,500 respectively.
(b) The business was looked after by a partner and his fair remuneration amounts to ₹ 1,500, per year. This amount was not considered in the calculation of the above profits.
Solution:
G/W = Average profit = Average profit x No.of years of purchase

G/W Average profit × No.of years of purchase
= 10,000 × 3yrs
G/W = Rs. 30,000
Answer:
Average profit: ₹ 10,000; Goodwill: ₹ 30,000

Question 4.
From the following information relating to Sridevi enterprises, cal te the value of goodwill on the basis of 4 years pruchase of the average profits of 3 years.
(a) Profits for the years ending 31st December 2016, 2017 and 2018 were ₹ 1,75,000, ₹ 1,50,0 and ₹ 2,00,000 respectively.
(b) A non-recurring-income of ₹ 45,000 is include in the profits of the year 2016.
(c) The closing stock of the year 2017 was overvalued by ₹ 30,000.
Solution:
Calculation of Adjusted profit

Note : Overvaluationof cl. Stock in 2017 will results in over valuation of opening stock in 2018
Average Profit.
Average Profit. = \(\frac{1,30,000+1,20,000+2,30,00}{3}\)
Average Profit = \(\frac{4,80,000}{3}\)
= Rs. 1,60,000
= Average profit x No.of years of purchase
= 1,60,000 x 4 yrs
= Rs. 6,40,000
Answer :
Average profit: ₹ 1,60,000; Goodwill: ₹ 6,40,000

Question 5.
The following particulars are available in respect of the business carried on by a partnership firm:
(i) Profit earned : 2016 ₹ 25,000; 2017: ₹ 23,000 and 2018: ₹26,000.
(ii) Profit of 2016 includes a non-recurring income of ₹ 2,500.
(iii) Profit of 2017 is reduced by ₹ 3,500 due to stock destroyed by fire.
(iv) The stock was not insured. But, it is decided to insure the stock in future. The insurance premium is estimated to be ₹ 250 per annum.
you are required to calculate the value of goodwill of the firm on the basis of 2 years purchase of average profits of the last three years.
Solution:
Calculation of Adjusted profit

G/W = Average profit × No.of years of purchase
G/W = 24,750 × 2 yrs
= Rs. 49,500
Answer:
Average profit: ₹ 24,750; Goodwill: ₹ 49,500

Weighted average profit method.

Question 6.
Find out the value of goodwill at three years pruchase of weighted average profit of last four years.

Solution:
Calculation of Weighted Average profit

Weighted profts = \(\frac{\text { Weighted profts }}{\text { Total of weights }}\)
Av.pft = \(\frac{1,54,000}{10}\) = Rs. 15,400
G/w = Weighted Average profit × No. of yrs of purchase
G/w = 15,400 × 3yrs
= Rs. 46,200
Answer:
Average profit: ₹15,400; Goodwill: ₹ 46,200

Purchase of super profit method

Question 7.
From the following details, calculate the value o oodwill at 2 years purchase of super profit:
(a) Total assets of a firm are ₹ 5,00,000;
(b) The liabilities of the firm are ₹ 2,00,000
(c) Normal rate of return in this class of business is 12.5%.
(d) Average profit of the firm is ₹ 60,000.
Solution:
G/W = Super Profit x No.of years of purchase

Cap. Emp = Total Assets – Total liabilities
= 5,00,000 – 2,00,000
= Rs. 3,00,000
Normal Profit = \(\frac{3,00,000 \times 12.5}{100}\)
Rs. 37,500
Average Profit = Rs. 60,00
Super profit = Average Profit – Normal profit
= 60,000 – 37,500
= Rs. 22,500
G/w = 22,500 x 2 yrs
= Rs. 45,000
Answer:
Super profit: ₹ 22,500; Goodwill: ₹ 45,000

Question 8.
A partnership firm earned net profits during the last three years as follows:
2016 ₹ 20,000; 2017: ₹ 17,000 and 2018: ₹ 23,000.
The capital investement of the firm throghout the above mentioned period has been ₹ 80,000. Having regard to the risk involved, 15% is considered to the a fair return on capital employed in the business. Calculate the value of goodwill on the basis of 2 years purchase of super profit.
Solution:
G/W = Super Profit x No.of years of purchase

Super profit = Average Profit – Normal profit
= 20,000 – 12,000
= Rs. 8,000
G/w = 8,000 x 2 yrs
= Rs. 16, 000
Answer :
Super profit: ₹ 8,000; Goodwill: ₹ 16,000

Annuity Method:

Question 9.
From the following information, calculate the value of goodwill under annuity method:

Present value of ₹ 1 for 5 years at 15% per annum as per the annuity table is 3.352
Solution:
G/W = Super Profit x Value of A nnuity
SuPer profit = Average Profit – Normal Profit
SuPer profit = 14,000- 4,000 = Rs. 10,000
G/w = 10.0 x 3.352 = Rs. 33,520
Answer :
Super profit: ₹ 10,000; Goodwill: ₹ 33,520

Capitalisation of super profit method:

Question 10.
Find out the value of goodwill by caitalising super profits:
(a) Normal Rate of Return 10%
(b) Profits for the last four years are ₹ 30,000, ₹ 40,000, ₹ 50,000, ₹ 45,000,
(c) A non-recurring income of ₹ 3,000 is included in the above mentioned profit of ₹ 30,000,
(d) Average capital employed is ₹ 3,00,000.
Solution:

= Rs. 30,000
Super profit – Average Profit – Normal Profit
Super Profit – 40,500 – 30,000
= Rs. 10,500

Answer :
Super Profit: ₹ 10,500; Good will: ₹ 1,05,000

Question 11.
From the following information, find out the value of goodwill by capitalisatio method:
(i) Average profit ₹ 20,000
(ii) Normal rate of return 10%
(iii) Tangible assets of the firm ₹ 2,20,000
(iv) Liabilities of the firm ₹ 70,000.
Solution:
G/W = Total capitalaised Value of the business – Actual cap.Employee Average Profit

Actual cap employed = Tangible Assets – Liabilities of the firm.
= 2,20,000 – 70,000
= Rs. 1,50,000
G/w = 2,00,000 – 1,50,000
= Rs. 50,000
Answer:
Total Capitalised value: ₹ 2,00,000; Net tangible assets: ₹ 1,50,000; Goodwill: ₹ 50,000

12th Accountancy Guide Goodwill in Partnership Accounts Additional Important Questions and Answers

Question 1.
The excess of average profit over normal profit is
(a) Super profit
(b) Goodwill
(c) Undistributed profit
Answer:
(a) Super profit

Question 2.
The goodwill that can be recoreded in the books of accounts and is shown on the assets side of the B/S is
(a) Acquired goodwill
(b) Self-Generated goodwil
(c) None of these two
Answer:
(a) Acquired goodwill

Question 3.
How many methods are there for valuing G/W?
(a) 3
(b) 2
(c) 5
Answer:
(a) 3

III Short Answer Questions

Question 4.
Describe the nature of G/W.
Answer:
The nature of goodwill can be described as follows:

• Goodwill is an intangible fixed asset. It is intangible because it has no physical existence. It cannot be seen or touched.
• It has a definite value depending on the profitability of the business enterprise.
• It cannot be separated from the business.
• It help in earning more profit and attracts more customers.
• It can be purchased or sold only when the business is purchased or sold in full or in part.

Question 5.
Explain any 5 factors that determine the value of G/W
Answer:
Factors determining the value of goodwill of a partnership firm
(i) Profitability of the firm
The profit earning capacity of the firm determines the value of its goodwill. A firm earning higher profits and having potential to generate higher profits in future will have higher value of goodwill.

(ii) Favourable location of the business enterprise
If the firm is located in a prominent place which is easily accessible to the customers, it can attract more customers. Its sales and profit will be higher when compared to a firm which is not located in a prominent place. Hence, it will ahve high value of goodwill.

(iii) Good equality of goods or services offered
If a firm is located in a prominent place which is easily accessible to the customers, it can attract more customers. Its sales and profit will be higher when compared to a firm which is not located in a prominent place. Hence, it will have high value of goodwill.

(iv) Tenure of the business enterprise
A firm which has caried on business for several years will have higher reputation among its customers as it is better known to the customers. Such a firm will have higher earnings and higher value of goodwill when compared to a new firm.

(v) Efficiency of management
A firm having efficient management will earn more profitsa and the value of its goodwill will be higher compared to a firm with less efficient managerial personnel.
(iii) It cannot be separated from the business.

Question 6.
What do you mean by self generated G/W?
Answer:
It is the goodwill which is self generated by a firm based on features of the business such as favourable location, loya customers, etc. Such self-generated goodwill cannot be recorded in the books of accounts.

Question 7.
How do you calculated Adjusted Profit?
Answer:
Adjusted profit = Actual profit
+ Past expanse not required in the future.
– Past revenues not likely to be earned in the future
+ Additional income expected in the future
– Additional expenses expected to be paid in the future

Question 8.
How do you calculate G/W under weighted average profit method?
Answer:
Weighted average profit method
Under this method, goodwill is calculated by mulitplying the weighted average profit by a certain number of years of purchase.
Goodwill = Weighted average profit × Number of years of purchase

In this method, weights are assigned to each year’s profit. Weighted profit is ascertained by multiplying the weights assigned with the respective years profit. The sum of the wieghted profits is divided by the sum of weights assigned to determine the weighted averagge profit.

Weighted average profit = \(\frac{\text { Total of weighted profits }}{\text { Total of weights }}\)

Question 9.
What do you mean by Annuity Method of valuing G/w?
Answer:
Under this method, value of goodwill is calculated by multiplying the super profit with the present value of annuity.
Goodwill = Super profit x Present value annuity factor
Present value annuity factor is the present value of annuity of rupee one at a given time. It can be found out from annuity table or by using formula.
Annuity factor = \(\frac{\mathrm{i}(1+\mathrm{i})^{\mathrm{n}}}{\mathrm{i}(1+\mathrm{i})^{\mathrm{n}-1}}\)
Where i = interest rate
n = estimated number of years.

Question 10.
Write the formula for calcultaing G/w under capitalisation of Super profit method.
Answer:

IV Additional Problems:

Question 1.
The Goodwill is to be valued at two years purchase of last four years average profit. The profit were Rs. 40,000, Rs. 32,000, Rs. 15,000 and Rs. 13,000 respectively. Find out the value of goodwill.
Solution :
a) Calculation of average profit:

Calculation of Goodwill:
Goodwill = Average Profit × two years purchase
= 25,000 × 2
= Rs. 50,000.

Question 2.
Three year’s purchase of the last four years average profits is eed the value of goodwill. The profits and losses for the last four years are: I year Rs. 50,000, II year Rs. 80,000; III year Rs. 30,000 (Loss): IV year Rs. 60,000.
Calculate the amount of goodwill.
Solution:
a) Calculation of average Profit:

Total profit No. of years 1,60,000

Calculation of Goodwill:
Goodwill = Average Profit × three year’s purchase
= 40,000 × 3
= Rs. 1,20,000

Question 3.
A finis’:-‘ net profits durL the last three years were Rs. 90,000, Rs. 1,00,000 and Rs. 1,10,000. The capital employed in the firm is Rs. 3,00,000. A normal return on the capital is 10%. Calculate the value of goodwill on the basis of two year’s pruchase of super p it.
Solution:
a) Calculation of average Profit:

Calculation of Super Profit:
Sceper Profit = Average Profit – Normal Profit
= 1,00,000 – 30,000
= Rs. 70,000

Goodwill at two years purchase of super profit:
Goodwill = Super Profit × No. of years of purchase
= 70,000 × 2
= Rs. 1,40,000

Question 4.
Goodwill is to be valued at three years purchase of five year’s average profits. The profits – for the last five years of the firm were: 2,000-Rs.4,200; 2001-Rs.4,500; 2002-Rs.4,700; 2003-Rs.4,600; and 2004-Rs.5,000.
Solution
G/W = Average Profits x No. of yrs of purchase
Average Profit = \(\frac{4200+4500+4700+4600+5000}{5}\)
Average Profit = \(\frac{23,000}{5}\)
= Rs. 4,600
G/W = 4,600 × 3yrs
= Rs. 13,800

Question 5.
Calculate the amount of goodwill on the basis of two year’s purchase of the last four years average profits. The profits for the last four year’s are:

Solution:
Average Profit = \(\frac{26,000+34,000+50,000-10,000}{4}\)
Average Profit = 25,000
G/W = Average Profits × No. of yrs of purchase
G/W = 25,000 × 2
= Rs. 50,000

Question 6.
A firm earned net profits during the last three years as follows:

The Capital inverstment of the firm is Rs. 1,20,000. A fair return on the capital having regard to the risk involved in 10% calculate the value of goodwill on the basis of three years purchase of super profits.
Solution:

1,20,000 x \(\frac{10}{100}\) = Rs. 12,000
Super Profit = Average profit – Normal profit
= 40,000 – 12,000
Rs. 28,000
G/w = Super profit × No. of. yrs of purchase
= 28,000 × 3 years.
G/w = Rs. 84,000

Question 7.
From the following information, calculate the value of goodwill at three year’s purchase of super profit.
Average Capital employed in the business Rs. 6,00,000
Net trading profits of the firm for the past three years were Rs. .1,07,600, Rs. 90,700 and Rs. 1,12,500.
Rate of interest expected from capital having to the risk involved is 12%.
Fair remuneration to the partners for their service Rs. 12,000 p.a.
Solution:
Average Profit = \(\frac{1,07,600+90,700+1,12,500}{3}\)
Average Profit = \(\frac{3,10,800}{3}\)= Rs. 1,03,600
Profit before adj of renumeration = Rs. 1,03,600
Less Remineration = 12,000
Avg. Profit = 91,600
Normal Profit = \(\frac{\text { Cap. Emp } \times \text { Normal rate }}{100}\)
= \(6,00,000 \times \frac{12}{100}\)
= Rs. 72,000

Super Profit = Average Profit – Normal Profit
= 91,600 – 72,000
= Rs. 19,600

G/w = Super profit x No. of yrs of purchase
= 19,600 × 3yrs
= Rs. 58,800

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Accountancy Guide Pdf Chapter 3 Accounts of Partnership Firms-Fundamentals Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 3 Accounts of Partnership Firms-Fundamentals

12th Accountancy Guide Accounts of Partnership Firms-Fundamentals Text Book Back Questions and Answers

I Multiple Choice Questions

Question 1.
In the absence of partnership deed, profits of the firm will be shared by the partners in
(a) Equal ratio
(b) Capital ration
(c) Both (a) and (b)
(d) None of these
Answer:
(a) Equal ratio

Question 2.
In the absence of an agreement among the partners, interest on capital is
(a) Not allowed
(b) Allowed at bank rate
(c) Allowed @ 5% per annum
(d) Allowed @ 6% per annum
Answer:
(a) Not allowed

Question 3.
As per the Indian Partnership Act, 1932, the rate of interest allowed on loans advanced by partners is
(a) 8% per annum
(b) 12% per annum
(c) 5% per annum
(d) 6% per annum
Answer:
(d) 6% per annum

Question 4.
Which of the following is shown in profit and loss appropriation account?
(a) Office Expenses
(b) Salary of staff
(c) Partner’s salary
(d) Interest on bank loan
Answer:
(c) Partner’s salary

Question 5.
When fixed capital method is adapted by a partnership firm, which of the following items will appear in capital account?
(a) Additional capital introduced
(b) Interest on capital
(c) Interest on drawings
(d) Share of profit
Answer:
(a) Additional capital introduced

Question 6.
When a partner withdraws regularly a fixed sum of money at the middle of every month period for which interest is to be calculated on the drawings on an average is
(a) 5.5 moths
(b) 6 months
(c) 12 months
(d) 6.5 months
Answer:
(b) 6 months

Question 7.
Which of the following is the incorrect pair?

Answer:
c

Question 8.
In the absence of an agreement, partners are entitled to
(a) Salary
(b) Commission
(c) Interest on loan
(d) Interest on capital
Answer:
(c) Interest on loan

Question 9.
Pick the odd one out
(a) Partners share profits and losses equally
(b) Interest on partners capital is allowed at 7% per annum
(c) No salary or remuneration allowed
(d) Interest on loan from partners is allowed at 6% per annum
Answer:
(b) Interest on partners capital is allowed at 7% per annum

Question 10.
Profit after interest on drawings, interest on capital and remuneration is ₹ 10,500. Geetha, a partner, is entitled to receive commission @ 5% on profits after charging such commission. Find out commission.
(a) ₹ 50
(b) ₹ 150
(c) ₹ 550
(d) ₹ 500
Hint:

Answer:
(d) ₹ 500

II Very Short Answer Questions

Question 1.
Define partnership.
Answer:
According to Section 4 of the Indian Partnership Act, 1932, the partnership is defined as, “the relation between person who has agreed to share the profits of a business carried on by all or any of them acting for all.

Question 2.
What is a partnership deed?
Answer:
A partnership deed is a document in writing that contains the terms of the agreement among the partners.

Question 3.
What is meant by the fixed capital method?
Answer:
Under the fixed capital method, the capital of the partners is not altered and it remains generally fixed. Two accounts are maintained for each partner namely:

1. Capital account and
2. Current account

The transactions relating to initial capital introduced, additional capital introduced, and capital permanently withdrawn are entered in the capital account and all other transactions are recorded in the current account.

Question 4.
What is the journal entry to be passed for providing interest on capital to a partner? For providing interest on capital.

Question 5.
Why profit and loss appropriation account prepared?
Answer:
The profit and loss appropriation account is an extension of the profit and loss account prepared for the purpose of adjusting the transactions relating to amounts due to and amounts due from partners. It is a nominal account in nature. The balance being the profit or loss is transferred to the partners’ capital or current account in the profit-sharing ratio.

III Short Answer Questions

Question 1.
State the features of partnership?
Answer:
Following are the essential features of a partnership.

1. A partnership is an association of two or more persons. The maximum number of partners is limited to 50.
2. There should be an agreement among the persons to share the profit or loss of the business. The agreement may be oral or written or implied.
3. The agreement must be to carry on a business and to share the profits of the business.
4. The business may be carried on by all the partners or any of them acting for all.

Question 2.
State any six contents of a partnership deed.
Answer:
Contents of partnership deed.
Generally, a partnership deed contains the following:

• Name of the firm and nature and place of business.
• Date of commencement and duration of business.
• Names and addresses of all partners.
• Capital contributed by each partner.
• profit sharing ratio

Question 3.
State the differences between fixed capital method and fluctuating capital method.
Differences between fixed capital method and fluctuating capital method.
Answer:

Question 4.
Write a brief note on the applications of the provisions of the Indian Partnership Act, 1932 in the absence of a partnership deed.
Answer:
Applications of the provisions of the Indian Partnership Act, 1932 in the absence of partnership deed.

1. Remuneration to partners: No salary or remuneration is allowed to any partner. [Section 13(a)]
2. Profit-sharing ratio: Profit and losses are to be shared by the partners equally. [Section 13(b)]
3. Interest on capital: No interest is allowed on the capital. When a partner is entitled to interest on capital contributed as per partnership deed, such interest on capital will be payable only out of profits [Section 13(c)]
4. Interest on loans advanced by partners to the firm: Interest on the loan is to be allowed at the rate of 6 percent per annum. [Section 13(d)]
5. Interest on drawings: No interest is charged on the drawings of the partners.

Question 5.
Jayaraman is a partner who withdrew 10,000 regularly in the middle of every month. Interest is charged on the drawings at 6% per annum. Calculate interest on drawings for the year ended 31st December 2018.
Answer:

IV Exercises

Question 1.
Akash, Bala, Chandru, and Daniel are partners in a firm. There is no partnership deed. How will you deal with the following?

1. Akash has contributed to maximum capital. He demands interest on capital at 10% per annum.
2. Bala has withdrawn ₹ 3,000 per month. Other partners ask Bala to pay interest on drawings @ 8% per annum to the firm. But, Bala did not agree to it.
3. Akash demands the profit to be shared in the capital ratio. But, others do not agree.
4. Daniel demands a salary at the rate of ₹ 10,000 per month as he spends full time for the business.
5. The loan advanced by Chandru to the firm is ₹ 50,000. He demands interest on loan@ 12% per annum

Solution:

1. No interest on capital is payable to any partner.
2. No interest is chargeable on drawings made by the partner.
3. Profits should be distributed equally
4. No remuneration is payable to any partner.
5. Interest on the loan is payable at 6% per annum.

Question 2.
From the following information, prepare capital accounts of partners Rooban and Deri, when their capitals are fixed.
Answer:

Solution :

Question 3.
Arun and Selvam are partners who maintain their capital accounts under fixed capital method. From the following particulars, prepare capital accounts of partners.

Solution :

Question 4.
From the following information, prepare capital accounts of partners Padmini and Padma, When their capitals are fluctuating.

Solution :

Question 5.
Mannan and Ramesh share profits and losses in the ratio of 3:2 and their capital on 1st April 2018 were Mannan ₹ 1,50,000 and Ramesh ₹ 1,00,000 respectively and their current accounts show a credit balance of ₹  25,000 and ₹ 20,000 respectively. Calculate interest on capital at 6% p.a for the year ending 31st March 2019 and show the journal entries.
Solution :
For providing interest on capital

Question 6.
Prakash and Supriya were partners who share profits and losses in the ratio of 5:3. Balance in their capital account on 1st April 2018 was Prakash ₹ 3,00,000 and Supriya ₹ 2,00,000. On 1st July 2018 Prakash introduced additional capital of ₹ 60,000. Supria introduced additional capital of ₹ 30,000 during the year. Calculate on capital at 6% p.a for the year ending 31st March 2019 and show journal entries.
Solution:

Note: Date of introduction of capital by Supria is not given, interest is calculated for an average period of 6 months.

Question 7.
The capital account of Begum and Fatima on 1st January 2018 showed a balance of ₹ 50,000 and ₹ 40,000 respectively. On 1st October 2018, Begum introduced an additional capital of ₹ 10,000 and On 1st May 2018 Fatima introduced an additional capital of ₹ 9,000. Calculate Interest on capital at 4% p.a. for the year ending 31st December 2018.
Solution :

Interest on capital
Begum Rs. 2,100
Fatima Rs. 1,840

Question 8.
From the following balance sheets of Subha and Sudha who share profits and losses in 2:3, Calculate interest on capital at 5% p.a. for the year ending 31st December 2018.

Drawings of subha and sudha during the year were ₹ 8,000 and ₹ 10,000 respectively. Profit earned during the year was ₹ 30,000.
Solution :

Question 9.
From the following balance sheets of Rajan and Devan who share profits and losses in 2:1, Calculate interest on capital at 6% p.a. for the year ending 31st December 2018.

on 1st April, Rajan introduced an additional capital of ₹ 40,000, and on 1st September 2018. Devan introduced ₹ 30,000. Drawings of Rajan and Devan during the year were ₹ 20,000 and ₹ 10,000 respectively. Profit earned during the year was ₹ 70,000
Solution :

Interest on Capital

Interest on capital
Rajan = Rs. 5,400
Devan = Rs. 3,600

Question 10.
Ahmad and Basheer Contribute ₹ 60,000 and ₹ 40,000 respectively as capital. Their respective share of profit is 2:1 and the profit before interest on capital for the year is ₹ 5,000. Computer the amount of interest on capital in each of the following situations:

1. If the partnership deed is silent as to the interest of capital
2. If interest on capital @ 4% is allowed as per the partnership deed
3. If the partnership deed allows interest on capital @ 6% per annum.

Solution
interest on capital @ 4%
Ahamed = 60,000 × \(\frac{4}{100}\) = 2,400
Basheer = 40,000 × \(\frac{4}{100}\) = 1,600

interest on capital @ 6%
Ahamed = 60,000 × \(\frac{6}{100}\) = 3,600
Basheer = 40,000 × \(\frac{6}{100}\) = 2,400

1. No Interest on capital is allowed
2. Since there is sufficient profit, interest on capital wifi be provided Ahamad. ₹ 2,400; Basheer. ₹ 1,600
3. Since the profit is insufficient, interest on capital will not be provided. profit of ₹ 5,000 will be distributed to the partners in their capital ratio of 3:2

Question 11.
Mani is a partner, who withdrew ₹ 30,000 on 1st September 2018. Interest on drawings is charged at 6% per annum. Calculate interest on drawings on 31st December 2018 and show the journal entries by assuming that the fluctuating capital method is followed.
Solution :

Question 12.
Santhosh is a partner in a partnership firm. As per the partnership deed, interest on drawings is charged per annum. During the year ended 31st December 2018 he withdrew as follows:

calculate the amount of interest on drawings.
Solution :

Question 13.
Kumar is a partner in a partnership firm. As per the partnership deed, interest on drawings is charged at 6% per annum. During the year ended 31st December 2018 he wi drew as follows:

calculate the amount of interest on drawings.
Solution :

Question 14.
Mathew is a partner who withdrew ₹ 20,000 during the year 2018. Interest on drawings is charged at 10% per annum. Calculate interest on drawings on 31st December 2018.
Solution :

Note: Date of withdrawal is not given, interest is calculated for an average period of 6 months.

Question 15.
Santhosh is a partner in a partnership firm. As per the partnership deed, interest on drawings is charged at 6% per During the ended 31st December 2018 he withdrew as follows:

calculate the amount of interest on drawings by using the product method.
Solution :

Question 16.
Kavitha is a partner in a firm. She withdraws ₹ 2,500 p.m. regularly interest on drawings is charged @ 4% p.a. Calculate the interest on drawings using average period, if she draws.
(i) at the beginning of every month
(ii) in the middle of every month
(iii) at the end of every month

Solution :
(i) at the beginning of every month
Amt withdrawn = 2,500 × 12 = Rs. 30,000

Question 17.
Kevin and Francis are partners. Kevin draws ₹ 5,000 at the end of each quarter. Interest on drawings is chargeable at 6% p.a. Calculate interest on drawings for the year ending 31st March 2019 using the average period.
Solution:
Kevin → Rs. 5,000 → at the end of each quarter
Average period = \(\frac{9+0}{2}=4 \frac{1}{2}\) Months
Amt withdrew = 5000 × 4 times = Rs. 20,000
IOD = 20,000 × \(\frac{6}{100} \times \frac{9}{24}\)= Rs. 450

Question 18.
Ram and Shyam were partners. Ram withdrew ₹ 18,000 at the beginning of each half year. Interest on drawings is chargeable @ 10% p.a. Calculate interest on the drawings for the year ending 31st December 2018 using the average period.
Solution :
Ram → Rs. 18,000 → beg of each half year.
Average period = \(\frac{12+6}{2}\) = 9 months
Amount withdrawn = 18,000 × 2 times = Rs. 36,000
IOD = 36,000 × \(\frac{10}{100} \times \frac{9}{12}\)= Rs. 2,700

Question 19.
Janani, Kamali, and Lakshmi are partners in the firm sharing profits and losses equally. As per the terms of the partnership deed, Kamali allows a monthly salary of ₹ 10,000, and Lakshmi is allowed a commission of ₹ 40,000 per annum for their contribution to the business of the firm. You are required to pass the necessary journal entry. Assume that their capitals are fluctuating.
Solution :

Question 20.
Sibi and Manoj are partners in a firm. Sibi is to get a commission of 20% of net profit before charging any commission. Manoj is to get a commission of 20% on net profit after charging all commission. Net profit for the year ended 31st Dec 2018 before ch y commission was ₹ 60,000. Find the commission of Sibi and Man Also show the distribution of profit.
Solution:

Note: Assumed to be equal partners (1:1)

Question 21.
Anand and Narayanan are partners in firm sharing profits and losses in the ratio of 5:3 On 1st January 2018, their capitals were ₹ 50,000 and ₹ 30,000 respectively. The partnership deed specifies the following:
(a) Interest on capital is to be allowed at 6% per annum.
(b) Interest on drawings charged to Anand and Narayanan are ₹ 1,000 and ₹ 800 respectively.
(c) The net profit of the firm before considering interest on capital and interest on drawings amounted to ₹ 35,000
Give necessary journal entries and prepare profit and loss appropriation account as on 31st December 2018. Assume that the capitals are fluctuating.
Solution :

Question 22.
Dinesh and Sugumar entered into a partnership agreement on 1st January 2018, Dinesh contributing ₹ 1,50,000 and Sugumar ₹ 1,20,000 as capital. The agreement provided that:
(a) Profits and losses shared in the ratio 2:1 as between Dinesh and Sugumar.
(b) Partners to be entitled to interest on capital @ 4% p.a
(c) Interest on drawings to be charged Dinesh: ₹ 3,600 and Sugumar: ₹ 2,200
(d) Dinesh to receive a salary of ₹ 60,000 for the year, and
(e) Sugumar to receive a commission of ₹  80,000
During the year ended on 31st December 2018, the firm made a profit of ₹ 2,20,000 before adjustment of interest, salary and commission.
Prepare the Profit and loss appropriation account.
Solution :

Question 23.
Antony and Ranjith Started a business on 1st April 2018 with capitals of ₹ 4,00,000 and ₹ 3,00,000 respect According to the Partnership Deed, Anton is to get the salary of ₹ 90,000 per annum, Ranj ith is got 25% commission on profit after allowing salary to Antony and interest on capital @ 5% p.a but before charging such commission. The profit-sharing ratio between the two partners is 1:1 During the year, the firm earned a profit of ₹ 3,65,000.
Prepare profit and loss appropriation account. The firm closes its accounts on 31st March every year.
Solution :

Calculation of profit before charging such commission
Net profit 3,65,000
Less: Ioc 20,000+15,000
salary 90,000 1,25,000
Profit before Commission 2,40,000
Commission = Net profit before commission × \(\frac{\text { Rate }}{100}\)
= 2,40,000 × \(\frac{25}{100}\)
= Rs.60,000

12th Accountancy Guide Accounts of Partnership Firms-Fundamentals Additional Important Questions and Answers

Question 1.
The minimum number of Reasons in a partnership firm is
(a) One
(b) Two
(c) Three
Answer:
(b) Two

Question 2.
In a partnership business, the agreement is
(a) Compulsory
(b) Optional
(c) Not necessary.
Answer:
(a) Compulsory

Question 3.
In a partnership, partners share their profits & losses in ‘ ratio
(a) their capital
(b) equal
(c) agreed
Answer:
(c) agreed

Question 4.
under fixed capital system, the profits and losses of partners will be transferred to their account
(a) Current
(b) Drawings
(c) Capital
Answer:
(a) Current

Question 5.
Interest on capital is calculated on the
(a) Opening capital
(b) Closing capital
(c) Average capital
Answer:
(a) Opening capital

Question 6.
Current accounts for partners will be opened under
(a) fixed capital method
(b) fluctuation capital method
(c) either fixed capital method or fluctuating capital method
Answer:
(a) fixed capital method

Question 7.
In the absence of agreement profits and losses are divided
(a) in the ratio of capitals
(b) in the ratio of time devoted by each partner
(c) equally
Answer:
(c) equally

Question 8.
x and y are partners sharing the profits and losses in the ratio of 2:3 with capitals of Rs. 1,20,000 and Rs. 60,000 respectively profits for the year are Rs. 9,000. If the partnership deed is silent as to interest on the capital show how profit is shared among x and y
(a) profit x – Rs.6,000 y – Rs.3,000
(b) profit x – Rs.3,600 y – Rs.5,400
(c) profit x – Rs.3,000 y – Rs. 6000
Hint:
Profit ₹ 9,000 Sharing ratio = 2:3
X Partner profit = 9,000 × 2/5 = ₹ 3,600
Y Partner profit = 9,000 × 3/5  = ₹ 5,400
Answer:
(b) profits – Rs.3,600 y – Rs.5,400

Question 9.
where a partner is entitled to interest on capital such interest will be payable
(a) Only out of profits
(b) Only out of Capital
(c) Out of profits or out of capital
Answer:
(a) Only out of profits

Question 10.
under fixed capital method, salary payable to a partner is recorded
(a) in the current account
(b) in capital account
(c) either in current account or capital account
Answer:
(a) in the current account

II Very Short Answer Questions

Question 11.
What is meant by the fluctuating capital method?
Answer:
Under this method, only a capital account is maintained for each partner. All the transactions between the partner and the firm are recorded in the capital account. This account is credited with the initial and additional capital introduced by the partner. The account is debited with capital withdrawn, drawings, interest on drawings, and share of loss of the partner. As a result, the balance in this account goes on fluctuating periodically. Under this method, the partner s capital account may show either credit balance or debit balance.

Question 12.
Mention the methods of calculating interest on drawings?
Answer:

1. Direct Method
2. Product Method
3. Average period Method.

Question 13.
How to calculate the average period under different circumstances?
Answer:
The average period is computed as follows:
The following table shows the average period in months for withdrawal made at the beginning. In the middle and at the end of every month, quarter and half- year of the year.

Additional problems:

Question 14.
Show how the following items will appear in the capital accounts of the partners, Anbu and Balu.

Solution :
a) When capital accounts are fixed:

b) when capital accounts are fluctuating:

capital Accounts

Question 15.
Write up the capital and. current accounts of the partners, Kala and Mala from the following and show how these will appear in the Balance Sheet.

Solution :

Question 16.
Ravi and Raghu started business on April 1, 2003, with a capital of 90,000 and Rs. 70,000 respectively. Ravi introduced Rs. 10,000 as additional capital on July 1, 2003. Interset on capital is to be allowed @ 10%. Calculate the interest payable to Ravi and Raghu for the year ending March 31, 2004.
Solution :

Question 17.
P, Q and R were partners sharing profits in the ratio of 3:2:1. P draws Rs.5,000 at the end of each quarter. Q draws Rs. 10,000 at the end of each half-year. R draws Rs.2,000 on 1.5.2004 Rs.3,000 on 31.10.2004, Rs.5,000 on 30.11.2004. Calculate interest on their drawings at 10% p.a for the year ending 31.3.2005.
Solution :
Calculation of interest on Drawings under Product Method:
a) Interest on drawing of P :

Interest on drawings at 10% p.a. on Rs.90,000
Interst on drawings = 90,000 × \(\frac{10}{100} \times \frac{1}{12}\) = Rs. 750
Interest on drawings of Q:

Interest on drawings at 10% p.a. on Rs.60,000
Interst on drawings = 60,000 × \(\frac{10}{100} \times \frac{1}{12}\) = Rs.500
Interst on drawings of R:

Interest on drawings at 10% p.a. on Rs.57,000
Interest on drawings = 57,000 x \(\frac{10}{100} \times \frac{1}{12}\) = Rs.475

Question 18.
Mahesh and Ramesh are partners sharing profits in the ratio of 3:2 with capitals of Rs.50,000 and Rs.40,000 respectively. Interest on capital is agreed at 8%p.a Interest on drawings is fixed at 10% p.a. The drawings of the partners were Rs.15,000 and Rs.10.000, the interest for Mahesh Rs.750 and for Ramesh Rs.500. Mahesh is entitled to a salary of Rs. 12,000 p.a. and Ramesh is entitled to get a commission of 10% on the Net Profit before charging such cc ion. The Net Profit of the firm before making the above adjustments was Rs.60,000 for the year ended 31st March 2005.
Prepare the profit and loss appropriation account.
Solution :
In the Books of the Firm Profit and Loss Appropriation Account

Note: Calculation of commission.
Step 1→ Total the credit side of the Profit and Loss Appropriation Account, i.e., Rs. 61,250
Step 2 → Total the debit side of the Profit and Loss Appropriation Account, i.e., Rs. 19,200
Step 3 → Find the balance, i.e., Rs. 42,050.
Step 4 → Apply the formula.
Commission = Net Profit before commission × \(\frac{\text { Rate of Commission }}{100} \text {\)
Commission = 42,050 × \(\frac{10}{100}\) = 4,205
The Balance of Rs. 37,845 (Rs.42,050 – Rs. 4,205) is transferred to Parnters Capital accounts in the ratio.

Question 19.
X and Y are partners in a firm, sharing profits losses equal. X is entitled to a salary of Rs. 5,000 p.m. Y is entitled to a con ion c 0% of Net Profit after charging such commission. Net Profit before charging commission and salary was Rs. 1,48,000. Show the Profit and loss appropriation account.
Solution:
a) Calculation of Salary to X:
5,0 p.m. for 12 months = 12 × 5,000 = Rs. 60,000
Calculation of Commission to Y:
Percentage of Net profit after charging the commission =
Net Profit before commission × \(\frac{\text { Rate of commission }}{100+\text { Rate of commision }}\)
Net Profit before commission = 1,48,000 – 60,000 = Rs.88,000
Commission = 88,000 × \(\frac{10}{100+10}\)
= 88,000 × \(\frac{10}{110}\)
= Rs.8,000

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.6 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.6

Question 1.
Evaluate the following
(i) \(\int_{0}^{π/2}\) sin¹⁰x dx
Solution:
Here n = 10, which is even

(ii) \(\int_{0}^{π/2}\) cos⁷x dx
Solution:
Here n = 7, which is odd

(iii) \(\int_{0}^{π/4}\) sin⁶ 2x dx
Solution:
put t = 2x
dt = 2 dx

(iv) \(\int_{0}^{π/6}\) sin⁵ 3x dx
Solution:
put t = 3x
dt = 3 dx

(v) \(\int_{0}^{π/2}\) sin² x cos⁴ x dx
Solution:
Here m = 2, which is even and n = 4, which is even

(vi) \(\int_{0}^{2π}\) sin⁷ \(\frac{x}{4}\) dx
Solution:

(vii) \(\int_{0}^{π/2}\) sin³ θ cos⁵ θ dθ
Solution:

Aliter Method

(viii) \(\int_{1}^{0}\) x² (1 – x)³ dx
Solution:
By applying reduction formula

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.4 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.4

Question 1.
Evaluate the following
\(\int_{0}^{1}\) x³e^-2x dx
Solution:
Bernoulli’s formula,
∫uv dx = uv₁ – u¹v₂ + u²v₃ – u³v₄ + ….

Question 2.
\(\int_{0}^{1}\) \(\frac{sin(3tan^{-1}x)tan^{-1}x dx}{1+x^2}\)
Solution:

Question 3.
\(\int_{0}^{1/√2}\) \(\frac{e^{sin^{-1}x}sin^{-1}x}{\sqrt{1-x^2}}\) dx
Solution:

Question 4.
\(\int_{0}^{\pi / 2}\) x² cos 2x dx
Solution:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.3 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.3

Question 1.
Evaluate the following definite integrals.
(i) \(\int_{3}^{4}\) \(\frac{dx}{x^2-4}\)
Solution:

(ii) \(\int_{-1}^{1}\) \(\frac{dx}{x^2+2x+5}\)
Solution:

(iii) \(\int_{0}^{1}\) \(\frac{\sqrt{1-x}}{\sqrt{1+x}}\) dx
Solution:

(iv) \(\int_{0}^{\pi / 2}\) (\(\frac{1+sin}{1+cosx}\))dx
Solution:

(v) \(\int_{0}^{\pi / 2}\) \(\sqrt{cos θ}\) sin³θ dθ
Solution:

(vi) \(\int_{0}^{1}\) \(\frac{1-x^2}{(1+x^2)^2}\) dx
Solution:

Question 2.
Evaluate the following integrals using properties of integration:
(i) \(\int_{-5}^{5}\) x cos (\(\frac{e^x-1}{e^x+1}\)) dx
Solution:

(ii) \(\int_{-\pi / 2}^{\pi / 2}\) (x⁵ + x cos x + tan³ x) dx
Solution:
\(\int_{-\pi / 2}^{\pi / 2}\) (x⁵ + x cos x + tan³ x) dx
= \(\int_{-\pi / 2}^{\pi / 2}\) (x⁵ + x cos x + tan³ x) dx + \(\int_{-\pi / 2}^{\pi / 2}\)
Let f(x) = x⁵ + x cos x + tan³x
f(-x) = -x⁵ – x cos x – tan³x
f(x) = -f(-x)
f(x) is an odd function

(iii) \(\int_{-\pi / 4}^{\pi / 4}\) sin² x dx
Solution:
I = \(\int_{-\pi / 4}^{\pi / 4}\) sin² x dx
f(x) = sin²x
f(-x) = sin²(-x) = sin²x
f(x) = f(-x)
f(x) is an even function

(iv) \(\int_{0}^{2π}\) x log(\(\frac{3+cos x}{3-cos x}\))dx
Solution:

(v) \(\int_{0}^{π}\) sin⁴ x cos³ x dx
Solution:
\(\int_{0}^{π}\) sin⁴ x cos³ x dx
f(x) = sin⁴x cos³x
f(2π – x) = sin⁴(2π – x) cos³ (2π – x)
= sin⁴x cos³x
f(2π – x) = f(x)

Limit from 0 to π tends to 0 to 0
∴ Integral value = 0
∴ \(\int_{0}^{π}\) sin⁴ x cos³ x dx = 0

(vi) \(\int_{0}^{1}\) |5x – 3|dx
Solution:

(vii) \(\int_{0}^{sin^2x}\) sin^-1 √t dt + \(\int_{0}^{cos^2x}\) cos^-1 √t dt
Solution:
I₁ = \(\int_{0}^{sin^2x}\) sin^-1 √t dt
Put sin^-1 √t = θ
√t = sin θ
\(\frac{1}{2√t}\) dt = cos θ dθ
dt = 2√t cos θ dθ
= 2 sin θ cos θ dθ
dt = sin 2θ dθ

I₁ = \(\int_{0}^{cos^2x}\) cos^-1 √t dt
Put cos^-1 √t = θ
√t = cos θ
\(\frac{1}{2√t}\) dt = -sin θ dθ
dt = -2√t sin θ dθ
= -2 cos θ sin θ dθ
dt = -sin 2θ dθ

(viii) \(\int_{0}^{1}\) \(\frac{log(1+x)}{1+x^2}\) dx
Solution:

(ix) \(\int_{0}^{π}\) \(\frac{x sin x}{1+sin x}\) dx
Solution:

(x) \(\int_{π/8}^{3π/8}\) \(\frac{1}{1+\sqrt{tan x}}\) dx
Solution:

(xi) \(\int_{0}^{π}\) x[sin²(sin x) + cos² (cos x)] dx
Solution:
Let I = \(\int_{0}^{π}\) x[sin²(sin x) + cos² (cos x)] dx
f(x) = sin² (sin x) + cos² (cos x)
f(π – x) = sin² (sin π – x)) + cos² (cos(π – x))
= sin² (sin x) + cos² (cos x)
f(x) = f(π – x)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Accountancy Guide Pdf Chapter 2 Accounts of Not-For-Profit Organisation Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 2 Accounts of Not-For-Profit Organisation

12th Accountancy Guide Accounts of Not-For-Profit Organisation Text Book Back Questions and Answers

I Multiple Choice Questions

Choose the correct answer

Question 1.
Receipts and payments account is a
(a) Nominal A/c
(b) Real A/c
(c) Personal A/c
(d) Representative personal account
Answer:
(b) Real A/c

Question 2.
Receipts and payments account records receipts and payments of
(a) Revenue nature only
(b) Captital nature only
(c) Both revenue and captial nature
(d) None of the above
Answer:
(c) Both revenue and captial nature

Question 3.
Balance of receipts and payments account indicates the
(a) Loss incurrd during the period
(b) Excess of income over expenditure of the period
(c) Total cash payments during the period
(d) Cash and bank balance as on the date
Answer:
(d) Cash and bank balance as on the date

Question 4.
Income and expenditure account is a
(a) Nominal A/c
(b) Real A/c
(c) Personal A/c
(d) Representative personal account
Answer:
(a) Nominal A/c

Question 5.
Income and expenditure Account is prepared to find out
(a) Profit or loss
(b) Cash and bank balance
(c) Surplus or deficit
(d) Financial position
Answer:
(c) Surplus or deficit

Question 6.
Which of the following should not be recorded in the income and expenditure account?
(a) Sale of old news papers
(b) Loss on sale of asset
(c) Honorarium paid to the secretary
(d) Sale proceeds of furniture
Answer:
(d) Sale proceeds of furniture

Question 7.
Subsceiption due but not received for the current year is
(a) An asset
(c) An expense
(b) A liability
(d) An item to be ignored
Answer:
(a) An asset

Question 8.
Legacy is a
(a) Revenue expenditure
(b) Capital expenditure
(c) Revenue receipt
(d) Capital receipt
Answer:
(d) Capital receipt

Question 9.
Donations received for a specific purpose is
(a) Revenue receipt
(b) Capital receipt
(c) Revenue expenditure
(d) Capital expenditure
Answer:
(b) Capital receipt

Question 10.
There are 500 members in a club each paying ₹ 100 as annual subscription. Subscription due but not received for the current year is ₹ 200; Subscription received in advance is ₹ 300. Find out the amount of subscription to be shown in the income and expend ; account.
(a) ₹ 50,500
(b) ₹ 50,200
(c) ₹ 49,900
(d) ₹ 49,800
Hint:

Answer:
(a) ₹ 50,500

II Very Short Answer Questions

Question 1.
State the meanting of not-for-profit organisation.
Answer:
Some organisations are established for the purpose of rendering servies to the public without any profit motive. They may be created for the promotion of art, culture, education, sports, etc. These organisations are called not-for-profit organisation.

Question 2.
What is receipts any payments account?
Answer:
Receipts and Payments account is a summary of cash and bank transactions of not-for- profit organisations prepared at the end of eac financial year.

It is a real account in nature. The receipts any payments account begins with the opening balances of cash and bank and ends with closing balances of cash and bank. All cash receipts are shown on the debit side and all cash payments are shown on the credit side of this account. All cash receipts and cash payments whether of capital or revenue nature will be recorded irrespective of the period for which the amoutn is received or paid, it is recorded if cash is received or paind during the year.

Question 3.
What is Legacy?
Answer:
A gift made to a not-for-profit organisation by a will, is called legacy. It is a capital receipt.

Question 4.
Write a short note on life membership fees.
Answer:
Amount received towards life membership fee from members is a capital receipt as it is non-recurring in nature.

Question 5.
Give four examples for Capital r pts of not-for-profit organisation.
Answer:
Life membership fees, Legacies Specifi, Specific donations, Sale of fixed assets, Special funds, Tournament fund prize

Question 6.
Give four examples for revenue receipts of not-for-profit organisation?
Answer:
Subscription, Interest on investment, Interest on fixed deposit
Sale of (old) sports materials, Sale of (old) newspapers

III Short Answer Questions

Question 1.
What is income and expenditure account?
Answer:
Income and expenditure account is a summary of income and expenditure of a not-for- profit organisation prepared at the end of an accounting year. It is prepared to find out the surplus or deficit pertaining to a particular year.
It is a nominal account in nature in which items of revenue receipts and revenue expenditure, relating to the current year alone are recorded. It is prepared following the accrual basis of accounting.

Question 2.
State the differences between Receipts and Payments Account and Income and Expenditure Account.
Answer:

Question 3.
How annual subscription is dealt with in the final accounts of not-for-profit organisation?
Answer:

Question 4.
How the following items are dealt with in the final accounts of not-for-profit organisation? (a) Sale of sports materials (b) Life membership fees (c) Tournament fund
Answer:

• Sale of sports materials: Shown on the credit side of Income & Edpenditure A/c.
• Life membership fees: being capital receipt, will be shown on the liability side of the B/s.
• Tournament Fund: being capital receipt, will be shown on the liability side o f the B/s.

IV Exercise

Question 1.
From the information given below, prepare Receipts and Payments account of Kurunji Sports Club for the year ended 31st December, 2018

Solution :
In the books of Kurunji Sports Club Receipts & Payments A/c for the year ended 31.12.2018
Answer:

Cash balance: ₹ 5,000

Question 2.
From the information given below, prepare Receipts and Payments account of Coimbatore Cricket club for the year ending 31st March, 2019.

Solution:
In the book of Coimbatore Cricket Club/Receipts & Payments A/c for the year ended 31.3.2019

Answer:
Bank balance: ₹ 2,400

(Hint: Wages yet to be paid is a non cash item. Hence, it is excluded in receipts and payments account)

Question 3.
From the information given below, prepare Receipts and Payments account of Madurai Mother Theresa Mahalir Madram for the year ended 31st December, 2018.

Solution: In the book of Mother Teresa Mahalir Mandram, Madurai Receipt & Payment A/c for the year ended 31.12.2018

Answer:
Bank Overdraft: ₹ 4,700
(Hint: As depreciation on furniture is a non cash item, it is excluded in receipts and payments account)

Question 4.
Mayiladuthrai Recreation Club gives you the following details. Prepare Receipts and Payments account for the year ended 31st March, 2019.

Solution:
In the books of Mayiladuthurai Recreation Club Receipts & Payments A/c for the year ended 31.3.2019


Answer:
Bank balance: ₹ 50,000

Question 5.
From the following information, prepare Receipts and Payments account of Cuddalore Kabaddi Association for the year ended 31st March, 2019

Solution :
In the books of Cuddalore Kabaddi Association
Receipts & Payments A/c for the year ended 31.3.2019

Answer:
Bank Overdraft: ₹ 4,500

Question 6.
From the following receipts and payments account of Tenkasi Ihiruvalluvar Mandram, Prepare income and expenditure account for the year ended 31st March, 2019.

In the books of Tenkasi Ihiruvalluvar Mandram Income &Expenditure A/c for the year ended 31.3.19


Answer:
Surplus: ₹14,000

Question 7.
From the following receipts and payment account, prepare income and expenditure account of Kumbakonam Basket Ball Association for the year ended 31st March, 2018.

Solution: In the books of Kumbakonam Basket Ball Association Income &Expenditure A/cfor the year ended 31.3.18

Answer:
Deficit: ₹ 26,000

Question 8.
From the following receipts and payments account and the additional information given below, calculate the amoumt of of subscription to be shown in Income and expenditure account for the year ending 31st December, 2018.

Additional information: Subscription outstanding for the year 2018 is Rs.8,000
Solution:
Income & Expenditure A/c for the year ended 31.12.2018

Answer:
Subscription for 2018: ₹ 1,80,000)

Question 9.
How the following items will appear in the final accounts of a club for the year ending 31st March 2019?

There are 200 members in the club each paying an annual subscription of ₹ 400 per annum. Subscription still outstanding for the year 2017-2018 is ₹ 2,000.
Solution :

Answer:
Income and Expenditure A/c: Subscription: ₹ 80,000 Balance Sheet: Assets: Subscription outstanding: ₹ 32,000; Liabilities: Subscription received in advance: ₹ 5,000

Question 10.
Ho w will the following items appear in the final accounts of a club for the year ending 31st March 2017? Received subscription of ₹ 40,000 during the year 2016-17. This includes subscription of ₹ 5,000 for 2015-16 and ₹ 3,000 for the year 2017-18. subscription of ₹ 1,000 is still outstanding for the year 2016-17.
Solution:
Income & Expenditure A/c for the year ended 31.3.17

Answer:
Income and Expenditure A/c: Subscription: ₹ 33,000
Balance Sheet: Assets: Subscription outstanding: ₹ 1,000;
Liabilities: Subscription received in advance: ₹ 3,000

Question 11.
Compute income from subscription for the year 2018 from the following particulars relating to a club.

Subscription received during in the year 2018: ₹ 45,000
Solution:
Calculation of income from subscription for the year 2018

Answer:
Income and Expenditure A/c: Subscription: ₹ 44,000

Question 12.
From the following particulars, show how the item ‘Subscription’ will appear in the Income and Expenditure Account for the year ended 31.12.2018?

Subscription received in 2018 is ₹ 50,000 which includes ₹ 5,000 for 2017 and ₹ 7,000 for 2019. Subscription outstanding for the year 2018 is ₹ 6,000. Subscription of ₹ 4,000 was received in advance for 2018 in the year 2017.
Solution:
Income & Expenditure A/c for three year ended 31.12.18

Answer:
Income and Expenditure A/c: Subscription:₹ 48,000

Question 13.
How the following items appear in the final accounts of Thoothukudi Young Pioneers Association?
There are one hundred members in the association each paying 25 as annual subscription. By the end of the year 10 members had not paid their subscription but four members had paid for the next year in advance.
Solution:
Calculation of income from subscription

Answer:
Income and Expenditure Account: Subscription: ₹ 2,500
Balance Sheet: Assets: Subscription outstanding: ₹ 250;
Liabilities: Subscription received in advance: ₹ 100

Question 14.
How will the following appear in the final accounts of Marthadam Women Cultural Association?

Solution :
Income & Expenditure A/c for the year ended 31.3.2019

Answer:
Income and Expenditure Account (Dr): ₹ 90,000;
Balance Sheet: Assets: Stock of sports materials: ₹ 10,000

Question 15.
How will the following appear in the final accounts of Vedaranyam Sports Club?

Solution:
Income & Expenditure A/c for the year ended

Answer:
Income and Expenditure Account: Bats and balls consumed : ₹ 16,000 (Dr);
Sale of old sports materials: ₹ 2,000 (Cr.)
Balance Sheet: Assets side: Stock of bats and balls: ₹ 4,000)

Question 16.
Show how the following items appear in the income and expenditure account of Sirkazhi Singers Association?

Solution:
Income & Expenditure A/c for the year ended 31.3.2018

Answer:
Stationery consumed : ₹ 6,900 (Dr.)

Question 17.
Chennai tennis club had Match fund showing credit balance of ₹ 24,000 on 1st April, 2018. Receipt to the fund during the year was ₹ 26,000. Match expensed incurred during the year was ₹ 33,000. How these items will appear in the final accounts of the club for the year ended 31st March, 2019?
Solution:
Balance Sheet As on 31.3.19

Answer:
Balance sheet: Liabilities: Match fund : ₹ 17,000

Question 18.
How will the following appear in the final accounts of Karaikudi sports club for the year ending 31st March, 2019?
Particulars ?

Solution:
Balance Sheet As on 31.3.2019

Answer:
Balance sheet: Liabilities: Tournament fund: ₹ 49,000
Assets: Tournament fund investment: ₹ 90,000

Question 19.
Compute capital fund of Salam Sports Club as on 01.4.2019?

Solution:
Balance Sheet As on 1.4.2019

Answer:
Capital fund: ₹ 80,000

Question 20.
From the follwoing Receipts and Paymentaccount and from the information given below of Ramanathapuram Sports Club, prepare Income and Expenditure account for the year ended 31st December, 2018 and the balance sheet as on that date.

Additional information:
(i) Capital fund as on 1st January 2018 ₹ 30,000
(ii) Opening stock of sports material ₹ 3,000 and closing stock of sports material ₹ 5,000
Solution
Dr. Income & Expenditure A/c for three year ended 31.12.18

Answer:
Surplus: ₹ 2,300; Balance Sheet total: ₹ 48,300

Question 21.
From the following Receipts and Payments account of Yarcaud Youth Association, prepare Income and expenditure account for the year ended 31st March, 2019 and the balance sheet as on the date.

Additional information:
(i) Opening capital fund ₹ 20,000.
(ii) Stock of books on 1.4.2018 ₹ 9,200.
(iii) Subscription due but not received ₹ 1,700.
(iv) Stock of stationery on 1.4.2018 ₹ 1,200 and stock of stationery on 31.3.2019, ₹ 2,000
Solution:
Income & Expenditure A/c for the year ended 31.3.19

Answer:
Surplus: ₹ 7,800; Balance Sheet total: ₹ 37,800

Question 22.
Following is the Receipts and Payments account of Neyveli Science Club for the year ended 31st December, 2018.
i
Additional information:
(i) Opening capital fund ₹ 6,400.
(ii) Subscription includes ₹ 600 for the year 2019.
(iii) Science equipment as on 1.1.2018 ₹ 5,000.
(iv) Surplus on account of exhibition should be kept in reserve for new auditorium. Prepare income and expenditure account for the year ended 31st December, 2018 and the balance sheet as on that date.
Solution:
Income & Expenditure A/c for the year ended 31.12.18

Answer:
Deficit: ₹ 1,400; Balance Sheet total: ₹ 10,600

Question 23.
From the following Receipts and Payments account of Sivasasi Pensioner’s Recreation Club, prepare income and expenditure account for the year ended 31st March, 2018 and the balance sheet as on that date.

Additional information:
(i) The club had 300 members each paying ₹ 100 as annual subscription.
(ii) The club had furniture ₹ 10,000 on 1.4.2017.
(iii) The subscription still due but not received for the year 2016 – 2017 is ₹ 1,000.
Solution:
Income & Expenditure A/c for thre year ended 31.3.18

Answer:
Surplus: ₹ 4,000; Opening capital fund ₹ 46,000; Balance sheet total: ₹ 99,000

Question 24.
Following is the Receipts and payments account ofVirudhunagar Volleyball Association for the year ended 31st December, 2018.

Additional information:
(i) On 1.1.2018, the association owned investments ₹ 10,000, premises and grounds ₹ 40,000, stock of bats and balls ₹ 5,000.
(ii) Subscription 15,000 related to 2017 is still due.
(iii) Subscription due for the year 2018, ₹ 6,000.
Prepare income and expenditure account for the year ended 31st December, 2018 and the balance sheet on that date.
Solution :
Balance Sheet As on 1.1.2018


Answer:
SurpIus: ₹ 52,000; Opening capital fund ₹ 75,000;
Closing balance sheet total: ₹ 1,32,000

12th Accountancy Guide Accounts of Not-For-Profit Organisation Additional Important Questions and Answers

I Multiple Choice Questions

Question 1.
Receipts & payments A/c records receipts & payments of
(a) Current year only
(b) Previous & future period only
(c) both a & b
Answer:
(c) both a & b

Question 2.
Non-cash items such as depreciation outstanding expenses and accrued income are not shown in
(a) Receipts & payments A/c
(b) Income & expenditure A/c
(c) Balance sheet
Answer:
(a) Receipts & payments A/c

Question 3.
Only revenue items are recorded in
(a) Income & Expenditure A/c
(b) Receipts & payments A/c
(c) Balance sheet
Answer:
(a) Income & Expenditure A/c

Question 4.
Interest on investment received is a
(a) Capital receipt
(b) revenue receipt
(c) Capital Expenditure
Answer:
(b) revenue receipt

Question 5.
Grants received towards construction of buildings acqisiton of assets etc are treated as
(a) Capital receipt
(b) Capital expenditure
(c) Revenue Receipt
Answer:
(a) Capital receipt

Question 6.
Grants received towards maintenance of assets, payments of salaries etc aire are treated as
(a) Capital receipt
(b) revenue receipt
(c) Capital Expenditure
Answer:
(b) revenue receipt

Question 7.
Revenue items of current year alone are recorded in
(a) Receipts & Payments A/c
(b) Income & Expenditure A/c
(c) Balance sheet
Answer:
(b) Income & Expenditure A/c

III. Short Answer Questions

Question 1.
What are the ffeatures of not-for-profit organisations.
Answer:
Following are the features of not-for-profit organisations:

1. Not-for-profit organisations are the organisations which function without any profit motive.
2. Their main aim is to provide services to a specific group or the public at large.
3. Generally, they do not undertake business or trading activities.
4. Their main sources of income include subscription from members, donations, grant-in¬aid and legacies.

Question 2.
What are the accounts prepared by not-for-profit organisations
Answer:

1. Receipts and payments account,
2. Income and expenditure account and
3. Balance sheet

Question 3.
What are the steps in the preparation of Receipts & Payments A/c?
Answer:
Following are the steps involved in the preparation of receipts and payments account:

• Record the opening balance of cash in hand and favourable bank balance on the debit side of receipts and payments account. If there is bank overdraft, it must be recorded on the credit side.
• Actual cash receipts during the year are recorded on the debit side and actual cash payments during the year are recorded on the credit side.
• While recording cash receipts and payments, no distinction needs to be made between revenue and capital items. Similarly, no distinction needs to be made whether the amount reveived or paid relates to the current period, previous period or future period.
• If the total of the debit side is more than the credit side, the balancing figure will appear on the credit side. It represents the closing balance of cash or bank.
• If the total of the credit side is more than the debit side, the balancing figure will appear on the debit side. It represents bank overdraft.

Question 4.
Name a few examples capital for expenditure & revenue expenditure.
Answer:

IV Additional Problems:

Question 1.
The following particulars of a club are available.

Prepare Receipts and Payments Account as on 31st March, 2002.
Solution:
Receipts and Payments A/c for the year ended 31st March, 2002

Question 2.
From the following particulars prepare receipts and payments account of a club for the year 2000:
Cash Balance (1st Jan. 2001) Rs. 4,000; Entry fee received Rs. 400; Subscription received Rs. 800; Last year’s arrears Rs. 80; Rent paid Rs. 1,060; Purchase of cricket balls Rs. 525; Purchase of cricket bats Rs. 580; Miscellaneous expenses Rs. 10; Purchase of Stationery Rs. 5; Salary paid Rs. 1,100.
Solution:
Receipts and Payments A/c for the year ended 31st December, 2001

Question 3.
Subscription received by the Health Club during the year 2014 were as under:

Calculate the amount of Subscription to be shown on the income side of Income and Expenditure A/c.
Solution:
Calculation of Subscription for 2014

Question 4.
During the year 2014, subscription received by a sports club were ₹ 80,000. These included ^ 3,000 for the year 2013 and 16,000 for the year 2015. On Decembrer 31, 2013 amount of subscription due but not reveived was ₹ 12,000. Calculate the ammount of subscription to be shown in Income and Expenditure Account as income from subscription?
Solution
Calculation of Subscription for 2014

Question 5.
The Receipts and Payments Account of Harimohan Charitable Institution is given:

Prepare the Income and Expenditure Account for the year ended on March 31, 2014 after considering the following:
(i) It was decided to treat 50% of the amount received on account of Legacies and Donations as income.
(ii) Liabilities to be provided for are: Rent’ 800; Salaries ‘ 1200; Advertisement’ 200
(iii) ‘2,000 due for interest on investment was not actually received.
Solution:
Income & Expenditure A/c the year ended 31.03.2018

Question 6.
From the following receipts and payments and information given below, prepare Income and Expenditure Account and opening Balance Sheet of Adult Literacy orgnization as on December 31, 2013: Receipts and Payments Account for the year ending as on December31,2013.


Information:
(i) Subscription outstanding as on 31.12.2012, ₹ 2,000 and for the year ending December 31,2013, ₹ 1,500
(ii) On December 31,2013 Salary outstanding’ 600, and one month Rent paid in advance.
(iii) On Jan. 01,2012 organization owned Furniture ₹ 12,000, Books ₹ 5,000.
Solution:

Question 7.
The following is the account of cash transactions of the Nari Kalayan Samittee for the year ended December 31, 2013:

You are required to prepare an Income and Expenditure Account after the following adjustments:
(i) Subscription still to be received are ‘ 750, but subscription include ‘ 500 for the year 2014.
(ii) In the beginning of the year the Sangh owned building ‘ 20,000 and furniture ‘ 3,000 and Books ‘2,000.
(iii) Provide depreciation of furniture @5% (including purchase), books @10% and building @5%.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.2 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.2

Question 1.
Evaluate the following integrals as the limits of sums:
(i) \(\int_{0}^{1}\) (5x + 4)dx
Solution:
(i) Here f(x) = 5x + 4, a = 0, b = 1
Hence we get

(ii) \(\int_{1}^{2}\) (4x² – 1)dx
Solution:
Here f(x) = 4x² – 1, a = 1, b = 2
We use the formula