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Students can Download Chapter 5 Dissolution of Partnership Questions and Answers, Plus Two Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Accountancy Chapter Wise Questions and Answers Chapter 5 Dissolution of Partnership

Plus Two Accountancy Dissolution of Partnership One Mark Questions and Answers

Question 1.
Which is not correct in the case of Dissolution of Partnership
(a) The original partnership agreement is terminated
(b) Some partners continue in the business
(c) No partner to continue in the business
(d) A new partnership comes in to existence
Answer:
(c) No partner to continue in the business

Question 2.
Dissolution of partnership does not lead to
(a) Termination of the original partnership agreement
(b) Dissolution of the existing partnership
(c) Coming in to existence of a new partnership
(d) Dissolution of the firm
Answer:
(d) Dissolution of the firm

Question 3.
Realisation Account is a
(a) Nominal Account
(b) Real Account
(c) Personal Account
(d) None of these
Answer:
(a) Nominal Account

Question 4.
The Account prepared at the time of dissolution of a partnership firm
(a) Revaluation Account
(b) P&L Adjustment A/c
(c) P&L Appropriation A/c
(d) Realisation Account
Answer:
(d) Realisation Account

Question 5.
The Realization account is closed by transferring the profit or loss to
(a) Partner’s Capital Accounts
(b) Partner’s Loan Account
(c) Bank Account
(d) Balance Sheet
Answer:
(a) Partner’s Capital Account

Question 6.
The Loan from Mrs.of a partner is credited to
(a) Her Capital Account
(b) Husband’s Capital Account
(c) Husband’s Loan Account
(d) Realisation Account
Answer:
(d) Realisation Account

Question 7.
On dissolution of partnership Firm, amount realised from unrecorded asset is credited to.
(a) Realisation A/c
(b) Re-Valuation A/c
(c) Capital A/c
(d) Goodwill A/c
Answer:
(a) Realisation A/c.

Question 8.
Entry for closing Provision for Bad debts at the time of dissolution of firm is_______.
Answer:
Provision for baddebt a/c Dr. To Realisation

Question 9.
Should you pass any entry for the payment of creditors worth Rs. 5000 on dissolution. If they accept stock of the same value? If yes, what is the journal entry?
Answer:
Creditors takes over stock of the same value. So no journal entry is need to be passed.

Question 10.
A firm is compulsorily dissolved when all partners or when all except one partner become insolvent – True or False
Answer:
True.

Question 11.
Unrecorded liabilities when paid by a partner are shown in_______.
Answer:
Debit of realisation a/c

Question 12.
On dissolution of a firm, bank overdraft is transferred.
(a) cash a/c
(b) bank a/c
(c) Realisation a/c
(d) capital a/c
Answer:
(c) Realisation a/c

Question 13.
On dissolution of the firm, partners capital accounts are closed through_______account.
Answer:
Cash/Bank Account

Plus Two Accountancy Dissolution of Partnership Two Mark Questions and Answers

Question 1.
What do you mean by Dissolution of partnership?
Answer:
Dissolution of partnership means termination of the existing partnership agreement between the partners. This may due to admission, retirement or death of a partner. In the case of dissolution of partnership, the firm continues to exist.

Question 2.
What is meant by Dissolution of firm?
Answer:
Termination of the partnership agreement between all the partners is known as dissolution of firm. In the case of dissolution of firm, the firm ceases to exist and the business of the firm is closed.

Question 3.
Why the balance at bank is not transferred to the Realisation A/c on the dissolution of a Partnership? Answer:
On the dissolution of a partnership the balance at bank is not transferred to the Realisation A/c be cause, there is no need to realise the same.

Question 4.
How will you settle firm’s debts and private debts of partner’s on the dissolution of a firm?
Answer:
Since the liability of partners is unlimited their private assets can be used to pay off the firm’s debts. But they will have the right to use their assets for paying their private debts first. They need to contribute only the remaining assets.

Question 5.
What is Realisation Account?
Answer:
Realisation Account is an account prepared at the time of dissolution of a partnership firm. It is to close the assets and liabilities and to find out the profit or loss and for the payment of liabilities.

Question 6.
What is the Accounting treatment of settlement with the creditors through transfer of an asset?
Answer:
Settlement with the creditors through transfer of assets require no entry. It is because the liability to the creditors has already been closed by transferring the same to realization account. The asset account also was closed by transferring to the same account.

Question 7.
How goodwill is treated on dissolution of the firm ?
Answer:
On dissolution of firm goodwill is treated like the other assets. It is transferred to realization account at its balance sheet amount. The amount realized for goodwill if any, is credited to realization account.

Question 8.
Complete the series

1. Sacrificing ratio: admission: Gainining ratio:?
2. Dissolution: Realisation A/c: Reconstitution: ?
3. Trading A/c: Profit and Loss A/c: Profit and Loss A/c:?
4. Balance of capital A/c: Balance sheet Balance of profit and loss appropriation a/c:?

Answer:

1. S.R. (Sacrificing Ratio) : Admission, G.R (Gaining Ratio) : Retirement
2. Dissolution : Realisation a/c Reconstitution: Revaluation a/c
3. Trading a/c : P&La/c, P & L a/c : P & L appropriation a/c
4. Balance of capital a/c : B/S, Balance of P & L appropciation a/c : Capital a/c.

Plus Two Accountancy Dissolution of Partnership Three Mark Questions and Answers

Question 1.
Which are the cases where a partnership is dissolved?
Answer:
Following are the cases in which a partnership is dissolved.

1. Change in the profit sharing ratio
2. Admission of a partner
3. Retirement, death of a partner
4. Insolvency of a partner
5. Completion of the venture for which it is formed
6. Expiry of the period

Question 2.
Toya and Soya are partners sharing profits and losses equally. They decided to dissolve the firm on 15^th March, 2005 which resulted in a loss of Rs. 30,000. The capital accounts of Toya and Soya was Rs. 20,000 and Rs. 30,000 respectively. The cash account showed a balance of Rs. 20,000. You are required to pass journal entries for

1. Transfer of loss to the capital accounts of partners.
2. Making final payments to the partners.

Answer:

Plus Two Accountancy Dissolution of Partnership Five Mark Questions and Answers

Question 1.
Distinguish between dissolution of partnership and dissolution of firm.
Answer:

Question 2.
How the accounts are settled on dissolution?
Answer:
On dissolution of a firm, the assets are realized (disposed) and the liabilities are paid off. Balance if any is shared among the partners. According to the Partnership Act, the following rules can be followed for the settlement of accounts.
1. Loss is to be paid first out of profits, next out of capital and last out of the private assets of partners in their ratio.

2. Amount realized from the assets of the firm shall be used in the following order

• Paying the realisation expenses
• Paying the liabilities to outsiders
• Paying the loans from partners
• Paying the capital of the partners
• Surplus if any is to be distributed to partners

Question 3.
Explain the different modes of dissolution of a partnership firm.
Answer:
Different modes of dissolution of a partnership firm are the following
(i) Dissolution by Agreement (section 40)
A firm is formed by an agreement between the partners. So it may be dissolved by the partnership agreement or with the consent of all the partners.

(ii) Compulsory Dissolution (Section 41)
A firm is dissolved compulsorily in the following cases

1. When all the partners or all except one become insane or insolvent.
2. When the business of the firm becomes illegal.
3. When all the partners except one retire.
4. When all the partners or all except one die.
   • When the period of partnership expires,
   • When the venture for which the partnership was formed becomes complete.

(iii) Dissolution on the happening of contingencies(42)
A firm may be dissolved on the happening of the following contingencies or events

1. On the death of a partner
2. On the insolvency of a partner

(iv) Dissolution by Notice (Sec 43)
If the partnership is a partnership at will, it can be dissolved by a partner by giving a notice to the other partners showing his will to dissolve the firm.

(v) Dissolution by Court (Sec 44)
A Court may issue an order to a partnership firm to dissolve the same on the suit of a partner in the following circumstances

1. If a partner becomes insane
2. If a partner becomes in capable of performing his duties
3. If a partner is found guilty of misconduct affecting the firm
4. If a partner intentionally and continuously commits breach of contract
5. If a partner transfers his interest in the firm to an outsider.
6. If the business of the firm cannot be carried on except at a loss
7. lf the court thinks it just and equitable to dissolve the firm

Question 4.
Differentiate between Realisation Account and Revaluation Account.
Answer:

Question 5.
What entry would you pass for the following transaction on the dissolution of a firm having partners Vishal and Rakesh.

1. An unrecorded asset realised Rs. 6200.
2. Dissolution expenses amounted to Rs. 3200.
3. Creditors already transferred to realisation account were paid Rs. 88,000.
4. Stock worth Rs. 5400 already transferred to realisation account was sold for Rs. 4100.
5. Profit on realisation Rs. 48000 to be distributed between partners, Vishal and Rakesh?

Answer:

Question 6.
Boby, Jestin, and Sudheer are in partnership in the ratio of 3:2:3. They have decided to dissolve the firm. On the date of dissolution total creditors were Rs.16,000; Bills discounted Rs. 2,650 during the year, has become a real liability which has to be paid, through this has not been recorded anywhere in the books of accounts. Their capital account balances were Boby Rs. 12000; Jestin Rs. 10000; Sudheer Rs. 8000 respectively. Boby advanced Rs. 14000 besides his capital account.
Find out

1. Total Sundry Assets
2. Realisation Account
3. Capital accounts of partners

Answer:
Balance Sheet

Realisation A/c

Capital A/cs

Plus Two Accountancy Dissolution of Partnership Eight Mark Questions and Answers

Question 1.
Appu and Chinku were partners in a firm sharing profits and losses in the ratio of 4: 3. Their Balance Sheet as on 31^st December 2005 was as follows.


The firm is dissolved as on the Balance sheet date. The assets were realized as follows.

Sundry Creditors were paid at a discount of 15%. The expenses on realisation amounted to Rs. 2,500. Pass journal entries and prepare ledger accounts on dissolution of the firm.
Answer:
Journal


Realisation Account

Partners’ Capital Account

Bank Account

General Reserve Account

Question 2.
The following is the Balance Sheet of Felix, Edwin, and Abel sharing profits and losses in the ratio of 2: 1: 1 as on 31^th March 2005.


The firm is dissolved. Sundry debtors realized Rs. 25.0 and stock Rs. 17,000. Trade mark and goodwill became valueless. Edwin agrees to discharge the bank loan. Creditors are paid Rs. 25,000 in full settlement, realisation expenses amounted to Rs. 6.0 paid by Felix. Pass journal entries and prepare ledger accounts.
Answer:
Journal

Realisation Account

Partner’s Capital Account

Bank Account

Notes:

• If nothing is given in the question, it is assumed that the assets(real) are realized and liabilities are paid off at their book value.
• Partners loan is not transferred to realisation account but paid directly.

Question 3.
Anu and Binu were partners sharing profits and losses in the ratio of 1/2 and 3/4. Their Balance Sheet as on 30^th June 2004 was as follows.

The firm is dissolved. Furniture and Machinery realized 10% less than their book values. Rs.20,000 is collected from debtors. Anu took over the stock at Rs. 25,000. The firm had an unrecorded liability on outstanding expenses Rs.2,500. Realisation expenses amounted to Rs. 2,000. Record journal entries and prepare ledger accounts.
Answer:
Journal

Realisation Account

Partner’s Capital Account

Bank Account

Question 4.
S.Raj, Narchison, and Boby are partners sharing profits in the ratio of 2:2:1, Whose ledger accounts on 31.03.06 reveal the following balances.

The firm dissolved on the above date.

1. During the years S.Raj withdraw from Bank Rs. 6,000 for his personal purpose which has not been brought into the records.
2. Rs. 10,000 was realised on account of unrecorded investments which was totally utilised for a liability on account of a claim payable to customers and the balance has been paid in cash.
3. Fixed Assets realised more than 10% of book value.
4. Sundry debtors could be collected only to the extend of 90% of the book value.

Prepare necessary accounts.
Answer:
1. Realisation A/c

Capital Account

Bank A/c

Question 5.
Mr. White and Mr. Black are partners sharing profits in the ratio of 3:2. They decided to close the firm and their Balance sheet is given below.
Balance sheet as on 31.03.2005

Assets realised as follows:
Building – 32,000, Debtors – 28,000, Furniture – 36,000 Liabilities settled as follows. Plant has been taken over by bank at Rs. 66,000 in respect of the loan granted by the bank and the rest has been paid in cash. Creditors are settled at Rs. 30,000. Realisation expenses came to Rs. 1,000 which have been met by Mr. Black. Prepare necessary accounts to dissolve the firm.
Answer:
Realisation A/c

Capital A/c

Bank A/c

Question 6.
Ali, Banu, and Cini were in partnership who have dissolved their firm on 31.10.2006 on which date their Balance sheet stood as follows.
Balance sheet as on 31.10.2006.

1. Bank passbook shows its balance to be Rs. 21,300. The difference is due to realisation of a claim directly credited to Bank Account.
2. Bills Receivable collected Rs. 300 less.
3. Stock has been utilized to settle the loan with Biju.
4. Unrecorded electronic equipments worth Rs.5000 has been utilised for settling the liability on account of Bills payable.
5. Land and Buildings was realised at Rs. 2,40,000.
6. All other assets were realised and liabilities were settled at book value.

Prepare necessary accounts.
Answer:
Realisation A/c

Capital A/cs

Cash A/c

Question 7.
Sam, Zen, and Jhony are in partnership sharing profits and losses in the ratio 3:2:1. Their Balance sheet as on 31^st December, 2004 was as follows.

The firm was dissolved on the above date with the following terms.

1. Building was taken over by Sam at book value and he agreed to discharge the creditors.
2. Accured interest was not collected, where as there was a contingent liability of Rs. 600 which was met.
3. Assets realised as follows: Plant – 25000, Stock – 5000, Debtors – 4600
4. Realisation expenses amounted to Rs. 600 You are required to prepare
   • Realisation account
   • Capital accounts
   • Cash account

Answer:
Realisation A/c

Capital A/cs

Cash A/c

Question 8.
Joe, Maggi, and Hassan were partners sharing profits and losses in the ratio of 1:2:2. Their Balance sheet as on 31 December 2004 was as follows.

The partners agreed to dissolve the firm on the following terms.

1. Assets realised as follows:
   Land and Building Rs. 1,20,000 Stock 40,000
   Accounts receivable 15,000
2. Expenses on dissolution is Rs. 3000
3. A creditor accepts office equipments for Rs. 7000 and the remaining creditors were paid in full by cheque.
4. The joint life insurance policy was surrendered for Rs.9000. Prepare realisation a/c, capital accounts and bank account.

Answer:
Realisation A/c

Capital A/c

Bank A/c

Students can Download Chapter 4 Reconstitution of a Partnership Firm – Retirement/Death of a Partner Questions and Answers, Plus Two Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Accountancy Chapter Wise Questions and Answers Chapter 4 Reconstitution of a Partnership Firm – Retirement/Death of a Partner

Plus Two Accountancy Reconstitution of a Partnership Firm – Retirement/Death of a Partner One Mark Questions and Answers

Question 1.
A partner who severs his connection with his firm is known as
(a) Retiring partner
(b) Outgoing partner
(c) Incoming partner
(d) None of these
Answer:
(b) Outgoing partner

Question 2.
On retirement of a partner
(a) The partnership is dissolved
(b) The firm is dissolved
(c) The business is dissolved
(d) None of these
Answer:
(a) The partnership is dissolved

Question 3.
The retiring partner is not liable for
(a) The losses of the firm
(b) The losses of the firm till the date of retirement
(c) The losses at the time of his retirement
(d) The losses after his retirement
Answer:
(d) The losses after his retirement

Question 4.
If the firm is not in a position to pay the amount due to the retiring partner, the amount is transferred to
(a) The retiring partner’s capital account
(b) The retiring partner’s loan account
(c) All the partner’s capital account
(d) None of these
Answer:
(b) The retiring partner’s loan account

Question 5.
The amount due to the deceased partner is transferred to
(a) His capital account
(b) His loan account
(c) His executor’s capital account
(d) His executor’s loan account
Answer:
(d) His executor’s loan account

Question 6.
When premium paid on Joint Life Policy is treated as an investment not as a business expense, it is transferred to
(a) Trading account
(b) Profit and Loss Account
(c) Joint Life Policy Account
(d) Joint Life Policy account and Balance Sheet
Answer:
(d) Joint Life Policy account and Balance Sheet

Question 7.
When partner retiring from the firm, the ratio relevant is_______.
(a) Sacrificing ratio
(b) Gaining ratio
(c) New ratio Gaining ratio
Answer:
(b) Gaining ratio

Question 8.
Write the narration of the given journal entry
Continuing Partners Capital A/c Dr.
To Retiring Partners capital A/c
(______________________)
Answer:
Give share of goodwil to retiring partners.

Question 9.
P/L Suspense A/c Dr.
To Deceased partners capital A/c What is the entry stands for?
Answer:
Credit of deceased partners share of profit in the in terim period.

Plus Two Accountancy Reconstitution of a Partnership Firm – Retirement/Death of a Partner Two Mark Questions and Answers

Question 1.
What do you mean by retirement of a partner?
Answer:
Withdrawal of a partner from a partnership firm either by giving a notice of retirement or with the consent of the other partners or as per the provisions of the partnership agreement is called retirement.

Question 2.
What do you mean by Gaining Ratio? How it is calculated?
Answer:
At the time of retirement of a partner, the ratio in which the continuing partners share the profit of out going partner’s profit is called gaining ratio.
Gaining ratio = New ratio – Old ratio.

Question 3.
How a retiring partner’s share of goodwill is compensated?
Answer:
A retiring partner has the right to get his share of goodwill, because the goodwill of the firm has been earned with his efforts too. So he should be compensated by the other partners in their gaining ratio.

Question 4.
What is the treatment of accumulated profits or losses on the retirement of a partner?
Answer:
The general reserve and accumulated profits or losses are transferred to all partners’ capital accounts in their profit sharing ratio. The general reserve and accumulated profits are transferred to the credit side of the account and the accumulated losses to the debit side.

Question 5.
What are the differences between retirement and death, from the accounting point of view?
Answer:

1. Retirement is a known thing, so usually takes place at the end of an accounting period, but death may take place at any time.
2. On retirement, a partner severs his connection with the firm Voluntarily. But in death, it is automatic.
3. On retirement, the amount due to the retiring partner is transferred to his Loan Account, while in death; the total amount due to the deceased partner is transferred to his Executor’s Loan Account.

Question 6.
X, Y, and Z were sharing profits in the ratio of 3:2:1. Z retires from the firm. X and Y decide to share future profits in the ratio of 7:5. Calculate the gaining ratio.
Answer:

Question 7.
A, B and C are partners sharing profits in the ratio of 5 : 3: 2. C retires and the goodwill is valued at Rs. 40,000. Give entries in the books of the firm regarding treatment of goodwill.
Answer:
C’s share of goodwill = 40,000 × \(\frac{2}{10}\) = Rs. 8,000

Question 8.
X, Y, Z are partners sharing profits in the ratio of 5:3:2. X retires and for this purpose goodwill is valued at Rs. 25,000. Continuing partners agree that their new profit sharing ratio shall be equal. Record necessary journal entry.
Answer:
Journal

NOTES:
1. Gain of partner = New share – Old share
Y = 1/2 – 3/10 = 5 – 3/10 = 2/10
Z = 1/2 – 2/10 = 5 – 2/10 = 3/10
Gaining ratio is 2 : 3

2. X’s share of goodwill = 25,000 × 5/10 = 12,500

Plus Two Accountancy Reconstitution of a Partnership Firm – Retirement/Death of a Partner Three Mark Questions and Answers

Question 1.
What are the problems that arise with regard to the accounting treatment on the retirement of a partner?
Answer:

1. Change in the Profit sharing ratio.
2. Adjustment of goodwill.
3. Treatment of accumulated profits and losses.
4. Revaluation of the assets and liabilities.
5. Calculation of the profit and loss up to the date of retirement.
6. Ascertainment of the total amount due to the retiring partner.
7. Payment of the amount due to the retiring partner.
8. Adjustment of the capitals of the continuing partners.

Question 2.
A, B and C were partners sharing profits in the ratio of 3: 5: 7. C retires and his share is taken up by A and B in the ratio of 3: 2. Find out the new profit sharing and gaining ratio of A and B.
Answer:
New share = Old share + Gain

A’s gain is 3/5 of 7/15 = 21/75
B’s gain is 2/5 of 7/15 = 14/75

Gaining ratio is the proportion in which they have acquired C’s share of profit, i.e., 3 : 2. This can be checked by working out in the following way.
Gaining ratio = New share – old share

Question 3.
Observe the following journal entry which has been passed at the time of retirement of Ganga Prasad. The other partners were Sheena and Rajani.

1. Which account would you prepare to share the profit on revaluation?
2. Prepare that account and give a journal entry to share the profit?

Answer:
1. Revaluation a/c

Journal Entry

Question 4.
X, Y, and Z are partners in a firm and they close their books on December 31 every year. They are sharing profits and losses in the ration of 3: 2: 1. The partnership deed provides that if a partner retires from the firm during the course of an accounting year, his share of profit from the date of last balance sheet to the date of retirement should be calculated on the basis of the average profits of the last three completed years.
On 1^st April 2004 Y retired from the firm. The profits of the firm during the years 2001, 2002 and 2003 were Rs. 12,500, Rs. 8,500 and Rs. 6,000 respectively. Write the journal entry to record the share of profit of the retiring partner for the year 2004.
Answer:

(Being Y’s share of profit for 2004 brought into A/c)
Notes: Profits for the last 3 years
= 12,500 + 8.500 + 6,000 = Rs. 27,000
Average profit = 27,000/3 = Rs. 9,000
Profit from the date of last balance sheet to the date . of retirement.
= 9,000 × 3/12 = Rs. 2,250
Y’s share thereof = 2,250 × 2/ 6 = Rs. 750.

Question 5.
Mr. Raj died on 25.08.2006 who was an active partner in a firm. The other partners were Mr. Das and Mrs.Das. The books of accounts reveal the following:

Interest on loan payable to Raj upto the date of death Rs. 1,000
Value of goodwill estimated at Rs. 24,000
Calculate the amount due to the legal heirs of Mr.Raj.
Answer:
Raj’s Capital A/c

R’s loan to the firm: 20000
R’s interest on loan: 1000
Total amount due to R’s legal heirs = His capital a/c
balance and R’s loan to the firm and interest on loan = 26000 + 20000 + 1000 = 47000.

Question 6.
X, Y, and Z are partners in a firm. Y retires from the firm on 1^st January, 2002. On his date of retirement, Rs. 60,000 is due to him. X and Z promise to pay in three equal annual instalments together with interest at 12% per annum. Prepare Y’s loan account for the three years.
Answer:
Dr. Y’s Loan Account Cr.

Amount of instalment each year = 60,000/3 = 20,000
Amount paid each year = Rs. 20,000 + Interest.

Question 7.
X,Y, and Z were partner sharing profit in proportion to 5:3:2. Good will does not appear in the books, but it is agreed to be worth Rs. 1,00,000. X retires from the firm and Y and Z decide to share future profits equally. You are required to make adjustment entry for good will without opening good will account at all. Show your working clearly.
Answer:
X’s share of goodwill adjusted through capital accounts in the gaining ratio.
Old ratio = 5:3:2
New ratio =1:1
Gaining ratio = New ratio – Old ratio
Gain of Y = 1/2 – 3/10 = 2/10
Gain of Z= 1/2 – 2/10 = 3/10
Gaining ratio of Y and Z = 2 : 3
Value of goodwill of the firm = 1,00,000
X’s share of goowill = 1,00,000 × 5/10 = 50,000
Journal Entry:

[X’s share of goodwill adjusted through the capital accounts of remaining partners in the gaining ratio of 2:3].

Question 8.
Aby, Suby and Minu are partners sharing profits in the ratio of 5:3:2. Minu retired on 31.09.06. The capital account balance and share of reserve due to Minu together amounted to Rs. 1,80,000. But Aby and Suby agreed to pay him Rs. 2,40,000. The new profit sharing ratio of Aby and Suby have been fixed at 3:2.

1. Why has Minu been paid over and above the actual amount due to him?
2. Give a journal entry to record this through capital a/c adjustments.

Answer:
A:S:M = 5:3:2
New ratio of A & S = 3:2
∴ Gaining ratio = New ratio – Old ratio

Gaining ratio = 1:1
Amount payable to Minu = 2,40,000
Capital + Reserve of Minu = 1,80,000
∴ Share of Goodwill due to Minu
= 240000 – 180000 = 60000

Question 9.
Debee, Sedee and Nedee are in partnership, who were sharing profits in the ratio of 3:2:1. On 31.03.05, Nedee left the firm as per agreement. The following details are available.
Balance Sheet:

1. Depreciate fixed assets @10%.
2. Only General Reserve is to be credited to the extent of Nedee’s share through capital adjustment of the partners.
3. Receivables are sold to a debt collection agency at Rs. 5,400/-

Nedee’s accounts are to be settled soon either by paying off or bringing in necessary cash as the case may be. Prepare the necessary a/c to show the amount due to Nedee.
Answer:
Capital A/cs

Amount to be brought in by Nedee is Rs. 5,800.

Plus Two Accountancy Reconstitution of a Partnership Firm – Retirement/Death of a Partner Five Mark Questions and Answers

Question 1.
How will you calculate the amount payable to a retiring partner?
Answer:
If retirement takes place on the closing date of the accounting year ascertainment of profit or loss is easy. But if the retirement occurs during an accounting year, the profit or loss from the date of last Balance Sheet to the date of retirement is also to be determined. Partnership deed provides the method of calculating the profit or loss of that period. It is calculated by any of the following methods.

1. On the basis of the last year’s profit.
2. On the basis of the average profit of a certain » number of past years.
3. By providing interest on the capital .of the retiring partner at. a certain rate.
4. By finding out the correct profit till the retirement date.
5. On any other basis as provided in the partnership agreement.

The profit or loss calculated above is only an estimate. So the journal entry for the same is; In case of profit.
Profit and Loss Suspense A/c Dr. To Retiring Partner’s Capital A/c:
P&L suspense A/c is shown on the asset side of the Balance Sheet prepared immediately after retirement. At the end of the accounting year; it is closed by transferring to Profit and Loss Account. Reverse is done in case of loss.

For calculating, the total amount payable to the retiring partner, his capital account is prepared.
It is started with the balance in the same on the last.
Balance Sheet and credited with:

• His share of goodwill.
• His share of revaluation profit.
• His share of accumulated profits and reserve.
• His share of profits upto the retirement date since the last Balance Sheet.
• Interest, salary or commission if any due to him.

The capital account is debited with:

• His share of accumulated losses.
• His share of revaluation loss.
• His drawings if any during the period.
• Interest on such drawings.
• His share of loss upto the retirement date since the last Balance Sheet.

If the account shows a credit balance, it is the total amount payable to him. On the other hand, if the account has a debit balance, it represents the amount payable by him to the firm.

Question 2.
X, Y, Z are partners in a firm sharing profits and losses equally. X retired from the firm on which date the balance sheet stood as follows:
Balance Sheet as on____

On the date of retirement it was found that

1. Patents have on value.
2. Furniture is to be depreciated by 15%.
3. Machinery is to be brought down to Rs. 10,000.

Pass the necessary journal entries to give effect to the revaluation of assets and liabilities at the time of retirement.
Answer:
Journal:

Question 3.
A, B and C were Partner’s sharing Profits and losses in the ratio of 3:2:1. Their capitals were as under as per the balance sheet as on 31-Dec-2010.
A-Rs. 30,000; B-Rs. 20,000; C-Rs. 15,000. On 31 March 2011, C died, and you are asked to prepare deceased partners Capital account after considering the following facts.

1. Capital carried interest at 12% p.a.
2. C’ drawings from 1^st Jan 2011 to the date of his death amounted to Rs. 4,500.
3. C’s share of Profits for the portion of current financial year for which he lived was to betaken at the sum calculated on the average Profit of the last three completed years and good will was to be raised on the basis of two years Purchase of the average Profit of those three years.

The annual Profits were Rs. 19,000, Rs. 16,000 and Rs. 19,000 respectively. Show C’s Capital account, Answer:
C’s Capital A/c

Working Note:
1. Share of profit to the date of death.
Average profit for past 3 years

Therefore, C’s share (1/6) for 3 months = 18.000 × 1/6 × 3/12 = 750.

2. Goodwill calculation
Average Profit = 18,000
Goodwill = Average profit × 2’years purchase = 18,000 × 2 = 36,000
C’s share of goodwill = 36,000 × 1/6 = 6,000
C’s share of goodwill adjusted through the capital account of remaining partners in the gaining ratio of 3:2.

Question 4.
X, Y, and Z were partners in a firm with capitals of Rs. 15,000, Rs. 9,500, Rs. 10,000 respectively and sharing profits in proportions of 1/2, 1/4 and 1/4. On 31st December 2005, Y retires and for the purpose of his retirement, the goodwill of the firm has been valued at Rs. 12,000. Pass the necessary entries assuming that ’Y’ has been paid and show the Capital Accounts of all partners.
Answer:
Capital A/c

Working Note:
Old ratio = 1/2 : 1/4 : 1/4
1 × 2/2 × 2 = 2/4

Goodwill of the form = 12,000
Y’s share of goodwill = 12,000 × 1/4 = 3,000
Gaining ratio = New ratio – Old ratio
X’s Gain = 2/3 – 2/4 = 8 – 6/12 = 2/12
Z’s Gain = 1/3 – 1/4 = 4 – 3 /12 = 1/12
Gaining ratio = 2 :1
Journal:

Question 5.
X, Y, and Z carried on business in partnership, shar¬ing profits in the ratio 3:2:1. The balance sheet on 31st December, 2003 showed their capitals to be Rs. 8,400; Rs. 6,800 and Rs. 7,400.
On 31^st March, 2004 X died. Write journal entries and prepare an account for presentation to his legal representatives having regard to the following facts:

1. Capital earned interest at 5 percent per annum.
2. X’s drawings from 1st January, 2004 to the date of his death amounted to Rs. 800; interest on drawings for the period Rs. 45.
3. X’s share of profit for the portion of current financial year for which he lived was to be taken a sum calculated on the average of the last three completed years.
4. Goodwill was to be raised on the basis of two year’s purchase of average profits of those three years.

The annual profits for the three years were Rs. 4,800; Rs. 3,500 and Rs. 4,300 respectively.
Answer:
Journal:



Notes
1. Share of profit to date of death:
Average profit for past 3 years
\(\frac{4,800+3,500+4,300}{3}\)
= Rs. 4,200
∴ X’s share (3/6) for 3 months 3
= 4,200 × 3/6 × \(\frac{3}{12}\) = RS. 525 12.

2. Calculation of goodwill:
Average profit for past 3 years = Rs. 4,200
2 years purchase = Rs. 4,200× 2 = Rs. 8,400
X’s share (3/6) of goodwill = Rs. 8,400 × 3/ 6 = Rs. 4,200.

Question 6.
Heisal, Roy and A Gomez are in partnership sharing profits in their capital ratio. The Balance Sheet on 15^th March, 2006 is given below.
Balance sheet

Further information on retirement of Roy on 15-6-06.

Interest on capital @ 5% p.a.
Salary to Roy Rs. 300 p.m.
The firm had a fixed deposit worth Rs. 3000 which has not accounted so far, has to be brought into the books. Marketable scrips were valued at Rs. 23,000. Prepare:

1. Profit and Loss appropriation a/c
2. Capital A/c’s
3. Balance sheet

Answer:
P/L Appropriation A/c


Revaluation A/c

Capital A/c

Balance Sheet

Plus Two Accountancy Reconstitution of a Partnership Firm – Retirement/Death of a Partner Eight Mark Questions and Answers

Question 1.
The Balance Sheet of X, Y. Z on 31st March 2003 is given below.
Balance sheet

They were sharing profits and losses in the ratio of 2:2:1. Y decided to retire from the firm. It was agreed that:

1. X and Z would share the profit in the ratio of 5:3
2. Goodwill was valued at Rs. 1,05,000
3. Machinery to be taken at Rs. 75,000
4. Buildings should be valued at Rs. 1,50,000
5. The value of stock should be Rs. 30,000
6. An amount of Rs. 1,500 should be written off as bad debt.

Pass the necessary journal entries and prepare the balance sheet of the new firm.
Answer:
Note: Calculation of gaining ratio
X = 5/8 – 2/5 – 25 – 16/40 = 9/40
Z = 3/8 – 1/5 = 15 – 8/40 = 7/40
Therefore, gaining ratio = 9:7

Journal



Balance Sheet of X and Z as on 1^st April 2003

Question 2.
Abey, Neha, and Anil are partners, who share profits and losses in 5:3:2 ratio. The following information is extracted from the books of accounts on 31.03.06.

On the above date Anil decided to retire from the firm as agreed upon. Fixed assets to be revalued at Rs. 86,000. Average profit calculated based on the past 5 year was Rs 15,000. Ascertain the amount due to the retiring partner.
Answer:
Profit and Loss Appropriation A/c

Anil’s Capital A/c

Working Note
Here, Goodwill is caculated on the basis of capitalization of Average profit method.
∴ Goodwill = Total value of business – Net tangible Asset.
Total value =

= 15,000 × \(\frac{100}{10}\) = 1,50,000
Net tangible asset = Fixed Asset + Current Asset
= 86,000 + 24,000 = 1,10,000
Goodwill = 1,50,000 – 1,10,000 = 40,000
Anil’s share of goodwill = 40,000 × 2/10 = 8,000
Anil’s share of goodwill adjusted through capital accounts in the gaining ratio.
Old ratio = 5:3:2
New ratio = 5:3
Gaining ration = 5 : 3
Abey’s share = 8,000 × 5/8 = 5,000
Neha’s share = 8,000 × 3/8 = 3,000
Journal entry:

Question 3.
P, Q and R partiners sharing profits and losses in the ratio 3:2:1. The Balance sheet as on 31^st December 2003 is given below:

On 31^st March 2004, Q decided to retire from the business due to ill-health subject to the following conditions.

1. That the goodwill should be valued at two year’s. purchase of the average profits of the preceding three years. The profits for the three preceding years were, 2001 – Rs. 9,000, 2000 – Rs. 15,000 and 2003-Rs. 12,000.
2. The profits for the three months ending 31 st March, 2004 be estimated on the basis of the profits for the year 2003.
3. That the motor car is to be given to Q, at a value of Rs. 16,000 and the balance due to him is to be paid immediately in cash by bringing the required amount by P and R in their profit sharing ratio which is 3: 1.

Answer:
Revaluation a/c

Calculation of Goodwill
Average profit of 3 years
\(=\frac{9,000+15,000+12,000}{3}\) = 12,000
Total goodwill = 12,000 × 2 = Rs. 24,000
Q’s share = 24,000 × 2/6 = Rs. 8,000
Q’s share of profit = 12,000 × 3/12 × 1/3 = 1,000
Balance Sheet as on 31^st March 2004

Question 4.
The balance sheet of X, Y, and Z on 31^st December, 2003, the date of X’s retirement was as follows:
Balance Sheet

The following terms have been agreed upon:

• Goodwill was valued at Rs. 18,000.
• The value of land and buildings should be appreciated by Rs. 10,000
• Plant and Machinery should be reduced to Rs. 23,000.
• Create provision @ 5% on debtors for bad and doubtful debts and Rs. 700 on creditors.
• The entire sum payable to X is to be brought by Y and Z in such a manner that their capital accounts are in proportion to their profit sharing ratio which is to be equal.

Prepare:

• Revaluation account.
• Partner’s capital accounts
• Bank account, and
• Balance sheet after X’s retirement.

Answer:



Balance Sheet of Y and Z as on 1^st January 2004

Notes:
Calculation of goodwill
Total goodwill ‘ = 18,000
X’s share 1/3 = 18,000 × 1/3 = Rs. 6,000
Working Notes:
(i) Gaining Ratio:
X: 3/4 – 3/6 = 9 – 6/12 = 3/12
Z: 1/4 – 1/6 = 3 – 2/12 = 1/12
Y’s share of goodwill of Rs. 12,000 (Rs. 36,000 × 2/6) will be contributed by X Rs. 9,000 and Z Rs. 3,000

(ii) Since the new profit sharing ratio between X and Z being 3:1, they will have to maintain their capitals at Rs. 90,000 and Rs. 30,000 respectively.

Question 5.
Anil, Bhanu, and Chandu were partners in a firm sharing profits in the ratio of 5:3:2. On March 31, 2007, their Balance Sheet was as under:
Books of Anil, Bhanu, and Chandu Balance Sheet as on March 31, 2007

Anil died on October 1, 2007. It was agreed between his executors and the remaining partners that:

1. Goodwill to be valued at 21/2 year’s purchase of the average profits of the previous four years which were:
   Year2003-04-Rs. 13,000, Year 2004-05-Rs. 12,000
   Year 2005-06-RS.20,000, Year 2006-07 – Rs. 15,000
2. Patents be valued at Rs. 8,000. Machinery at Rs. 28,000 and Building at Rs. 25,000
3. Profit for the year 2007-08 be taken as having accrued at the same rate as that of the previous year.
4. Interest on capital be provided at 10% p.a.
5. Half of the amount due to Anil be paid immediately.

Prepare Revaluation Account, Anil’s Capital Account and Anil’s Executor’s Account as on October 1, 2007.
Answer:
Revaluation Account

Anil’s Capital Account

Anil’s Executor’s Account

Working Note:
1. Goodwill = 21/2 years purchase × Average Profit
Average Profit

Anil’s share of Goodwill = \(\frac{5}{10}\) × Rs.37,500 = 18750

2. Profit from the date of last balance sheet to date of death (April 1,2007 to October 1,2007) = 6 months
Profit for 6 months = Rs.15,000 × \(\frac{6}{12}\) = Rs.7,500
Anil’s share of profit = Rs.7,500 × \(\frac{5}{10}\) = Rs.3,750.

3. Interest on Capital
(April 1, 2007 to October 1, 2007)
= Rs. 30000 × \(\frac{10}{100}\) × \(\frac{6}{12}\) = Rs. 1,500.

Students can Download Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner Questions and Answers, Plus Two Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Accountancy Chapter Wise Questions and Answers Chapter 3 Reconstitution of a Partnership Firm – Admission of Partner

Plus Two Accountancy Reconstitution of a Partnership Firm – Admission of Partner One Mark Questions and Answers

Question 1.
Total value of business-Net tangible assets’ is the value of goodwill under.
(a) Superprofit method
(b) Present value of super profit method
(c) Capitalization of average profit method
(d) Weighted average profit method
Answer:
(c) Capitalization of average profit method.

Question 2.
Which of the following does not lead to reconstitution of a partnership firm?
(a) Admission of a new partner
(b) Retirement of a partner
(c) Death of a partner
(d) Dissolution of a partnership
Answer:
(d) Dissolution of a partnership

Question 3.
Change in agreement (relationship among the partners) lead to
(a) Reconstitution
(b) Dissolution
(c) Reconstruction
(d) Amalgamation
Answer:
(d) Reconstitution

Question 4.
Change in profit sharing ratio of the existing partners result in
(a) Gain to all partners
(b) Sacrifice to ail partners
(c) Gain to some partners and sacrifice to others
(d) None of these
Answer:
(c) Gain to some partners and sacrifice to others.

Question 5.
Unless otherwise mentioned, sacrificing ratio will be
(a) Equal ratio
(b) New ratio
(c) Old ratio
(d) None of these
Answer:
(c) Old ratio

Question 6.
The profit or loss arising from revaluation of assets and liabilities is transferred to
(a) Old partners’ capital account
(b) All partners’ capital account
(c) New partners’ capital account
(d) Profit and Loss Appropriation account
Answer:
(a) Old partners’ capital account

Question 7.
The profit or loss on revaluation is transferred to the old partners capital a/c in
(a) Old ratio
(b) Sacrificing ratio
(c) New ratio
(d) In the ratio of capital
Answer:
(a) Old ratio

Question 8.
At the time of admission of a new partner the re-serves and accumulated profits in P & L account is transferred to
(a) Profit and Loss appropriation account
(b) Profit and Loss Adjustment Account
(c) Old Partner’s capital account
(d) Revaluation account
Answer:
(c) Old partner’s capital account

Question 9.
The amount of goodwill brought in by the new partner is shared among the old partners in
(a) Old ratio
(b) Sacrificing ratio
(c) New ratio
(d) None of these
Answer:
(b) Sacrificing ratio

Question 10.
The goodwill brought in kind (assets) by the new partner is transferred to
(a) Revaluation account
(b) Profit and Loss Account
(c) Sacrificing partners’ capital account
(d) All partners’ capital account
Answer:
(c) Sacrificing partners’ capital account

Question 11.
When the new partner is not able to bring his share of goodwill, his account will be
(a) Debited
(b) Credited
(c) Omitted
(d) Closed
Answer:
(a) Debited

Question 12.
In partnership, a minor
(a) Cannot be a partner
(b) Can be a partner
(c) Can be admitted only to the benefits of a partnership.
(d) Can be a partner and share profit & losses along with other partners.
Answer:
(c) Can be admitted only to the benefits of a partnership.

Question 13.
Complete the following on the basis of the hint given

1. Premium – Sacrificing ratio.
2. Revaluation profit – _______.

Answer:
Old profit sharing.ratio

Question 14.
Joy’s capital A/c
Dr. Saju’s capital A/c Dr
To Profit and Loss A/c.
What is the entry stands for?
Answer:
Accumulated losses transferred to old partners capital a/c.

Plus Two Accountancy Reconstitution of a Partnership Firm – Admission of Partner Two Mark Questions and Answers

Question 1.
‘Goodwill is an asset, but is not visible’. Describe.
Answer:
Goodwill is the value of the reputation of a firm. As such it is an asset to the firm. But it is an intangible asset and is not visible.

Question 2.
A firm has an average profit of Rs. 50,000 during the last certain years. The normal rate of return is 10%. The firm has net tangible assets of Rs. 3,00,000. Calculate the value of goodwill using capitalization method.
Answer:
Average profit = Rs. 50,000
Normal rate of return = 10%
Capitalised value of
average profit = \(\frac{50,000 \times 100}{10}\) = Rs. 5,00,000
Goodwill = Capitalised value of average profit – Total of net tangible assets.
= Rs. 5,00,000 – 3,00,000 = Rs. 2,00,000.

Question 3.
When a new partner is admitted into a firm?
Answer:
Inclusion of a new partner into an existing firm is called admission of a partner. A new partner is admitted, when a firm needs more capital, managerial skill, etc.

Question 4.
What are the rights acquired by a new partner?
Answer:

1. Right to share the assets of the firm – For this the new partner has to bring a certain amount of capital.
2. Right to share the profits of the firm – For this he has to bring his share of goodwill.

Question 5.
What do you mean by sacrificing ratio?
Answer:
At the time of admission of a new partner, the old partners have to sacrifice a certain portion of their profits for the incoming partner. The ratio in which they give up or sacrifice their profit is called sacrificing ratio.

Question 6.
What is a revaluation account?
Answer:
Revaluation account is a nominal account prepared at the time of admission of a new partner. This is prepared to find out the profit or loss on revaluing the assets, if they are overstated or understated.

Question 7.
What is meant by premium or goodwill?
Answer:
At the time of admission, the new partner has to bring in a certain amount for getting a share in future profit. This amount is called premium or goodwill.

Question 8.
What treatment is made of accumulated profits and losses on the admission of a new partner?
Answer:
Accumulated profits and losses are distributed amongst the old partner’s in their old profit sharing ratio. The new partner should not share such profits or losses because these arose before his admission.

Question 9.
What is Memorandum Revaluation Account?
Answer:
A memorandum revaluation account is prepared when the partners decide to record the effect of revaluation of assets and liabilities without affecting the old figures of assets and liabilities in the balance sheet.

Question 10.
A new partner is admitted into a firm; but he is not in a position to bring his share of goodwill. What the firm will do?
Answer:
When the new partner is not able to bring his share of goodwill, his account is debited and the sacrificing partners capital account is credited. The following is the journal entry.
New Partners’ Capital A/c Dr.
To Sacrificing Partners’Capital A/c.

Question 11.
An equipment having a book value of Rs.2,600 was sold at Rs. 3,000 on the date of admission of a partner.

1. How much amount will be credited to Revaluation A/c?
2. Give journal entry for the above.

Answer:

Question 12.
In connection with the admission of Mr.Santhosh Kumar as equal partner, one of the existing partners
of the firm Mrs. Sreema has taken over the plant and equipments worth Rs. 15000 at Rs. 18000 on the date of admission. Give a journal entry to this effect.
Answer:

Question 13.
X and Y are partners sharing profits and loses in the ratio of 2:1. They admit Z into the firm for a fourth, share. Calculate new ratio and sacrificing ratio.
Answer:
Old ratio = 2:1
New ratio =2:1:1

Here old ratio and sacrificing ratio are the same.

Question 14.
P and Q are partners in a firm sharing profits in the ratio of 5 : 3. They admit R for 1/6 share. The total goodwill of the firm is Rs. 50,000. Goodwill existing in the books is Rs. 25,000. Pass the journal entry for the share of goodwill to be brought in by R.
Answer:
Amount of goodwill to be brought in by R
= 1/6 of (50,000 – 25,000)
= 1/6 of 25,000
= 1/6 × 25,000 = Rs. 4,167
The Journal entry is

Plus Two Accountancy Reconstitution of a Partnership Firm – Admission of Partner Three Mark Questions and Answers

Question 1.
Calculate the value of goodwill at 2 years, purchase from the following 3 years average profits.

Answer:
Average profit = \(\frac{27,000+28,000+29,000}{3}\)
= 28,000
Goodwill = 2 yeas purchase of the average profit = 2 × 28,000 = 56,000.

Question 2.
A business has earned average profits of Rs. 1,00,000 during the last few years and the normal rate of return in a similar business is 10%. Ascertain the value of goodwill by capitalisation of superprofits method, given that the value of net assets of the business is Rs.8,20,000.
Answer:
Goodwill = superprofit × 100/ Normal rate of return Super profit = Actual/Average profit – Normal profit Normal profit = Capital employed × Normal rate of return = 820000 × 10/100 = 82000
Super profit = 100000 – 82000 = 18000
Goodwill = 18000 × 100/10 = Rs. 180000.

Question 3.
Ram and Rahim are partners in a firm sharing profits in the ratio of 3:2. Their capitals were Rs.80000 and Rs.50000 respectively. They admitted Syam on January 1^st 2014 as a new partner for 1/5 share in the future profits. Syam bought Rs.60,000 as his capital. Calculate the value of goodwill of the firm.
Answer:
Syam’s capital = 60000
Syam’s share of capital = 1/5
Total capital of new firm = 60000 × 5/1 = 300000
Total capital of Ram, Rahim & Syam
= 80000 + 50000 + 60000 = 190000
Goodwill of the firm = 300000 – 190000 = 110000
Syam’s share of goodwill = 110000 × 1/5 = Rs.22,000.

Question 4.
L and M are partners sharing profits in the ratio of 5: 4. On 1^st July 2005 they admit N into the firm for 1/10 share in future profits. N contributed the following assets for his capital and share of goodwill. Stock-in-trade Rs. 50,000, Furniture Rs. 25,000 and Land and Buildings Rs. 75,000 and machinery Rs. 50,000. Goodwill of the firm was valued at Rs. 45,000. Give the journal entries.
Answer:
Journal

Question 5.
P and Q are partners sharing profits and losses in the ratio of 3:2. They admit R into the firm with 2/5 share which he gets equally from P & Q. Calculate the new ratio and sacrificing ratio.
Answer:
Old ratio = 3 : 2 = 3/5 : 2/5
R’s share = 2/5 ie. (1/5 from P 1/5 from Q)

Here the sacrificing ratio is equal (1: 1) as R gets equally from P&Q.

Question 6.
Which are the matters on which accounting adjustments are required at the time of the admission of a new partner?
Answer:
At the time of the admission of a new partner, accounting adjustments are required on the following

1. Capital of the new partner
2. Ascertainment of profit sharing ratios – new and sacrificing
3. Revaluation of assets and liabilities
4. Adjustment of accumulated profits (including reserves) or losses.
5. Calculation of goodwill
6. Adjustment of capital accounts of partners.

Question 7.
Explain the premium method of treatment of goodwill.
Answer:
Under premium method, the new partner brings his share of goodwill in cash. The amount so brought in by him is shared among the old partners in the sacrificing ratio.
The journal entries here are:

1. Cash a/c Dr.
   To premium for goodwill a/c (cash brought in by the new partner for goodwill)
2. Premium for good will a/c
   To old partners capital a/c (Goodwill shared among the old partners)

“If the amount of premium is paid privately to the old partners, no need of entering the same in the books.”

Question 8.
A new partner instead of bringing his share of good¬will in cash brought the same as assets. How will you treat it?
Answer:
When an incoming partner brings his share of goodwill in kind (as assets), the assets account will be debited. Credit is given to premium for good will account with the share of goodwill and new partner’s capital account with the share of capital.
The journal entries here are:

1. Assets a/c Dr. To New partner’s Capital A/c
   To Premium (goodwill) A/c (Assets brought in by the new Partner)
2. New partners’ capital a/c Dr. To Sacrificing partners Capital A/c
   (Share of goodwill brought in by the new partner transferred to old partners capital).

Question 9.
A and B are partners sharing profits and losses equally (1:1). They admit C for 1/6 share in future profits. Calculate the new ratio and sacrificing ratio.
Answer:
Old Ratio = 1:1 C’s
Share = 1/6
Remaining portion = 1 – 1/6 = 5/6
This 5/6 is to share among A & B in their old ratio.
So their new shares will be
A’s share 1/2 of 5/6 = 1/2 × 5/6 = 5/12
B’s share 1/2 of 5/6 = 1/2 × 5/6 = 5/12
The new ratio between

Old ratio and sacrificing ratio are the same here.

Question 10.
Roshi and Riya are partners sharing profits and losses in the ratio of 5 : 3. Maria is admitted into the firm. Roshi sacrifices 1 /5 of her share and Riya sacrifices 1/6 in favour of Maria. Calculate the new ratio.
Answer:
Old ratio = 5:3

Question 11.
Ansa and Valsa are partners in a firm sharing profits and losses in the ratio of 3:2. They admit Sona into the partnership fora sixth share for which she brings in Rs. 35000 as capital and Rs. 20,000 for good will. Pass journal entries in the books of the firm.
Answer:
Journal

Note: Old ratio itself is the sacrificing ratio here as the ratio between old partners is not changed.

Question 12.
X and Y are partners sharing profits and losses equally. They admit Z for a third share for which he brings Rs. 15,000 for good will. X and Y withdrew the full amount of goodwill immediately. Pass journal entries in the books of the firm.
Answer:
Journal

Question 13.
P and Q are partners sharing profits and losses in the ratio of 2:1. They admit R for a third share. He brings in Rs. 30,000 for goodwill, half of which is withdrawn by the old partners. Pass journal entries.
Answer:
Journal

Question 14.
X and Y are partners sharing profits in the ratio of 4:3. Z is admitted for 1/6 share in profits. Their capitals were Rs. 50,000 and 40,000 respectively. It is also agreed that Z’s capital should be proportionate to her profit sharing ratio. Find out the amount to be brought in by Z as capital.
Answer:
Share of Profit of Z = \(\frac{1}{6}\)
Share of profit of X and Y = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
Total capital of X & Y= 50,000+ 40,000 = 90,000
Capital of X and Y for 5/6 share = 90,000
∴ Total capital of X,
Y and Z = 90,000×6/5 = 1,08,000
∴ Capital to be brought in by Z = 1,08,000 – 90,000
or 1,08,000 × 1/6 = 18,000.

Question 15.
A trading firm has in its ledger book, an accumulated profit balance of Rs. 30,000 in general reserve. The partners Smitha, Neha, and Anila who share profits in the ratio of 3:2:1. They have decided to become equal partners. Show journal entry to adjust the existing general reserve through capital accounts.
Answer:
Old ratio = 3:2:1 = 3/6 : 2/6: 1/6
New ratio = 1:1:1 = 1/3:1/3:1/3


Share of general reserve = 30,000 × \(\frac{1}{6}\) = 5000
Anila’s capital A/c Dr. 5000
To Smitha’s capital 5000
(Being goodwill adjusted between capital a/cs of Smitha and Anila, Neha’s profit sharing ratio remains
unchanged (Neha’s old ratio = \(\frac{2}{6}\) – \(\frac{1}{3}\), New ratio = \(\frac{1}{3}\)).

Question 16.
The Profits of firm for the last five years were as follows

Calculate the value of goodwill on the basis of 3 year’s Purchase of weighted average profit based on weights 1, 2, 3, 4, & 5 respectively to the profits for 2002, 2003, 2004, 2005 and 2006.
Answer:

Weighted Average Profit = \(\frac{3,48,000}{15}\) = 23,200
Goodwill = 23,200 × 3 = 69,600

Plus Two Accountancy Reconstitution of a Partnership Firm – Admission of Partner Five Mark Questions and Answers

Question 1.
List the factors affecting goodwill.
Answer:
Following are the important factors affecting the goodwill of a firm

1. Nature of business -A firm producing goods having constant demand will have more goodwill.
2. Suitable location – A firm which is situated in a favourable locality will have more goodwill.
3. Efficiency of management – if the management of a firm is efficient, it wil have high goodwill.
4. Running period – a firm which is running for a long period of time, will have more goodwill.
5. Requirement of capital – if a firm requires a lesser amount of capital, it will have high goodwill.
6. Market situation – if competition in the market is limited, it helps a firm to have more goodwill.

Question 2.
Describe the methods of valuing goodwill.
Answer:
The following are the common methods used for valuing good will

1. Average profit method/simple average profit method.
2. Super profit method
3. capitalisation method.

1. Average Profit method:
Under this method, the goodwill is valued at agreed number of years purchase of the average profits of the past few years.
Goodwill = Average profits × No. of years purchased.
Average profit = \(\frac{\text { Total profits }}{\text { No. of years }}\)
Weighted Average Profit method:
Goodwill = weighted Average Profit × No. of years purchase. Weighted average is based on specified weights like 1, 2, 3, 4 for respective year’s profit.

2. Superprofit Method:
Under this method, goodwill is calculated by multiplying the super profit with the agreed num-ber of years.
Goodwill = Super Profit × No.of years purchase
Super Profit = Actual or Average Profit – Normal Profit
Normal profit = Capital employed × Normal Rate of Return
Capital employed = Total Assets – Total Liabilities or outside liabilities
Average profit = \(\frac{\text { Total profits }}{\text { No. of years }}\).

3. Capitalisation Method:
Under this method, the goodwill can be calcu-lated in two ways

• by capitalising the average profits.
• by capitalising the super profits.

(i) Capitalisation of average profits
Under this method, the value of goodwill is calculated by deducting the capital employed (net assets) in the business from the capitalized value of average profits on the basis of normal rate of return.
Good will = capitalised value – capital employed (net asset)
Capitalised value of average profit
Average Profit × \(\frac{100}{\text { Normal Rate of Return }}\)

(ii) Capitalisation of Super Profits
Under this method the goodwill can be ascer tained by capitalising the super profit directly.
Goodwill = Super Profits × \(\frac{100}{\text { Normal Rate of Return }}\).

Question 3.
The following are the particulars in respect of two partnership firms.

Manu wishes to join in any one of the above firm which can make better profit. He seeks your advice as to which firm is more worth while and reputed.
Answer:
Capital Exployed = Assets – Liabilities = (10,000 + 15,000 + 20,000 + 20,000) – 5,000 = 60,000
Normal profit = Capital employed × Normal rate of return
60,000 × 10/100 = 6,000
Actual profit = 5,500
Super profit = Actual profit – Normal Profit
= 5,500 – 6,000 = -500
Firm Y
Capital Employed= (2,000 + 8,000 + 10,000 + 20,000) – 5,000 = 35,000
Normal profit = 35.000 × 10/100 = 3,500
Actual profit = 4,000
Superprofit = 4,000 – 3,500 = 500
Conclusion: Firm ‘Y’ earns Rs. 4,000 which is above normal profit.
Firm Y’s performance is better. So select Firm Y’.

Question 4.
How the capital accounts of the partners are adjusted at the time of admission of a new partner?
Answer:
At the time of admission of a new partner the capital accounts of the partners may be adjusted in the following ways.
1. Asking the new partner to bring in the capital on the basis of the existing partner’s capitals. Here new partners’ capital is calculated as follows.

• Totalling the capitals of the existing partners left after making all adjustments.
• Totalling the new profit sharing rights of the old partners.
• Total capital as per (a) is treated as the capital for the total rights as per (b)
• On the basis of the above, calculating the amount of capital to be brought in by the new partner.

2. Adjusting the capital of the old partners on the basis of the capital brought in by the new partner. This is done as follows.

• Comparing the capital of the incoming partner with the capitals of old partners.
• Asking the partner to bring in the required amount whose capital is less.
• Allowing the partner to withdraw the surplus amount whose capital is more.

Question 5.
A and B are partners sharing profits in the ratio of 3:2. On 1^st April 2005 they admit C into the firm. C brought in Rs. 1,00,000 for his capital but he was not in a position to bring his share of goodwill. The goodwill of the firm was valued at Rs. 1,50,000. Goodwill existing in the books of the firm is Rs. 2,75,000. The new profit sharing ratio is 2: 1: 1. Pass the journal entries.
Answer:
Working Note:
Sacrificing Ratio = Old Ratio – New ratio

C’s share of goodwill = 1,50,000 × 1/4 = 37,500 This Rs. 37,500 is to be debited to the new partner’s capital account and credited to old partners’ capital as C (new partner) cannot bring in the same.
Journal

Question 6.
R & S are partners sharing profits and losses in the ratio of 3:2. They admitted T into the firm. They have agreed to share the future profits – equally. T brought in Rs. 45,000 as his capital and Rs. 40,000 for his share of goodwill. The goodwill of the firm as in the books is Rs. 12,500. Write the journal entries.
Answer:
Notes
Sacrificing Ratio = Old ratio – New ratio
Old ratio = 3:2
New ratio = 1:1:1

Journal

Question 7.
Terry and C.L. Stephen are in partnership engaged in software development on accounting packages to various companies.
Ledger balances as shown by the books of accounts are:

• They have decided to admit Francis who is the son of Mr. Terry on the following terms.
• Fixed Assets valued 10% more than the book value.
• Interest payable on Bank loan Rs 16,000
• Office Fixtures was taken over by C.L. Stephen.

The new partner’s capital A/c is to be credited with half of Mr. Terry’s capital A/c before making any adjustments. Prepare capital accounts of the partners.
Answer:
Revaluation A/c

Capital Accounts

Note: The new partners capital A/c is to be credited with half of Mr. Terry’s capital a/c before making Adjustment.
Journal entry is:

Question 8.
A and B are partners sharing profit & losses in the ratio of 3:2. They admit C as a partner who is unable to bring goodwill in cash, but pays Rs. 16,000 as his capital. A Goodwill Account is raised in the books of the firm. Goodwill of the firm is valued at two year’s purchase of average three year’s profits. The profits for the three years were Rs. 10,000, Rs. 8,000 and Rs. 9,000. The net profit sharing ratio will be 5:2:2. The partners decided to write off goodwill after C’s admission. Make the Journal Entries, write up the Capital A/c. of partners & Goodwill calculation.
Answer:
Journal

Capital A/c

Calculation of Goodwill
Average profit = \(\frac{10,000+8,000+9,000}{3}\) = 9,ooo
Goodwill = Average profit × No. of years purchase
= 9,000 × 2 = 18,000
C’s share of Goodwill = 18,000 × 2/9 = 4,000
Sacrificing ratio = Old ratio – New ratio
Old ratio = 3:2
New ratio = 5:2:2
Sacrifice of A = 3/5 – 5/9 = 27 – 25 / 45 = 2/45
Sacrifice of B = 2/5 – 2/9 = 18 – 10 / 45 = 8/45
Sacrificing ratio of A and B = 2 : 8 = 1 : 4

Question 9.
You are given the following information on a reconstitution of a firm.
Partners capital

• Ammu – Rs. 20,000
• Beena – Rs. 30,000
• Ceema – Rs. 20,000
• Old profit sharing ratio – 2:3:1
• New Ratio – 1:2:3
• Revaluation profit – 22,500

1. State the reason for reconstitution.
2. Give a journal entry to adjust the revaluation profit through capital accounts of partners.

Answer:
Reconstitution refers to a change in the nature of relationship among partners due to

1. Change in profit sharing ratio
2. Admission
3. Retirement
4. Death or Amalgamation of two partnership firms.

In the firm of Ammu, Beena and Ceema reconstitution of firm takes place because they decided to change their profit sharing ratio:
Old ratio = 2:3:1 = 2/6 : 3/6: 1/6
New ratio = 1:2:3 = 1/6 : 2/6: 3/6
Ammu’s sacrifice = old ratio – new ratio
= 2/6 – 1/6 = 1/6
Beena’s sacrifice = old ratio-new ratio
= 3/6 – 2/6 = 1/6
Ceema’s gain = new ratio-old ratio = 3/6 – 1/6 = 2/6
Ammu’s revaluation profit = 22,500 × \(\frac{1}{6}\) = 3750
Beena’s revaluation profit = 22,500 × \(\frac{1}{6}\) = 3750
Ceema’s revaluation profit = 22,500 × \(\frac{1}{6}\) = 7500.

Question 10.
Ann and Gopu were doing sole proprietorship business of same nature. Both are close friends. On 1^st April 2005, they have decided to start a partnership business and have brought their existing assets into the new firm. They share profits in the ratio of 3:2. Details of existing assets and liabilities.

On 31^st December, 2006, they have changed their profit sharing ratio and become equal partners. The assets were then revalued as follows:

• Building is up by 10%
• Plant is down by 10%
• Furniture is up by 10%
• Stock is valued at 1,50,000
• Goodwill valued at Rs. 10,000

1. Give journal entries at to bring capital into the records on 01.04.2005.
2. Prepare :
   • Revaluation A/c
   • Capital A/c
3. New Balance Sheet

Answer:
Revaluation A/c

Capital A/c

New B/S

Question 11.
Haridas and Sudheer Raj are partners in a firm sharing profits in the ratio 3:2. On 1.04.2004, they admit Ramdas into the firm for a 5^th share in profits. Ramdas contributed the following in respect of his capital and goodwill.

Goodwill has been valued at 2 years purchase of super profit of past 3 years.

Capital employed is Rs. 2,00,000 and normal rate of return is 10%.
Give journal entries in respect of:

1. Capital contributed by Ramdas.
2. Goodwill brought in by him.

Answer:
Super profit = (Actual profit) Average profit – Normal profit
Average profit =

Normal Profit = Capital employed × Normal rate of
return =200000 × \(\frac{10}{100}\) = 20000
Super Profit = 25000 – 20000 = 5000
Value of Goodwill = Super profit × No. of years purchase = 5000 × 2 = 10000
Ramdas’s (New Partner) Share of goodwill = Total goodwill of firm × Ramdas’ share
= 10000 × \(\frac{1}{5}\) = 2000 5

Plus Two Accountancy Reconstitution of a Partnership Firm – Admission of Partner Eight Mark Questions and Answers

Question 1.
Jo and Sony are partners sharing profits and losses in the raito of 2: 1. Their Balance Sheet as on 31^st December 2004 was as follows.
Balance Sheet As on 31^st December 2004

Ebo is admitted into the partnership on the Balance Sheet date on the basis of the following.

1. Ebo will bring Rs. 50,000 as his capital.
2. Stock in trade should be decreased by Rs. 5,000
3. Plant and Machinery should be increased to Rs. 35,000 and Land and Building should be appreciated by 10%.
4. Bills payable and creditors be decreased by 5% and 10% respectively.

Record necessary journal entries, prepare revaluation account and partner’s capital account and also prepare the Balance sheet after Ebo’s admission.
Answer:
Journal

Revaluation A/c

Partner’s Capital A/c

Balance Sheet as on 1^st January 2005

Question 2.
Sunu and Jinu are partners in a firm sharing profits and losses equally. The following is their Balance Sheet as on 31.12.2005.

On the balance sheet date Tinu is admitted into the partnership on the following terms.

1. Tinu should bring in Rs. 60,000 as his capital
2. Furniture should be revalued at Rs. 50,000 and machinery at 25% less.
3. Bank overdraft should be decreased to Rs. 75,000
4. A provision of 10% is to be made for bad debts.
5. An unrecorded liability of Rs. 5,000 on rent is to be recorded.

Give journal entries, prepare revaluation account, capital accounts of partners and the Balance Sheet after Tinu’s admission.
Answer:
Journal

Revaluation A/c

Partner’s capital A/c

Balance Sheet as on 1^st January 2006

Question 3.
The following is the Balance sheet of L & M as on 30^th June 2005.

L & M were sharing profits and losses in the ratio of 2:1. N is admitted into the firm for a fourth share. The following are the conditions agreed upon.

1. Provision for bad and doubtful debts be increased to Rs. 2,500
2. Land and Buildings was to be depreciated by Rs. 10,000
3. The firm had an unrecorded machinery of Rs. 10,000 which is to be recorded.
4. N is asked to bring Rs. 50,000 as his capital and Rs. 15,000 for good will.
5. L & M had to withdraw half of the goodwill brought in by N.

Record journal entries, prepare Profit Loss adjustment account, capital accounts, and the New Balance Sheet
Answer:
Journal

Profit & Loss Adjustment (Revaluation) a/c

Partner’s Capital A/c

Balance Sheet as on 1^st July 2005

Question 4.
Edwin and Abel are partners sharing profits and losses in the ratio of 4:3. Their Balance sheet as on 30^th June 2005 is as follows.

Jerin is admitted into the firm with 2/7^th share. The following are the terms and conditions.

1. Jerin should bring in Rs. 40,000 as his capital and share of goodwill. The value of the goodwill of the firm is fixed at Rs. 35,000.
2. The amount of furniture and fittings should be written down by Rs. 5,000.
3. The full amount of goodwill should be withdrawn by old partners.
4. Creditors should be reduced by Rs. 2,000.
5. The new profit sharing ratio should be 3: 2:2.

Prepare necessary accounts and the Balance Sheet after the admission of Jerin.
Answer:
Working notes :
1. Jerin’sshare of goodwill = Goodwill of the firm × 2/7
= 35,000 × 2/7 = 10,000

2. Sacrificing ratio = Old share – New share
Edwin’s sacrifice = 4/7 – 3/7 = 1 / 7
Abel’s sacrifice = 3/7 – 2/7 = 1 / 7
Ratio = 1/7 : 1/7 ie. = 1 :1
Revaluation A/c

Partner’s capital a/c

Balance Sheet as on 1^s July 2005

Question 5.
Ram and Gopal are partners in a firm sharing profit and loss in the ratio of 3:1 respectively. The following is their Balance Sheet as on 31 /12/2008.
Balance Sheet

They admit Menon into partnership on 1-1-2009 on the following terms.

1. Menon should bring Rs. 10,000 as his capital for 1/5^th share and 18,000 as her share of good will.
2. Liability for workmen compensation estimated at Rs. 1500.
3. Value of land and building be appreciated by Rs. 5000.
4. Stock reduced by 5%
5. A provision of 5% should be made for doubtful debts in debtors.

Prepare revaluation a/c, capital account of partners and the Balance sheet of new firm.
Answer:
Revaluation A/c

Capital A/c

Balance Sheet

Working Note:
Menon’s share of goodwill = 18,000 Sacrificing ratio = 3:1
Goodwill credited in Ram = 18,000 × 3/4 = 13,500 Gopal = 18,000 × 1/4 = 4,500.

Question 6.
The following is the Balance Sheet of A and B as on 31^st Dec. 2004.

A & B share profits and losses in the ratio of 5 : 3. They admit C into the partnership for equal share. C brings in Rs. 50,000 as capital and Rs. 5,000 for his share of goodwill. Goodwill of the firm is valued at Rs. 30,000. The following conditions were agreed upon.

1. Land and buildings are appreciated by 20%.
2. Stock is decreased by Rs. 5,000/-
3. Creditors include Rs. 2,500/- not become payable.
4. Unexpired insurance or insurance paid in advance Rs. 2,500 is to be recorded.

Record journal entries, prepare ledger accounts and the new Balance Sheet.
Answer:
Notes:
Calculation of sacrificing ratio
Sacrificing ratio = old ratio – new ratio

Journal

Revaluation A/c

Partner’s Capital A/c

Balance sheet as on 1^st January 2005

Question 7.
P and Q are partners sharing profits and losses in the ratio of 4:3. Their balance sheet as on 30^th June 2004 is as follows.

R is admitted into the firm on the basis of the follow-ing conditions.

1. Sundry debtors should be revalued at Rs. 1,00,000.
2. R should bring in Rs. 15000 as capital and Rs. 10000 as his share of goodwill. He will get 1/6 share in future profits.
3. The capital accounts of all partners should be adjusted on the basis of their profit sharing ratio by bringing in or paying off the cash as the case may be.

Prepare necessary ledger accounts and the new Balance sheet of the firm.
Answer:
Working notes:
Calculation of new ratio:
Old Ratio between P & Q = 4 : 3
R’s share = 1/6
Remaining portion = 1 – 1/6 = 5/6
P’s new share = 4/7 of 5/6 = 4/7 × 5/6 = 20/42
Q’s new share = 3/7 of 5/6 = 3/7 × 5/6 = 15/42
R’s share = 1/6 = 7/42

Calculation of capital required:
R’s capital for
1/6 share in profits = 15,000
Total capital of the firm= 15,000 × 6/1 = 90,000
P’s capital = 90,000 × 20/42 = 42,857
Q’s capital = 90,000 × 15/42 = 32,143
Revaluation A/c

Partners capital a/c

Cash A/c

Balance sheet as on 1^st July 2004

Question 8.
A and B are partners in a firm sharing profits and losses as 3/5 and 2/5.C, comes in for 1/5^Th share of profit. He pays Rs. 8,000 as goodwill premium and 50% of the adjusted capitals of A and B. Balance Sheet of A and B on the date of Cs’ admission stand as follows:
Balance Sheet

Land and Buildings are to be valued at Rs. 40,000. Plant is to be depreciated by 10% and stock by Rs. 500.Sundry Debtors is worth Rs. 31,750. A liability of Rs. 1,750 for outstanding expenses has been omitted to be recorded in the books. A and B have a joint life policy of Rs. 15,000 not shown in the books, the premium for which has been charged to Profit & Loss Account. The surrender value of the policy on the date of admissions is Rs. 2,000, and is agreed to raise a life policy account in the books at this value. Give Journal Entries.
Answer:
Journal

Revaluation A/c

Dr. Partner’s Capital A/c Cr.


Capital to be brought in by C = 50% of the adjusted capital of A and B.
i.e., = 50% of 57,500+ 40,000 = 50% of 97,500 = 48,750.

Question 9.
Below given the details related to a firm.


Can you analyse the adjustment on admission of a new partner and show the balance sheet after admission.
Answer:
Revaluation A/c

Capital Accounts

Balance Sheet

Working Note
Goodwill Calculation: Here, Goodwill is calculated on the basis of super profit method.
Goodwill = Superprofit × No. years purchase
Superprofit = Actual or Average profit – Normal profit
Normal profit = Capital Employed × Normal rate of return
Capital employed = Asset – Liabilities
= (10,000 + 20,000 + 31,500 + 30,000+ 20,000) – 11,500
= 1,00,000
Normal profit = 1,00,000 × 10/100 = 10,000
Superprofit =40,000 – 10,000 = 30,000
Goodwill = 30,000 × 2 = 60,000
New partner’s share of Goodwill – 60,000 × 1/3
= 20,000
Sacrifacina ratio
Manu = 3/5 – 1/3 = 9 – 5/15 = 4/15
Raju = 2/5 – 1/3 = 6 – 5/15 = 1/15
Sacrificing ratio = 4:1
Manu = 3/5 – 1/3 = 9 – 5/15 = 4/15
Manu’s Sacrifice = 20,000 × 4/5 = 16,000
Raju’s Sacrifice = 20,000 × 1/5 = 4,000 [Hint: New ratio 1:1:1].

Question 10.
Observe the following table.

Furniture was sold at Rs. 2700 on the date of admission. You are required:

1. Revaluation A/c
2. Capital Accounts
3. Balance sheet of the new firm

Answer:
Revaluation A/c

Capital A/c

Cash A/c

Balance Sheet

Question 11.
A and B are partners in a firm sharing profits in the ratio of 2:1 ‘C’ is admitted into the firm with 1/4 share in Profits. He will bring in Rs. 30,000 as capital and capital of A and B are to be adjusted in the profit sharing ratio. The balance sheet of A and B as on 31/03/2017 (before c’s admission) was as under.
Balance sheet of A and B as on 31/03/2017

The terms of agreement are as follows:

1. ‘C’ will bring in Rs. 12,000ashisshareofgoodwill.
2. Building was valued at Rs. 45,000 and Machinery at Rs. 23,000
3. A provision for bad debts is to be created @ 6% on debtors
4. The capital accounts of A and B are to be adjusted by opening current accounts.

Prepare necessary ledger accounts and new balance of the firm.
Answer:
Revaluation A/c

Capital A/c

Balance Sheet

Working Note:
1. New Profit Sharing ratio C’s share of Profit = 1 /4

2. New capital of A and B on the basis of c’s capital
Total capital of the new firm = 30,000 × \(\frac{4}{1}\) = 1,20,000
As new capital = 1,20,000 × \(\frac{2}{4}\) = 60,000
The Existing capital of A = 63,680 Excess (A) = 3,680
B’s new capital = 1,20,000 × \(\frac{1}{4}\) = 30,000
The existing capital of B = 38,840
Excess (B) – 8,840
The current accounts can be opened and the amount to be withdrawn by A and B will be transferred to their respective current accounts.

Students can Download Chapter 4 Chemical Kinetics Notes, Plus Two Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Notes Chapter 4 Chemical Kinetics

Chemical kinetics is the branch of chemistry which deals with the study of the velocity of chemical reactions and their mechanism.

Rate of a Chemical Reaction :
amount of chemical change per unit time.

Average Rate of Reaction:
change in concentration of any one of the reactants or products per unit time. Unit of rate of a reaction mol L^-1 s^-1 Fora reaction, R → P

Instantaneous Rate of Reaction:
the rate of change in concentration of any one of the reactants or products at a particular instant of time for a gven temperature. It may be expressed as \(\frac{dx}{dt}\) where dx is the change in concentration at the instant dt.
For the reaction aA + bB → cC + dD

Graphically,- instantaneous rate = slope of the tangent drawn to the concentration vs time graph

corresponding to the time t. i.e., r_inst = \(\frac{dx}{dt}\) , where dx and dt are the intercepts.

Factors affecting rate of reaction:
Concentration of reactants, Nature of reactants and products, Temperature, Pressure (for gaseous reactants), Presence of catalyst, Presence of light (radiation)

Rate Expression and Rate Constant:
According to law of mass action, the rate of a chemical reaction is proportional to the product of molar concentrations of the reactants.
Consider a general reaction.
aA + bB → cC + dD
Rate α [A]^x [B]^y

where exponents ‘x’ and ‘y’ may or may not be equal to ‘a’ and ‘b’ respectively.
The above equation is also written as.
Rate = k[A]^x [B]^v
or \(\frac{-\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}\) = k[A]^x [B]^v
where ‘k’ is a proportionality constant called rate constant. The equation is known as rate expression or rate law.

Rate law:
expression in which reaction rate is given in temis of molar concentration of reactants with each term raised to some power, which may or may not be same as the stiochiometric coefficient of the reacting species in a balanced chemical equation.

Order of Reaction :
sum of powers of the concentration of the reactants in the rate law expression. Considers general reaction,
aA + bB → cC + dD
Rate = k[A]^x [B]^v
Order = x + y

Example: H₂ + l₂ → 2 HI
Rate = k[H₂]¹ [l₂]¹, Order = 1 + 1 = 2

Order of a reaction is an experimentally determined quantity. It may be zero, whole number, fractional and even negative.
Elementary reactions –
reactions taking place in one step.

Complex reactions –
reactions involving a sequence of elementary reactions. These may be consecutive reactions, reverse reactions and side reactions.

Some example of reactions of different orders: First Order:
i) Decomposition of N₂O₅
N₂O₂ → 2NO₂ + ½ O₂
Or 2N₂O₅ → NO₂ + O₂
Rate = k[N₂O₅]¹

ii) Decomposition of NH₄NO₂ in aqueous solution.
NH₄NO₂ → N₂ + 2H₂O
Rate = k[NH₄NO₂]¹

Second order:
i. 2NO₂ → 2NO + O₂ Rate = k[NO₂]²
ii. H₂ + l₂ → 2Hl Rate = k[H₂]¹[l₂]¹

Third order:
i. 2NO + O₂ → 2NO₂
Rate = k[NO]² [O₂]¹
ii. 2NO₂ + Cl₂ → 2NOCl + O₂
Rate = k[NO₂]² [Cl₂]¹

Units of Rate Constant:
For an n^th order reaction, the unit of rate constant is given by the formula, mol^1-n L^n-1 s^-1

Molecularity of a Reaction :
number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which collide simultaneously in order to bring about a chemical reaction. It is always a whole number.

Reactions which involve simultaneous collision between two species are bimolecular.

When one reacting species is involved in the reaction, it is unimolecular.
Example:
NH₄NO₂ → N₂ +2H₂O
O₃ → O₂ + O

Reactions which involve simultaneous collision between two species are bimolecular.
Example:
2 Hl → H₂ + l₂

Reactions which involve simultaneous collision between three species are trimolecular or termolecular.
Example :
2 NO + O₂ → 2 NO₂

The probability that more than 3 molecules can collide and react simultaneously is very small. Hence, molecularity greaterthan 3 is not observed.
In a complex reaction, the slowest step in a reaction determine the rate of reaction, i.e., slowest step is the rate determining step.

Difference between order and molecularity

Integrated Rate Equation :
Integrated rate equation gives a relation between concentrations at different times and rate constant.

Zero Order Reaction :
The rate of reaction is independent of the concentration of the reactants.

For a zero order reaction, R → P,
d[R] = – kdt
[R] = – kt + [R]₀ ………….. (1)
or \(k=\frac{[R]_{0}-[R]}{t}\)

Equation (1) is of the form y = mx + c, equation for a straight line. If we plot [R] versus t, we get a straight line with slope = -k and intercept equal to [R]₀

Note:
R0 initial concentration of reacting species (i.e., at time = 0)
R → concentration of reacting species (i.e., at time = t)

First Order Reaction
Fora reaction, R → P

If we plot a graph between log [R]₀/[R] vs ‘t’ we get a straight line with slope = k/2.303

All natural and artificial radioactive decay take place by first order kinetics.

Half-Life of a Reaction (t_½):
time required to reduce the concentration of a reactant to half of its initial concentration.
Forzero order reaction,
\(t_{1 / 2}=\frac{[R]_{0}}{2 k}\)
Derivation.
For a zero order reaction R → P, the rate constant is given by the equation,

Derivation:
For a first order reaction R → P, the rate constant is given by the equation,

For first order reactio t_½ is independent of [R]0.

Pseudo First Order Reaction :
Reaction which appear to be of higher order but actually follow lower order kinetics.

Example:
Acid hydrolysis of ethylacetate.

Rate = k[CH₃-COOC₂H₅]

Since the concentration of H₂O is quite large and does not change appreciably, it does not appear in the rate law.
Another example: Inversion of cane sugar in presence of dilute acids.

Temperature Dependence of the Rate of a Reaction :
The rate of the reaction increases considerably with increase in temperature. For a chemical reaction with rise in temperature by 10°, the rate constant is nearly doubled.

Temperature Coefficient –
The ratio between the rate constant of a reaction at two temperatures differing by 10°.

Arrhenius Equation –
The temperature dependence of the rate of a chemical reaction can be explained by Arrhenius equation.
k = A e^-Ea/RT
A → Arrhenius factor or frequency factor or pre-exponential factor
E_a → Activation energy in J mol^-1
R → Gas constant
T → Temperature in kelvin

Activation energy (E_a)-
The energy required to form activated complex or intermediate. Some energy is released when the complex decomposes to form products.

Most probable kinetic energy –
kinetic energy of maximum fraction of molecules. The peak of the Boltzmann-Maxwell curve corresponds to this.

From the Arrhenius equation,
In k = In A \(\frac{E_{a}}{R T}\)
A polt of In k vs. \(\frac{1}{T}\)

If k₁ and k₂ are the rate constants at temperatures T₁ and T₂ respectively, Arrhenius equation can be written in the form,

Effect of Catalyst :
A catalyst is a substance which alters the rate of a reaction without itself undergoing any permanent chemical change. The function of a catalyst is to provide an alternate path of reaction with a lower energy of activation.

A small amount of the catalyst can catalyse a large amount of reactants. A catalyst does not alter Gibbs energy ∆ G of a reaction. It does not change the equilibrium constant but helps in attaining the equilibrium faster.

Collision Theory of Chemical Reactions :
It is based on kinetic theory of gases.
1. According to collision theory, the reactant molecules are assumed to be hard spheres and a chemical reaction takes place when reactant molecules collide with one another.

2. All collisions are not effective collisions. An effective collision is that collision which results into chemical reaction.

3. For effective collision, the molecule possess a certain minimum amount of energy called threshold energy and should have proper orientation.

Threshold energy – the minimum amount of energy which the colliding molecules must possess to make an effective collision.

4. Collision frequency (Z) – The number of collisions per second per unit volume of the reaction mixture.

5. To account for effective collisions, the probability or steric factor (P) is considered. It accounts for the fact that in a collision, molecules must be properly oriented.
Rate = PZ_ABe-Ea/RT

Thus, in collision theory activation energy and proper orientation of the molecules together determine the criteria for effective collision and hence the rate of ’ reaction.

Students can Download Chapter 2 Accounting for Partnership – Basic Concepts Questions and Answers, Plus Two Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts

Plus Two Accountancy Accounting for Partnership – Basic Concepts One Mark Questions and Answers

Question 1.
A partner is entitled to get 6% per annum as
(a) Profit
(b) Interest on capital
(c) Interest on loan
(d) Remuneration
Answer:
(c) Interest on loan

Question 2.
Profit and Loss Appropriation is an extension of
(a) Capital Account
(b) Current Account
(c) Trading Account
(d) Profit and Loss Account
Answer:
(d) Profit and Loss Account

Question 3.
Find odd one and state the reason
(a) Interest on capital
(b) Interest on drawings
(c) Salary
(d) Commission
Answer:
(b) Interest on drawings

Question 4.
Complete the following

• Interest on loan – Charge against profit.
• Interest on Partners capital – _______.

Answer:

• Appropriation of profit
• Reason: All others are increase to capital A/c.

Question 5.
Find the odd one and state reason.
(a) Interest on partner’s capital
(b) Interest on partner’s loan.
(c) Interest on partner’s drawings
(d) Borrowings from the firm
Answer:
(d) Borrowings from the firm. Others are P/L appropriation A/c items.

Question 6.
Rugma, Neha and Lekshmi are partners in a firm sharing profits and loses in the ratio of 3:3:4. Their fixed capitals were Rs. 1,00,000, Rs. 2,00,000 and Rs. 3,00,000 respectively. For the year 2005, interest on capital was credited to them @ 10% instead of 9% per annum. You are required to rectify the mistake by passing an adjustment entry.
Answer:

Plus Two Accountancy Accounting for Partnership – Basic Concepts Two Mark Questions and Answers

Question 1.
Define Partnership.
Answer:
According to Section 4 of the Indian Partnership Act 1932, a partnership is “the relation between persons who have agreed to share the profits of a business carried on by all or any of them acting for all”. The persons entered into agreement are individually known as ‘partners’ and collectively as ‘firm’.

Question 2.
What is Partnership Deed?
Answer:
Partnership agreement may be oral or written. When the agreement is written, it is called Partnership Deed. It is also called ‘Articles of Partnership’.

Question 3.
What is meant by Profit and Loss Appropriation Account?
Answer:
Profit and Loss Appropriation Account is an extension of the Profit and Loss Account. It is prepared to show the appropriation (distribution) of profit among partners.

Question 4.
What are the circumstances in which goodwill is need to be valued?
Answer:

1. When a new partner is admitted.
2. When a partner is retired ordied.
3. When two or more firms are amalgamated.
4. When a firm is dissolved or its business is sold.

Question 5.
What is divisible profit?
Answer:
Divisible profit is the balance net profit that remains after making all adjustments to net profit regarding interest on capital, salary to partners, interest on partners loan, interest on drawings, etc. and which is distributed among partners in their profit sharing ratio.

Question 6.
Edwin and Abel are partners sharing profits and losses in the ratio of 1:1. Edwin drew Rs. 1000/- at the beginning of every month for the year ending 31st Dec. 2004. Calculate interest on drawings at 6% perannum.
Answer:
Total amount withdrawn by Edwin = 1,000 × 12 = 12,000
1,000 × 12 = 12,000
\(Average period =\frac{\text { Total period (months) }+1}{2}\)
\(=\frac{12+1}{2}\) = 6 : 5 months
Interest on drawings = 12,000 × 6/100 × 6.5/12 = Rs. 390.

Question 7.
Christy and Fiyona are partners in a firm with equal profit sharing ratio. Christy drew regularly Rs. 15000 at the end of every month.
Answer:
Total amount withdrawn by Christy = Rs. 1,500 × 2 = 18,000
\(Average period =\frac{\text { Total period (months) }-1}{2}\)
\(=\frac{12-1}{2}\) = 5.5 months
Interest ondrawings= 18, 000 × 5/100 × 5.5/12 = Rs. 412.5

Question 8.
After closing the books of accounts, it was discovered that an item, interest on capital was omitted to be recorded in the books of accounts. Even then, there was no difference in the closing balance of capital account, before and after the treatment of the item. What do you infer from this?
Answer:
Partner’s Share of Capital and their profit sharing ratio are in accordance with their capital account balances.

Question 9.
Paul, Kumar, and Lakshman are partners in a firm, sharing profits and losses in the ratio of 3:2:1. After the preparation of final accounts, it was discovered that interest on drawing had not been taken into consideration. The interest on drawings of partners amounted to Rs. 600, Rs. 400 and Rs. 200. Give necessary adjustment journal entry.
Answer:
Items which are omitted while preparing P&L Appropriation A/c can be brought into accounts through P&L adjustment a/c by passing the following entry.

Question 10.
Give your suggestions to the following arguments. “Under capitalisation method, the firm will have good will only if the value of net tangible assets are more than the capitalised value of profit.”
Answer:
Under capitalisation method
Value of goodwill = Total value of business – Net as-sets
Total value of business

Net assets = Assets – Liabilities The above equation proves that, a firm will have goodwill only if the value of net tangible assets are less than the capitalised value of profit.

Question 11.
Sanu and Manu are in partnership, who have not made any written agreement. Sanu has given a loan of Rs. 12,000/- to the firm in addition to his capital contribution. During the year the firm made a net loss of Rs. 40,000. Regarding the interest on loan, Manu is of the opinion that no interest be paid being the loan was not external one. Is Mr.Manu right in his stand? State your views.
Answer:
Normally partnership deed contains rules and regulations regarding the conduct of partnership business. In such cases partnership may not have a written agreement. In some other firms, partnership deed may be silent on some matter.

Then relevant discussion in the IPA 1932 becomes applicable. As per IPA 1932, interest on loan is payable at 6% p.a. on Partner’s loan. So Manu’s opinion that interest on loan is not payable, is wrong.

Question 12.
A business has been purchased by a firm for Rs. 1,00,000. But its net tangible assets were worth Rs. 92,000.

1. What does this difference in value indicate?
2. Where is it shown in the Balance sheet of the firm?

Answer:
Total value of business-Net Assets = Value of goodwill 1,00,000 – 92,000 = 8,000
So ‘Rs. 8000’ implies the value of goodwill of the firm. Goodwill refers to the value of reputation of a business. It is an intangible asset. So it is shown on the asset side of B/S.

Plus Two Accountancy Accounting for Partnership – Basic Concepts Three Mark Questions and Answers

Question 1.
Match the following.

Answer:

Question 2.
Ameer and Sudheer are partners sharing profits equally. They are entitled to salaries as follows, Ameer Rs. 6000, Sudheer Rs.4000. The partnership has made a profit of Rs. 15,000. How much is the increase in capital of Mr.Ameer?
(a) 3500
(b) 3900
(c) 8500
(d) 9300
Answer:
(c) 8500
Notes: A:S = 1:1
A’s salary 6000
S’s salary 4000
Out of Net profit of Rs. 15,000.
P/L Appropriation A/c:

A will get salary 6000 and share of profit 2500. So increase in capital is Rs. 8500.

Question 3.
Gomez and Arun Gomez are partners sharing profits and losses in the ratio of 2:1. They are allowed interest at 10% per annum on capitals and loans to the partnership.

The partnership has made a net profit of Rs. 40,000 for the year. How much is the total increase in the net worth of A.Gomez?
(a) 24,800
(b) 25,000
(c) 26,800
(d) 27,100
Answer:
(c) 26,800
Notes:
Capital A/c of A.Gomez:

P/L Appropriation A/c:

A Gomez’s opening capital (capital in the begining of the year) is Rs. 20,000 and closing capital (capital at the end of the year) is Rs. 46,800. So total increase in the net worth of A.Gomez is Rs. 26,800. (46,800-20,000).

Question 4.
Lalu and Beena are in partnership. He is also entitled to a salary of Rs. 12,000 per annum. Profits and losses are shared equally. The partnership has made a net profit of Rs. 30,000. How much is Lalu’s total increase in his Capital A/c?
(a) 18,000
(b) 15,000
(c) 21,000
(d) 42,000
Answer:
(c) 21,000
Notes: P&L Appropriation A/c:

Lalu’s capital A/c:

Increase in capital = 12000 + 9000 = 21000.

Question 5.
“Partnership deed must be in writing.” Do you agree with the statement? Give reasons in favour of having partnership deed in writing.
Answer:
Partnership is the result of agreement between two or more persons. The agreement may be oral or written. The written agreement is called Partnership Deed. It is always advisable to put the partnership agreement in writing because of the reasons given below:

1. To avoid disputes, quarrels, and misunderstanding among the partners.
2. To remind the partners about their rights, duties, and liabilities.
3. To maintain healthy atmosphere to carry on business smoothly.

Question 6.
X, Y, and Z are partners in a firm sharing profits and losses in the ratio of 4:3:2. During 2005, their fixed capital and drawings were as follows:

Partners are entitled to a salary of Rs. 12,000 p.a. and interest on capital @ 5% p.a. You are required to prepare the Current Accounts of partners.
Answer:
Current Account:

Plus Two Accountancy Accounting for Partnership – Basic Concepts Five Mark Questions and Answers

Question 1.
Define Partnership Deed. Mention some of its contents.
Answer:
Partnership deed is a written document which contains the rules and regulations regarding the conduct of business.
Contents of Partnership deed:

1. Name and address of the firm.
2. Name and Address of partners
3. Nature of business
4. Duration of partnership
5. Capital introduced by partners
6. Interest on capital
7. Drawing made by partner
8. Interest on partner drawings
9. Salary, commission and other remunerations payable to partners
10. Rights, Duties, and Liabilities of Partners.

Question 2.
What are the rules applicable as per the Partnership Act in the absence of an agreement?
Answer:
In India Partnerships are governed by the Indian Partnership Act 1932. The following are the rules applicable as per the Partnership Act in the absence of an agreement.

1. Profits and Losses – Profits and losses are to be shared equally among partners.
2. Salary or remuneration – Partners are not entitled to salary or any other remuneration.
3. Interest on capital- Partners are not entitled to interest on capital.
4. Interest on drawings – No interest is charged on drawings made by partners.
5. Interest on loan – partners are entitled to get an interest at 6% per annum for any loans they have given to the firm.

Question 3.
Distinguish between Fluctuating and Fixed Capital methods.
Answer:

Question 4.
What are the general characteristics/ features of a partnership? Explain.
Answer:
1. Number of members:
The minimum member of persons required for a partnership is two (2). Maximum number is ten (10) in case of banking business and twenty (20) in other partnerships.

2. Business Purpose:
The purpose of forming a partnership should be to carry out some business. The business must not be illegal.

3. Agreement:
For the formation of a partnership an agreement is must. The agreement may be oral or written. Only competent persons can enter into a partnership agreement.

4. Profit sharing:
The profits and losses of a partnership business must be shared among the partners. Profits must be shared in an agreed ratio or equally.

5. Mutual agency:
Mutual agency is there in partnership. Every partner is an agent and a principal at a time. He is an agent when he acts for others and a principal when the others act for him.

6. Unlimited liability:
The liability of the partners . in a firm is unlimited. Every partner is individually and jointly liable for all the debts of the firm.

7. No legal existence:
A partnership has no legal existence. It has no existence different from its members.

8. No transfer of share:
A partner cannot transfer his share in the firm to outsiders without the consent of the other partners.

Question 5.
Sabu and Sekhar commenced business in partnership on 1^st January, 2005. No written agreement was in force between them. They contributed Rs. 40,000 and Rs. 10,000 respectively as capital. In addition, Sabu advanced Rs. 20,000 on 1^st July, 2005 as loan to the firm. Sabu met with an accident on 1^st April, 2005 and could not attend the partnership business upto 30^th June, 2005. The . profits for the year ended on 31^st December, 2005 amounted to Rs. 50,600. Dispute arise between them for sharing profits.
Sabu claims:

• He should get an interest @ 10% p.a. on capital. Sekhar claims:
• Net Profit should be shared equally
• He should be allowed remuneration of Rs. 1,000 p.m. during the period of Sabu’s illness.

You are required to:

1. In your opinion how much profit will each partner get?
2. State your reason.

Answer:
In the absence of agreement, partners are not entitled to interest on capital contributed by them. So Sabu’s claim is not admitted.

1. In the absense of agreement, partners are not entitled to any salary or other remuneration.
2. In the absence of written agreement, partners are entitled to share profits equally, Here, net profit is divided equally among Sabu and Sekhar.

Profit for the year = 50,600
Less:Interest on Sabus loan (20000 × 6/100 × 6/12) = 600
The actual profit = 50,000
In the absence of agreement, partners are entitled to interest on loan (to the firm) at the rate of 6% p.a. Sabu’s share of profit = 50,000 × 1/2 = 25000
Stephen’s share of profit = 50000 × 1/2 = 25000.

Question 6.
Sona and Jerin started a partnership business on 1^st January 2004. Sona contributed Rs. 50,000 and Jerin Rs. 25,000 as capital. They decided to share profits and losses in the ratio of 2:1. Sona was entitled to a salary of Rs. 2000 per month. Partners are entitled to interest on their capitals at 5% per annum. The drawings of Sona and Jerin during the year are Rs. 9,000 and Rs. 6,000 respectively. The profit of the firm after making all the adjustment was Rs. 15,000. Prepare the capital accounts of the partners under fluctuating capital method.
Answer:
Partners Capital Account:

Question 7.
A and B are partners sharing profits in the ratio of 2: 3. On 1^st January 2001, they admitted C into the firm for a sixth share of profits with a guaranteed minimum of Rs. 25000. A & B continue to share profits as before but agrees to suffer any excess over 1/6 of profit going to C equally. The profits of the firm forthe year was Rs. 75,000. Prepare Profit and Loss appropriation account.
Answer:
Notes:

The ratio in which this difference is to be borne 1: 1 (equally)
Dr. Profit and Loss Appropriation Account Cr.

Question 8.
From the following Balance Sheet of Aneesh and Jaya, calculate interest on capital @ 5% per annum, payable to Jaya for the year ending 31.12.05.
Balance Sheet as on 31.12.2005:

During the year, Jaya’s drawings were Rs. 3,000 and the firm made a profit of Rs. 4,000.
Answer:
Interest on capital is payable onthe opening capital (ie. capital on 1.1.05)
Opening capital = Closing capital + drawings – Net Profit
Closing capital = Rs. 6000
Drawings = 3000
Net Profit during the year = 4000
P&L appropriation shown in B/S = 2000
∴ Net profit credited to partners capital = 4000 – 2000 = 2000
Net profit credited to Jaya = 2000 × 1/2 (ratio being 1:1) = 1000
Opening capital = 6000 + 3000 – 1000 = 8000
Interest on capital = 8000 × 5/100 = 400.

Question 9.
Aby and Anu are partners sharing profits in the ratio of 4:1. Their capital a/c balances are
Aby : 4,00,000
Anu : 5,00,000
Profit made during the year was Rs. 1,00,000.
Anu is of the opinion that their agreement must include a provision for interest on capital @10% p.a. Otherwise the profit sharing ratio must be made equal. Why did Anu put forward such an opinion? Will it be worthwhile to her if such changes are made. Which of the above condition is more advantageous to her. Give your advice.
Answer:
Profit sharing ratio of Aby and Anu is 4:1.
Net Profit = 1,00,000
As per this ratio Aby will get Rs. 80,000 (1,00,000 × 4/5) and Anu will get 20,000 (1,00,000 × 1/5) as their share of profit. But Anu has contributed Rs. 1,00,000 more than Aby’s capital. Now Anu is in a disadvantageous position.
Conditions:
1. If there is a provision for interest on capital @10%.
Anu’s interest on capital = 5,00,000 × \(\frac{10}{100}\) = 50,000
Her share of profit = (Net Profit – Interest on capital of Aby and Anu) × 1/5
=[1,00,000 – (40,000 + 50,000)] × \(\frac{1}{5}\)
= 1,00,000 – 90,000 × 1/5
=10,000 × \(\frac{1}{5}\) = 2,000
Anu will get interest on capital = 50,000+ share of profit 2000 = 52,000.

2. Profit sharing ratio made equal:
Abu’s share of profit = 1,00,000 × 1/2 = 50,000
Anu’s share of profit = 1,00,000 × 1/2 = 50,000
So First condition is more advantageous to Anu.

Question 10.
The partner’s capital account prepared by Mr.Jose, an accountant in Gokul and Co, where Mr. Raman and Mrs.Seetha are partners, is given below. Rectify the errors, if any in the capital accounts prepared by him and show the partners capital accounts under fixed capital method. What should have been the profit of the firm as per profit and loss account?
Partners Capital A/c

Answer:
Calculation of NP:

Capital A/c:

Question 11.
T and S are partners with equal profit sharing ratio. T withdrew the following amounts during the year 2005.

The interest on drawings charted is at 6% p.a. Assuming that the accounting year ends on 31^st December. Calculate the interest on drawings under product method.
Answer:
Calculation of Interest on Drawings:

Interest on drawings = 73,500 × 6/100 = Rs. 4,410
Interest for one month= 4,410 × 1/12 = Rs. 367.5.

Question 12.
On 1^st January 2006, X&Y entered into partnership contributing Rs. 60,000 and Rs. 40,000 respectively. They agreed to share profits & losses in the ratio of 3 : 2. Y is allowed a salary of Rs. 15,000 per year. Interest on capital is to be allowed at 10% per annum. During the year X withdrew Rs. 9,000 and Y Rs. 8,000 as drawings. The interest on drawings paid by X and Y was Rs.150 and Rs. 130 respectively. Profits as on 31^st December, 2006 before the above mentioned adjustments were Rs. 65,000. Show the distribution of Profits by preparing Profit and Loss Appropriation A/c & Prepare Partner’s Capital Accounts.
Answer:
Profit and Loss Appropriation Account:

Capital A/c:

Plus Two Accountancy Accounting for Partnership – Basic Concepts Eight Mark Questions and Answers

Question 1.
On 1^st January 2005 Sneha and Surya started partnership business by contributing capitals of Rs. 50,000 and Rs. 60,000 respectively. They share profits in the ratio of 2 : 3. Sneha is entitled to a salary of Rs. 12,000 per annum. Interest on capital allowed is at 6% p.a. Surya is entitled to a commission of Rs. 3,000. During this year Sneha withdrew Rs. 3,000 and Surya Rs. 2,000. Interest on drawings charged is Rs. 100 and Rs. 150 respectively. Profit in the year before making the adjustments was Rs. 25,000. Pass necessary journal entries, Prepare Profit and Loss Appropriation Account and Partners Capital Accounts.
Answer:
Journal:


Dr. Profit & Loss Appropriation Account Cr.

Dr. Partners Capital Account Cr.

Students can Download Chapter 1 The Solid State Notes, Plus Two Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Notes Chapter 1 The Solid State

General Characteristics of Solid State
Definite mass, volume and shape; short inter molecular distances; strong intermolecular forces; constituent particles have fixed position and can only oscillate about their mean positions; incompressible and rigid.

Classification of Solids:
1. Crystalline Solids :
The constituent particles are well orderly arranged. It has long range orcfe/-which means that there is a regular pattern of arrangement of particles which repeats itself periodically overthe entire crystal, e.g. NaCl, Quartz.

2. Amorphous Solids (Greek amorphos = no form):
The constituent particles are irregularly arranged; i.e. they have irregular shape, the arrangement of constituent particles in this solid has only short range order, e.g. rubber, quartz, glass, plastic.

Amorphous solids have a tendency to flow slowly, therefore these are called pseudo solids or supercooled liquids, e.g. Glass – Glass flows down very slowly and makes the bottom portion slightly thicker.

Crystalline solids are anisotropic in nature (i.e. their physical properties like electrical resistance or refractive index show different values in different directions). Amorphous solids on the other hand are isotropic, (i.e. their physical properties would be same along any direction)

Difference between Crystalline and Amorphous Solids

Classification of Crystalline Solids:
Crystalline solids can be classified on the basis of nature of intermolecular forces operating in them into four categories.

Molecular Solids :
In this, molecules are the constituent particles, they are of three categories:

a. Non-polar Molecular Solids :
They are formed by the regular arrangement of either atoms or non-polar molecules held by weak dispersion forces or London forces. They have low melting points and are liquids or gases at room temperature and pressure, e.g. solid forms of argon, He, H₂ and Cl₂, l₂.

b. Polar Molecular Solids :
In these polar molecules are held by relatively stronger dipole-dipole interaction. They are soft and non-conductors of electricity e.g. solid forms of HCl, SO₂, NH₃.

c. Hydrogen bonded Molecular Solids :
Strong H- bonding binds the molecules of these solids. They are non-conductors of electricity and are volatile liquids or soft solids.
e.g. Ice (Solid H₂O)

Ionic Solids :
In these ions are held together by strong coulombic (electrostatic) forces. They are hard and brittle in nature and have high melting point and boiling points. They are electrical insulators in the solid stale, but in molten state/solutions conduct electricity because ions become free to move.
e.g. NaCl, KCl, KNO₃ etc.

Metallic Solids :
In these positive ions are surrounded by and held together by a sea of free electrons. These free and mobile electrons are responsible for high electrical and thermal conductivity. Due to this mobile electrons metals have lustre and colour, e.g. Gold, Silver etc.

Covalent or Network Solids:
They have covalent bonds between adjacent non-metal atoms which are held strongly in their positions. They have high melting points and are insulators, e.g. Diamond, SiC Graphite is an exceptional case – it is soft and conducts electricity due to its typical layer structure.

Crystal Lattice and Unit Cells
The three dimensional arrangement of constituent particles in a crystal, represented by points is called crystal lattice/space lattice.

Unit Cell:
The smallest repeating portion of space lattice/crystal lattice. A unit cell is characterised by,
1. The distance along with three edges: a, b & c
2. Angle between the edges:
α (between b & c)
β (between a & c)
γ (between a & b)

Primitive and Centred Unit Cells
a. Primitive Unit Cell :
the constituent particles are present only on the corner positions of unit cell.

b. Centred Unit Cells
i. Body-Centred Unit Cell (bcc) :
The constituent particles at all the corners as well as at the centre of the unit cell.
ii. Face-Centred Unit Cell (fee) :
The constituent particles at all the corners as well as at the centre of the each face.
iii. End-Centred Unit Cell:
One constituent particle is present at the centre of any two opposite faces besides the one present at its corners.

Seven Primitive Unit Cells and their Possible Variations as Centred Unit Cells:

Number of Atoms in a Unit Cell
1. Primitive Cubic Unit Cell or Simple Cubic Unit Cell:
Primitive cubic unit cell has atoms only at its corner. Each atoms at a corner is shared between eight adjacent unit cells. Therefore the contribution of an atom/molecule to a particular unit cell is 1/8

Total number of atoms in unit cell = 8 × 1/8 = 1

2. Body Centred Cubic Unit Cell (bcc):
In bcc unit cell, particles are present not only at the eight corners but also at the centre of the cube.

In a bcc unit cell:
i. 8 corners x 1/8 per corner atom = 8 × 1/8 = 1 atom

ii. 1 body centre atom = 1×1 = 1 atom
∴ Total number of atoms per unit cell = 1 + 1=2 atoms

3. Face Centred Cubic Unit Cell :
A fee unit cell contains atoms at all the corners as well as at the centre of 6 faces of the cube. The atoms at the face centre is shared between two adjacent unit cells and the contribution is only ½ to the unit cell.

In a fcc unit cell:
i. 8 comers x 1/8 per corner atom = 8 × 1/8 = 1 atom
ii. 6 face centre x ½ atom per unit cell = 6 × ½ = 3 atoms
∴ Total number of atoms per unit cell = 1+3 = 4 atoms

Close Packed Structures:
In solids, the constituent particles (considered as identical hard spheres) are close-packed, leaving the minimum vaccnt space.

Coordination number (C.N.) :
The number of nearest neighbouring particles (spheres) in a crystal lattice is called coordination number.

a. Close Packing in One Dimension :
In this arrangment, each sphere is in contact with two of its neighbours. The cordination number is two.

b. Close Packing in Two Dimensions :
Two dimensional close packing can be done in two different ways:

i. Square Close Packing (scp):
The second row is placed in contact with the first one such that the spheres are exactly above those of the first row. The spheres of the two rows are aligned horizontally as well as vertically. Each sphere is in contact with four of its neighbours. Thus coordination number is 4. This is also known as AAA type arrangment.

ii. Hexagonal Close Packing (hep) :
The spheres in every second row are placed in the depressions of the first row. The spheres in the third row are seated in the depressions of the second row and so on. The fourth layer is aligned with the second layer. Hence this type of arrangement is of ABAB type. Each sphere is in contact with 6 neighbouring spheres. Thus the C.N. is 6.

In hexagonal close packing, particles are more closely packed than in square close packing. Hence, it is more efficient than square close packing.

c. Close Packing in Three Dimensions :
i. Three-dimensional close packing from two-dimensional square close-packed layers :
the second layer is placed over the first layer such that the spheres of the upper layers are exactly above those of the first layer. Here, spheres of both the layers are perfectly aligned horizontally as well as vertically. This type of arrangement has AAA type pattern. This arrangement generates the simple cubic lattice.

ii. Three dimensional close packing from two-dimensional hexagonal close-packed layers
a. Placing second layer over the first layer :
The spheres of the second layer are placed in the depressions of the first 2 dimensional hep layer. Wherever a sphere of the second layer is above the triangular void of the first layer (or vice versa) a tetrahedral void is formed.|

The vaccant space/sites in close packed structure are called voids/interstitial voids.

Tetrahedral void (td void) :
The vacant space surrounded by 4 spheres are called tetrahedral voids.

Octahedral void (oh void):
The vaccant space surrounded by six spheres in a close packed arrangement.

If ‘N’ is the number of spheres in close packed structure,

b. Placing third layer over the second layer :
If particles in the third layer are arranger) in the tetrahedral voids, the spheres of the third layer are exactly aligned with those of the first layer. Thus, the pattern of spheres is repeated in alternate layers. This is known as ABAB or hexagonal close packed (hep) structure, e.g. Mg, Zn, Cd etc.

If particles in the third layer are arranged in the octahedral voids, spheres of the fourth layer are aligned with those of the first layer. Thus, ABCABC … arrangement is obtained. This is known as cubic close packed (ccp) structure or face-centred cubic (fee) structure, e.g. Cu, Ag, Au etc.

Both the hep and ccp are highly efficient and 74% space in the crystal is filled. Coordination number is 12 in either of these two structures.

Packing Efficiency
It is the percentage of total space filled by the particles.
Let ‘a’ be the edge length of a unit cell and ‘r’ the radius of sphere.

Packing Efficiency in ccp and hep Structures:
In the case of ccp and hep, the edge length,

Packing Efficiency of Body Centred Cubic Structures :
In this case radius of a sphere,

Packing Efficiency in Simple Cubic Lattice :
In simple cubic lattice edge length, ‘a’ and radius of the sphere ‘r’ are related as,
a = 2r
r = a/2
We know that a simple cubic unit cell contains only one sphere.

Calculations Involving Unit Cell Dimensions :
Edge length of unit cell = a
Volume of the unit cell = a³
Mass of unit cell = Number of atoms in unit cell × mass of each atom = z × m
Mass of an atom in unit cell m = \(\frac{M}{N_{A}}\) (M-molar mass)

Note:
The density of the unit cell is the same as the density of the substance.

Imperfection in Solids :
The crystal defects are irregularities in the arrangement of constituent particles. There are two types of defects:
1. Point Defects:-
Irregularities from ideal arrangement around a point or an atom.
2. Line Defects:-
Irregularities/deviations from ideal arrangement in entire rows of lattice points.

Point Defects:
They can be classified into three types – stoichiometric defects, impurity defects and non-stoichiometric defects.

a. Stoichiometric Defects:
These are the point defects that do not disturb the stoichiometry of the solid. They are also called intrinsic or thermodynamic defects. These are of two types:
i. Vacancy Defect:
When some of the lattice sites are vacant (missing of constituent particles), the crystal is said to have vacancy defect. As a result density of the substance decreases.

ii. Interstitial Defect:
When some constituent particles occupy an interstitial site, the crystal is said to have interstitial defect.

There are two types of stoichiometric defects in ionic solids : Schottky Defect and Frenkel Defect.

Schottky Defect:
It arises due to the missing of equal number of cations and anions from their normal positions leaving behind a pair of holes. It is observed in ionic compounds having high coordination numberwith ions of almost similar size. Since equal number of cations and anions are missing they maintain electrical neutrality. Density of the substance decreases.
e.g. NaCl, KCl, CsCl and AgBr.

Frenkel Defect:
It arises due to an ion, usually cation which is dislocated from its normal site to an interstitial site. It creates a vaccancy defect at its original site and an interstitial defect at its new location. This is also called dislocation defect. It is usually observed in ionic compounds having low coordination number and crystals with anions much larger in size than the cations. Since no ions are missing, density of the solid does not change, e.g. ZnS, AgCI, AgBrandAgl.
Note: AgBr shows both Schottky as well as Frenkel defects.

Note: AgBr shows both Schottky as well as Frenkel defects.

b. Impurity Defects :
Defect caused by foreign ions. e.g. If molten NaCl containing a little amount of SrCl₂ is crystallised, some of the sites of Na⁺ ions are occupied by Sr²⁺. Each Sr²⁺ replaces two Na⁺ ions. It occupies the site of one Na⁺ ion and the othersite remains vaccant. The cationic vaccancies thus pro duced are equal in number to that of Sr²⁺ ions. Another example is the solid solution of CdCl₂ and AgCl.

c. Non-Stoichiometric Defects:
The stoichiometry of the crystal is altered due to defects. These defects are of two types:
i. Metal Excess Defect
ii. Metal Deficiency Defect

i. Metal Excess Defect
1. Due to anionic vacancies :
A compound may have excess of metal ion if a negative ion is absent from its lattice site leaving a hole which is occupied by electron. The anionic sites occupied by unpaired electrons are called F-centres. (German word Farbenzenter means colour centre). These can impart colourto the crystal by excitation of electrons when they absorb energy from visible light, e.g. When crystals of NaCl are heated in an atmosphere of Na vapour, F-centres are formed. As a result, the crystal has an excess of sodium which imparts yellow colour.
Excess Li in LiCl – imparts pink colour
Excess K in KCl – imparts violet(lilac) colour

2. Metal excess defect due to extra cations at interstitial sites :
Here, electrical neutrality is maintained by the presence of an electron in the interstitial site. e.g. Zinc oxide (ZnO) is white in colour at room temperature. On heating it loses O₂ and turns yellow.

Now there is excess of zinc in the crystal and its formula becomes Zn_1+x O. The excessZn²⁺ ions move to interstitial sites and electrons move to neighbouring interstitial sites.

ii. Metal Deficiency Defect
1. Due to cation vaccancies:
A positive ion is missing from its lattice postion and the extra negative charge thus created is balanced by the adjacent metal ion which attains one additional positive charge.
e.g. FeO. It is mostly found with a composition of Fe_0.95O. In crystal of FeO some Fe²⁺ cations are missing and the loss of positive charge is made up by the presence of required number of Fe³⁺ ions. As a result, the crystal has metallic lustre. Other example is FeS (fool’s gold).

2. Due to the presence of anions in interstitial site:
An extra negative ion would occupy an interstitial site and the extra negative charge thus formed is balanced by an adjacent cation possessing additonal positive change. This defect is not common because anions are bigger than cations and cannot be occupied in interstitial sites.

Properties of Solids
Electrical Properties :
Solids can be classified into 3 types on the basis of their conductivities.

1. Conductors :
Solids with conductivities of the order of 10⁴ to 10⁷ ohm^-1 m^-1. Metals are good conductors of electricity and the conductivity is in the order of 10⁷ ohm^-1 m^-1.

ii. Insulators :
The solids which almost do not allow the passage of electricity, e.g. S, P, wood, paper, rubber. Conductivity order 10^-20– 10^-10 ohm^-1 m^-1

iii. Semiconductors :
The solids whose conductivity lies between metallic conductors and insulators. Conductivity range from 10^-6 to 10⁴ ohm^-1 m^-1 (or 10^-8 to 10² ohm^-1cm^-1).

Oxides like TiO, VO, ReO₃ etc. are good conductors. But oxides like Ti₂O₃, V₂O₃ etc. behave as insulators at centain termperature. TiO₂, V₂O₅ etc. are perfect insulators. The conductivity of semiconductors increases with temperature while that of metals decreases with temperature.

Conduction of Electricity in Metals :
Metals conduct electricity in solid as well as molten state through movement of electrons. The conductivity of metals depend upon the number of valence electrons available per atom. The atomic orbitals of metal atoms form molecular orbitals which are so close in energy to each other as to form a band. If this band is partially filled or if overlap with a higher energy unoccupied conduction band, then the electrons can flow easily under an applied electric field.

Conduction of Electricity in Semiconductors :
Here, the energy gap between the valence band and conduction band is small. Therefore some electrons may jump to conduction band and show some conductivity. Electrical conductivity of semi conductors increases with rise in temperature, since more electrons can jump to conduction band. This type of semiconductors are known as intrinsic semiconductors, e.g. Silicon, Germanium.

The conductivity of intrinsic semi conductors is increased by adding appropriate amount of suitable impurities, which introduce electronic defects. This process is called dopping. It can be done in two ways:

a. By adding electron rich impurities (n-type semiconductors) :
If Si and Ge are dopped with group 15 elements like P or As, the 5th electron is extra and becomes delocalised which is responsible for electrical conductivity. Since conductivity is due to the negatively charged electron it is called n-type semiconductor.

b. By adding electron deficit impurities (p-type semi conductors):
If Si and Ge are dopped with group 13 elements like B, Al or Ga an electron deficient bond or electron hole is produced. Under the influence of electric field, electrons move towards the positively charged plate through electronic holes. It appears as if electron holes are positively charged and are moving towards negatively charged plate. This type semi conductors are called p-type semi conductors.

Application of n-type and p-type semi conductors:

1. Diode is a combination of n-type and p-type semi conductors and is used as a rectifier.
2. The npn and pnp types of transistors are used to amplify radio or audio signals.
3. Photo-diode is used for conversion of light energy into electrical energy.

Magnetic Properties :
Every substance has some magnetic properties associated with it. Each electron in an atom behaves like a tiny magnet. Its magnetic moment originates from two types of motions.
i. Electron’s orbital motion around the nucleus
ii. Electron’s spin around its own axis
Magnetic moment is measured in Bohr magneton
(B.M), µ_B. 1 B.M. = 9.27 × 10^-24 A m²
Magnetic properties are of 5 types:

i. Paramagnetism :
It is due to the presence of one or more unpaired electrons. Paramagnetic substances are weakly attracted by a magnetic field.
e.g. O₂, Cu²⁺, Fe³⁺, Cr³⁺ Ni²⁺, VO, VO₂, CuO, NO

ii. Diamagnetism :
It is due to paired electrons in the substance. Diamagnetic substances are weakly repelled by a magnetic field.
e.g. H₂O, NaCl, C₆H₆, TiO₂, N₂

iii. Ferromagnetism :
It is considered as an extreme case of paramagentism and is caused by spontaneous alignment of magnetic domains (metal ions grouped into small regions) in the direction of the magnetic field. Ferromagnetic substances are strongly attracted by magnetic field. They retain a permanent magnetism even when the field is removed, e.g. Fe, Co, Ni, Alloys of Fe, Co and Ni, CrO₂ Once such a material is magnetised, it remains permanently magnetised.
Alignment of magnetic moments: ↑↑↑↑↑↑

iv. Antiferromagnetism:
It arises due to the alignment of magnetic domains in opposite direction and the resulting moment is zero. Antiferromagnetic substances are expected to possess paramagnetism or ferromagnetism on the basis of unpaired electrons but actually they possess zero net magnetic moment.
e.g. MnO, MnO₂, Mn₂O₃, FeO, NiO, CuO
Alignment of magnetic moments: ↑↓↑↓↑↓

v. Ferrimagnetism :
It is due to the alignment of magnetic moments in opposite directions in unequal numbers resulting in a net magnetic moment. These substances are expected to possess large magnetism on the basis of unpaired electrons but actually have small net magnetic moment and are weakly attracted by magnetic field as compared to ferromagnetic substances.
e.g. Fe₃O₄ (magnetite), MgFe₂O₄ & ZnFe₂O₄.
Alignment of magnetic moments: ↑↑↓↑↑↓

Students can Download Chapter 10 The s Block Elements Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements

Plus One Chemistry The s Block Elements One Mark Questions and Answers

Question 1.
The element placed at the bottom of the alkali metal family is expected to
a) Have maximum ionisation enthalpy
b) Be the least reducing agent
c) Be the least electropositive element
d) Be the most easily ionisable
Answer:
d) Be the most easily ionisable

Question 2.
Which of the following have lowest thermal stability?
a) Li₂CO₃
b) Na₂CO₃
c) K₂O₃
d) Rb₂CO₃
Answer:
a) Li₂CO₃

Question 3.
The chemical formula of Plaster of Paris is
a) Ca₂SiO₄.½H₂O
b) Ca₂SiO₄.2H₂O
c) CaSO₄.½H₂O
d) CaSO₄.2H₂O
Answer:
c) CaSO₄.½H₂O

Question 4.
A mixture of NaOH and CaO is known as _________ .
Answer:
Soda lime

Question 5.
Which of the following is least soluble in water?
a) BeSO₄
b) BaSO₄
c) CaSO₄
d) SrSO₄
Answer:
BaSO₄

Question 6.
Pick out the odd one and write reason for it.
Ca(OH)₂, Mg(OH)₂, Ba(OH)₂, Be(OH)₂ Sp
Answer:
Be(OH)₂. The others hydroxides are basic but Be(OH)₂ is amphoteric.

Question 7.
The stability of alkaline earth metal carbonate
Answer:
Increases from Be to Ba

Question 8.
Mg₂C₃ on hydrolysis gives _________
Answer:
Propyne

Question 9.
The raw materials required for the manufacture of cement clinker are
Answer:
limestone & clay

Question 10.
A sodium potassium alloy is used as a _________ .
Answer:
coolant in nuclear reactor.

Question 11.
Li₂CO₃ decomposes at a lower temperature whereas Na₂CO₃ at highertemperature _________ .
Answer:
Li⁺ is very small

Plus One Chemistry The s Block Elements Two Mark Questions and Answers

Question 1.
Choose the false statements and rewrite them.
a) Alkali metals possess both +1 & +2 oxidation states.
b) Lithium is found to be the strongest reducing agents among the alkali metals.
c) Manufacturing of rayon is known as viscose process.
d) Washing soda is used to remove temporary hardness of water.
Answer:
a) False. Alkali metals possess only +1 oxidation state.
b) True
c) True
d) False. Washing soda is used to remove the permanent hardness.

Question 2.
a) Write the scientific name of slaked lime.
b) What is its role in white wash?
c) What do you mean by slaking of lime?
d) What happens when slaked lime is treated with dry chlorine?
Answer:
a) Calcium hydroxide
b) It is disinfectant hence it is used in white wash.
c) Calcium hydroxide is obtained by adding water to quick lime. This process is called slaking of lime.
d) Calcium hydroxide reacts with dry chlorine to form calcium hypochlorite, a constituent of bleaching powder.
2Ca(OH)₂ + 2Cl₂ → CaCl₂ + CaOCl₂ + 2H₂O

Question 3.
1. Which metal is present in chlorophyll?
2. What is the role of calcium in our body?
Answer:
1. Mg
2. About 90% of body calcium is present in bones and teeth. It also plays important roles in neuromuscular function, intemeuronal transmission, cell membrane integrity and blood coagulation.

Question 4.
A compound is used as drying agent as such or as soda lime with NaOH and it is used on a very large scale as a building material.
a) Which is this compound?
b) How can we prepare this compound?
c) What are the properties of this compound?
Answer:
a) Calcium oxide
b) It can be prepared by heating lime stone in a rotary kiln at 1070-1270 K.
c) Pure calcium oxide is an amorphous white solid of very high melting point. It is sparingly soluble in water. Calcium oxide readily absorbs moisture and carbon dioxide.

Question 5.
Listen the chemical reaction,

1. Write the common name of the reactant.
2. What is the temperature corresponding to A?
3. What is B (product)? Write its chemical formula.

Answer:

1. Gypsum
2. 393 K
3. Plaster of Paris (CaSO₄. ½H₂O) or (CaSO₄)₂.H₂O

Question 6.
Explain with the help of chemical equations what happens when

1. lime stone is heated?
2. water is dropped on quick lime?
3. gypsum is heated to 393 K?

Answer:

1. When lime stone is heated, it is converted into CaO and CO₂.

2. When water is dropped on quick lime calcium hydroxide is formed.
CaO + H₂O → Ca(OH)₂
3. When gypsum is heated to 393 K, Plaster of Paris is obtained.

Question 7.
Alkali metal halides are all high melting, colourless crystalline solids.

1. Write any other physical property of alkalimetal halides.
2. How alkali metal halides are prepared?

Answer:
1. All the alkali metal halides are soluble in water except LiF. They have high negative enthalpies of formation. The melting and boiling points always follow the trend: fluoride > chloride > bromide > iodide.

2. Alkali metal halides are prepared by the reaction of the appropriate oxide, hydroxide or carbonate with aqueous hydrohalic acid.

Question 8.
What is Plaster of Paris? How is it prepared? What is its use?
Answer:
The chemical formula of plaster of Paris is (CaSO₄)₂.H₂O.
It is prepared by heating gypsum at 393 K.

It is a hemihydrate of calcium sulphate.
It is a white powder. On mixing with an adequate quantity of water it forms a plastic mass that gets into a hard solid in 5 to 10 minutes.

Plaster of Paris is used

• for making moulds for pottery, ceramics, etc.
• for making models, statues and decorative materials.
• for immobilising the affected part of organ where there is a bone fracture or sprain.

Question 9.
Give the biological importance of Na and K.
Answer:
Biological importance of sodium: The Na⁺ ions participate in the transmission of nerve signals, in regulating the flow of water across cell membranes and in the transport of sugars and amino acids into cells. Biological importance of potassium: The K⁺ ions activate many enzymes, participate in the oxidation of glucose to produce ATP and, with sodium, are responsible for the transmission of nerve signals.

Question 10.
1. Name the element showing anomalous behaviour in group 2.
2. Give reason forthis anomalous behaviour,
3. List any two similarities between Be and Al.
Answer:
1. Be

2. Be has exceptionally small atomic and ionic size, it has high ionisation enthalpy, it cannot exhibit coordination number more than four as in its valence shell there are only four orbitals. There are no d-orbitals.

2. i) Both Be and Al are not readily attacked by acids because of the presence of an oxide film on the surface of the metal.
ii) The chlorides of both Be and Al have Cl^– bridged chloride structure in vapour phase.

Plus One Chemistry The s Block Elements Three Mark Questions and Answers

Question 1.
A compound of calcium is used for immobilising the fractured bones of body.

1. Write down the common name and molecular formula of the compound.
2. Which property of the compound helps to make plaster?
3. What do you mean by dead burnt plaster? How does it form?

Answer:
1. Plaster of Paris (CaSO₄. ½H₂O) or (CaSO₄)₂.H₂O

2. When we add water to Plaster of Paris, there is slight increase in volume and this helps Plaster of Paris to take the shape of any mould in which it is present.

3. When Plaster of Paris is heated above 393 K, Plaster of Paris loses water of crystallization to form anhydrous calcium sulphate, CaSO₄. It is known as dead burnt plaster which does not set in presence of water.

Question 2.
Lithium is group 1 element. It shows some similarities with group 2 element magnesium.
1. Write the name of the relationship.
2. Explain this relationship.
3. What is the reason for this relationship?
4. Give other example for this relationship.
Answer:
1. Diagonal relationship.

2. The first element of alkali group shows some similarities with the second element of the second group, which is diagonally present. This relation is called diagonal relationship.

3. Diagonal relationship is due to the following reasons:

• Similarity in their polarising powers due to similar charge/radius ratio.
• Almost similar electronegativity of the 2 elements.

4.

• Berylium and Aluminium
• Boron and Silicon

Question 3.
Mentionafewimportarrtusesofthefollowing compounds.

1. Epsom salt
2. Marbl
3. Sodium hydroxide

Answer:
1. Epsom salt:
It is used as a mordant in dyeing. Used in medicine in purgative. Used in tanning industry.

2. Marble:
It is used to make quick lime, cement etc. Used as a raw material in the Solvay process. Used as a building material.

3. Sodium hydroxide:
It is used as a laboratory reagent. Used in petroleum refining. Used in viscose process.

Question 4.
Cement is a complex mixture of silicates and aluminates.
1. What is the function of gypsum in cement?
2. Explain setting of cement.
Answer:
1. The purpose of adding gypsum is only to slow down the process of setting of the cement so that it gets sufficiently hardened.

2. When the cement is mixed with water, it forms a gelatinous mass which sets slowly to a hard mass having 3 dimensional network structure with -Si- O-Si and -Si-O-AI- chains. This is an exothermic process and is called setting of cement.

Question 5.
The alkali metals and their salts impart characteristic colour to an oxidising flame.
1. What is the reason for this?
2. Give the flame colour of Na and K.
3. Name the alkali metal which imparts crimson red colour to the oxidizing flame.
Answer:
1. When the heat energy is supplied to alkali metal or its salt, the electrons are excited to higher energy levels. When these excited electrons come back to their original energy levels, they emit radiations which fall in the visible region of the electromagnetic spectrum and they appear coloured.

2. Na-Yellow, K-Violet

3. Li

Question 6.
a) How will you prepare Ca(OH)₂ and CaCO₃ from quicklime.
b) Give any two uses of quick lime.
Answer:
a) By adding water to quick lime we can prepare Ca(OH)₂. This process is called slaking.
CaO + H₂O → Ca(OH)₂
When CO₂ is passed thruogh lime water we can prepare CaCO₃.
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

b) 1. As an important primary material for manufacturing cement.
2. In the manufacture of Na₂CO₃ from NaOH.

Question 7.
Lithium shows similarities in properties with Magnesium.
a) Namethe above phenomenon.
b) Give any two similarities of Lithium with Magnesium.
c) How is bleaching powder prepared?
Answer:
a) Diagonal relationship

b) 1) Both Lithium and Magnesium are harder and lighter than other elements in the respective groups.
2) The oxides of both (Li₂O and MgO) do not combine with excess O₂ to give any superoxide.

c) It is prepared by passing Cl₂ gas through dry slaked lime.
2Ca(OH)₂ + 2Cl₂ → CaCl₂ + CaOCl₂ + 2H₂O

Plus One Chemistry The s Block Elements Four Mark Questions and Answers

Question 1.
1. What is the name of the method that is used in manufacturing of sodium hydroxide?
2. Explain the method.
3. Write the equations of the reactions involved in this process.
Answer:
1. Castner-Kellner method

2. The Castner-Kellner cell consists of a large tank. The bottom of the tank is filled with mercury. The tank is made into 3 compartments by 2 partitions. The outer compartments are filled with NaCl solution and the middle compartment is filled with a very dilute solution of NaOH. The mercury layer of the bottom of cell in the outer compartments acts as cathode and the mercury in the middle compartments acts as anode. As a result of electrolysis Na and Cl₂ are produced in the outer compartments. Na gets coated with mercury, and sodium amalgam is produced in the outer compartments. Due to the action of eccentric wheels at the bottom of the tank, the tank can be tilted. As a result of this Na is carried into the middle compartment and is then allowed to react with water in the middle compartment. As a result of this reaction NaOH is produced. When the concentration of NaOH in the middle compartment is raised to 40%, it is taken out from the tank. On cooling, crystals of NaOH separate out.

3. NaCl → Na⁺ + Cl^–
At cathode: Na⁺ + e^– → Na
Na + Hg → Na/Hg (Sodium amalgam)
At anode: Cl^– → Cl + e^–
Cl + Cl → Cl₂ (g)

Question 2.
Match the following:

Answer:

Question 3.
A piece of metallic sodium is added to liquid ammonia.

1. What is the observation?
2. What is the reason for this?
3. What happens when the solution is kept for some time?
4. What happens if the solution is concentrated?

Answer:

1. The solution turns into blue.
2. The blue colour of the solution is due to the presence of solvated electrons.
3. On standing, the solution slowly liberates hydrogen resulting in the formation of amide.
4. When the concentration of the solution increases the colour turns bronze and the solution becomes diamagnetic.

Question 4.
1. Some compounds of sodium and calcium are given below. Match them.

2. On passing CO₂ through lime water, milkiness appears. On further passing CO₂ milkiness disappears. What is the Chemistry behind it?
Answer:
1.

2. When CO₂ is passed through lime water it turns milky due to the formation of insoluble calcium carbonate.
Ca(OH)₂ + CO₂ → CaCO₃ +H₂O
When CO₂ is passed through lime water for a long time, it becomes colourless due to the formation of water soluble calcium bicarbonate.
CaCO₃ + H₂O + CO₂ → CaOHCO₃)₂

Question 5.
Cement is an important building material employed in different kinds of construction works.

1. What are the major raw materials for making cement?
2. How is cement prepared?
3. What are the important ingredients of Portland cement?
4. Explain the Chemistry involved in the setting of cement.

Answer:
1. Limestone, Clay

2. When clay and lime are strongly heated together they fuse and react to form cement clinker. This clinker is mixed with 2-3% by weight of gypsum to form cement.

3. Dicalcium silicate (Ca₂SiO₄) – 26%, tricalcium silicate (Ca₃SiO₅) – 51% and tricalcium aluminate (Ca₃Al₂O₆) – 11%

4. When mixed with water, the setting of cement takes place to give a hard mass. This is due to the hydration of the molecules of the constituents and their rearrangement.

Question 6.
1. Match the following :

2. Study the list of elements given below:
Na, Mg, Li, B, C
Select the pairs showing diagonal relationship.
3. Write the chemical equation showing the reaction of sodium metal with water.
Answer:
1.

2. Li, Mg
3. 2Na + 2H₂O → 2NaOH + H₂

Question 7.
a) Arrange the following compounds in the increasing order of solubility in water:
i) Mg(OH)₂, Be(OH)₂, Ba(OH)₂, Ca(OH)₂, Sr(OH)₂
ii) SrSO₄, MgSO₄, BeSO₄, BaSO₄, CaSO₄,
b) i) Write the difference between lime water and milk of lime.
ii) Complete the following reaction.

Answer:
a) i) Be(OH)₂ < Mg(OH)₂< Ca(OH)₂< Sr(OH)₂< Ba(OH)₂
ii) BaSO₄ < SrSO₄< CaSO₄< MgSO₄< BeSO₄

b) i) A clear solution of calcium hydroxide in water is called lime water. A suspension of calcium hydroxide in water is called milk of lime,

Plus One Chemistry The s Block Elements NCERT Questions and Answers

Question 1.
Explain why alkali and alkaline earth metals cannot be obtained by chemical reduction methods? (2)
Answer:
Alkali and alkaline earth metals are themselves very strong reducing agents and reducing agents stronger than them are not easily available. Therefore, these metals cannot be obtained by chemical reduction methods.

Question 2.
Why are potassium and caesium, rather than lithium used in photoelectric cells? (2)
Answer:
Potassium and caesium have much lower ionisation enthalpy than that of lithium. As a result, these metals on exposure to light, lose electrons much more easily than lithium. Hence, potassium and caesium rather than lithium are used in photoelectric cells.

Question 3.
When an alkali metal dissolves in liquid ammonia the solution can acquire different colours. Explain the reasons for this type of colour change. (2) Answer:
When an alkali metal is dissolved in liqiud ammonia it produces a blue coloured conducting solution due to formation of ammoniated cation and ammoniated electron as given below:

When the concentration is above 3 M, the colour of solution is copper-bronze. This colour change is because the concentrated solution contains clusters of metal ions and hence possess metallic lustre.

Question 4.
Beryllium and magnesium do not give colour to flame whereas other alkaline earth metals do so, why? (2)
Answer:
Due to the small size, the ionisation enthalpies of Be and Mg are much higher than those of other alkaline earth metals. This means that the valence electrons in beryllium and magnesium are more tightly held by the nucleus. Therefore, they need large amount of energy for excitation of electrons to higher energy levels. Since such a large amount of energy is not available in bunsen flame, therefore, these metals do not impart any colour to the flame.

Question 5.
Potassium carbonate cannot be prepared by Solvay process. Why? (2)
Answer:
Potassium carbonate cannot be prepared by Solvay process because potassium biocarbonate being highly soluble in water does not get precipitated in carbonation tower when CO₂ is passed through a concentrated solution of KCI saturated with ammonia.

Question 6.
Why Li₂CO₃ decomposes at lower temperature whereas Na₂CO₃ at higher temperature? (2)
Answer:
Li⁺ ion is smaller in size. It forms more stable lattice with smaller anion oxide, O^2- than with CO₃^2- ion. Therefore, Li₂CO₃decomposes into Li₂O at lower temperature on the other hand, Na⁺ ion being larger in size forms more stable lattice with larger anion CO₃^2- than with O^2- ion. Therefore, Na₂CO₃ is quite stable and decomposes into Na₂0 at very high temperature.

Question 7.
How would you explain? (3)
1. BeO is insoluble but BeSO₄ is soluble in water.
2. BaO is soluble but BaSO₄ is insoluble in water.
3. Lil is more soluble than Kl in ethanol.
Answer:
1. BeO has higher lattice enthalpy than hydration enthalpy and hence is insoluble in water. BeSO₄ on the contrary, is soluble because the lattice enthalpy is less due to bigger sulphate ion.

2. In barium oxide, BaO, due to larger size of barium ion the lattice enthalpy is less than hydration enthalpy and hence it is soluble in water. On the other hand, in BaSO₄ dueto largersize of barium as well as sulphate ion, the magnitude of hydration enthalpy is much smallerthan the lattice enthalpy and hence it is insoluble in water.

3. In Kl the chemical bond is ionic in character. On the other hand, due to small size of lithium ion and its high polarising power, the bond in Li is predominantly covalent in character. Hence, Li is more soluble than Kl in ethanol.

Question 8.
Which of the alkali metals is having least melting point? Justify. (2)
a) Na
b) K
c) Rb
d) Cs
Answer:
d) Cs.
As the atomic size of the metal increases, the strength of metallic bonding decreases and hence its melting point decreases. Since the size of Cs is the largest, therefore its melting point is the lowest.

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Kerala Plus Two Computer Application Chapter Wise Questions and Answers Chapter 11 Trends and Issues in ICT

Plus Two Computer Application Trends and Issues in ICT One Mark Questions and Answers

Question 1.
SIM is_____.
(a) Subscriber Identity Module
(b) Subscriber Identity Mobile
(c) Subscription Identity Module
(d) Subscription Identity Mobile
Answer:
(a) Subscriber Identity Module

Question 2.
The protocol used to send SMS message is______.
Answer:
SS7 (Signalling System No.7)

Question 3.
_____is a standard way to send and receive short text message using mobile phone.
Answer:
Short Message Service(SMS)

Question 4.
_____is a standard way to send and receive message with multimedia content using mobile phone.
Answer:
Multimedia Messaging Service (MMS)

Question 5.
From the following which generation network has more speed?
(a) 1G
(b) 3G
(c) 2G
(d) 4G
Answer:
(d) 4G

Question 6.
GSM stands for______.
Answer:
Global System for Mobiles.

Question 7.
CDMA stands for_____.
Answer:
Code Division Multiple Access

Question 8.
GPRS stands for______.
Answer:
Global Packet Radio Sysetm.

Question 9.
GPS stands for______.
Answer:
Global Positioning System.

Question 10.
EDGE stands for____.
Answer:
Enhanced Data rates for GSM Evolution

Question 11.
_____technology provides speed, voice quality, wider coverage, better security to the mobile network than GSM network.
Answer:
CDMA

Question 12.
Write down the popular standards introduced by 2G network.
Answer:
GSM and CDMA

Question 13.
From the following which one is used analog signals instead of digital signals.
(a) 2G
(b) 3G
(c) 1G
(d) 4G
Answer:
(c) 1G

Question 14.
Consider a person while using internet through mobile phone, it shows G on the network coverage icon. Which type of network is he using and name the communication feature.
Answer:
The network is 2G and the communication feature is GPRS.

Question 15.
Consider a person while using internet through mobile phone, it shows E on the network coverage icon. Which type of network is he using and name the communication feature.
Answer:
The network is 2G and the communication feature is EDGE.

Question 16.
Consider a person while using internet through mobile phone, it shows H on the network coverage icon. Which type of network is he using and name the technology.
Answer:
The network is 3G and the technology is WCDMA

Question 17.
_____is the technology used in 3G.
Answer:
WCDMA (Wideband Code Division Multiple Access)

Question 18.
______is the technology used in 4G.
Answer:
OFDMA(Orthogonal Frequency Division Multiple Access)

Question 19.
From the following which generation network provides good quality images and videos than TV.
(a) 1G
(b) 2G
(c) 3G
(d) 4G
Answer:
(d) 4G

Question 20.
SMS stands for_______.
Answer:
Short Message Service

Question 21.
MMS stands for_______.
Answer:
Multimedia Messaging Service

Question 22.
______is a plastic card with small chip.
Answer:
Smart card

Question 23.
MOS stands for_______.
Answer:
Mobile Operating systems

Question 24.
Name a Mobile Operating System.
Answer:
Android.

Question 25.
KitKat, Jelly Bean, Donut, Cupcake are the different versions of______OperatingSystem.
KitKat, Jelly Bean, Donut, Cupcake 4rmloi
Answer:
Android.

Question 26.
RFID stands for______.
Answer:
Radio Frequency Identification

Question 27.
_______technology helped in Business Logistics to identify, track, sort or detect objects.
Answer:
RFID technology

Question 28.
IPR stands for_______.
Answer:
Intellectual Property Right

Question 29.
WIPO stands for_______.
Answer:
World Intellectual Property Organisation

Question 30.
_______is the exclusive rights to prevent unauthorized copying of inventions by a Creator from the Unauthorised person or company.
Answer:
Patent

Question 31.
______is a unique, simple and memorable sign to promote a brand and hence increase the business and goodwill of a company.
Answer:
Trademark

Question 32.
A product or article is designed so beautifully to attract the customers. This type of design is called______.
Answer:
Industrial Design.

Question 33.
Aranmuia Kannadi, Palakkadan Matta, Marayoor Sarkkara, etc are example of______.
Answer:
Geographical indications.

Question 34.
_______is the property right that arises automatically when a person creates a new work by his own and by Law it prevents the others from the unauthorized or intentional copying of this without the permission of the creator.
Answer:
Copyright

Question 35.
From the following which is the symbol for copyright.
(a)  $
(b) ©
(c) ®
(d) ™
Answer:
(b) ©

Question 36.
From the following which is the symbol for Unregistered trademark.
(a)  $
(b) ©
(c) ®
(d) ™
Answer:
(d) ™

Question 37.
From the following which is the symbol for Registered trademark.
(a)  $
(b) ©
(c) ®
(d) ™
Answer:
(c) ®

Question 38.
Unauthorized copying or use of Intellectual property rights such as Patents, Copy rights and Trademarks are called_______.
Answer:
Intellectual Property Infringement.

Question 39.
______prevents others from the unauthorized or intentional copying or use of Patent without the permission of the creator.
Answer:
Patent Infringement

Question 40.
______is the illegal copying, distribution, or use of software.
Answer:
Piracy

Question 41.
_______prevents others from the unauthorized or intentional copying or use of Trademark without the permission of the creator.
Answer:
Trademark Infringement

Question 42.
_______prevents others from the unauthorized or intentional copying or use of Copy right without the permission of the creator.
Answer:
Copy right Infringement

Question 43.
______is a virtual environment created by computer systems connected to the internet.
Answer:
Cyber space

Question 44.
A person committing crimes and illegal activities with the use of computer over Internet. This crime is included as______crime.
Answer:
Cyber crime

Question 45.
State True or False.
Cyber crimes can be classified into three categories such as against individual, property, and Government.
Answer:
True

Question 46.
Phishing, hacking, denial of service attacks, etc are______crimes.
Answer:
Cyber

Question 47.
Odd one out
(a) Identity theft
(b) Harrassment
(c) vilation of privacy
(d) credit card fraud
Answer:
(d) credit card fraud, it is a cyber crime against individual others are cyber crimes against property

Question 48.
Odd one out
(a) Credit card theft
(b) Intellectual property theft
(c) Internet time theft
(d) Dissemination of obscene material
Answer:
(d) Dissemination of obscene material, It is cyber crime against individual, the others are cyber against property

Question 49.
Odd one out
(a) cyber terrorism
(b) Attacks against e-Governance websites
(c) Impersonation and cheating
(d) Website defacement
Answer:
(c) Impersonation and cheating, it is cyber crime against individual others are cyber crimes against Government

Question 50.
IT Act amended in_______
(a) 2015
(b) 2008
(c) 1900
(d) 1998
Answer:
(b) 2008

Question 51.
IT Act passed in Indian parliament is_______.
Answer:
2000

Question 52.
The laws to prevent cyber crimes is termed as______.
Answer:
Cyber law

Question 53.
_______ is the excessive enthusiasm for acquiring knowledge.
Answer:
Infomania

Question 54.
Phishing is an example of_______.
Answer:
Cyber crime

Question 55.
Expand the term GPRS.
Answer:
General Packet Radio Services.

Question 56.
SMS messages are exchanged using the protocol called_______.
Answer:
SS7(Signaling System No.7)

Question 57.
Pick the odd one out and justify your answer.
(A) SMS
(B) MMS
(C) GPS
(D) Android
Answer:
(D) Android. It is a mobile operating System. Others are mobile communication services.

Question 58.
Which among the following statements is wrong in relation with 2G network?
(A) It support MMS.
(B) It can provide only voice Service.
(C) GSM and CDMA are two popular standards introduced in 2G.
(D) 2G Network were later expanded to include GPRS and EDGE.
Answer:
(B) It can provide only voice service.

Question 59.
Pick the odd one out.
(A) Windows
(B) Android
(C) IOS
(D) Linux
Answer:
(D) Linux. Others are mobile Operating Systems.

Question 60.
_____refers to the exclusive right given to a person over the creation of his/her mind fora period of time.
Answer:
Patent / Intellectual Property Right

Question 61.
Which among the following are considered as violation to privacy?

1. Keeping hidden cameras in private places
2. Publishing private photos of individual in social media without their permission
3. Use of unauthorized software
4. Using simple password

(A) All the above are correct
(B) 1, 2 and 3 only
(C) 1 and 4 only
(D) 1 and 2 only
Answer:
(D) 1 and 2 only

Plus Two Computer Application Trends and Issues in ICT Two Mark Questions and Answers

Question 1.
What is mobile computing?
Answer:
The drawbacks of Desk computers are, it is heavy and power consumption rate is high and it is not portable (not mobile).

The advancements in computing technology, light weight, and low power consumption have led to the developments of more computing power in hand held devices like laptops, tablets, smart phones, etc.

Nowadays instead of desktops, light weight and low power consumption devices are used because they are cheap and common. Moreover, people are able to connect to others through internet even when they are in move.

Question 2.
Write short notes on SMS.
Answer:
It allows transferring short text messages containing up to 160 characters between mobile phones. The sent message reaches a Short Message Service Center(SMSC), that allows ‘store and forward’ systems.

It uses the protocol SS7(Signaling System No7). The first SMS message ‘Merry Christmas’ was sent on 03/12/1992 from a PC to a mobile phone on the Vodafone GSM network in UK.

Question 3.
Expand GPS? Explain?
Answer:
It is a space-based satellite navigation system that provides location and time information in all weather conditions, anywhere on or near the Earth where there is an unobstructed line of sight to four or more GPS satellites.

The system provides critical capabilities to military, civil and commercial users around the world. It is maintained by the United States government and is freely accessible to anyone with a GPS receiver. GPS was created and realized by the U.S.

Department of Defense (DoD) and was originally run with 24 satellites. It is used for vehicle navigation, aircraft navigation, ship navigation, oil exploration, Fishing, etc. GPS receivers are now integrated with mobile phones.

Question 4.
What is a smart card? How is it useful?
Answer:
A smart card is a plastic card with a computer chip or memory that stores and transacts data. A smart card (may be like your ATM card) reader used to store and transmit data. The advantages are it is secure, intelligent and convenient. The smart card technology is used in SIM for GSM phones. A SIM card is used as an identification proof.

Question 5.
How do trademark and industrial design differ?
Answer:
1. Trademark:
This is a unique, simple and memorable sign to promote a brand and hence increase the business and goodwill of a company. It must be registered. The period of registration is for 10 years and can be renewed. The registered trademark under Controller General of Patents Design and Trademarks cannot use or copy by anybody else.

2. Industrial designs:
A product or article is designed so beautifully to attract the customers. This type of designs is called industrial design. This is a prototype and used as a model for large scale production.

Question 6.
Explain the exclusive right given to the owner by IPR?
Answer:
The exclusive right given to the owner by IPR is owner can disclose their creations for money.

Question 7.
What is piracy?
Answer:
It is the unauthorized copying, distribution, and use of a creation without the permission of the creator. It is against the copy right act and hence the person committed deserve the punishment.

Question 8.
What do you meant by infringement?
Answer:
Unauthorized copying or use of Intellectual property rights such as Patents, Copy rights and Trademarks are called intellectual property lnfringement(violation). It is a punishable offence.

Question 9.
Match the following.

Answer:

• (a) – (iii)
• (b) – (ii)
• (c) – (iv)
• (d) – (i)

Plus Two Computer Application Trends and Issues in ICT Three Mark Questions and Answers

Question 1.
Compare GSM and CDMA standards.
Answer:
1. Global System for Mobile (GSM):
It is the most successful standard. It uses narrow band TDMA(Time Division Multiple Access), allows simultaneous calls on the same frequency range of 900 MHz to 1800 MHz. The network is identified using the SIM(Subscriber Identity Module).

2. GPRS (General Packet Radio Services):
It is a packet oriented mobile data service on the 2G on GSM. GPRS was originally standardized by European Telecommunications Standards Institute (ETSlJ GPRS usage is typically charged based on volume of data transferred. Usage above the bundle cap is either charged per megabyte or disallowed.

3. EDGE(Enhanced Data rates for GSM Evolution):
It is three times faster than GPRS. It is used for voice communication as well as an internet connection.

4. Code Division Multiple Access (CDMA):
It is a channel access method used by various radio communication technologies. CDMA is an example of multiple access, which is where several transmitters can send information simultaneously over a single communication channel.

This allows several users to share a band of frequencies To permit this to be achieved without undue interference between the users, and provide better security.

Question 2.
Differentiate GPRS and EDGE?
Answer:
1. GPRS(General Packet Radio Services):
It is a packet oriented mobile data service on the 2G on GSM. GPRS was originally standardized by European Telecommunications Standards Institute (ETSI) GPRS usage is typically charged based on volume of data transferred. Usage above the bundle cap is either charged per megabyte or disallowed.

2. EDGE(Enhanced Data rates for GSM Evolution):
It is three times faster than GPRS. It is used for voice communication as well as an internet connection.

Question 3.
Explain the features of Android OS.
Answer:
It is a Linux based OS for Touch screen devices such as smart phones and tablets.lt was developed by Android Inc. founded in Palo Alto, California in 2003 by Andy Rubin and his friends. In 2005, Google acquired this.

Ateam led by Rubin developed a mobile device platform powered by the Linux Kernel. The interface of Android OS is based on touch inputs like swiping, tapping, pinching in and out to manipulate on screen objects.

In 2007 onwards this OS is used in many mobile phones and tablets. Android SDK(Software Development Kit) is available to create applications(apps) like Google Maps, FB, What’s App.etc. It is of open source nature and many Apps are available for free download from the Android Play Store hence increase the popularity.

Question 4.
What is cyberspace?
Answer:
Earlier Traditional communication services such as postal service(Snail mail) are used for communication. It is a low speed and not reliable service. In order to increase the speed Telegram Services were used. Its speed was high but it has lot of limitations and expensive too. Latertelephoneswere used for voice communication.

Nowadays telephone system and computer system are integrated and create a virtual(unreal) environment. This is called cyber space. The result for this integration is that tremendous speed and it is very cheap.

Question 5.
Why is cyberspace called a virtual world?
Answer:
The telephone system and computer system are integrated and create a virtual(unreal) environment. This is called cyber space. The result for this integration is that tremendous speed and it is very cheap.

This is an imaginary world. We can see persons with different behaviour. Because of good and bad people we can’t believe blindly. If we search a solution fora problem thousands of answers will get instantly and may confused us.

Question 6.
What is copyright? How does it differ from patent?
Answer:
1. Copyright:
The trade mark is ©, copy right is the property right that arises automatically when a person creates a new work by his own and by Law it prevents the others from the unauthorized or intentional copying of this without the permission of the creator for 60 years after the death of the author.

2. Patents:
A person or organization invented a product or a creation can be protected from unauthorized copying or creation without the permission of the creator by law. This right is called Patent. In India the validity of the right is up to 20 years. After this anybody can use freely.

Question 7.
Why is Cyber law important?
Answer:
Just like normal crimes (theft, trespassing private . area, destroy, etc.) Cyber crimes (Virus, Trojan Horse, Phishing, Denial of Service, Pornography, etc.) also increased significantly. Due to cyber crime, the victims lose money, reputation, etc. and some of them commit suicide.

Cyber law ensures the use of computers and Internet by the people safely and legally. It consists of rules and regulations like Indian Penal Code (IPC) to stop crimes and for the smooth functions of Cyber world. Two Acts are IT Act 2000 and IT Act Amended in 2008.

Question 8.
“Infomania has became a psychological problem”. Write your opinion.
Answer:
Infomania is the excessive desire(lnfatuation) for acquiring knowledge from various modern sources like Internet, Email, Social media, Instant Message Application(WhatsApp) and Smart Phones. Due to this the person may neglect daily routine such as family, friends, food, sleep, etc. hence they get tired.

They give first preference to Internet than others. They create their own Cyber World and no interaction to the surroundings and the family. They are more anxious and afraid that they will be out from the cyber world unless they updated.

Question 9.
What do you mean by big data in business? Explain big data analytics.
Answer:
Earlier before buying a product people may consult two or three shop keepers or local friends and take decisions. But nowadays before taking decisions people search shopping sites, social network groups(Facebook, WhatsApp, Instagram, twitter, etc), web portals etc.forthe best prices. Almost all online sites have product comparison menus.

By this we can compare the price, features, etc. Earlier a product is created and customers are forced to buy. But today customer is the King of the market, so products are created for the choices of the customers.

So companies gathering information about the customers from various sources such as social medias like Internet forums, social blogs, Micro blogs, etc. The volume of such data is very large and considered as big data in business.

With the help of a s/w analysis this big data and generate a report that contain all the information such as choices, taste, needs, status, etc of a customer.

Question 10.
What do you mean by business logistics?
Answer:
It is the management of the flow(transportation) of resources such as food, consumer goods, services, animals, etc in a business between the point of origin (source) and the point of consumption (destination) in order to meet the needs of companies and customers.

Business logistics consists of many more complexities. The effective use of hardware and software reduces the complexities faced in Business logistics. For this the hardware used is RFID(Radio Frequency Identification) tag and the reader.

It is like the barcode. The RFID tag contains all the details of a product and it consists of a combination of transmitter and a receiver. The data stored in the RFID tag can be accessed by a special reader and to read the data no need of RFID tag and reader in a line of site instead both are within a range.

This tag is used in Vehicles as a prepaid tag and makes the payments easier in Toll booths. Similarly, it is useful to take the Census of wild animals also.

Question 11.
How does RFID improve the way business is done?
Answer:
The data stored in the RFID tag can be accessed by a special reader and to read the data no need of RFID tag and reader in a line of site instead both are within a range.

This tag is used in Vehicles as a prepaid tag and makes the payments easier in Toll booths. Similarly, it is useful to take the Census of wild animals also.

In business the tag(contains details about Product code, Price, Batch no., Manufacturing Date, Expiry date, etc.) is stick on the cartons and by using RFID reader process it speedily.

Question 12.
Define Mobile computing.
Answer:
1. Mobile computing:
The drawbacks of Desk computers are, it is heavy and power consumption rate is high and it is not portable(not mobile).

The advancements in computing technology, light weight, and low power consumption have led to the developments of more computing power in hand held devices like laptops, tablets, smart phones, etc.

Nowadays instead of desktops, light weight and low power consumption devices are used because they are cheap and common. Moreover, people are able to connect to others through internet even when they are in move.

Question 13.
“2G networks introduced data services for the mobile. Two popular standards of 2G systems are GSM and CDMA.”

1. Expand CDMA.
2. Discuss the different technologies that are used to enhance data communication features of GSM.

Answer:
1. Code Division Multiple Access.

2. different technologies:

a. GPRS (General Packet Radio Services):
It is a packet oriented mobile data service on the 2G on GSM. GPRS was originally standardized by European Telecommunications Standards Institute (ETSI) GPRS usage is typically charged based on volume of data transferred. Usage above the bundle cap is either charged per megabyte or disallowed.

b. EDGE(Enhanced Data rates for GSM Evolution):
It is three times faster than GPRS. It is used for voice communication as well as an internet connection.

Question 14.
“Mobile communication offers many services apart from the basic voice calling facility”

1. Name the service which helps to locate a geographical position anywhere on the earth.
2. Differentiate SMS and MMS.

Answer:
1. GPS

2. Differentiate SMS and MMS:

a. Short Message Service(SMS):
It allows transferring short text messages containing up to 160 characters between mobile phones. The sent message reaches a Short Message Service Center(SMSC), that allows ‘store and forward’ systems. It uses the protocol SS7(Signaling System No7). The first SMS message ‘Merry Christmas’ was sent on 03/ 12/1992 from a PC to a mobile phone on the Vodafone GSM network in UK.

b. Multimedia Messaging Service (MMS):
It allows sending Multi-Media(text, picture, audio and video file) content using mobile phones. It is an extension of SMS.

Question 15.
Write a short note on GPS.
Answer:
Global Positioning System(GPS):
It is a space-based satellite navigation system that provides location and time information in all weather conditions, anywhere on or near the Earth where there is an unobstructed line of sight to four or more GPS satellites.

The system provides critical capabilities to military, civil and commercial users around the world. It is maintained by the United States government and is freely accessible to anyone with a GPS receiver.

GPS was created and realized by the U.S. Department of Defense (DoD) and was originally run with 24 satellites. It is used for vehicle navigation, aircraft navigation, ship navigation, oil exploration, Fishing, etc. GPS receivers are now integrated with mobile phones.

Question 16.
Write a short note on Android OS.
Answer:
1. Android OS:
It is a Linux based OS for touch screen devices such as smart phones and tablets.lt was developed by Android Inc. founded in Palo Alto, California in 2003 by Andy Rubin and his friends.

In 2005, Google acquired this. A team led by Rubin developed a mobile device platform powered by the Linux Kernel. The interface of Android OS is based on touch inputs like swiping, tapping, pinching in and out to manipulate on screen objects. In 2007 onwards this OS is used in many mobile phones and tablets.

Question 17.
“In some of the states ration cards have been replaced by smart cards”. What is a smart card? List any two advantages of replacing ration card using smart cards?
Answer:
Smart Cards:
A smart card is a plastic card with a computer chip or memory that stores and transacts data. A smart card (may be like your ATM card) reader used to store and transmit data.

The advantages are it is secure, intelligent and convenient. The smart card technology is used in SIM for GSM phones. A SIM card is used as an identification proof.

Question 18.
Business firms have started watching the conversation and opinions posted in social media. Why?
Answer:
Business firms gathering information about the customers from various sources such as social medias like Internet forums, social blogs, Micro blogs, etc. The volume of such data is very large and considered as big data in business.

With the help of a s/w, analyses this big data and generate a report that contain all the information such as choices, taste, needs, status, opinions, suggestions etc of a customer.

Question 19.
“The following is a figure of a tag which helps in business logistics”

1. Name the tag shown in the figure.
2. How it helps business logistics?

Answer:
1. RFID tag

2. The data stored in the RFID tag can be accessed by a special reader and to read the data no need of RFID tag and reader in a line of site instead both are within a range.

This tag is used in Vehicles as a prepaid tag and makes the payments easier in Toll booths. Similarly, it is useful to take the Census of wild animals also.

In business the tag(contains details about Product code, Price, Batch no., Manufacturing Date, Expiry date, etc.) is stick on the cartons and by using RFID reader process it speedily.

Question 20.
“IPR (Intellectual Property Right) encourages innovation” Justify.
Answer:
Some people spend lots of money, time body and mental power to create some products such as a classical movie, album, artistic work, discoveries, invention, software, etc.

These type of Intellectual properties must be protected from unauthorized access by law. This is called Intellectual Property right(IPR). It enables to earn recognition, financial benefit, can sell the innovation, etc. It motivates further innovation.

Question 21.
Write a short note on

1. Trademark
2. Industrial design

Answer:
1. Trademark:
This is a unique, simple and memorable sign to promote a brand and hence increase the business and goodwill of a company. It must be registered. The period of registration is for 10 years and can be renewed. The registered trademark under Controller General of Patents Design and Trademarks cannot use or copy by anybody else.

2. Industrial designs:
A product or article is designed so beautifully to attract the customers. This type of designs is called industrial design. This is a prototype and used as a model for large scale production.

Question 22.
Compare patent and Trade mark.
Answer:
1. Patents:
A person or organization invented a product or a creation can be protected from unauthorized copying or creation without the permission of the creator by law. This right is called Patent. In India the validity of the right is up to 20 years. After this anybody can use freely.

2. Trademark:
This is a unique, simple and memorable sign to promote a brand and hence increase the business and goodwill of a company. It must be registered. The period of registration is for 10 years and can be renewed. The registered trademark under Controller General of Patents Design and Trademarks cannot use or copy by anybody else.

Question 23.
Write a short note on intellectual property theft.
Answer:
Intellectual property theft: The violation of Intellectual Property Right of Copy right, Trademark, Patent, etc. In film industry crores of investment is needed to create a movie. Intellectual Property thieves upload the movies on the Releasing day itself. Hence the revenue from the theatres are less significantly and undergoes huge loss. (Eg: Premam, Bahubali, etc).

Question 24.
What is cyberspace?
Answer:
a. CyberSpace:
Earlier Traditional communication services such as postal service (Snail mail) are used for communication. It is a low speed and not reliable service. In order to increase the speed Telegram Services were used.

Its speed was high but it has lot of limitations and expensive too. Later telephones were used for voice communication. Nowadays telephone system and computer system are integrated and create a virtual (unreal) environment. This is called cyber space. The result for this integration is that tremendous speed and it is very cheap.

Question 25.
Write a short note on the importance of IT Act 2000.
Answer:
Information Technology Act 2000 (amended in 2008):
IT Act 2000 controls the use of Computer(client), Server, Computer Networks, data and Information in Electronic format and provide legal infrastructure for E-commerce, in India.

This is developed to promote IT industry, control e-commerce, also ensures the smooth functioning of E-Governance and it prevents cyber crimes.

The person those who violate this will be prosecuted. In India IT bill introduced in the May 2000 Parliament Session and it is known as Information Technology Act 2000. Some exclusions and inclusions are introduced in December 2008.

Question 26.
“Informania affects people’s lives and their loved ones” Justify.
Answer:
Infomania is the excessive desire(lnfatuation) for acquiring knowledge from various modern sources like Internet, Email, Social media, Instant Message Application(WhatsApp) and Smart Phones. Due to this the person may neglect daily routine such as family, friends, food, sleep, etc.

Hence they get tired. They give first preference to Internet than others. They create their own Cyber World and no interaction to the surroundings and the family. They are more anxious and afraid that they will be out from the cyber world unless they updated.

Plus Two Computer Application Trends and Issues in ICT Five Mark Questions and Answers

Question 1.
Explain generations in mobile communication?
Answer:
The mobile phone was introduced in the year 1946. Early stage it was expensive and limited services hence its growth was very slow. To solve this problem. Cellular communication concept was developed in 1960’ satBell Lab. 1990’s onwards cellular technology became a common standard in our country.
The various generations in mobile communication are:

a. First Generation networks (1 G):
It was developed around 1980, based on analog system and only voice transmission was allowed.

b. Second Generation networks (2G):
This is the next generation network that was allowed voice and data transmission. Picture message and MMS(Multimedia Messaging Service) were introduced. GSM and CDMA standards were introduced by 2G.

1. Global System for Mobile (GSM):
It is the most successful standard. It uses narrow band TDMA(Time Division Multiple Access), allows simultaneous calls on the same frequency range of 900 MHz to 1800 MHz. The network is identified using the SIM(Subscriber Identity Module).

(i) GPRS (General Packet Radio Services):
It is a packet oriented mobile data seivice on the 2G on GSM. GPRS was originally standardized by European Telecommunications Standards Institute (ETSI) GPRS usage is typically charged based on volume of data transferred. Usage above the bundle cap is either charged per megabyte or disallowed.

(ii) EDGE (Enhanced Data rates for GSM Evolution):
It isthree times faster than GPRS. It is used for voice communication as well as an internet connection.

2. Code Division Multiple Access (CDMA):
It is a channel access method used by various radio communication technologies. CDMA is an example of multiple access, which is where several transmitters can send information simultaneously over a single communication channel. This allows several users to share a band of frequencies To permit this to be achieved without undue interference between the users, and provide better security.

c. Third Generation networks(3G):
It allows high data transfer rate for mobile devices and offers high speed wireless broadband services combining voice and data. To enjoy this service 3G enabled mobile towers and hand sets required.

d. Fourth Generation networks(4G):
It is also called Long Term Evolution(LTE) and also offers ultra broadband Internet facility such as high quality streaming video. It also offers good quality image and videos than TV.

e. Fifth Generation networks:
This is the next generation network and expected to come in practice in 2020. It is more faster and cost effective than other four generations. More connections can be provided and more energy efficient.

Question 2.
Explain different categories of cyber crimes in detail.
Answer:
Just like normal crimes(theft, trespassing private area, destroy, etc,) Cyber crimes(Virus, Trojan Horse, Phishing, Denial of Service, Pornography, etc) also increased significantly. Due to cyber crime, the victims lose money, reputation, etc and some of them commit suicide.

A. Cyber crimes against individuals:

1. Identity theft:
The various information such as personal details(name, Date of Birth, Address, Phone number, etc.), Credit / Debit Card details(Card number, PIN, Expiry Date, CW, etc), Bank details, etc. are the identity of a person. Stealing these information by acting as the authorized person without the permission of a person is called Identity theft. The misuse of this information is a punishable offence.

2. Harassment:
Commenting badly about a particular person’s gender, colour, race, religion, nationality, in Social Media is considered as harassment. This is done with the help of Internet is called Cyber stalking (Nuisance). This is a kind of torturing and it may lead to spoil friendship, career, self image and confidence. Sometimes may lead to a big tragedy of a whole family or a group of persons.

3. Impersonation and cheating:
Fake accounts are created in Social Medias and act as the original one for the purpose of cheating or misleading others. Eg: Fake accounts in Social Medias (Facebook, Twitter, etc), fake sms, fake emails, etc.

4. Violation of privacy:
Trespassing into another person’s life and try to spoil the life. It is a punishable offence. Hidden camera is used to capture the video or picture and black mailing them.

5. Dissemination of obscene material:
With the help of hidden camera capture unwanted video or picture. Distribute or publish this obscene clips on Internet without the consent of the victims may mislead the people specifically the younger ones.

B. Cyber crimes against property:
Stealing credit card details, hacking passwords of social media accounts or mail account or Net banking, uploading latest movies, etc, are considered as cyber crimes against property.

1. Credit card fraud:
Stealing the details such as credit card number, company name, expiry date, cvv number, password, etc. and use these details to make payment for purchasing goods or transfer funds also.

2. Intellectual property theft:
The violation of Intellectual Property Right of Copy right, Trademark, Patent, etc. In film industry crores of investment is needed to create a movie. Intellectual Property thieves upload the movies on the Releasing day itself.

Hence the revenue from the theatres are less significantly and undergoes huge loss. (Eg: Premam, Bahubali, etc). Copying a person’s creation and present as a new creation is called plagiarism. This can be identified some tools (programs) available in the Internet.

3. Internet time theft:
This is deals with the misuse of WiFi Internet facility. If it is not protected by good password there is a chance of misuse our devices (Modem/Router) to access Internet without our consent by unauthorized persons. Hence our money and volume of data (Package) will lose and we may face the consequences if others make any crimes.

C. Cyber crimes against government:
The cyber crimes against Govt, websites is increased significantly. For example in 2015 the
website of Registration Department of Kerala is hacked and destroys data from 2012 onwards.

1. Cyber terrorism:
It is deals with the attacks against very sensitive computer networks like computer-controlled atomic energy power plants, air traffic controls, Gas line controls, telecom, Metro rail controls, Satellites, etc. This is a very serious matter and may lead to huge loss (money and life of citizens). So Govt is very conscious and give tight security mechanism for their services.

2. Website defacement:
It means spoil or hacking websites and posting bad comments about the Govt.

3. Attacks against e-governance websites:
Its main target is a Web server. Due to this attack the Web server/ computer forced to restart and this results refusal of service to the genuine users.

If we want to access a website first you have to type the web site address in the URL and press Enter key, the browser requests that page from the web server. Dos attacks send huge number of requests to the web server until it collapses due to the load and stops functioning.

Question 3.
Discuss various generations of mobile communication.
Answer:
The various generations in mobile communication are:

a. First Generation networks(1 G):
It was developed around 1980, based on analog system and only voice transmission was allowed.

b. Second Generation networks (2G):
This is the next generation network that was allowed voice and data transmission. Picture message and MMS(Multimedia Messaging Service) were introduced. GSM and CDMA standards were introduced by 2G.

1. Global System for Mobile(GSM):
It is the most successful standard. It uses narrow band TDMA (Time Division Multiple Access), allows simultaneous calls on the same frequency range of 900 MHz to 1800 MHz. The network is identified using the SIM(Subscriber Identity Module).

(i) GPRS (General Packet Radio Services):
It is a packet oriented mobile data service on the 2G on GSM. GPRS was originally standardized by European Telecommunications Standards Institute (ETSI) GPRS usage is typically charged based on volume of data transferred. Usage above the bundle cap is either charged per megabyte or disallowed.

(ii) EDGE(Enhanced Data rates for GSM Evolution):
It is three times faster than GPRS. It is used for voice communication as well as an internet connection.

2. Code Division Multiple Access (CDMA):
It is a channel access method used by various radio communication technologies. CDMA is an example of multiple access, which is where several transmitters can send information simultaneously over a single communication channel. This allows several users to share a band of frequencies To permit this to be achieved without undue interference between the users, and provide better security.

c. Third Generation networks (3G):
It allows high data transfer rate for mobile devices and offers high speed wireless broadband services combining voice and data. To enjoy this service 3G enabled mobile towers and hand sets required.

d. Fourth Generation networks (4G):
It is also called Long Term Evolution(LTE) and also offers ultra broadband Internet facility such as high quality streaming video. It also offers good quality image and videos than TV.

e. Fifth Generation networks (5G):
This is the next generation network and expected to come in practice in 2020. It is more faster and cost effective than other four generations. More connections can be provided and more energy efficient.

Question 4.
“Due to anonymous nature of Internet it is possible for the people to engage in variety of criminal activities”. Justify the statement with special reference to cyber crimes taking place against Individual.
Answer:
Cyber crimes against individuals:
1. Identity theft:
The various information such as personal details(name, Date of Birth, Address, Phone number, etc), Credit / Debit Card details(Card number, PIN, Expiry Date, CW, etc), Bank details, etc. are the identity of a person. Stealing these information by acting as the authorized person without the permission of a person is called Identity theft. The misuse of this information is a punishable offence.

2. Harassment:
Commenting badly about a particular person’s gender, colour, race, religion, nationality, in Social Media is considered as harassment. This is done with the help of Internet is called Cyber stalking (Nuisance). This is a kind of torturing and it may lead to spoil friendship, career, self image and confidence. Sometimes may lead to a big tragedy of a whole family or a group of persons.

3. Impersonation and cheating:
Fake accounts are created in Social Medias and act as the original one for the purpose of cheating or misleading others. Eg: Fake accounts in Social Medias(Facebook,Twitter,etc), fake sms, fake emails etc.

4. Violation of privacy:
Trespassing into another person’s life and try to spoil the life. It is a punishable offence. Hidden camera is used to capture the video or picture and black mailing them.

5. Dissemination of obscene material:
With the help of hidden camera capture unwanted video or picture. Distribute or publish this obscene clips on Internet without the consent of the victims may mislead the people specifically the younger ones.

Students can Download Chapter 10 Enterprise Resource Planning Questions and Answers, Plus Two Computer Application Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Computer Application Chapter Wise Questions and Answers Chapter 10 Enterprise Resource Planning

Plus Two Computer Application Enterprise Resource Planning One Mark Questions and Answers

Question 1.
______is a group of people and other resources working together for a common goal.
Answer:
An enterprise

Question 2.
ERP stands for_______.
Answer:
Enterprise Resource Planning

Question 3.
The 4 M’s related to resources are
Answer:
Man, Material, Money, and Machine.

Question 4.
_____consists of single database and a collection of programs to handle the database hence handle the enterprise efficiently and hence enhance the productivity.
Answer:
ERP

Question 5.
state True or False
An ERP package consists of only one module
Answer:
False. It consists of many modules.

Question 6.
_____module is the core of ERP package
Answer:
Financial

Question 7.
ERP stands for_____.
(a) Entertain Resource Package
(b) Enterprise Retain Plan
(c) Enterprise Resource Planning
(d) None of these
Answer:
(c) Enterprise Resource Planning

Question 8.
_____module of ERP contains rules to manage production process.
Answer:
Manufacturing Module

Question 9.
_____module of ERP contains rules to plan the production process.
Answer:
Production Planning Module

Question 10.
_____module of ERP focus on human resource and human capital.
Answer:
HR module

Question 11.
______module of ERP is used to get the raw materials in the right time and right price.
Answer:
Purchasing Module

Question 12.
______module of ERP is used for monitoring and tracking customer oriented activities.
Answer:
Marketing Module

Question 13.
______module of ERP deals with important parts of sales cycle.
Answer:
Sales and distribution module

Question 14.
______module of ERP is used for managing the quality.
Answer:
Quality management module

Question 15.
BPR stands for ______.
Answer:
Business Process Re-engineering (BPR)

Question 16.
In general_____is the series of activities such as rethinking and redesign of the business process to enhance the enterprise’s performance such as reducing the cost (expenses), improve the quality, prompt and speed (time bound) service.
Answer:
BPR (Business Process Re-engineering)

Question 17.
Odd one out
(a) Inputs
(b) Processing
(c) Outcome
(d) SAP
Answer:
(d) SAP is a ERP package the others are 3 elements of business process.

Question 18.
Odd one out
(a) Oracle
(b) SAP
(c) Odoo
(d) C++
Answer:
(d) C++ is a general purpose programming language. The others are ERP packages.

Question 19.
Odd one out
(а) html
(b) C++
(c) JavaScript
(d) Tally ERP
Answer:
(d) Tally ERP is a ERP package.

Question 20.
From the following select the open source ERP.
(a) Microsoft Dynamics
(b) Tally ERP
(c) Odoo
(d) SAP
Answer:
(c) Odoo

Question 21.
Odd one out. Give the reason.
(a) Microsoft Dynamics
(b) Tally ERP
(c) Odoo
(d) SAP
Answer:
(c) Odoo is an open source ERP. The others are not.

Question 22.
PLM stands for_____.
Answer:
Product Life cycle Management

Question 23.
CRM stands for______.
Answer:
Customer Relationship Management

Question 24.
MIS stands for______.
Answer:
Management Information System

Question 25.
SCM stands for______.
Answer:
Supply Chain Management

Question 26.
DSS stands for_____.
(a) Decision Signal System
(b) Decision Support System
(c) Decision Support Scheme
(d) Design Support System
Answer:
(b) Decision Support System

Question 27.
Pick the odd one out and justify.
(a) SAP
(b) Oracle
(c) C++
(d) Tally
Answer:
(c) C++. This is a programming Language. The others are ERP packages

Question 28.
SAP stands for______.
Answer:
Systems, Applications, and Products for data processing

Question 29.
Pick the odd one out from the following list and justify the selection.
CRM, MIS, SCM, SAP
Answer:
SAP others are ERP related technologies. SAP is an ERP package.

Question 30.
Consider the following two statements.
Statement 1: “The number of functional modules in an ERP vary with nature of enterprise” Statement 2: “There is no connection between BPR and ERP”
Then choose the correct one from the following.
(i) Both statements are true
(ii) Both statements are false
(iii) Statement 1 is true and statement 2 is false
(iv) Statement 1 is false and statement 2 is true
Answer:
(iii) Statement 1 is true and Statement 2 is false.

Question 31.
Choose the correct answer from the following: Implementation of ERP in an enterprise ______.
(a) minimize planning risks.
(b) integrates different functional units of an enter-prise.
(c) uses centralized data base.
(d) All the above.
Answer:
(d) All the above.

Question 32.
State True or False

1. Every ERP package can manage all the functional units of an enterprise.
2. In ERP, a centralized database is used for integrating functional units.

Answer:

1. False
2. True

Question 33.
ERP stands for_____.
(a) Enterprise Resource Project
(b) Enterprise Resource Processing
(c) Enterprise Resource Planning
(d) Enterprise Requirement Planning
Answer:
(c) Enterprise Resource Planning

Plus Two Computer Application Enterprise Resource Planning Two Mark Questions and Answers

Question 1.
Define an Enterprise.
Answer:
An Enterprise is a group of people and other resources working together for a common goal. It is also an example for System.

Question 2.
Define Enterprise Resource Planning ERP.
Answer:
An enterprise(organization) is considered as a system (A system is an orderly grouping of interdependent components linked together to achieve an objective, according to a plan. Human body is an example for System).

All the departments of an enterprise are connected to a centralized data base. ERP consists of single database and a collection of programs to handle the database hence handle the enterprise efficiently and hence enhance the productivity.

Question 3.
Give the significance of HR module in an ERP package.
Answer:
This model ensures the effective use of Human resources and Human capital and enhance the productivity of the Enterprise hence increase the profit.

Question 4.
Give the relation between business process reengineering (BPR) and enterprise resource planning.
Answer:
ERP and BPR will not make much change if they are in stand alone. To improve the efficiency of an enterprise integrate both ERP and BPR because they are the two sides of a coin.

For the better results conducting BPR before implementing ERP, this will help an enterprise to avoid unnecessary modules from the software.

Question 5.
“The key concept of ERP is a centralised database management system”. Justify.
Answer:
ERP is an integrated business management system which uses a single database to store and communicate information of various departments of an enterprise.

Question 6.
Match the following.

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

Question 7.
Briefly explain the importance of Business Process Re-engineering (BPR) in the implementation of ERP in an enterprise.
Answer:
Business Process Reengineering: In general BPR is the series of activities such as rethinking and redesign of the business process to enhance the enterprise’s performance such as reducing the cost(expences), improve the quality, prompt and speed(time bound) service.

Plus Two Computer Application Enterprise Resource Planning Three Mark Questions and Answers

Question 1.
Write short notes about BPR.
Answer:
In this world, tight competition is based on price, quality, wide variety of selection and quick service. To increase the business and hence increase the profit of a Business firm various activities are involved.

IT and Re-engineering plays major roles to increase the productivity. In general BPR is the series of activities such as rethinking and redesign of the business process to enhance the enterprise’s performance such as reducing the cost(expenses), improve the quality, prompt and speed(time bound) service. BPR enhances the productivity and profit of an enterprise.

Question 2.
Give an example for an enterprise from your real life and identify different departments/functional units in it.
Answer:
Indian Oil Corporation is an example for an enterprise. The activities involved are planning, purchasing raw material, production, storing finished goods (warehouse), sales, finance, etc. These activities are performed by different departments and their duties are interlinked.

Question 3.
The first five phases of ERP implementation are listed below. Rearrange them in correct order.
Package selection, BPR, Gap analysis, pre evaluation screening, project planning.
Answer:
The correct order is as follows

1. Pre evaluation screening
2. Package selection
3. Project planning
4. Gap analysis
5. Business Process Reengineering

Question 4.
Write a short note about the following terms:

1. DSS
2. MIS

Answer:
1. Decision Support System (DSS):
It is a computer based system that takes inputs as business data and after processing it produces good decisions as output that will make the business easier. Management Information.

System (MIS):
Management is the decision and policy makers. A good management can take good decision and that will help to do the business well. The good relationship between Management and employees is a key to success.

MIS will collect relevant data from inside and outside of a company. Based upon this information produce reports and take appropriate decisions.

Plus Two Computer Application Enterprise Resource Planning Five Mark Questions and Answers

Question 1.
Explain functional units of ERP in detail.
Answer:
Different modules are given below:

1. Financial Module:
It is the core. This is used to generate financial report such as balance sheet, general ledger, trial balance, financial statement, etc.

2. Manufacturing Module:
It provides information for the production and capable to change the methods in manufacturing sector.

3. Production planning Module:
This module ensures the effective use of resources and helps the enterprise to enhance the productive hence increase the profit.

4. HR (Human Resource) Module:
This model ensures the effective use of Human resources and Human capital.

5. Inventory control Module:
This model is useful to maintain the appropriate level of stock(includes raw material, work in progress and finished goods)

6. Purchasing Module:
This module is useful to make available the required raw materials in good condition and in the right time and price.

7. Marketing Module:
It is used for handle the orders of customers.

8. Sales and distribution Module:
Existence of a company is based on the income from sales. This module will help to handle the sales enquiries, order placement, and scheduling, dispatching and invoicing.

9. Quality (Ql & QC) management module:
Quality of a product or service is very much important to a company. This module helps to maintain the quality of the product. Quality planning, inspection, and control are the main activities involved in this module.

Question 2.
Explain the different phases of ERP implementation.
Answer:
The different phases of ERP implementation are given below

1. Pre evaluation screening:
Many ERP packages are available in the markets. At most care should be taken before implementing a ERP. Select a few from from the available ERP packages.

2. Package selection:
Selection of right ERP to our enterprise is a laborious task and it needs huge investment. Various factors should be keep in mind before you purchase an ERP that should meet our complete needs.

3. Project planning:
A good planning is essential to implement an ERP. From the beginning to the end activities are depicted in this phase.

4. Gap analysis:
A cent percent(100%) problem solving ERP is not available in the market. Most of them solve a maximum of 70% to 80% problems. The rest (30% to 20%) of the problems and their solutions are mentioned here.

5. Business Process Reengineering:
In general BPR is the series of activities such as rethinking and redesign of the business process to enhance the enterprise’s performance such as reducing the cost(expences), improve the quality, prompt and speed(time bound) service. BPR enhances the productivity and profit of an enterprise.

6. Installation and configuration:
In this phase the new system are installing, before implementing the whole system a miniature of the actual system is going to be implemented as a test dose. Then check the reactions if it is good it is the time to install the whole system completely.

7. Implementation team training:
In this phase the company trains its employees to implement and run the system.

8. Testing:
This phase is very important. It determines whether the system produces proper result. Errors in design and logic are identified.

9. Going live:
Here a change over is taken place to new system from old system. It is not an easy process without the support and service from the ERP vendors.

10. End user training:
This phase will start familiarising the users with the procedures to be used in the new system. It is very important.

11. Post implementation:
Once the system is implemented maintenance and review begin. In this phase repairing or correcting previously ill defined problems and upgrade or adjust the performance according to the company needs.

Question 3.
Write short notes regarding ERP package Companies.
Answer:
Popular ERP packages are given below
1. Oracle:
American based company famous in database(Oracle 9i-SQL) packages situated in Redwood shores, California. Their ERP packages is a solution for finance and accounting problems. Their other products are

• Customer Relationship Management(CRM)
• Supply Chain Management (SCM)Software SAP

2. SAP:
stands for Systems, Applications, and Products for data processing. It is a German MNC in Walldorf and founded in 1972. Earlier they developed ERP packages for large MNC. But nowadays they developed for small scale industries also.
The other software products they developed are

• Customer Relationship Management(CRM)
• Supply Chain Management(SCM)
• Product Life cycle Management(PLM)

3. Odoo:
Formerly known as OpenERP.
It is an open source code ERP. Unlike other companies their source code is available and can be modified as and when need arises.

4. Microsoft Dynamics:

• American MNC in Redmond, Washington
• ERP for midsized companies.
• This ERP is more user friendly
• Other s/w is Customer Relationship Management(CRM)

5. Tally ERP:

• Indian company situated in Bangalore.
• This ERP provides total solution for accounting, inventory, and Payroll.

Question 4.
Explain the merits and demerits of ERP packages.
Answer:
ERP packages have a lot of advantages as well as many drawbacks also.
Benefits of ERP system:

1. Improved resource utilization:
Resources such as Men, Money, Material, and Machine are utilized maximum hence increase the productivity and profit.

2. Better customer satisfaction:
Without spending more money and time all the customer’s needs are considered well. Because customer is the king of the market. Nowadays a customer can track the status of an order by using the docket number through Internet.

3. Provides accurate information:
Right information at the right time will help the company to plan and manage the future cunningly. Acompany can increase or reduce the production based upon the right information hence increase the productivity and profit.

4. Decision making capability:
Right information at the right time will help the company to take good decision.

5. Increased flexibility:
A good ERP will help the company to adopt good things as well as avoid bad things rapidly. It denotes the flexibility.

6. Information integrity: A good ERP integrates various departments into a single unit. Hence reduce the redundancy, inconsistency, etc.

Risks of ERP implementation:

1. High cost:
Very huge investment is required to purchase and configure an ERP. Moreover, it requires up gradation or. replacement of hardware(Man, computer or machine) is an additional investment. So small scale enterprise cannot afford this.

2. Time consuming:
The full fledge implementation of ERP package needs one or two years. That is highly time consuming.

3. Requirement of additional trained staff:
The existing staffs may not capable to work with ERP. To overcome this give proper training to them otherwise appoint trained and experienced employees to Cop up.

4. Operational and maintenance issues:
The first major problem is that the resistance from the existing employees. To overcome this give awareness to the existing employees. The second problem is that ERP package is a cyclic process oriented package. It is a continuous process and should be maintained well otherwise the correct output will not available.

Question 5.
Explain in detail the ERP packages and related technologies.
Answer:
It is an all in one system. It integrates various functions such as raw material purchase, production planning, marketing, financial, etc., into a single application.

1. Product Life Cycle Management (PLM):
It manages the entire life cycle of a product. PLM consists of programs to increase the quality and reduce the price by the efficient use of resources.

2. Customer Relationship Management (CRM):
As we know that customer is the king of the market. The existence of a company mainly the customers. CRM consists of programs to enhance the customer’s relationship with the company.

3. Management Information System (MIS):
Management is the decision and policy makers. A good management can take good decision and that will help to do the business well. The good relationship between Management and employees is a key to success. MIS will collect relevant data from inside and outside of a company. Based upon this information produce reports and take appropriate decisions.

4. Supply Chain Management (SCM):
This is deals with moving raw materials from suppliers to the company as well as finished goods from company to customers. The activities includes are inventory(raw materials, work in progress and finished goods) management, warehouse management, transportation management, etc.

5. Decision Support System (DSS):
It is a computer based system that takes inputs as business data and after processing it produces good decisions as output that will make the business easier.

Question 6.
“The number and functioning of modules vary with the nature of enterprise and the type of ERP package.” List any six common modules of an enterprise.
Answer:
Different modules are given below

1. Financial Module:
It is the core. This is used to generate financial report such as balance sheet, general ledger, trial balance, financial statement, etc.

2. Manufacturing Module:
It provides information for the production and capable to change the methods in manufacturing sector.

3. Production planning Module:
This module ensures the effective use of resources and helps the enterprise to enhance the productive hence increase the profit.

4. HR (Human Resource) Module:
This model ensures the effective use of Human resources and Human capital.

5. Inventory control Module:
This model is useful to maintain the appropriate level of stock(includes raw material, work in progress and finished goods)

6. Purchasing Module:
This module is useful to make available the required raw materials in good condition and in the right time and price.

7. Marketing Module:
It is used for handle the orders of customers.

8. Sales and distribution Module:
Existence of a company is based on the income from sales. This module will help to handle the sales enquiries, order placement, and scheduling, dispatching and invoicing.

9. Quality (Ql & QC) management module:
Quality of a product or service is very much important to a company. This module helps to maintain the quality of the product. Quality planning, inspection, and control are the main activities involved in this module.

Question 7.
Mr. Suresh uses separate software for managing different functional units of an enterprise and Mr. Saleem uses an integrated software package formanaging overall functioning of the enterprise. Compare the benefits and risks of above two methods of an enterprise management.
Answer:
Benefits of ERP system:

1. Improved resource utilization:
Resources such as Men, Money, Material, and Machine are utilized maximum hence increase the productivity and profit.

2. Better customer satisfaction:
Without spending more money and time all the customer’s needs are considered well. Because customer is the king of the market. Nowadays a customer can track the status of an order by using the docket number through Internet.

3. Provides accurate information:
Right information at the right time will help the company to plan and manage the future cunningly. A company can increase or reduce the production based upon the right information hence increase the productivity and profit.

4. Decision making/apability:
Right information at the right time will help the company to take good decision.

5. Increased flexibility:
A good ERP will help the company to adopt good things as well as avoid bad things rapidly. It denotes the flexibility.

6. Information integrity:
A good ERP integrates various departments into a single unit. Hence reduce the redundancy, inconsistency, etc.

Risks of ERP implementation:

1. High cost:
Very huge investment is required to purchase and configure an ERP. Moreover, it requires up gradation or replacement of hardware(Man, computer or machine) is an additional investment. So small scale enterprise cannot afford this.

2. Time consuming:
The full fledge implementation of ERP package needs one or two years. That is highly time consuming.

3. Requirement of additional trained staff:
The existing staffs may not capable to work with ERP. To overcome this give proper training to them otherwise appoint trained and experienced employees to cop up.

4. Operational and maintenance issues:
The first major problem is that the resistance from the existing employees. To overcome this give awareness to the existing employees. The second problem is that ERP package is a cyclic process oriented package. It is a continuous process and should be maintained well otherwise the correct output will not available.

Question 8.
“Implementation of an ERP system in an enterprise is not a single step action”. Justify this statement by listing all the phases of ERP implementation in correct order.
Answer:
The different phases of ERP implementation are given below:

1. Pre evaluation screening:
Many ERP packages are available in the marketfs. At most care should be taken before implementing a ERP. Select a few from from the available ERP packages.

2. Package selection:
Selection of right ERP to our enterprise is a laborious task and it needs huge investment. Various factors should be keep in mind before you purchase an ERP that should meet our complete needs.

3. Project planning:
A good planning is essential to implement an ERP. From the beginning to the end activities are depicted in this phase.

4. Gap analysis:
A cent percents 00%) problem solving ERP is not available in the market. Most of them solve a maximum of 70% to 80% problems. The rest (30% to 20%) of the problems and their solutions are mentioned here.

5. Business Process Reengineering:
In general BPR is the series of activities such as rethinking and redesign of the business process to enhance the enterprise’s performance such as reducing the cost(expences), improve the quality, prompt and speed(time bound) service.

BPR enhances the productivity and profit of an enterprise:

1. Installation and configuration:
In this phase the new system are installing, before implementing the whole system a miniature of the actual system is going to be implemented as a test dose. Then check the reactions if it is good it is the time to install the whole system completely.

2. Implementation team training:
In this phase the company trains its employees to implement and run the system.

3. Testing:
This phase is very important. It determines whether the system produces proper result. Errors in design and logic are identified.

4. Going live:
Here a change over is taken place to new system from old system. It is not an easy process without the support and service from the ERP vendors.

5. End user training:
This phase will start familiarising the users with the procedures to be used in the new system. It is very important.

6. Post implementation:
Once the system is implemented maintenance and review begin. In this phase repairing or correcting previously ill defined problems and upgrade or adjust the performance according to the company needs.

Question 9.
“Selection of ERP package is one of the important phases of ERP implementation”.
Write a short note about any of the ERP packages.
Answer:
Popular ERP packages are given below:

1. Oracle:
American based company famous in database (Oracle 9i-SQL) packages situated in Redwood shores, California. Their ERP packages is a solution for finance and accounting problems. Their other products are

• Customer Relationship Management(CRM)
• Supply Chain Management (SCM)Software

2. SAP:
SAP stands for Systems, Applications, and Products for data processing. It is a German MNC in Walldorf and founded in 1972. Earlier they developed ERP packages for large MNC. But nowadays they developed for small scale industries also.
The other software products they developed are

• Customer Relationship Management(CRM)
• Supply Chain Management(SCM)
• Product Life cycle Management(PLM)

3. Odoo:
Formerly known as OpenERP.
It is an open source code ERP. Unlike other companies their source code is available and can be modified as and when need arises.

4. Microsoft Dynamics:

• American MNC in Redmond, Washington
• ERP for midsized companies.
• This ERP is more user friendly
• Other s/w is Customer Relationship Management(CRM)

5. Tally ERP:

• Indian company situated in’Bangalore.
• This ERP provides total solution for accounting, inventory, and Payroll.

Students can Download Chapter 9 Structured Query Language Questions and Answers, Plus Two Computer Application Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Computer Application Chapter Wise Questions and Answers Chapter 9 Structured Query Language

Plus Two Computer Application Structured Query Language One Mark Questions and Answers

Question 1.
______form of SQL is designed for use with in general purpose programming languages such as COBOL, C, etc.
Answer:
Embedded SQL.

Question 2.
What is TCL?
Answer:
Transaction Control Language- component of SQL includes commands for specifying transactions.

Question 3.
Data dictionary is a special file in DBMS. What is it used for?
Answer:
Table details are stored in this file

Question 4.
What does the following statement mean?
Name VARCHAR(30)
Answer:
Field Name can store up to 30 characters. It is a column definition.

Question 5.
Is there any data type available in SQL to store your date of birth information?
Answer:
Ip Yes. DATE data type

Question 6.
Pick the odd one out.
(DEC, NUMBER, INT, DATE)
Answer:
DATE

Question 7.
How do you ensure that the field ‘ name’ will have some value always?
Answer:
using the constraint NOT NULL

Question 8.
How to set the default value of column District in a table to ‘Thrissur’?
Answer:
District VARCHAR(30) DEFAULT ‘ Thrissur’

Question 9.
_______symbol is used as substitution operator in SQL
Answer:
&

Question 10.
Is there any method to find the strings starting with letter ‘a’ from a field in SQL?
Answer:
Use LIKE operator, LIKE ‘a%’

Question 11.
Howto check whether a particular field contains null values or not?
Answer:
Use operator IS NULL

Question 12.
“ORDER BY” clause is used for_______
Answer:
Sorting the results of a query by ascending(Asc) or Descending (Desc).

Question 13.
The built-in functions in SQL that return just a single value for a group of rows in a table are called______
Answer:
Summary functions or aggregate functions.

Question 14.
How to find the number of values in a column in the table?
Answer:
Using the function COUNT()

Question 15.
______is the clause in SQL used for categorization.
Answer:
GROUP BY

Question 16.
Thomas wants to remove a table that he were using. How could you help him doing this?
Answer:
Using DROP TABLE command

Question 17.
Pick the odd one out.
(SELECT, UPDATE, DELETE, DROP TABLE)
Answer:
DROP TABLE

Question 18.
Name the aggregate function that can be used to find the total number of records.
Answer:
COUNT()

Question 19.
Which of the following is an essential clause used with SELECT command?
(GROUP BY, ORDER BY, WHERE, FROM)
Answer:
FROM

Question 20.
Which of the following is not a column constraint?
(CHECK, DISTINCT, UNIQUE, DEFAULT)
Answer:
DISTINCT

Question 21.
Which of the following is a DDL command?
(SELECT, UPDATE, CREATE TABLE, INSERT INTO)
Answer:
CREATETABLE

Question 22.
What are the logical operators used in SQL?
Answer:
And, Or, Not

Question 23.
The______operator of SQL is used to match a pattern with the help of %. February 2009
(a) BETWEEN
(b) AND
(c) LIKE
(d) OR
Answer:
(c) LIKE

Question 24.
The structure of a table Book is given below.

Write SQL query to Insert an additional column Purchase date of date type to the. Books table.
Answer:
Alter table Book add(PurchaseDate Date);

Question 25.
Suppose we want to include a column in a table in which serial numbers are to be stored automatically on adding new records. Which constraint is to be used for that column during table creation?
Answer:
The constraint Autojncrement is used.

Question 26.
Which of the following cannot be used to name a table in SQL? Give the reason.
(a) Studnt50
(b) Table
(c) $Employee
(d) Stock_123
Answer:
(b) Table. This is a keyword hence it cannot be used.

Question 27.
Which of the following commands is used to view the strucutre of a table?
(a) SHOW TABLES
(b) DESC
(C) SELECT
(d) DISPLAY
Answer:
(b) DESC

Question 28.
The command to eliminate the table CUSTOMER from a database is:
(a) REMOVE TABLE CUSTOMER
(b) DROP TABLE CUSTOMER
(c) DELETE TABLE CUSTOMER
(d) UPDATE TABLE CUSTOMER
Answer:
(b) DROP TABLE CUSTOMER

Question 29.
Which SQL command is used to open a database?
(a) OPEN
(b) SHOW
(c) USE
(d) CREATE
Answer:
(c) USE

Question 30.
Which is the keyword used with SELECT command to avoid duplication of rows in the selction?
Answer:
DISTINCT

Question 31.
Pick odd one out and write reason:
(a) WHERE
(b) ORDER BY
(c) UPDATE
(d) GROUP BY
Answer:
(c) UPDATE. It is a command and others are clauses.

Question 32.
Which of the following clause is not used with SELECT command in SQL?
(a) GROUP BY
(b) WHERE
(c) SET
(d) ORDER BY
Answer:
(c) SET. This clause is used with UPDATE.

Question 33.
______operator in SQL is used with wildcard characters for selection of records,
(a) LIKE
(b) IN
(c) NOT IN
(d) IN and NOT IN
Answer:
(a) LIKE

Plus Two Computer Application Structured Query Language Two Mark Questions and Answers

Question 1.
How is SQL different from other computer high level languages?
Answer:
SQL means Structured Query Language. It is a relational database language, not a programming language like other high level languages. It provides facilities to create a table, insert data into a table, retrieve information from a table, modify data in the table, etc.

Question 2.
Distinguish the SQL keywords UNIQUE and DISTINCT.
Answer:

1. Unique: It ensures that no two rows have the same value in a column while storing data. It is used with create command.
2. Distinct: This keyword is used to avoid duplicate values in a column of a table while retrieving data. It is used with select command.

Question 3.
Identify errors in the following SQL statement and rewrite it correctly. Underline the corrections.
CREATE student TABLE
(admno PRIMARY KEY,
roll no INT,
name CHAR);
Answer:
The correct SQL statement is as follows.
CREATE TABLE student(
admno INT PRIMARY KEY,
roll_no INT,
name CHAR(3));

Question 4.
Suppose a column named Fee does not contain any value for some records in the table named STUDENT. Write SQL statement to fill these blanks with 1000.
Answer:
UPDATE STUDENT SET Fee = 1000 WHERE Fee IS NULL;

Question 5.
Identify the errors in the following SQL statement and give reason for the error.
SELECT FROM STUDENT
ORDER BY Group
WHERE Marks above 50;
Answer:
In this query Group is the key word hence it cannot be used. The correct query is as follows.
SELECT * FROM STUDENT WHERE Marks > 50 ORDER BY Marks;

Question 6.
Differentiate CHAR and VARCHAR data types of SQL.
Answer:

• Char: It is used to store fixed number of characters. It is declared as char(size).
• Varchar: It is used to store characters but it uses only enough memory.

Question 7.
Assume that CUSTOMER is a table with columns Cust_code, Cust_name, Mob_No and Email. Write an SQL statement to add the details of a customer who has no e-mail id.
Answer:
INSERT INTO CUSTOMER VALUES(1001, ‘ALVIS’, 9447024365, NULL);

Question 8.
Find the correct clause from the 2^nd column for each SQL command in the1 st column.

Answer:

Question 9.
Is ALL a keyword in SQL? Explain.
Answer:
Yes, ALL retains all the duplicate values of a field

Question 10.
Differentiate LIKE ‘a%’ and LIKE ‘a_’
Answer:
%, it replaces a string but _ replaces only one character at a time.LIKE ‘a%’ retrieve all strings of any length that start with letter ‘a.’ while LIKE ‘a_’ retrieve all two letter strings that start with letter ‘a’.

Question 11.
Is multiple sorting possible using ORDER BY clause? Explain.
Answer:
Yes, by giving multiple field names separated by comma in the ORDER BY clause.

Question 12.
While inserting records into a table, Raju finds it is not possible to give more than 20 characters in the ‘name’ field. How can you help Raju solve this problem?
Answer:
Using ALTER TABLE command he can modify the width of field so that it can accommodate more than 20 characters.

Question 13.
After executing a query Mohan gets a message like ‘Table Altered’. What would have he done? Give the syntax.
Answer:
He would have used the ALTER TABLE command.
Syntax, ALTER TABLE <name> ADD/MODIFY (<column>);

Question 14.
During discussion one student said that DELETE and DROP TABLE commands are same. Do you agree with that? Justify.
Answer:
No, DROP TABLE remove a table entirely from the database whereas DELETE command deletes only records from an existing table.

Question 15.

1. Pick odd one out .
   DROP TABLE, DELETE, ALTER TABLE, CREATE VIEW
2. Justify your answer

Answer:

1. DELETE
2. All others are DDL commands

Question 16.

1. Pick odd one out from the following.
   NOT NULL, Group By, Check, Unique, Default
2. Justify your answer

Answer:

1. Group By
2. Group By is not a Constraint.

Question 17.
Consider the following SQL statements.
DELETE Name
From Student
Where name = “Raju”
Find the error in the SQL if any and correct it.
Answer:
Delete from student where name = ‘Raju’

Question 18.
Consider the following Query in SQL.
SELECT Department, avg(salary)
From Employee
Where branch = ‘Kannur’
Group By Department
Having avg (salary) > 7000,
Compare the ‘where’ clause and ‘Having’ Clause using the above query.
Answer:
‘where’ clause applies on single rows but Having clause applies on a group.

Question 19.
Some constraints in SQL are called column constraints. Some constraints are called table constraints. How do they differ?
Answer:
Column constraints are specified while defining each column, table constraints are specified once for the entire table at the end of table definition.

Question 20.

1. Choose the odd one from the following.
   Primary key, Unique, Distinct, Default, Check
2. Justify your choice

Answer:

1. Distinct
2. Distinct is used with select command while others are column constraints of create table command.

Question 21.
Name the most appropriate SQL data type required to store the following data.

1. Name of a student (maximum 70 characters)
2. Date of Birth of a student.
3. RollNo. of a student (in the range 1 to 50)
4. Percentage of marks obtained (correct to 2 decimal places)

Answer:

1. Varchar(70)
2. Date
3. Smallint OR Integer Or decimal(2)
4. Dec(5, 2)

Question 22.

1. From the list given below select the names that cannot be used as a table name.
   Adm_No, Date, Salary 2006, Table, Column_Name, Address.
2. Justify your selection.

Answer:

1. Date, Table
2. There is a data type Date and Table is a key word used to create a table. So these two are not used.

Question 23.
Classify the following SQL elements into two and give proper title for each category.
NOT NULL. AVG. COUNT. CHECK. SUM. DEFAULT
Answer:
1. Aggregate functions:
AVG
COUNT
SUM

2. Column Constraints:
NOT NULL
CHECK
DEFAULT

Plus Two Computer Application Structured Query Language Three Mark Questions and Answers

Question 1.
As a part of your school project you are asked to create a table Student with the fields RollNo, Name, Date of Birth and Score in IT. The constraints required are RollNo. is the primary key. Name cannot be empty.
Answer:
Create table student(RollNo decimal(2) not null primary key, Name varchar(20) not null, DOB date, Score number(2));

Question 2.
A table Employee consists of fields EmployeeNo, Name, Designation and Salary. Consider that you are forced to give access to this table for an engineer from another company. But for security reasons you need to hide Salary from him.

1. Name the concept that provides this engineer, a facility to work on this table without viewing salary.
2. Write SQL query for implementing this.

Answer:

1. Views
2. Create view Empview as select employee no, Name, Designation from Employee.

Question 3.
Explain the keywords in the following query.
SELECT DISTINCT course FROM Student;
Answer:
SELECT allows to retrieve a subset of rows from the table, DISTINCT avoids duplication of courses from the table STUDENT, FROM is a key word used to specify the name of the table

Question 4.
A table named student is given below.

Write answers for the questions based on the above table.

1. SQL statement to display the different courses available without duplication.
2. SQL statement to display the Name and Batch of the students whose percentage has a null value.
3. Output of the query select count (percentage) from Student.

Answer:

1. Select distinct batch from Student;
2. Select name, batch from Student where percent is null
3. 5

Question 5.
Once the creation of a table is over, one can perform two changes in the schema of the table. What are they? Give syntax.
Answer:
We can alter the table in two ways.

• We can add a new column to the existing table using the following syntax,
  ALTER TABLE <tablename>ADD(<cloumnname> <type> <constraint>);
• We can also change or modify the existing column in terms of type or size using the following syntax, ALTER TABLE<tablename>MODIFY(<column> <newtype>);

Question 6.
Explain how pattern matching can be done in SQL with an example.
Answer:
Pattern matching can be done using the operator LIKE while setting the condition with pattern matching.

for eg., to display the names of all students whose name begins with letters ‘ma’, we can write the following query, SELECT name FROM STUDENT WHERE name LIKE ‘ma%’; here the character ‘%’ substitutes any number of characters in the value of the specified column. Another character substitutes only one character of the specified column.

Question 7.
During the discussion of study your friend say that table and view are the same. How can you correct him?
Answer:
Tables and views are different. A view is a single virtual table that is derived from other physically existing tables. When we access a view we actually access the base tables.

We can use all the DML commands with the views but care should be taken as operation actually reflects in the base tables. The advantage of view is that without sparing extra storage space, we can use same table as different virtual tables. It also implements sharing along with privacy).

Question 8.
Find the errors if any of the following code. Create table emp (name char, R0IIN0 int(20));
Answer:
Here the argument size of data type char of field name is missing. So we can’t store a character only. The second error is the data type int has no argument. The correct statement is as follows Create table emp (name char(20), RollNo int);

Question 9.
Consider the following variable Declaration in SQL.

• name char (25)
• name Varchar (25)
  1. Considering the utilisation of memory, which variable declaration is more suitable.
  2. Justify your answer

Answer:
1. name varchar (25)

2. Because char data type is fixed length. It allocates maximum memory i.e, here it allocates memory for 25 characters may be there is a chance of memory wastage. But Varchar allocates only enough memory to store the actual size.

Question 10.
Mr. Dilip wants to construct a table and implement some restrictions on the table.

1. Column can never have empty values.
2. One of the columns must be a key to identify the rows etc.

Can you help him to create that table satisfying the above restrictions with an example.
Answer:

1. Use of create table, Not NULL,
2. Primary key

Eg: Create table employee(RollNo decimal(3) not null primary key, Name Varchar(70) not null, Desgn Varchar(30), DOB Date, Salary Dec(7,2));

Question 11.
Write SQL query to construct a table student with fields Reg No, Stud Name, Sex, Course, grade, etc. As per the following conditions.

1. Reg No should not be empty and using this column any rows in that table can be identified.
2. Student name should have some value.
3. Default value of sex be Female.
4. Grade should be any of the values: A+, A, B+, B, C+, C, D+, D, E

Answer:
Create table Student(RollNo decimal(3) not null primary key,Name Varchar(70) not null, sex char default ‘Female’, Grade char(2));

Question 12.
Ramu create Table employee with fields empld, empname, Designation, salary using SQL statements. Later he found that data type of emPId is typed as Integer instead of character and missed a field ‘Department’. Can you help him to solve this problem without recreating ‘employee’ table.
Answer:

• Alter Table Employee Modify(Empld char(4));
• Alter Table Employee Add(Dept char(15));

Question 13.
While creating a table Alvis give “Emp Details” for table name. Is it Possible. Write down the rules for naming a Table.
Answer:
It is not possible because there is a space between Emp and Details. The rules are given below:

1. It must not be a keywond(Key words are reserved words and have predefined meaning)
2. It must begin with alphabets
3. Digits can be used followed by alphabets
4. Special characters cannot be used except under score
5. We cannot give a name of another table

Plus Two Computer Application Structured Query Language Five Mark Questions and Answers

Question 1.
Consider the table ITEMS.

1. SELECT ITEMCODE, NAME FROM ITEMS WHERE CATEGORY = ‘Stationery’;
2. SELECT * FROM ITEMS WHERE SALES_ PRICE < UNIT_PRICE;
3. SELECT CATEGORY, COUNT(*) FROM ITEMS GROUP BY CATEGORY;’

Answer:
1.

2.

3.

Question 2.
Write SQL queries to
1. Create a table Employee with the fields given below.

2. List the name of all employees whose name’s second letter is ‘a’.
3. List the Name and Designation of employees whose Designation is not ‘Manager’.
4. Increase the salary of all employees by 10 percent.
5. Remove all managers whose salary is less than Rs. 10,000 from the table.
Answer:

1. Create table employee(Name Varchar(70), Desgn Varchar(30), DOB Date, Salary Dec(7,2));
2. Select Name from Employee where Name like ‘_a%’;
3. Select Name, Desgn from Employee where Desgn<> ’Manager’;
4. Update Employee set Salary = Salary + Salary * .01;
5. Delete from Employee where Desgn = ‘Manager’ and Salary < 10000;

Question 3.
A table Student consists of fields Roll No, Name, Batch and Percent. Write SQL statements to

1. Display RollNo and Name of students whose percentage is Iessthan90 and greater than 70.
2. Display RollNo and Name of all students in science batch whose percentage is more than
3. Display Names of all students in commerce and science batches.
4. Display the average Percent of students in each batch.
5. Display RollNo and Name in the ascending order of Batch and descending order of Percent.

Answer:

1. Select RollNo.Name from Student where percent < 90 and percent > 70;
2. Select RollNo.Name from Student where Batch = ’Science’ and percent > 90;
3. Select Name from Student where batch = ’Science’ Or batch = ’Commerce’;
4. Select batch, Avg(percent) from Student group by batch;
5. Select RollNo.Name from Student order by batch, percent desc;

Question 4.

1. Classify the following SQL commands. Create table, Insert Into, AlterTable, Delete, Update, Drop Table, Select.
2. List the features of each category.

Answer:
1. DDL – Create Table, Alter Table, Drop Table DML- Insert Into, Delete, Update, Select.

2. DDL – DDL means Data Definition Language. It is used to create the structure of a table, modify the structure of a table and delete the structure of a table.

DML -DML means Data Manipulation Language. It is used to insert records into a table, modify the records of a table, delete the records of a table and retrieve the records from a table.

Question 5.
Explain the available database integrity constraints.
Answer:

1. NOT NULL: it specifies that a column can never have null values, i.e., not empty
2. UNIQUE: it ensures that no two rows have same value in the specified column.
3. PRIMARY KEY: it declares a column or a set of columns as the primary key of the table. This constraint makes a column NOT NULL and UNIQUE.
4. DEFAULT: it sets a default value for a column when the user does not enter a value for that column.
5. Auto_increment: This constraint is used to perform auto_increment the values in a column. That is automatically generate serial numbers. Only one auto_increment column per table is allowed.

Question 6.
Consider a table student with fields Reg No, Stud Name, Sex, Course, total score.
Write Sql queries for the following:

1. Enter a Record
2. List the Detail’s of all students
3. Display Details of the student whose name ends with ‘Kumar’.
4. List all Female students who got more than 50 marks.
5. List all students who are studying either science or in Commerce group.
6. List details of those students who are studying Humanities in the Descending order of their names.

Answer:

1. Insert into Student values. (52, ‘JOSE’, ‘Male’, ‘Science’, 500);
2. Select * from Student;
3. Select * from Student where name like ‘%Kumar’;
4. Select * from Student where sex=’Female’ and Total > 50;
5. Select * from Student where course = ‘science’ or course = ‘commerce’;
6. Select * from Student where course=’Humanities’ order by name desc;

Question 7.
Construct a Table Product with the following fields using SQL.
Product code – Consist of Alphanumeric code
Product name – Consist of maximum of 30 Alphabets
Unit price – Numeric values
Quantity – Numeric values
Product price – Numeric Values
Write Query for the following:

1. Enter 5 records in the table (not give values for product prices)
2. Calculate product price (unit price x quantity)
3. List all product whose unit price ranging from 10 to 20
4. Calculate total price of all product.

Answer:
Create table Product(ProdCode Varchar(20) not null primary key, ProdName char(30), UPrice decimal(7, 2), Qty decimal(6), ProdPricedecimal(7, 2));

1. Insert into Product(Product Code,Product Name, Unit Price, Quantity) values (‘101’, ‘LUX’, 29.50, 500); Similarly insert four more records
2. Update Product set ProdPrice = Qty * UPrice;
3. Select ProdName from Product where UPrice between 10 and 20;
4. Select sum(ProdPrice) from Product;

Question 8.
Mr. Wilson wants to store the details of students. The details consists of different types of data. Explain different data types used in SQL to store data.
Answer:
The different data types are:

1. Char (Fixed):
It is declared as char (size). This data type is used to store alpha numeric characters. We have to specify the size. If no size is specified, by default we can store only one character. Eg: name char(20).

2. Variable Character:
It is declared as varchar (size). This data type is also used to store alpha numeric characters. But there is a slight difference.lt allocates only enough memory to store the actual size.
Eg: namevarchar(20)

3. Decimal:
It is declared as Dec(size, scale), where size is the number of digits and scale is the maximum number of digits to the right of the decimal point.
Eg: weight dec(3,2)

4. Integer:
It is declared as int. It does not have any arguments. It takes more memory.
Eg: RollNo int.

5. Small Integer:
It is declared as small int. It takes less memory RollNo int.

6. Date:
It is used to store date.
Eg: DOB date.

7. Time:
It is used to store time.
Eg: Joining_Time Time.

Question 9.
A company wants to create a table to store its employees details. Write SQL Commands for the following :

1. Create a table with EMP table having fields EMPNO primary key varchar(10), Name varchar(20), Salary number(6), Department varchar(3)
2. Insert values to table
3. List all employees whose salary > 10,000
4. Display name and salary in the order of name.

Answer:

1. Create table EMP (EMPNO varchar(10) not null primary key, Name varchar(20), Salary decimal(6), Dept varchar(3));
2. Insert into EMP values (‘1001’, ‘ALVIS’, 50000, ‘Sales’);
3. Select * from EMP where salary > 10000;
4. Select name,salary from EMP order by name;

Question 10.
Create a table ‘Savings’ with the following fields.
Acc No (Integer), Name (char), age (Integer) and balance (Number) where Acc No is the primary key. Write SQL commands for the following.

1. Insert data in all the fields for 3 records
2. Display the list of account holders in the ascending order of their names.
3. Display the list of all account holders having age between 20 and 30
4. Display the name and Acc No of customers having a balance > 10 lakhs

Answer:
Create table Savings(Accno decimal(4) not null primary key, name char(2), age decimal(3), Balance decimal(8,2));

1. Insert into savings values (501, ‘Andrea’, 18,45000)
2. Select * from Savings order by name
3. Select * from Savings where age between 20 and 30
4. Select name.accno from Savings where balance > 100000

Question 11.
Which are the components of SQL? How do they help to manage database?
Answer:
The components of SQL are given below.
DDL commands (3 commands)

• Create table: Used to create a table.
• Alter table: Used to modify existing column or add new column to an existing table. There are 2 keywords used ADD and MODIFY.
• Drop table: Used to remove a table from the memory.

DML commands (4 commands)

• Select: Used to select rows from a table. The keyword From is used with this. Where clause is used to secify the conition.
• Insert: Used to insert new records into a table. So the keyword used is INTO.
• Delete: Used to delete records in a table.
• Update: Used to modify the records in a table the keyword used is set.

DCL (Data Control Language) commands

• Grant: It grants permission to the users to the database
• Revoke: It withdraws user’s rights given by using Grant command.

Plus Two Computer Application Structured Query Language Let Us Practice Questions and Answers

Question 1.
The structure of a table is given to store the details of marks scored by students in an examination. (5 Mark)

Write SQL statements for the creation of the table and the following requirements:

1. Insert data into the fields (at least 10 records).
2. Display the details of all students.
3. List the details of Science group students.
4. Count the number of students in each course.
5. Add a new column named Total to store the total marks.
6. Fill the column Total with the sum of the six marks of each student.
7. Display the highest total in each group.
8. Find the highest, lowest and average score in Subject 6 in Commerce group.
9. Display the names in the alphabetical order in each course.
10. Display the name of the student with the highest total.

Answer:

1.

2.

3. mysql>select * from student1 where course = ’Science’;

4. mysql>select course,count(*) from → student1 group by course;

5. mysql>altertable student1 add(total int);

6. mysql>update studentl set total=mark1 + mark2 + mark3 + mark4 + mark5 + mark6;

7. mysql>select course,max(total) from student1 group by course;

8. mysqt>select max(mark6), min(mark6), avg(mark6) from student1;

9. mysql>select course, name from studentl order by course, name;

10. mysql>select name from studentl where total=(select max(total) from student1);

Question 2.
The structure of a table is given to store the details of items in a computer shop. (5 Mark)

Write SQL statements for the creation of the table and the following requirements.

1. Insert data into the fields (at least 10 records).
2. Display the details of all items in the table.
3. Display the names of items and total price of each.
4. List the items manufactured by a company (specify the name) available in the table.
5. Find the number of items from each manufacturer.
6. Display the details of items with the highest price.
7. List the names of items whose price is more than the average price of all the items.
8. Display the names of items purchased after 1 -1 -2015.
9. Get the details of items manufactured by two or three companies (specify the names) available in the table.
10. Display the details of items from a company (specify the name) with a stock of more than 20 pieces.

Answer:
mysql>create table shop (ItemNo int primary key, name char(30) not null,
DOP date,
UnitPrice float(8, 2),
Qty int,
mfrer char(30));
1. mysql>insert into shop values(1,’Keyboard’, ’2014-08-21’, 300.00, 100, ’Tech Com’);
mysql>insert into shop values(2,’Mouse’, ‘2014-08-21 ’, 300.00, 100, ‘Tech Com’);
mysql>insert into shop values(3,’Speaker’,’2015-08-21’, 550.00, 100, ’I Ball’);
mysql>insert into shop values(4,’CPU’,’2015-07-21’, 3500.00, 100, ’AMD’);
mysql>insert into shop values(5,’RAM’, ‘2015-08-1’, 1300.00, 100, ’Hynix’);

2. mysql>select * from shop;

3. mysql>select name, UnitPrice*Qty from shop;

4. mysql>select name from shop where mfrer=’Tech Com’;

5. mysql>select mfrer,count(*) from shop group by mfrer;

6. mysql>select * from shop where UnitPrice = (select max(UnitPrice) from shop);

7. mysql>select name from shop where UnitPrice > (select avg(UnitPrice) from shop);

8. mysql> select “from shop where DOP>’2015-1-1’;

9. mysql> select name frogi shop where mfrer=’l Ball’ and Qty>20;

Question 3.
The structure of a table is given to store the details of higher secondary school teachers. (5 Mark)

Write SQL statements for the creation of the table and the following requirements:

1. Insert data into the fields (at least 10 records).
2. Display the details of all female teachers in the table.
3. List the details of male teachers in the Science department.
4. Display the names and basic pay of teachers in the Language department whose basic pay is Rs. 21000/-or more.
5. Display the names and 71 % of basic pay of the teachers.
6. Find the number of teachers in each department.
7. Display the details of teachers whose basic pay is less than the average basic pay.
8. List the male teachers who joined before 1-1-2010.
9. Increment the basic pay of all teachers by Rs. 1000/-.
10. Delete the details of teachers from the Language department.

Answer:
mysql>create table hsst(Teacherld int primary key, name varchar(30) not null,
gender char,
DOJ date,
Dept varchar(15),
BP float(8, 2));
1. mysql>insert into hsst values(1 ,’Jose’,’M’, ‘2002-01-01’,’Science’, 25660);
mysql> insert into hsst values(2,’Christy’,’F’, ‘2012-01-01′,’Commerce’, 20740);
mysql>insert into hsst values(3,’Geejo George’,’M’, ‘2007-01-01’,’Humanities’, 22360);

2. mysql>select * from hsst where gender=’F’;

3. mysql>select * from hsst where gender=’M’ and Dept=’Science’;

4. mysql>select name, BP from hsst where dept=’Language’ and BP >21000;

5. mysql>select name, BP*.71 from hsst;

6. mysql>select Dept,count(*) from hsst group by Dept;

7. mysql>select * from hsst where BP < (select avg(BP) from hsst);

8. mysql>select * from hsst where gender=’M’ and DOJ<’2010-01-01’;

9. mysql>update hsst set BP=BP+1000;

10. mysql>delete from hsst where Dept= ’Language’;

Question 4.
The structure of a table is given to store the details of customers in a bank. (5 Mark)

Write SQL statements for the creation of the table and the following requirements:

1. Insert data into the fields (at least 10 records).
2. Display the details of customers having SB account.
3. Display the names of customers with a balance among greater than Rs. 5000/-.
4. Display the details of female customers with a balance amount greater than Rs. 10000/-
5. Count the number of male and female customers.
6. Display the names of customers with the highest balance amount.
7. Display the names of customers whose names end with ‘kumar’.
8. Update the balance amount of a particular customer with a deposit amount of Rs. 2000/-.
9. Display the details of customers with a tax deduction of 2% of the balance amount for those who have Rs. 2,00,000/- in their account.
10. Delete the details of customers with current account.

Answer:
mysql>create table customer
AccNo int primary key,
name varchar(30),
gender char,
DOJ date,
TypeOfAcc char(8),
Balance double(10, 2));
1. mysql>insert into customer values (1001,’Adeline’,’F’,’2008-11-26’,’SB’, 50000.00);
mysql>insert into customer values (1002,’Aivis’.’M’,’2007-05-19’,’Current’, 500000.00);
mysql>insert into customer values (1003,’Andrea’,’F’,’2012-07-29’,’SB’, 450000.00);

2. mysql>select * from customer where TypeOfAcc=’SB’;

3. mysql>select name from customer where Balance>5000;

4. mysql>select name from customer where gender=’F’ and Balance>10000;

5. mysql>select gender, count(*) from customer group by gender;

6. mysql>select name from customer where Balance=(select max(Balance) from customer);

7. mysql> select name from customerwhere name like “%kumar”;

8. mysql>update customer set Balance= Balance+ 2000 where Accno= 1001;

9. mysql>select Accno.name, Balance*.02 from customerwhere Balance>=200000;

10. mysql > delete from customerwhere Type Of Acc = ‘Current’;

Plus Two Computer Application Structured Query Language Let Us Assess Questions and Answers

Question 1.
The command to remove rows from a table ‘CUSTOMER’ is: (1 Mark)
(a) REMOVE FROM CUSTOMER
(b) DROP TABLE CUSTOMER
(c) DELETE FROM CUSTOMER
(d) UPDATE CUSTOMER
Answer:
(c) DELETE FROM CUSTOMER

Question 2.
If values for some columns are unknown, how is a row inserted? (2 Mark)
Answer:
In this occasion the column list must be included, following the table name.
Eg. INSERT INTO <TABLE NAME> (COLUMN NAME1, COLUMN NAME2,….) VALUES (VALUE1, VALUE2, );

Question 3.
Distinguish between CHAR and VARCHAR data types of SQL. (2 Mark)
Answer:

• Char: It is used to store fixed number of characters. It is declared as char(size).
• Varchar: It is used to store characters but it uses only enough memory. It is declared as varchar(size).

Question 4.
What is the difference between PRIMARY KEY and UNIQUE constraints? (2 Mark)
Answer:

• Unique: It ensures that no two rows have the same value in a column.
• Primary key: Similar to unique but it can be used only once in a table.

The strings (i) and (iv) only

Question 5.
What do you mean by NULL value in SQL? (1 Mark)
Answer:
Null is a key word in SQL that represents an empty value.

Question 6.
Which of the following is the correct order of clauses for the SELECT statements? (1 Mark)
(a) SELECT, FROM, WHERE, ORDER BY
(b) SELECT, FROM, ORDER BY, WHERE
(c) SELECT, WHERE, FROM, ORDER BY
(d) SELECT, WHERE, ORDER BY, FROM
Answer:
(a) SELECT, FROM, WHERE, ORDER BY

Question 7.
The SQL operator______is used with pattern matching. (1 Mark)
Answer:
LIKE OPERATOR

Question 8.
Read the following strings : (1 Mark)
(i) ‘Sree Kumar’
(ii) ‘Kumaran’
(iii) ‘Kumar Shanu’
(iv) ‘Sreekumar’
Choose the correct option that matches with the pattern ‘%Kumar’, when used with LIKE operator in a SELECT statement.

1. Strings (i) and (ii) only
2. Strings (i), (iii) and (iv) only ,
3. Strings (i) and (iii) only
4. All the strings

Answer:

Question 9.
List any five built-in functions of SQL and the value returned by each. (2 Mark)
Answer:
Aggregate functions:

1. Sum()- find the total of a column.
2. Avg()- find the average of a column.
3. Min() – find the smallest value of a column.
4. Max() – find the largest value of the column.
5. Count() – find the number of values in a column.

Question 10.
Distinguish between WHERE clause and HAVING clause. (2 Mark)
Answer:
Where clause is used to specify the condition.
Syntax: Select * from student where roll=1;

Having clause is used with Group By to give to form groups of records, not conditions and individual rows.

Question 11.
Write any four DML commands in SQL. (2 Mark)
Answer:
The four DML commands are:

1. INSERT
2. UPDATE
3. SELECT
4. DELETE

Question 12.
Write the essential clause required for each of the following SQL command. (2 Mark)

1. INSERT INTO
2. SELECT
3. UPDATE

Answer:

1. INSERT INTO – VALUES
2. SELECT – FROM
3. UPDATE – SET

Question 13.
Consider the given table Customer and write the output of the following SQL queries: (5 Mark)

1. SELECT * FROM customer WHERE Amount>25000;
2. SELECT Name FROM customer
   WHERE Branch IN (‘Calicut, ‘Kannur’);
3. SELECT COUNT (*) FROM customer WHERE Amount < 20000;
4. SELECT Name FROM customer WHERE Name like “%m%”;
5. SELECT * FROM customer ORDER BY Amount DESC;

Answer:
1.

2.

3.

4.

5.

Question 14.
Distinguish between COUNT (*) and COUNT (column-name). (2 Mark)
Answer:

• Count(): find the number of non null values in a column.
• Count(*): This is used to find the number of records with at least one field.

Question 15.
Considerthe given table ITEMS. (5 Mark)

1. Suggest a suitable primary key for the above table. Give justification.
2. Write SQL statements for the following:
   • To list all stationery items.
   • To list item code, name, and profit of all items.
   • To count the number of items in each category.
   • To list all stationery items in the descending order of their unit price.
   • To find the item with the highest selling price.
   • To create a view that contains the details of all stationery items.

Answer:
1. Item code is the primary key for the table

2. SQL statements:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Question 16.
What are the different modifications that can be made on the structure of a table? Which is the SQL command required for this? Specify the clauses needs for each type of modification. (3 Mark)
Answer:
Alter table command is used to modify existing column or add new column to an existing table. There are 2 keywords used ADD and MODIFY.
We can alter the table in two ways.

• We can add a new column to the existing table using the following syntax,
  ALTER TABLE <tablename>ADD(<cloumnname> <type> <constraint>);
• We can also change or modify the existing column in terms of type or size using the following syntax,
  ALTER TABLE<tablename>MODIFY(<column> <newtype>);

Question 17.
A table is created in SQL with 10 records. Which SQL command is used to change the values in a column of specified rows? Write the format. (2 Mark)
Answer:
UPDATE command is used for this.
Syntax : UPDATE <table name> set <column name>=value where condition.

Question 18.
Name the keyword used with SELECT command to avoid duplication of values in a column. (1 Mark)
Answer:
DISTINCT.

Question 19.
Distinguish between DISTINCT and UNIQUE in SQL. (2 Mark)
Answer:

• DISTINCT: This keyword is used to avoid duplicate values in a column of a table.
• Unique: It ensures that no two rows have the same value in a column.

Question 20.
Pick the odd one out and give reason: (1 Mark)
(a) CREATE
(b) SELECT
(c) UPDATE
(d) INSERT
Answer:
(a) CREATE, It is a DDL command the others are DML commands.

Students can Download Chapter 8 Database Management System Questions and Answers, Plus Two Computer Application Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Computer Application Chapter Wise Questions and Answers Chapter 8 Database Management System

Plus Two Computer Application Database Management System One Mark Questions and Answers

Question 1.
Select the property which is desirable for a database.
(a) Redundancy
(b) Inconsistency
(c) Integrity
(d) Complexity
Answer:
(c) Integrity

Question 2.
Pick the odd man out.
(a) Create
(b) Select
(c) Update
(d) Insert
Answer:
(a) Create

Question 3.
______is the ability to modify a schema definition in one level without affecting the schema definition in the next higher level.
Answer:
Data Independance.

Question 4.
For accessing data from a database, provides an interface with programming languages.
Answer:
SQL (or DML)

Question 5.
Give an example for RDBMS package.
Answer:
Packages such as Oracle, My SQL, etc.

Question 6.
Name the key that acts as a candidate key but not a primary key.
Answer:
Alternate key

Question 7.
With the help of_______the process of storing, retrieving and modifying date are greatly simplified.
Answer:
DBMS

Question 8.
If______is controlled , DBMS can guarantee that database is never inconsistent.
Answer:
Redundancy.

Question 9.
The property of a DBMS that guarantees that the database is never inconsistent.
Answer:
Redundancy.

Question 10.
______and_____are the two types of integrity checks.
Answer:
Range checks, Value checks.

Question 11.
Which component of DBMS provides interfaces with programming Languages.
Answer:
DML

Question 12.
The level of database abstraction that describes how the data is actually stored in the storage medium.
Answer:
Physical level.

Question 13.
The level of database abstraction that describes what data are stored in the database.
Answer:
Logical level.

Question 14.
Data Base Administrators (DBA) are more concerned with level______of Abstraction.
Answer:
Logical level.

Question 15.
The programmers are connected with______level of abstraction.
Answer:
Logical level.

Question 16.
The teller at a bank sees only that part of the database that has information on customer accounts. Which level of Database abstraction he is at?
Answer:
View level.

Question 17.
As part of project work, Ashish defines the type of data and the relationship among them. He is at_______level of database abstraction.
Answer:
Logical.

Question 18.
_______is the other name for logical level.
Answer:
Conceptual level

Question 19.
If the modifications made on storage format does not affect the structure of data, then we achieve______data independence.
Answer:
Physical data independence.

Question 20.
Pick the odd one out.
(a) net work model
(b) hybrid model
(c) relational model
(d) hierarchical model
Answer:
(b) Hybrid model.

Question 21.
“I am a data model. My records can have more than one parent record” Who am I?
Answer:
Network model.

Question 22.
Name the language that enables user to access or manipulate data as organized by RDBMS.
Answer:
DBML

Question 23.
ADatabaseAdministratorisableto modify the structure a programmer changes data types and length of a database without affecting certain fields in a database of a bank and the program. Identify the data independence associated in it.
Answer:
Logical data Independence.

Question 24.
Name the language that used to define a database scheme.
Answer:
DDL

Question 25.
Name the person who has central control over the database and programs in DBMS
(a) Naive user
(b) Programmer
(c) Database Administrator
(d) System Analyst
Answer:
(c) Database Administrator.

Question 26.
Oracle DBMS package is based on______model.
Answer:
Relational

Question 27.
Match the following.

Answer:
(a) – (ii), (b) – (iii), (c) – (i).

Question 28.
Name the Relational operation which selects certain columns from the table while discarding others.
Answer:
Project.

Question 29.
How to define the Domain of a column ‘subject 1’ of MARKS relation.
Answer:
Range of values from 0 to 100.

Question 30.
State whetherTrue or False. A view is a kind of table whose contents are taken from other tables.
Answer:
True

Question 31.
State True or False. A view can be queried, inserted into, updated and deleted from.
Answer:
True.

Question 32.
Name an efficient way to provide only required data to users hiding other data from the database.
Answer:
View

Question 33.
Pick the key which can not be used to uniquely identify a tuple on a relation:
(Candidate Key, Primary Key, Alternate Key, Super Key, None of these)
Answer:
None of these.

Question 34.
π(pi) Greek letter is used to denote________operation in relational algebra.
Answer:
Project.

Question 35.
Which relational Algebra operation returns all possible combinations of tuples from two relations.
Answer:
Cartesian product.

Question 36.
Pick the odd one out.
(Select, Cartesian product, Union, intersection)
Answer:
Select – unary operator.

Question 37.
What will be the cardinality of the resultant table if after the following operation if the cardinality of STUDENT is 5 and INSTRUCTOR is 3?
Answer:
15

Question 38.
Consider two relations FOOTBALL AND CRICKET, How to get the names of players play only cricket not also FOOTBALL.
Answer:
CRICKET – FOOTBALL

Question 39.
Why we call a Foreign Key so?
Answer:
It is a candidate Key in another table, A foreigner.

Question 40.
_______is range of values from which actual values are appearing in a given columns are drawn.
Answer:
Domain

Question 41.
Name the table that does not contain data of its own, but is derived from a base table.
Answer:
view

Question 42.
Name a way to uniquely identify a tuple in a relation.
Answer:
By using primary key.

Question 43.
Give two Unary operations performed on a relation in Relational Algebra.
Answer:
Select, Project

Question 44.
Data redundancy is not a desirable property. But All redundancy can not or should not be eliminated. Do you agree with this statement. Justify.
Answer:
Yes. Because sometimes there can be technical or business reasons for maintaining several distinct copies of same data.

Question 45.
Anju is able to do all the internal operations in a DBMS. What type of user is she? What are the other type of users?
Answer:
DBA, Other type users are Appl Prograammer and Naive users.

Question 46.
Which of the following statements are true?

1. DBMS facilitates storage, retrieval, and management of databases.
2. We must keep more copies of the same data in databases.
3. Data inconsistency is eliminated in DBMS.
4. DBMS allows sharing of data, but does not ensure security.

Choose the correct option from the following:
(a) Both 1 and 3 are true
(b) Statements 1, 3 and 4 are true
(c) Statements 1, 2 and 4 are true
(d) All statements are true
Answer:
(a) Both 1 and 3 are true

Question 47.
Which of the following refers to duplication of data in files?
(a) Data redundancy
(b) Data inconsistency
(c) Data integrity
(d) Data security
Answer:
(a) Data redundancy

Question 48.
The following are some responsibilities of database users. Which of them belong to Database Administrator?

1. Design the conceptual schema of the database.
2. Develops programs to interact with the database.
3. Interacts with the database through queries.
4. Ensures authorised and secured access of data

(a) Both 1 and 3
(b) Except 2 and 3
(c) 1, 2 and 4
(d) All the four
Answer:
(b) Except 2 and 3

Question 49.
Which of the keys in a relation do not allow null values? Choose the most appropriate option from the following.
(a) Primary key
(b) Candidate key
(c) Both primary key and candidate key
(d) Either primary key or candidate key
Answer:
(c) Both primary key and candidate key

Question 50.
Choose the level of database abstraction that describes what data is stored in the database and what relationships exist among them.
(a) External
(b) Logical
(c) Physical
(d) View
Answer:
(b) Logical

Question 51.
Which of the following operations can extract the specified columns of a table?
(a) Selection
(b) Projection
(c) Intersection
(d) Set Difference
Answer:
(b) Projection

Plus Two Computer Application Database Management System Two Mark Questions and Answers

Question 1.
The schema of a table is EMPLOYEE(emp_code, emp_name, designation, salary). Write down the relational expressions for the following:

1. To get the name and designation of all employees.
2. To get the details of employees whose salary is above 25000.
3. To get the names of employees who designation is Manager.
4. To get the details of Managers with salary less than 25000.

Answer:

1. π_{name, designation} (EMPLOYEE)
2. σ_salary>25000 (EMLOYEE)
3. π_name( σ_{designation=”Manager”}(EMPLOYEE))
4. σ_{designation = “Manager” and salary < 25000} (EMLOYEE).

Question 2.
Data sharing is an essential feature of DBMS. How data sharing reduces the data inconsistency in a database? Data sharing is an essential feature of DBMS. How data sharing reduces the data inconsistency in a database?
Answer:
Instead of storing more than one copy of the same data, it stores only one copy. This can be shared by several users. If redundancy occurs there is a chance to inconsistency. If redundancy is removed then inconsistency cannot occur.

Question 3.
Pick the odd one out and justify your answer:
(a) Column
(b) Attribute
(c) Field
(d) Tuple
Answer:
(d) Tuple. The other three terminologies indicate the same characteristic of a table.

Question 4.
Suppose a table (relation) contains the details of customers in a bank. Which attribute of the customer will be set as primary key for the table? Give reason for your opinion.
Answer:
Account number can be set as primary key since account number is different for different customers. That is it is unique hence it can be set as primary key.

Question 5.
How many distinct tuples and attributes are there in a relation with cardinality 22 and degree 7. Answer:

• Cardinality is the number of rows (tuples)
• Hence number of tuples is 22
• Degree is the number of coloums (attributes)
• Hence number of attributes 7

Question 6.
Distinguish primary key and alternate key.
Answer:

1. Primary key: It is a set of one or more attributes used to uniquely identify a row Alternate
2. key: A candidate key other than the primary key.

Question 7.
Write an example for relational data model.
Answer:

Question 8.
Observe the following table and choose the correct match from the following options:

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

Question 9.
Consider the table with the following fields Name, RollNumberand Mark for a set of students. Suggest a field among them, which is suitable for primary key. Justify your answer.
Answer:
field RollNumberis suitable for the primary key. The name and mark can have same values so they are not suitable for the primary key.

Question 10.
Raju is confused with the statement ‘logical data independence is more difficult to achieve than physical data independence’. How can you help Raju to understand the statement.
Answer:
Because Appl. Programs heavily dependent on the logical structure of data. So any change in structure means chance of rewriting Appl. Programs.

Question 11.
Match the following.

Answer:
(a) – (ii), (b) – (iii), (c) – (i)

Question 12.
Match the following.

Answer:
(a) – (iv), (b) – (iii), (c) – (i)

Question 13.
Your friend tells you that only relational model is used nowadays as DBMS. Will you agree with that? Justify.
Answer:
Yes. Other two models are complex. In RDMS, no redundancy and relationships can be formed easily.

Question 14.
The telephone number of Gokul is entered in Library file as 802111 and in admission register file as 802171.

1. Can you correlate this problem with a concept in DBMS
2. Can you propose a solution to avoid this.

Answer:

1. Consistency problem
2. Remove data redundancy.

Question 15.
What is a DBMS?
Answer:
A DBMS is used to store large volume of data and it is used to retrieve data whenever needed, edit the existing data, update the data and it is possible to delete also.

Question 16.
“View provides an excellent way to access data from data.” Do you agree with this statement? Justify your answer.
Answer:
Yes. Views can have data from more than one table, view can be queried, inserted into, deleted from and updated like a normal table.

Question 17.
A relation is given below.

Mark the following:
Tuple, Attributes, Cardinality, Degree
Answer:

• Tuple- It is the Rows
• Attributes – It is the columns
• Cardinality – 3(Number of Rows)
• Degree – 4 (Number of Columns)

Question 18.
State whetherTrue or False.

1. Primary key cannot be composite key.
2. Only a candidate key can become a Primary Key.
3. Foreign key of a table is a candidate key in another table.
4. Super key uniquely identifies a row in a relation.

Answer:

1. False
2. True
3. True
4. True

Question 19.
Explain the meaning of following operations.
Answer:
select the tuples whose department is sales and who have salary >5000.

Question 20.
Two relations are given below.
FOOTBALL

CRICKET

Is it possible to find the players those who play both FOOTBALL AND CRICKET by applying any of the Relational Algebra Operations? Explain.
Answer:
Intersection operation.
FOOTBALL ∩ CRICKET.

Question 21.
How will you differentiate Primary key and Super Key? Answer:
Answer:

• primary key- one of the candidate keys chosen to uniquely identifies the rows of a table.
• Super key – Combination of a Primary key with any other attribute or group of attributes.

Question 22.
Cardinality of a table T1 is 10 and of table, T2 is 8 and the two relations are union compatible. If the cardinality of result T1 ∪ T2 is 13, then what is the cardinality of T1 ∩ T2? Justify your answer.
Answer:
Cardinally of table T1 is 10 means it has 10 rows Cardinally of table T2 is 8 means it has 8 rows Normally T1 ∪ T2 is 10 + 8 = 18 But Here T1 ∪ T2 is 13 means after eliminating duplication of 5 rows this happened. This means 5 rows are common. That is T1 ∩ T2 is 5.

Question 23.
Cardinality of a table T1 is 10 and of table, T2 is 8 and the two relations are union compatible.

1. What will be the maximum possible cardinality of T1 ∪ T2?
2. What will be the minimum possible cardinality of T1 ∩ T2?

Answer:
1. Degree(CD) -the number of Columns is the Degree
Cardinality (RC)-: the number of Rows is the Cardinality
T1 ∪ T2 = Sum of cardinalities of Table 1 and Table 2
i.e. T1 ∪ T2 = 10 + 8 = 18.

2. T1 ∩ T2 is the common rows(tuples) in T1 and T2 If there is no common tuples then T1 n T2 is 0 hence the cardinality is 0.

Plus Two Computer Application Database Management System Three Mark Questions and Answers

Question 1.
For catering to the needs of users, a database is implemented through three general levels. Name the three levels and discuss them.
Answer:

1. Physical Level is the lowest level.lt describes how the data is actually stored in the storage medium. At physical level complex low-level data structures are described in detail.
2. Logical level describes what data are stored in the database and what relationships exist among data. Here database is described in terms of simple structures. Records are defined in this level. Programmers work at this level.
3. View level is the highest level of data abstraction. It is concerned with the way in which the users view the database. It describes only part of the database.

Question 2.
Consider the following table and write relational algebra operations for the following DEPOSIT.

1. To display those tuples from DEPOSIT relation where amount is greater than 25,000.
2. To display only AccNo. and Amoun to fall depositors.

Answer:

1. σ_amount >25000 (Deposit)
2. π_{AccNo, amount} (deposit)

Question 3.

Show the output of the following relational operations.

1. R₁ – R₂
2. R₁ ∩ R₂
3. R₁ ∪ R₂

Answer:
1.

2.

3.

Question 4.
Developers hide the complexity of Database system three several levels of abstraction. What are they?
Answer:

1. Physical – (how data)
2. logical – (what data)
3. view level – (view data)

1. Physical Level is the lowest level. It describes how the data is actually stored in the storage medium. At physical level complex low-level data structures are described in detail.

2. Logical level describes what data are stored in the database and what relationships exist among data. Here database is described in terms of simple structure. Records are defined in this level. Programmers work at this level.

3. View level is the highest level of data abstraction. It is concerned with the way in which the users view the database. It describes only part of the database.

Question 5.
How data are organized in a database.
Answer:

1. Field: Smallest unit of data. Eg: RolINo, Name.
2. Record: Collection of related fields. Eg: The information of a particular student.
3. File: Collection of related records. Eg: The informations of 10 students.

Question 6.
Salih check his account details using an ATM machine.

1. Identify the levels of abstraction associated with this?
2. Specify other levels.

Answer:

1. View level
2. Logical level, Physical level

Question 7.
Match the following.

Answer:
(1) – (b), (2) – (c), (3) – (a)

Question 8.
Match the following.

Answer:

Question 9.
Explain the major components of DBMS.
Answer:
Components of DBMS:

1. Databases – It is the main component.
2. Data Definition Language (DDL) – It is used to define the structure of a table.
3. Data Manipulation Language (DML) – It is used to add, retrieve, modify and delete records in a database.
4. Users – With the help of programs users interact with the DBMS.

Question 10.
Categorise the users of DBMS and write their functions.
Answer:
Users of Database:

1. Database Administrator: It is a person who has a central control over the DBMS.
2. Application Programmer: These are computer professionals who interact with the DBMS through programs.
3. Naive users: He is an end user. He does not know the details of DBMS.

Question 11.
A table with three columns is given below. For each relational operation given in the 1^st column find the best matches from 2^nd and 3^rd columns.

Answer:

1. Select → c) σ → (ii)
2. Union → d) ∪ → (iv)
3. Set difference → b) → (i)

Question 12.
Observe the given table BOOK and write down the outputs of the following relational expressions:

Answer:
1. This query returns all the tuples(rows) that contain BPB in column Publisher.

2. This query returns the column Book_Title with price<200.
Book Title

• Computer Fundamentals
• C++ Programming
• Mystery of Chemistry

Plus Two Computer Application Database Management System Five Mark Questions and Answers

Question 1.
Cardinality of a table A is 10 and of table, B is 8 and the two relations are union compatible.

1. What will be the maximum possible cardinality of (A ∪ B) and (A ∩ B)?
2. What will be the minimum possible cardinality of (A ∪ B) and (A ∩ B)?

Give justifications for your answers.
Answer:
There are two possibilities:

1. Both relations contain different tuples (rows).
2. The 8 tuples (rows) of table B are same as that of table A.

Case 1.
If both relations contain different tuples then the maximum possible cardinality of A ∪ B is 10 + 8 = 18.

Case 2.
If 8 tuples of table B are same as that of table A then the maximum possible cardinality of A ⊂ B is 8.

Case 3
If 8 tuples of table B are same as that of table then the minimum possible cardinality of A ∪ B is 10.

Case 4.
If both relations contain different tuples then the minimum possible cardinality of A ∩ B is 0.

Question 2.
There are different data models of which one is currently used in all business transactions. Specify it and discuss in detail.
Answer:
Relational data model is currently used in all business transactions. It is based on the concept introduced by E F Codd. It is composed of one or more tables. Tables are made up of rows and columns.

Here tables are called relations, rows are called tuples and the columns are called attributes. The advantages of this model is neither data redundancy nor complexity. Eg:

Question 3.
You have to present a seminar on the topic “Keys in RDBMS”. Prepare the seminar report.
Answer:
1. Candidate Key:
It is a set of attributes that uniquely identifies a row. There may be more than candidate key and may be a combination of more than one attribute.

2. Primary Key:
A primary key is one of the Candidate Keys. It is a set of one or more attributes that can uniquely identify tuples in a relation.

3. Alternate Key:
The Candidate key that is not the primary key is called the alternate key.

4. Super Key:
A combination of a primary key with any other attribute or group of attributes is called a super key.

5. Foreign Key:
A single attribute or a set of attributes, which is a candidate key in another table, is called foreign key.

Question 4.

Consider the relations STUDENT and GRADE given above and predict the output of the following relational operations in table format.

1. σ (STUDENT)
   Score > 80
2. π (STUDENT)
   RegNo, Score
3. σ (GRADE)
   RegNo <103
4. σ (GRADE)
   Grade = ‘B+’
5. π_{RegNo, Name}( σ_{Scores >70}(STUDENT))

Answer:
1.

2.

3.

4.

5.

Question 5.
Explain any 4 Relational Algebra Operations.
Answer:
A. SELECT operation:
SELECT operation is used to select tuples in a relation that satisfy a selection condition. Greek letter σ (sigma) is used to denote the operation. Syntax,
σ_condition (relation)
eg. σ_{salary < 10000} (EMPLOYEE) – selects tuple whose salary is less than 10000 from EMPLOYEE relation.

B. PROJECT operation:
PROJECT operation selects certain columns from the table and discards the other columns. Greek letter π(pi) is used to denote PROJECT operation. Syntax,
π_condition (relation)
eg. π_{name, salary} (EMPLOYEE) displays only the name and salary of all employees.

C. UNION operation:
This operation returns a relation consisting of all tuples appearing in either or both of the two specified relations. It is denoted by U. duplicate tuples are eliminated.

Union operation can take place between compatible relations only, i.e., the number and type of attributes in both the relations should be the same and also their order.

e.g. SCIENCE ∪ COMMERCE gives all the tuples in both COMMERCE and SCIENCE.

D. INTERSECTION operation:
This operation returns a relation consisting of all the tuples appearing in both of the specified relations. It is denoted by n. It can takes place only on compatible relations, e.g. FOOTBALL ∩ CRICKET returns the players who are in both football and cricket teams.

Question 6.
Why should you choose a database system instead of simply storing data in conventional files?
Answer:
Advantages of DBMS over conventional files:
1. Data Redundancy:
It means duplication of data. DBMS eliminates redundancy. DBMS does not store more than one copy of the same data.

2. Inconsistency can be avoided:
If redundancy occurs there is a chance to inconsistency. If redundancy is removed then inconsistency cannot occur.

3. Efficient data access:
It stored huge amount of data. efficiently and can be retrieved whenever a need arise.

4. Data can be shared:
The data stored in the database can be shared by the users or programs.

5. Standards can be enforced:
The data in the database follows some standards. Eg: a field ‘Name’ should have 40 characters long. Some standards are ANSI, ISO, etc.

6. Security restrictions can be applied:
The data is of great value so it must be kept secure and private. Data security means the protection of data against accidental or intentional disclosure or unauthorized destruction or modification by unauthorized person.

7. Integrity can be maintained:
It ensures that the data is to be entered in the database is correct.

8. Crash recovery:
Some times all or a portion of the data is lost when a system crashes. A good DBMS helps to recover data after the system crashed.

Question 7.
We have admission register, attendance register, marks register, etc. in our school to keep various details of students. Briefly describe how DBMS can replace these registers by stating any five merits.
Answer:
Advantages of DBMS:
1. Data Redundancy:
It means duplication of data. DBMS eliminates redundancy. DBMS does not store more than one copy of the same data.

2. Inconsistency can be avoided:
If redundancy occurs there is a chance to inconsistency. If redundancy is removed then inconsistency cannot occur.

3. Data can be shared:
The data stored in the database can be shared by the users or programs.

4. Standards can be enforced:
The data in the database follows some standards. Eg : a field ‘Name’ should have 40 characters long. Some standards are ANSI, ISO, etc.

5. Security restrictions can be applied:
The data is of great value so it must be kept secure and private. Data security means the protection of data against accidental or intentional disclosure or unauthorized destruction or modification by unauthorized person.

6. Integrity can be maintained:
It ensures that the data is to be entered in the databse is correct.

7. Efficient data access:
It stored huge amount of data efficiently and can be retrieved whenever a need arise.

8. Crash recovery:
Some times all or a portion of the data is lost when a system crashes. A good DBMS helps to recover data after the system crashed.

Plus Two Computer Application Database Management System Let Us Assess Questions and Answers

Question 1.
Who is responsible for managing and controlling the activities associated with the database? (1 Mark)
(a) Database administrator
(b) Programmer
(c) Native user
(d) End user
Answer:
(a) Database administrator

Question 2.
In the relational model, cardinality is the (1 Mark)
(a) number of tuples
(b) number of attributes
(c) number of tables
Answer:
(a) number of tuples

Question 3.
Cartesian product in relational algebra is (1 Mark)
(a) a Unary operator
(b) a Binary operator
(c) a Ternary operator
(d) not defined
Answer:
(b) a Binary operator

Question 4.
Abstraction of the database can be viewed as (1 Mark)
(a) two levels
(b) four levels
(c) three levels
(d) one level
Answer:
(c) three level

Question 5.
In a relational model, relations are termed as (1 Mark)
(a) tuples
(b) attributes
(c) tables
(d) rows
Answer:
(c) tables

Question 6.
In the abstraction of a database system the external level is the (1 Mark)
(a) physical level
(b) logical level
(c) conceptual level
(d) view level
Answer:
(d) view level

Question 7.
Related fields in a database are grouped to form a (1 Mark)
(a) datafile
(b) data record
(c) menu
(d) bank
Answer:
(b) data record

Question 8.
A relational database developer refers to a record as (1 Mark)
(a) criteria
(b) relation
(c) tuple
(d) attribute
Answer:
(c) tuple

Question 9.
An advantage of the database management approach is (1 Mark)
(a) data is dependent on programs
(b) data redundancy increases
(c) data is integrated and can be accessed by multiple programs
(d) none of the above
Answer:
(c) data is integrated and can be accessed by multiple programs

Question 10.
Data independence means (1 Mark)
(a) data is defined separately and not included in programs
(b) programs are not dependent on the physical attributes of data
(c) programs are not dependent on the logical attributes of data
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

Question 11.
Key to represent relationship between tables is called (1 Mark)
(a) primary key
(b) candidate Key
(c) foreign Key
(d) alternate Key
Answer:
(c) foreign key

Question 12.
Which of the folowing operations is used if we are interested only in certain columns of a table? (1 Mark)
(a) PROJECTION
(b) SELECTION
(c) UNION
(d) SELECT
Answer:
(a) PROJECTION

Question 13.
Which of the following operations need the participating relations to be union compatible? (1 Mark)
(a) UNION
(b) INTERSECTION
(c) SET DIFFERENCE
(d) All of the above
Answer:
(d) All of the above

Question 14.
Which database level is closest to the users? (1 Mark)
(a) External
(b) Irttemal
(c) Physical
(d) Conceptual
Answer:
(a) View level (External)

Question 15.
The result of the UNION operation between R1 and R2 is a relation that includes (1 Mark)
(a) all the tuples of R1
(b) all the tuples of R2’
(c) all the tuples of R1 and R2
(d) all the tuples of R1 and R2 which have common columns
Answer:
(c) All the tuples of R1 and R2 (eliminating the duplication)

Question 16.
A file manipulation command that extracts some of the records from a file is called (1 Mark)
(a) SELECT
(b) PROJECT
(c) JOIN
(d) PRODUCT
Answer:
(a) SELECT

Question 17.
An instance of relational schema R (A, B, C) has distinct values of A including NULL values. Which one of the following is true? (1 Mark)
(a) A is a candidate key
(b) A is not a candidate key
(c) A is a primary key
(d) Both (a) and (c)
Answer:
(a) A is a candidate key

Question 18.
How many distinct tuples are there in relation instance with cardinality 22? (1 Mark)
(a) 22
(b) 11
(c) 1
(d) none
Answer:
(a) 22

Question 19.
A set of possible data values is called (1 Mark)
(a) Attribute
(b) Degree
(c) Tuple
(d) Domain
Answer:
(d) Domain

Question 20.
Why should you choose a database system instead of simply storing data in conventional files? (5 Mark)
Answer:
Advantages of DBMS over conventional files:
1. Data Redundancy:
It means duplication of data. DBMS eliminates redundancy. DBMS does not store more than one copy of the same data.

2. Inconsistency can be avoided:
If redundancy occurs there is a chance to inconsistency. If redundancy is removed then inconsistency cannot occur.

3. Efficient data access:
It stored huge amount of data efficiently and can be retrieved whenever a need arise.

4. Data can be shared:
The data stored in the database can be shared by the users or programs.

5. Standards can be enforced:
The data in the database follows some standards. Eg: a field ‘Name’ should have 40 characters long. Some standards are ANSI, ISO, etc.

6. Security restrictions can be applied:
The data is of great value so it must be kept secure and private. Data security means the protection of data against accidental or intentional disclosure or unauthorized destruction or modification by unauthorized person.

7. Integrity can be maintained:
It ensures that the data is to be entered in the database is correct.

8. Crash recovery:
Some times all or a portion of the data is lost when a system crashes. A good DBMS helps to recover data after the system crashed.

Question 21.
Explain the different levels of data abstraction in DBMS? (3 Mark)
Answer:
Levels of Database Abstraction:

1. Physical Level (Lowest Level): It describes how the data is actually stored in the storage medium.
2. Logical Level (Next Higher Level): It describes what data are stored in the database.
3. View Level (Highest level): It is closest to the users. It is concerned with the way in which the individual users view the data.

Question 22.
How are schema layers related to the concepts of logical and physical data independence? (3 Mark)
Answer:
Data Independence:
It is the ability to modify the scheme definition in one level without affecting the scheme definition at the next higher level.

1. Physical Data Independence: It is the ability to modify the physical scheme without causing application programs to be rewritten.
2. Logical Data Independence: It is the ability to modify the logical scheme without causing application programs to be rewritten.

Question 23.
Consider the instance of the EMPLOYEE relation shown in the following table. Identify the attributes, degree, cardinality, and domain of Name and Emp_code. (3 Mark)

Answer:

1. Attributes: These are column names, i.e, Emp_Code, Name, Department, Designation, and Salary.
2. Degree(CD): The number of Columns is the Degree i.e Degree is 5(Here 5 columns).
3. Cardinality (RC): the number of Rows is the Cardinality.

i.e. Cardinality is 4(Here 4 rows)
Domain is the pool of possible values. Domain of Name is a String(Sudheesh, Dhanya,
Fathima, Shajan.etc). Domain of Emp_Code is a number (1000,1001,1002, 1003, etc).

Question 24.
Identify primary key, candidate keys and alternate keys in the instance of EMPLOYEE relation in Question 23. (3 Mark)
Answer:
Candidate key: It is used to uniquely identify the row.

1. Emp_code and Emp_Code + Department (Composite) are the candidate keys:
   Primary key: It is a set of one or more attributes used to uniquely identify a row.
2. Emp_code is the primary key:
   Alternate key: A candidate key other than the primary key.

We set Emp_code as the primary key then Emp_code+ Department is the alternate key.

Question 25.
Consider the instance of the STUDENT relation shown in the following table Assume Reg_no as the primary key. (3 Mark)

1. Identify the candidate keys and alternate keys in the STUDENT relation
2. How are the primary key and the candidate key related?

Answer:

1. Reg_no and Reg_no+Batch are the candidate keys. We set Reg_no as the primary key hence Reg_no+Batch is the alternate key
2. Candidate Key: It is a set of attributes that uniquely identifies a row. There may be more than candidate key and may be a combination of more than one attribute.
3. Primary Key: A primary key is one of the Candidate Keys. It is a set of one or more attributes that can uniquely identify tuples in a relation.

Question 26.
What is a database? Describe the advantages and disadvantages of using DBMS. (5 Mark)
Answer:
A Database is a collection of large volume of data.
Advantages of DBMS:
1. Data Redundancy: It means duplication of data. DBMS eliminates redundancy. DBMS does not store more than one copy of the same data.

2. Inconsistency can be avoided: If redundancy occurs there is a chance to inconsistency. If redundancy is removed then inconsistency cannot occur.

3. Efficient data access: It stored huge amount of data efficiently and can be retrieved whenever a need arise.

4. Data can be shared: The data stored in the database can be shared by the users or programs.

5. Standards can be enforced: The data in the database follows some standards. Eg: a field ‘Name’ should have 40 characters long. Some standards are ANSI, ISO, etc.

6. Security restrictions can be applied: The data is of great value so it must be kept secure and private. Data security means the protection of data against accidental or intentional disclosure or unauthorized destruction or modification by unauthorized person.

7. Integrity can be maintained: It ensures that the data is to be entered in the database is correct.

8. Crash recovery: Some times all ora portion of the data is lost when a system crashes. A good DBMS helps to recover data after the system crashed.

Question 27.
What is data independence? Explain the difference between physical and logical data independence. (3 Mark)
Answer:
Data Independence: It is the ability to modify the scheme definition in one level without affecting the scheme definition at the next higher level.

1. Physical Data Independence: It is the ability to modify the physical scheme without causing application programs to be rewritten.
2. Logical Data Independence: It is the ability to modify the logical scheme without causing application programs to be rewritten.

Question 28.
Enforcement of standard is an essential feature of DBMS. How are these standards applicable in a database? (3 Mark)
Answer:
There is a standard BIS (Bureau of Indian Standards) in the field of Gold and ISBN (International Standard Book Number) in the field of publication. Similarly here is also some standards like ANSI(American National Standards Institute), ISO (International Organization for standardization), etc.. For example a filed “Name” should have 40 characters is a standard.

Question 29.
Cardinality of a table T1 is 10 and of table, T2 is 8 and the two relations are union compatible. If the cardinality of result T1 ∪ T2 is 13, then what is the cardinality of T1 ∩ T2? Justify your answer. (3 Mark)
Answer:
Cardinalty of table T1 is 10 means it has 10 rows Cardinalty of table T2 is 8 means it has 8 rows Normally T1 ∪ T2 is 10 + 8 = 18 But Here T1 ∪ T2 is 13 means after eliminating duplication of 5 rows this happened. This means 5 rows are common. That is T1 ∩ T2 is 5.

Question 30.
Cardinality of a table T1 is 10 and of table, T2 is 8 and the two relations are union compatible.

1. What will be the maximum possible cardinality of T1 ∪ T2?
2. What will be the minimum possible cardinality of T1 ∩ T2? (3 Mark)

Answer:
1. Degree(CD): the number of Columns is the Degree.
Cardinality (RC)-: the number of Rows is the Cardinality
T1 ∪ T2 = Sum of cardinalities of Table1 and Table2
i.e. T1 ∪ T2 = 10 + 8 = 18.

2. T1 ∩ T2 is the common rows(tuples) in T1 and T2 If there is no common tuples then T1 ∩ T2 is 0 hence the cardinality is 0.

Question 31.
Consider the relations, City (city_name, state) and Hotel (name, address, city_name). Answer the following queries in relational algebra (5 Mark)

1. Find the names and address of hotels in Kochi.
2. List the details of cities in Kerala state.
3. List the names of the hotels in Thrissur.
4. Find the names of different hotels.
5. Find the names of hotels in Kozhikode or Munnar.

Answer:

1. π_{name, address} (σCity_name = “Kochi” (Hotel))
2. π_{city_name} (σstate = “Kerala” (City))
3. π_name (σcity_name = “Thrissur” (Hotel))
4. π_name (Hotel)
5. π_name(σcity_name = “Kozhikode” V city_name = “Munnar” (Hotel)).

Question 32.
Using the instance of the EMPLOYEE relation shown in question 23, write the result of the following relational algebra expressions. (5 Mark)

1. σ_{Departments=“Sales”}(EMPLOYEE).
2. σ_{salary> 20000 ∧ Department = “Sales”} (EMPLOYEE).
3. σ_{Salary>20000 ∨ Department = “Sales”} (EMPLOYEE).
4. π_{name, salary} (EMPLOYEE).
5. π_{name, salary} (σ_{Designations=“Manager”} (EMPLOYEE)).
6. π_{name, Department} (σ_{Designation = “Clerk” salary > 20000}(EMPLOYEE)).

Answer:
1.

2.

3.

4.

5.

6. No rows selected.

Question 33.
Consider the instance of the BORROWER and DEPOSITOR relations shown in following figure which stores the details of customers in a Bank. Answer the following queries in relational algebra. (5 Mark)

1. Display the details of the customers who are either a depositor or a borrower.
2. Display the name of customers who are both a depositor and a borrower.
3. Display the details of the customers who are d positors but not borrowers.
4. Display the name and amount of customer who is a borrower but not depositor.

Answer:
1.

2.

3.

4.

Question 34.
Consider the instance of the CUSTOMER and BRANCH relations shown in the following table. Write the Cartesian Product of the two relations. (3 Mark)


Answer:

Students can Download Chapter 7 Web Hosting Questions and Answers, Plus Two Computer Application Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Computer Application Chapter Wise Questions and Answers Chapter 7 Web Hosting

Plus Two Computer Application Web Hosting One Mark Questions and Answers

Question 1.
The companies that provide web hosting services are called_____.
Answer:
Web Hosts

Question 2.
The service of providing storage space in a web server to make a website available on Internet is called______.
Answer:
web hosting

Question 3.
Which of the following is true in the case of dedicated hosting?
(a) It shares server with other websites.
(b) It is usually inexpensive.
(c) It does not guarantee performance.
(d) It offers freedom for the clients to choose the hardware and the software.
Answer:
(d) It offers freedom for the clients to choose the hardware and the software.

Question 4.
Choose the odd one out, and justify your answer,
(a) Shared hosting
(b) Dedicated hosting
(c) DNS
(d) Virtual Private Server
Answer:
(c) DNS others are types of web hosting.

Question 5.
______is the service of providing storage space in a Webserver.

OR

Storing the web pages of a website in a server is popularly known as_____.
Answer:
Web hosting

Question 6.
The companies that provides web hosting services are called______.
Answer:
Web hosts

Question 7.
Odd one out.
(a) Shared
(b) Dedicated
(c) VPS
(d) DNS
Answer:
(d) DNS, means Domain Name System. The others are types of web hosting.

Question 8.
VPS stands for______.
Answer:
Virtual Private Server

Question 9.
From the following select the most commonly used web hosting.
(a) Shared
(b) Dedicated
(c) VPS
(d) DNS
Answer:
(a) Shared hosting

Question 10.
DNS stands for_____.
Answer:
Domain Name System

Question 11.
‘A record’ means_____.
Answer:
Address record

Question 12.
FTP stands for______.
Answer:
File Transfer Protocol

Question 13.
Odd one out.
(a) Mozilla Firefox
(b) FileZilla
(c) CuteFTP
(d) SmartFTP
Answer:
(a) Mozilla Firefox, it is a web browser, the others are popular FTP client software.

Question 14.
______hosting provides web hosting services free of change.
Answer:
Free hosting

Question 15.
Give an example for Free web hosting services with sub domain website address.
Answer:
www.bvmhsskaiparamba.facebook.com

Question 16.
Give an example for Free web hosting services with directory service website address.
Answer:
www.facebook.com/bvm hss kalparamba

Question 17.
CMS stands for______.
Answer:
Content Management System

Question 18.
The term ‘responsive web designing’ was introduced by______.
Answer:
Ethan Marcotte.

Question 19.
The method of Flexible designing of web pages to suit the various types of screens is called______.
Answer:
Responsive web design method

Question 20.
In dedicated hosting, if the client is allowed to place his own purchased web server in the service
provider’s facility, then it is called______.
Answer:
Co-location

Question 21.
What are the information contain in a ICANN database?
Answer:
Registered domain names/name, address, telephone number and e-mail address of the registrants.

Question 22.
What is ‘A record’?
Answer:
‘A record’ is used to store the IP address and the corresponding domain name.

Question 23.
Joomla is an example for______.
(a) CMS
(b) ISP
(c) DNS
(d) None of the above
Answer:
(a) CMS