Accountancy 12th Previous Year Question Paper 2018 (CBSE)

Accountancy

Q.1 Amit and Beena were partners in a firm sharing profits and losses in the ratio of 3 : 1. Chaman was admitted as a new partner for 1⁄6 th share in the profits. Chaman acquired 2⁄5th of his share from Amit. How much share did Chaman acquire from Beena ? 

Answer: Share of profit acquired by Chaman from Aman=  1⁄6 × 2⁄5 = 2⁄30

Therefore, share of profit acquired by Chaman from Beena =  1⁄6 ⁻ 2⁄30 = 3⁄30 = 1⁄10

                                                                 OR

Answer: Share of profit acquired by Chaman from Beena= 3⁄5 × 1⁄6 = 3⁄30 = 1⁄10

 

Q. 2. Neetu, Meetu and Teetu were partners in a firm. On 1st January, 2018, Meetu retired. On Meetu’s retirement the goodwill of the firm was valued at ₹. 4,20,000. Pass necessary journal entry for the treatment of goodwill on Meetu’s retirement.

Answer: 

 

Q.3. Distinguish between ‘Dissolution of partnership’ and ‘Dissolution of partnership firm’ on the basis of settlement of assets and liabilities. 

Answer: 

Basic Dissolution of partnership Dissolution of partnership firm
Settlement of assetsand liabilities Assets and liabilities are revaluedand new balance sheet is drawn Assets are sold andliabilities are paid off

 

Q.4. Ritesh and Hitesh are childhood friends. Ritesh is a consultant whereas Hitesh is an architect. They contributed equal amounts and purchased a building for ₹. 2 crores. After a year, they sold it for ₹. 3 crores and shared the profits equally. Are they doing the business in partnership ? Give reason in support of your answer. 

Answer: No, they are not doing business in partnership because they are not involved in doing sale and purchase of land/ plot on a regular basis/ Mere co-ownership of a property does not amount to partnership.

 

Q.5. Is ‘Reserve Capital’ a part of ‘Unsubscribed Capital’ or ‘Uncalled Capital’ ? 

Answer: Reserve Capital is a part of Uncalled Capital.

 

Q.6. Give the meaning of ‘Debentures issued as Collateral Security’. 

Answer: When the company issues debentures to the lenders as an additonal/ secondary security, in addition to other assets already pledged/ some primary security. Such issue of debentures is called debentures issued as a collateral security.

 

Q.7. Jayant, Kartik and Leena were partners in a firm sharing profits and losses in the ratio of 5 : 2 : 3. Kartik died and Jayant and Leena decided to continue the business. Their gaining ratio was 2 : 3. 

Calculate the new profit sharing ratio of Jayant and Leena. 

Answer: Jayant’s gain= 2⁄5 ×  2⁄10 = 4⁄50 

Leena’s gain = 3⁄5 ×  2⁄10 = 6⁄50  

Jayant’s new share= 5⁄10 + 4⁄50 = 29⁄50 

Leena’s new share = 3⁄10 + 6⁄50 = 21⁄50 

New profit sharing ratio of Jayant and Leena = 29:21 or 29⁄50  : 21⁄50

 

Q.8. What is meant by a ‘Share’ ? Give any two differences between ‘Preference Shares’ and ‘Equity Shares’. 

Answer:  A Share refers to the unit into which the total share capital of the company is divided. 

OR 

               A share means a share in the share capital of the company and includes stock.

Differences between ‘Preference Shares’ and ‘Equity Shares’:

(i) Preference Shares are shares which carry a prefrential right at the time of payment of

dividend and at the time of repayment of capital.

(ii) Equity shares are shares which do not carry a prefrential right at the time of payment of dividend and at the time of repayment of capital.

OR

Differences between ‘Preference Shares’ and ‘Equity Shares’: (Any two)

# Preference Shares  Equity Shares
(i) Share which enjoys preferential right at the time of payment of dividend/Dividend is paid on preference sharesbefore it is paid on equity shares. Shares which do not enjoy preferentialright at the time of payment ofdividend/Dividend is paid on equity shares after it is paid on preference shares.
(ii) Enjoy preferential right at the time ofrepayment of capital. Do not enjoy preferential right at thetime of repayment of capital.
(iii) Rate of dividend may be fixed. Rate of dividend is proposed every yearby the directors and approved by theshareholders.
(iv) Preference shares may be converted into equity shares if the terms of issueprovide for it. Equity shares are not convertible.
(v) Preference shareholders have votingrights in special circumstances. Equity shareholders have voting rightsin all circumstances.
(vi) Preference shareholders do not have the right to participate in the management of the company. Equity shareholders have the right toparticipate in the management of thecompany.
(vii) Arrears on cumulative preference sharesare paid before dividend is paid onequity shares. If dividend is not declared during theyear, it is not accumulated to be paidthe coming years.

 

Q.9. NK Ltd., a truck manufacturing company, is registered with an authorised capital of ₹. 1,00,00,000 divided into equity shares of ₹. 100 each. The subscribed and paid up capital of the company is ₹. 50,00,000. 

The company decided to open technical schools in the Jhalawar district of Rajasthan to train the specially abled children of the area. It is planning to provide them employment in its various production units and industries in the neighbourhood area. To meet the capital expenditure requirements of the project, the company offered 20,000 shares to the public for subscription. The shares were fully subscribed and paid. Present the share capital in the Balance Sheet of the company as per the provisions of Schedule III of the Companies Act, 2013. 

Also identify any two values that the company wants to communicate. 

Answer: 

Balance Sheet of NK Ltd.

As at ………………..(As per revised schedule III)

Notes to Accounts :

Values (Any two):

(i) Concern for the specially abled.

(ii) Creation of job opportunities.

(iii) Development of backward regions.

          (Or any other suitable value)

 

Q.10. Complete the following journal entries left blank in the books of VK Ltd. : VK Ltd. 

Answer:

VK Ltd.

Journal

 

Q.11. Banwari, Girdhari and Murari are partners in a firm sharing profits and losses in the ratio of 4 : 5 : 6. On 31st March, 2014, Girdhari retired. On that date the capitals of Banwari, Girdhari and Murari before the necessary adjustments stood at ₹. 2,00,000, ₹. 1,00,000 and ₹. 50,000 respectively. On Girdhari’s retirement, goodwill of the firm was valued at ₹. 1,14,000. Revaluation of assets and reassessment of liabilities resulted in a profit of ₹. 6,000. General Reserve stood in the books of the firm at ₹. 30,000. 

The amount payable to Girdhari was transferred to his loan account. Banwari and Murari agreed to pay Girdhari two yearly instalments of ₹. 75,000 each including interest @ 10% p.a. on the outstanding balance during the first two years and the balance including interest in the third year. The firm closes its books on 31st March every year. Prepare Girdhari’s loan account till it is finally paid showing the working notes clearly. 

Answer: 

Working Notes:

Calculation of amount payable to Girdhari:          ₹

Girdhari’s Capital                                            1,00,000

Share of goodwill                                                38,000

Share of Revaluation profit                                  2,000

Share of General reserve                                     10,000

                                                                                  1,50,000

(WORKING NOTES MAY BE SHOWN IN ANY FORM)

 

Q.12. Asha and Aditi are partners in a firm sharing profits and losses in the ratio of 3 : 2. They admit Raghav as a partner for ¼ th share in the profits of the firm. Raghav brings ₹. 6,00,000 as his capital and his share of goodwill in cash. Goodwill of the firm is to be valued at two years’ purchase of average profits of the last four years. 

The profits of the firm during the last four years are given below : 

Year Profit (₹)
2013-14 3,50,000
2014-15 4,75,000
2015-16 6,70,000
2016-17 7,45,000

The following additional information is given : 

(i) To cover management cost an annual charge of ₹. 56,250 should be made for the purpose of valuation of goodwill.

(ii) The closing stock for the year ended 31.3.2017 was overvalued by ₹. 15,000. Pass necessary journal entries on Raghav’s admission showing the working notes clearly. 

Answer: 

Working Notes:

Calculation of goodwill:

Profits

2013-14 ₹3,50,000 – ₹56,250 = ₹2,93,750

2014-15 ₹4,75,000 – ₹56,250 = ₹4,18,750

2015-16 ₹6,70,000 – ₹56,250 = ₹6,13,750

2016-17 ₹7,45,000 – ₹56,250 – ₹15,000 = ₹6,73,750

Goodwill of the firm = (₹2,93,750 + ₹4,18,750 + ₹6,13, 750 + ₹6,73,750)/4 x 2 = ₹10,00,000

Raghav’s share of goodwill = 1⁄4 x ₹10,00,000 = ₹2,50,000

OR

Answer: Calculation of goodwill:

Total Profits of four years = ₹3,50,000 + ₹4,75,000 + ₹6,70,000 + ₹7,30,000 = ₹22,25,000

Average Profits = ₹ 5,56,250 – ₹ 56,250 = ₹ 5,00,000

Goodwill of the firm = ₹ 5,00,000 x 2 = ₹10,00,000

Raghav’s share of goodwill = 1⁄4 x ₹10,00,000 = ₹2,50,000

 

Q.13. Pranav, Karan and Rahim were partners in a firm sharing profits and losses in the ratio of 2 : 2 : 1.

 On 31st March, 2017 their Balance Sheet was as follows : 

Balance Sheet of Pranav, Karan and Rahim as on 31.3.2017 

Liabilities Amount ₹ Assets Amount ₹
Creditors 3,00,000 Fixed Assets 4,50,000
General Reserve  1,50,000 Stock 1,50,000
Capitals
Pranav – 2,00,000Karan – 2,00,000
Rahim – 1,00,000



5,00,000
DebtorsBank 1,50,000
Total 9,50,000 Total 9,50,000

Karan died on 12.6.2017. According to the partnership deed, the legal representatives of the deceased partner were entitled to the following : 

(i) Balance in his Capital Account. 

(ii) Interest on Capital @ 12% p.a. 

(iii) Share of goodwill. Goodwill of the firm on Karan’s death was valued at ₹. 60,000. 

(iv) Share in the profits of the firm till the date of his death, calculated on the basis of last year’s profit. The profit of the firm for the year ended 31.3.2017 was ₹. 5,00,000. 

Prepare Karan’s Capital Account to be presented to his representatives. 

Answer: 

Working Notes:

Interest on Capital = 12/100 x 73/365 x ₹2,00,000 = ₹4,800

Share of Profits = 2/5 x 5,00,000 x 73/365 = ₹40,000

Share of goodwill = 2/5 X ₹60,000 = ₹24,000

Share of General Reserve = 2/5 x ₹1,50,000 = ₹60,000

 

Q.14. Chander and Damini were partners in a firm sharing profits and losses equally. On 31st March, 2017 their Balance Sheet was as follows : 

On 1.4.2017, they admitted Elina as a new partner for 31 rd share in the profits on the following conditions : 

(i) Elina will bring ₹. 3,00,000 as her capital and ₹. 50,000 as her share of goodwill premium, half of which will be withdrawn by Chander and Damini. 

(ii) Debtors to the extent of ₹. 5,000 were unrecorded. 

(iii) Furniture will be reduced by 10% and 5% provision for bad and doubtful debts will be created on bills receivables and debtors. 

(iv) Value of land and building will be appreciated by 20%. 

(v) There being a claim against the firm for damages, a liability to the extent of ₹. 8,000 will be created for the same. 

Prepare Revaluation Account and Partners’ Capital Accounts. 

Answer:

 

Q.15. On 1st April, 2014, KK Ltd. invited applications for issuing 5,000 10% debentures of ₹. 1,000 each at a discount of 6%. These debentures were repayable at the end of 3rd year at a premium of 10%. Applications for 6,000 debentures were received and the debentures were allotted on pro-rata basis to all the applicants. Excess money received with applications was refunded. 

The directors decided to transfer the minimum amount to Debenture Redemption Reserve on 31.3.2016. On 1.4.2016, the company invested the necessary amount in 9% bank fixed deposit as per the provisions of the Companies Act, 2013. Tax was deducted at source by bank on interest @ 10% p.a. 

Pass the necessary journal entries for issue and redemption of debentures. Ignore entries relating to writing off loss on issue of ebentures and interest paid on debentures. 

Answer:

Q.16. Srijan, Raman and Manan were partners in a firm sharing profits and losses in the ratio of 2 : 2 : 1. On 31st March, 2017 their Balance Sheet was as follows : 

On the above date they decided to dissolve the firm. 

(i) Srijan was appointed to realise the assets and discharge the liabilities. Srijan was to receive 5% commission on sale of assets (except cash) and was to bear all expenses of realisation. 

(ii) Assets were realised as follows : 

Plant 85,000 

Stock 33,000 

Debtors 47,000 

(iii) Investments were realised at 95% of the book value. 

(iv) The firm had to pay ₹. 7,500 for an outstanding repair bill not provided for earlier. 

(v) A contingent liability in respect of bills receivable, discounted with the bank had also materialised and had to be discharged for ₹. 15,000. 

(vi) Expenses of realisation amounting to ₹. 3,000 were paid by Srijan. Prepare Realisation Account, Partners’ Capital Accounts and Bank Account. 

OR 

Q.16. Moli, Bhola and Raj were partners in a firm sharing profits and losses in the ratio of 3 : 3 : 4. Their partnership deed provided for the following : 

(i) Interest on capital @ 5% p.a. 

(ii) Interest on drawing @ 12% p.a. 

(iii) Interest on partners’ loan @ 6% p.a. 

(iv) Moli was allowed an annual salary of ₹. 4,000; 

Bhola was allowed a commission of 10% of net profit as shown by Profit and Loss Account and Raj was guaranteed a profit of ₹. 1,50,000 after making all the adjustments as provided in the partnership agreement. 

Their fixed capitals were Moli : ₹. 5,00,000; 

Bhola : ₹. 8,00,000 and Raj : ₹. 4,00,000. 

On 1st April, 2016 Bhola extended a loan of ₹. 1,00,000 to the firm. The net profit of the firm for the year ended 31st March, 2017 before interest on Bhola’s loan was ₹. 3,06,000. Prepare Profit and Loss Appropriation Account of Moli, Bhola and Raj for the year ended 31st March, 2017 and their Current Accounts assuming that Bhola withdrew ₹. 5,000 at the end of each month, Moli withdrew ₹. 10,000 at the end of each quarter and Raj withdrew ₹. 40,000 at the end of each half year. 

Answer: 

OR

Answer:

 

Q.17. X Ltd. invited applications for issuing 50,000 equity shares of ₹. 10 each. The amount was payable as follows : 

On Application : ₹. 2 per share 

On Allotment : ₹. 2 per share 

On First Call : ₹. 3 per share 

On Second and Final Call : Balance amount Applications for 70,000 shares were received. Applications for 10,000 shares were rejected and the application money was refunded. Shares were allotted to the remaining applicants on a pro-rata basis and excess money received with applications was transferred towards sums due on allotment and calls, if any. Gopal, who applied for 600 shares, paid his entire share money with application. Ghosh, who had applied for 6,000 shares, failed to pay the allotment money and his shares were immediately forfeited. These forfeited shares were re-issued to Sultan for ₹. 20,000; ₹. 4 per share paid up. The first call money and the second and final call money was called and duly received. Pass necessary journal entries for the above transactions in the books of X Ltd. Open Calls-in-Advance Account and Calls-in-Arrears Account wherever necessary. 

OR 

Q.17. A Ltd. invited applications for issuing 1,00,000 shares of ₹. 10 each at a premium of ₹. 1 per share. The amount was payable as follows : 

On Application : ₹. 3 per share 

On Allotment : ₹. 3 per share (including premium) 

On First Call : ₹. 3 per share 

On Second and Final Call : Balance amount 

Applications for 1,60,000 shares were received. 

Allotment was made on the following basis : 

(i) To applicants for 90,000 shares : 40,000 shares 

(ii) To applicants for 50,000 shares : 40,000 shares 

(iii) To applicants for 20,000 shares : full shares 

Excess money paid on application is to be adjusted against the amount due on allotment and calls. 

Rishabh, a shareholder, who applied for 1,500 shares and belonged to category (ii), did not pay allotment, first and second and final call money. 

Another shareholder, Sudha, who applied for 1,800 shares and belonged to category (i), did not pay the first and second and final call money. 

All the shares of Rishabh and Sudha were forfeited and were subsequently re-issued at ₹. 7 per share fully paid. 

Pass the necessary journal entries in the books of A Ltd. Open Calls-in-Arrears Account and Calls-in-Advance Account wherever required.

Answer

OR

Answer:

 

PART B

(Analysis of Financial Statements) 

Q.18. State the primary objective of preparing a Cash Flow Statement.

Answer:The primary objective of Cash Flow Statement is to provide useful information about cash flows (inflows and outflows) of an enterprise during a particular period underoperating, investing and financing activities.

 

Q.19.‘Interest received and paid’ is considered as which type of activity by a finance company while preparing a Cash Flow Statement ?

Answer: Interest received – Operating activity.

Interest paid – Operating activity.

OR

Answer: Interest received and paid – Operating activity.

 

Q.20. Prepare a common size Balance Sheet of KJ Ltd. from the following information : 

Answer:

In case the examinee has prepared only columns (i) and (ii) in the correct order, one mark may be awarded.

 

Q.21. From the following information obtained from the books of Kundan Ltd., calculate the inventory turnover ratio for the years 2015 − 16 and 2016 – 17 :

                                                     2015 − 16(₹)              2016 − 17(₹)

Inventory on 31st March              7,00,000                17,00,000

Revenue from operations           50,00,000               75,00,000

(Gross profit is 25% on cost of revenue from operations)

In the year 2015 − 16, inventory increased by 2,00,000.

Answer:

Inventory turnover ratio = Cost of Revenue from operations/Average inventory

2015-16 :

Cost of Revenue from operations= ₹50,00,000 -₹10,00,000 = ₹40,00,000

Average inventory = Opening inventory + Closing inventory /2

= (₹5,00,000 + ₹7,00,000)/2 

= ₹6,00,000

Inventory turnover ratio = ₹40,00,000/₹6,00,000 = 6.67 times

2016-17 :

Cost of Revenue from operations= ₹75,00,000 – ₹15,00,000 = ₹60,00,000

Average inventory = Opening inventory + Closing inventory/2

= (₹7,00,000 + ₹17,00,000)/2

= ₹12,00,000

Inventory turnover ratio = ₹60,00,000/₹12,00,000 = 5 times

 

Q.22. JW Ltd. was a company manufacturing geysers. As a part of its long term goal for expansion, the company decided to identify the opportunity in rural areas. Initial plan was rolled out for Bhiwani village in Haryana. Since the village did not have regular supply of electricity, the company decided to manufacture solar geysers. The core team consisting of the Regional Manager, Accountant and the Marketing Manager was taken from the Head Office and the remaining employees were selected from the village and neighbourhood areas. At the time of preparation of financial statements, the accountant of the company fell sick and the company deputed a junior accountant temporarily from the village for two months. The Balance Sheet prepared by the junior accountant showed the following items against the Major Heads and Sub-heads mentioned which were not as per Schedule III of the Companies Act, 2013. 

Item Major Head/Sub-Head
Loose Tools  Trade Receivables
Cheques in Hand Current Investments
Term Loan from Bank Other Long-term Liabilities
Computer Software Tangible Fixed Assets

Identify any two values that the company wants to communicate to the society. Also present the above items under the correct major heads and sub-heads as per Schedule III of the Companies Act, 2013. 

Answer:

Values (Any two):

(i) Development of rural areas.

(ii) Sensitivity towards the environment.

(iii) Generation of employment.

(Or any other suitable value)

Item Head Sub-Head
Loose Tools  Current assets Inventories
Cheques in Hand Current assets Cash and Cash Equivalents
Term Loan from Bank Non Current Liabilities Long Term Borrowings
Computer Software Non Current Assets Fixed – Intangible Assets

 

Q.23. From the following Balance Sheet of JY Ltd. as at 31st March 2017, prepare a Cash Flow Statement : 

Balance Sheet of JY Ltd. as at 31.3.2017 

Notes to Accounts :

Additional Information :

₹ 1,00,000, 10% debentures were issued on 31.3.2017.

Answer: 

Working Notes:

Calculation of Net profit before tax:

                                                         ₹

Net Profit for the year           1,25,000

Add Proposed dividend            75,000

Add Provision for tax            1,25,000

                                                    3,25,000

FULL CREDIT IS TO BE GIVEN IF AN EXAMINEE HAS TAKEN ‘SHORT TERM LOANS AND ADVANCES’ AS INCREASE IN CURRENT ASSETS UNDER OPERATING ACTIVITIES.

In that case,

CASH FROM OPERATIONS = ₹2,52,000

CASH GENERATED FROM OPERATING ACTIVITIES = ₹1,77,500

CASH USED IN INVESTING ACTIVITIES = ₹2,12,500

 

PART B 

(Computerised Accounting) 

Q.18. How does the usage of computer sharpen the competitive edge and enhance the profitability of a business ?

Answer: The quick, accurate and timely access to the information, helps decision making fast and correct, hence it helps the business to earn better.

 

Q.19. Give an example to explain the meaning of ‘stored’ and ‘derived’ attribute.

Answer: The information which is stored e.g. date of birth of a person is an example of stored attribute where as when his/her age is calculated automatically is derived attribute.

 

Q.20. Name the value which represents absence of data. Also state the situations which may require the use of these values.

Answer: The value is called “Null value” The three situations in which these can be used are

1. When a particular attribute does not apply to an entry.

2. Value of an attribute is unknown.

3. Unknown because it does not exit.

 

Q.21. Differentiate between desktop database and server database.

Answer: (Any four)

1. Application : Desktop database can be used by a single user server data base can

be used by many users at the same time.

2. Additional provision for reliability : Desktop database

Doesn’t present this but these provisions are available in server based database.

3. Cost : Desktop database tend to cost less than the server database.

4. Flexibility regarding the performance in front end applications : It is not present in desktop database but server database provide this flexibility.

5. Suitability : Desktop database are suitable for small/home offices and server

database are more suitable for large business organisations.

Q.22. Give four limitations of computerised accounting system.

Answer: Following are the limitations of computerised accounting softwares :

1. Faster obsolescence of technology necessitates investment in shorter period of time.

2. Data may be lost or corrupted due to power interruptions.

3. Data are prone to hacking.

4. Un-programmed and un-specified reports cannot be granted .

 

Q.23. ABC Ltd. operates in two cities  Bengaluru and Mangaluru. House Rent Allowance for Bengaluru is ₹ 5,000 and for Mangaluru is ₹ 4,000. Dearness Allowance is calculated on Basic Pay as follows :

15% of Basic Pay if basic pay is less than ₹ 15,000.

10% of Basic Pay if basic pay is greater than ₹ 15,000.

Standard number of days are taken as 30 days in a month.

Calculate the amount using Excel :

(i) Gross Salary of Mr. Mahesh, who is working in Bengaluru. He has availed leave without pay for 3 days and his Basic Pay is ₹ 25,000.

(ii) Gross Salary of Mr. Ranjan, who is working in Mangaluru. Basic Pay of Mr. Ranjan is ₹ 14,000.

Answer: 

Gross salary of Mr. Mahesh and Ranjan

Basic pay of Mahesh Column A1 = 25000

Basic pay of Ranjan column A2 = 14000

Basic pay earned for Mahesh column B1 = A1* 27/30 = 22500

Basic pay earned for Ranjan column B2 = A2 = 14000

HRA for Mahesh Column C1 = 5000

HRA for Ranjan Column C2 = 4000

DA for Mahesh Column D1 = IF (A1>15000, 10/100*B1, 15/100*B1)

DA for Ranjan Column D2 = IF (A2 > 15000, 10/100*B2, 5/100*B2)

D1 = 2250

D2 = 2100

Gross salary for Mahesh = Column E1 = SUM (B1,C1,D1)

Gross salary for Ranjan = Column E2 = SUM (B2,C2,D2)

Mr. Mahesh’s Salary E1 = 22500 + 5000 + 2250 = ₹29750

Mr. Ranjan’s Salary E2 = 14000 + 4000 + 2100 = ₹20100

Maths 12th Previous Year Question Paper 2018 (CBSE)

Maths

Section – A

Q.1. Find the value of tan-1 -cot-1 .

Solution: We have 

Q.2. If the matrix A = is skew-symmetric, find the values of’a’ and ‘b’. 

Solution:

Q.3. Find the magnitude of each of the two vectors  , having the same magnitude such that the angle between them is 60° and their scalar product is 9/2 

Solution: Let be two such vectors. 

Q.4. If a * b denotes the larger of ‘a’ and ‘b’ and if a o b = (a * b) + 3, then write the value of (5) o (10), where * and o are binary operations.** 

 

Section – B

Q.5. Prove that

Solution: R.H.S = sin-1(3x – 4x3)

Putting x = sin θ in R.H.S, we get

R.H.S = sin-1(3 sin θ – 4 sin3 θ)

= sin-1(sin 3θ)

 

Q.6. Given A =   compute A-1 and show that 2A-1 = 9I – A. [2]

Solution: 

 

Q.7. Differentiate  with respect to x. 

Solution: 

 

Q.8. The total cost C(x) associated with the pro-duction of x units of an item is given by C(x) = 0.005x3 – 0.02x2 + 30s + 5000. Find the marginal cost when 3 units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output.

Solution : Cost function is given as

C(s) = 0.005x3 – 0.02x2 + 30s + 5000

Marginal cost (MC) =

= 0.005(3x2) – 0.02 (2x) + 30

= 0.015x2 – 0.04s+ 30

When x = 3,MC = 0.015(3)2 – 0.04(3) + 30

= 0.135 – 0.12 + 30

= 30.015

 

Q.9. Evaluate, 

Solution: We have, 

 

Q.10. Find the differential equation representing the family of curves y = aebx + 5, where a and b are arbitrary constants. 

Solution: Given curve is

y = aebx + 5 …(i)

Differentiating (i) w.r.t. s, we get

Q.11. If θ is the angle between two vectors   and , find sin θ.

Solution: 

 

Q.12. A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Solution : The sample space has 36 Outcomes. Let A be event that the sum of observations is 8.

∴ A = {(2, 6), (3, 5), (5, 3), (4, 4), (6, 2)}

 

Section – C

Q.13. Using properties of determinants, prove that:

Solution:

 

Q.14. If (x2 + y2)2 = xy, find dy/dx.

Solution: We have,

(x2 + y2)2 = xy …(i)

OR

Q.14. If x = a(2θ – sin 2θ) and y = a(1 – cos 2θ), find dy/dx when θ =π ⁄3

Solution: Given, x = a (2θ – sin 2θ)  and y = a (1 – cos 2θ)

Differentiating x* and y w.r.t. θ, we get 

 

Q.15. If y = sin(sin x), prove that:

Solution: Given, y = sin (sin x)

Differentiating y w.r.t. x, we get

 = cos (sin x) • cos x

Differentiating again w.r.t. x, we get

 = cos (sin x) (- sin x) + cos x [-sin (sin x)-cos x]

 

Q.16. Find the equations of the tangent and the normal to the curve 16x2 + 9y2 = 145 at the point (x1, y1), where x1 = 2 and y1 > 0. 

Solution: Given curve is

16x2 + 9y2 = 145 ….(i)

Since (x1, y1) lies on equation (i),

OR

Q.16. Find the intervals in which the function

(a) strictly increasing,

(b) strictly decreasing.

Solution: Given function is

 

Q.17. An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of material will be least when depth of the tank is half of its width. If the cost is to be borne by nearby settled lower income families, for whom water will be provided, what kind of value is hidden in this question ? 

Solution: Let the length, breadth and height of the open tank be x, x and y units respectively.

Then, Volume (V) = x2y …(i)

Area is minimum, thus cost is minimum when x = 2y.

i.e., depth of tank is half of the width.

Value : Any relevant value.

 

Q.18. Find

Solution: 

 

Q.19. Find the particular solution of the differential equation ex tan ydx + (2 – ex) sec2 ydy = 0, given that y = 

when x = 0.

Solution: Given differential equation is,

OR

Q.19. Find the particular solution of the differential equation   + 2y tan x = sin x, given than y = 0 when x = .

Solution: Given differential equation is,

Hence, particular solution is

y sec2 x = sec x – 2

or y = cos x – 2 cos2x

 

Q.20. Let Find a vector d  which is perpendicular to both .

Solution: Given

Q.21. Find the shortest distance between the lines and .

Solution: Gives lines are,

Q.22. Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3, 4, 5 or 6, she tosses a coin once and notes whether a ‘head’ or ‘tail’, is obtained. If she obtained exactly one “tail7, what is the probability that she threw 3, 4, 5 or 6 with the dice? 

Solution : Let E1 be the event that girl gets 1 or 2 on the roll and E2 be the event that girl gets 3, 4, 5, or 6 on the roll of a die.

Let A be the event that she gets exactly one tail. If she tossed coin 3 times and gets exactly one tail then possible outcomes are HTH, HHT, THH.

                                      = 8/11

 

Q.23. Two numbers are selected at random (without replacement) from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and variance of X.

Solution: First five positive integers are 1, 2, 3, 4, 5. We select two positive numbers in 5 × 4 = 20 ways. Out of these, two numbers are selected at random. Let X denote larger of the two selected numbers. Then, X can have values 2, 3, 4 or 5.

P(X = 2) – P (larger no. is 2) = {(1, 2) and (2, 1)}

Therefore, mean and variance are 4 and 1 respectively.

 

Q.24. Let A = {x ϵ Z : 0 ≤ x ≤ 12}. Show that R = {(a, b): a, b ϵ A, | a – b | is divisible by 4} is an equivalence relation. Find the set of all elements related to 1. Also write the equivalence class . 

Solution: Given, R = {(a, b): a, b ϵ A, | a – b | is divisible by 4}

Reflexivity: For any a ϵ A

| a – a | = 0, which is divisible by 4

(a, a) ϵ R

So, R is reflexive.

Symmetry: Let (a, b) ϵ R

⇒ | a – b | is divisible by 4

⇒ | b – a | is divisible by 4 [ ∵ | a – b | = | b – a | ]

⇒ (b, a) ϵ R

So, R is symmetric.

Transitive : Let (a, b) ϵ R and (b, c) ϵ R

⇒ | a – b | is divisible by 4<br> ⇒ | a – b | = 4k

∴ a – b = ± 4k, k ϵ Z …(i)

Also, |b – c| is divisible by 4

⇒ |b – c| = 4m

∴ b – c = ±4m, m ϵ Z …(ii)

Adding equations (i) and (ii)

a – b + b – c = ±4 (k + m)

⇒ a – c = ±4 (k + m)

|a – c | is divisible by 4,

⇒ (a, c) ϵ R

So, R is symmetric.

⇒ R is reflexive, symmetric and transitive.

∴ R is an equivalence relation.

Let x be an element of R such that (x, 1) ϵ R

Then | x – 1| is divisible by 4

x – 1 = 0, 4, 8, 12,

⇒ -x = 1, 5, 9 (∵ x ≤ 12)

∴ Set of all elements of A which are related to 1 are {1, 5, 9}.

Equivalence class of 2 i.e.

[2] = {(a, 2): a ϵ A, |a – 2| is divisible by 4}

⇒ | a – 2| = 4k(k is whole number, k ≤ 3)

⇒ a = 2, 6, 10

Therefore, equivalence class [2] is {2, 6, 10]

OR

Q.24. Show that the function f : R → R defined by f(x) = , ∀x ϵ R is neither one-one nor onto. Also,if g: R → R is defined as g(x) = 2x – 1, find fog(x).

Solution: 

Q.25. If A = , find A-1. Use it to solve the system of equation: 

2x – 3y + 5z = 11

3x + 2y – 4z = -5

x + y – 2z = -3

Solution:

OR

Q.25. Using elementary row transformations, find the inverse of the matrix:

We know that, A = IA

Q.26. Using integration, find the area of the region in the first quadrant enclosed by the X-axis, the line y = x and the circle x2 + y2 = 32. 

Solution: Given curve is

Q.27. Evalute

Solution:

OR

Q.27. Evaluate  as the limit of the sum.

Solution:

 

Q.28. Find the distance of the point (-1, -5, -10) from the point of intersection of the line 

 and the plane 

Solution:  Equation of line is

 

Q.29. A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic and a hand-operated. It takes 4 minutes on the automatic and 6 minutes on the hand-operated machines to manufacture a packet of screws ‘A’ while it takes 6 minutes on the automatic and 3 minutes on the hand- operated machine to manufacture a packet of screws ‘B’. Each machine is available for at most 4 hours on any day. The manufacturer can sell a packet of screws ‘A’ at a profit of 70 paise and screws ‘B’ at a profit of ₹1. Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory owner produce in a day in order to maximize his profit ? Formulate the above LPP and solve it graphically and find the maximum profit 

Solution: Let the number of packets of screw ‘A’ manufactured in a day be x and that of screw B be y.

Plotting the points on the graph, we get the feasible region OABC as shown (Shaded).

Hence, profit will be maximum if company produces 30 packets of screw A and 20 packets of screw B and maximum profit = ₹ 41.

Maths 12th Previous Year Question Paper 2019 (CBSE)

Previous Paper (SET-1) Click Here!

Maths

Section – A

Q.1. Find | AB |, if A = and B = .

Solution: 

Q.2. Differentiate  , with respect to x.

Solution:

Section – B

Q.6. If A = and A3 = | = 125, the find the value of p.

Solution: 

Q.12. Find the general solution of the differential equation  

Solution: 

Section – C

Q.21. If (a + bx)ey/x = x, then prove that

Solution: We have

Q.22. The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of its edge is 12 cm? 

Solution: Let x be the length of side, V be the volume and S be the surface area of cube.

Then, V = x3 and S = 6x2, where x is a function of time t

Q.23. Find the cartesian and vector equations of the plane passing through the point A(2, 5, – 3), B(- 2, 3, 5) and C(5, 3, – 3). 

Solution: We know that the general equation of the plane passing through three points (x1, y1, z1) (x1, y1, Z1), (x3, y3, z3)

Section – D

Q.24. Find the point on the curve y2 = 4x, which is nearest to the point (2, – 8). 

Solution: Given curve is of the form, y2 = Ax and let p(x, y) is a point on the curve which is nearest to the point (2, – 8).

Q.25. Find as the limit of sums  .

Solution: We have

OR

Q.24. Using integration, find the area of the triangular region whose sides have the equation y = 2x + 1, y = 3x + 1 and x = 4.

Solution: The equations of sides of triangle are

y = 2x + 1, …(i)

y = 3x + 1 …(ii)

and x = 4 …(iii)

The equation y = 2x + 1 meets x and y axes at  and (0, 1). By joining these two points we obtain the graph of x + 2y = 2. Similarly, graphs of other equations are drawn.

Solving equation (i), (ii) and (iii) in pairs, we obtain the coordinates of vertices of ∆ABC are A(0, 1), B (4, 13) and C (4, 9).

Then, area of ∆ABC = Area (OLBAO) – Area (OLCAO)

Maths

SET-III

Section – A

Q.1.Find the differenital equation representing the family of curves y = ae2x + 5, where a is an arbitrary constant. 

Solution:

Given, y = ae2x + 5

On differentiating w.r.t. x, we get

 

Q.2. If y = cos√3x, then find dy/dx

Solution:

Section – B

Q.5. Show that the points  ,  and   are collinear. 

Solution:

Q.6. Find:

Solution:

 

Section – C

Q.13. Solve for x: tan-1 + tan-1(x – 1) = tan-1(8/31).

Solution:

 

Q.14. If x = aet (sin t + cost t) and y = aet (sin t – cos t), then prove that 

dy/dx= x+y/x-y.

Solution:

                              dy/dx= x+y/x-y                  Hence proved.

 OR

Q.14. Differentiate xsinx + (sinx)cosx with respect to x.

 Solution:

 

Q.15. Find:

Solution:

 

Section – D

Q.24. 

Show that for the matrix A = , A3 – 6A2 + 5A + 111 = 0. Hence, find A -1.

Solution:

OR

Q.24. Using matrix method, solve the following system of equations:

3x – 2y + 3z = 8

2x + y – z = 1

4x – 3y + 2z = 4

Solution:

The given equations are

3x – 2y + 3z = 8

2x + y – z = 1

4x – 3y + 2z = 4

These equations can be written in the form AX = B, where

 

Q.26. A bag contains 5 red and 4 black balls, a second bag contains 3 red and 6 black balls. One of the two bags is selected at random and two balls are drawn at random (without replacement) both of which are found to be red. Find the probability that the balls are drawn from the second bag. 

Solution:

 Let E1 be the event of choosing the bag I, E2 be the event of choosing the bag II and A be the event of drawing a red ball.

 Then,

Maths 12th Previous Year Question Paper 2019 (CBSE)

Maths

SET-I

Section – A

Q.1. If A is a square matrix satisfying A’ A = I, write the value of |A|.

Solution: Given, A’ A= I

 

Q.2. If y = x | x |, find for x &lt; 0. 

Solution: If y = x | x |

 

Q.3. Find the order and degree (if defined) of the differential equation 

Solution:

Order of this equation is 2.

Degree of this equation is not defined.

 

Q.4. Find the direction cosines of a line which makes equal angles with the coordinate axes. 

Solution : Let the direction cosines of the line make an angle with each of the coordinate axes and direction cosines be l, m and n.

OR

Q.4 A line passes through the point with position vector  and is in the direction of the vector  . Find the equation of the line in cartesian form.

Solution: The line passes through a point (2, -1, 4) and has direction ratios proportional to (1, 1, – 2).

Cartesian equation of the line

Section – B

Q.5. Examine whether the operation * defined on R, the set of all real numbers, by   is a binary operation or not, and if it is a  binary operation, find out whether it is associative or not.** 

Q.6. If A = , show that (A – 2I)(A – 3I) = 0. 

Solution: Given,

 

Q.7. Find

Solution:

 

Q.8. Find 

Solution:

OR

Q.8. Find

Solution:

 

Q.9. Find the differential equation of the family of curves y = Ac2x + Be-2x, where A and B are arbitrary constants. 

Solution: Given, y = Ae2x + Be-2x         ………(i)

On differentiating equation (i) w.r.t. x, we get

 

Q.10. If  and , find the angle between and .

Solution: Given,

OR

Q.10. Find the volume of a cuboid whose edges are given by  and .

Solution: If a, b, c are edges of a cuboid.

 

Q.11. If P(not A) = 0·7, P(B) = 0·7 and P(B/A) = 0·5, then find P(A∆B).

Solution:

 

Q.12. A coin is tossed 5 times. What is the probability of getting (i) 3 heads, (ii) at most 3 heads? 

Solution:

OR

Q.12. Find the probability distribution of X, the number of heads in a simultaneous toss of two coins.

Solution: If we toss two coins simultaneously then sample space is given by (HH, HT, TH, IT)

Then probability distribution is,

 

Section – C

Q.13. Check whether the relation R defined on the set A = {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive. 

Solution: Here, R = {(a, b): b = a +1}

 ∴ R = {(a, a + 1): a, a + 1 ϵ (1, 2, 3, 4, 5 ,6)}

 ⇒ R = {(1, 2,) (2, 3), (3, 4), (4, 5), (5, 6)}

(i) R is not reflexive as {a, a} ∉ R∀a

(ii) R is not symmetric as (1, 2) ϵ R but (2, 1) ∉ R

(iii) R is not transitive as (1, 2) ϵ R, (2, 3) ϵ R but (1, 3) ϵ R

OR

Q.13. Let f: N → Y be a function defined as

f(x) = 4x + 3,

where Y = {y ϵ N : y = 4x + 3, for some x ϵ N}.

Show that f is invertible. Find its inverse.

Solution: Consider an arbitrary element of Y. By the definition of y, y = 4x + 3, for some x in the domain N.

Q.14. Find the value

Solution:

Q.15. Using properties of determinants, show that 

Solution :

Q.16. If = 0 and x ≠ y, prove that 

Solution:

Solution:

Q.17. If If (x – a)2 + (y – b)2 = c2, for some c > 0, prove that  is a constant independent of a and b. 

Solution:

Q.18. Find the equation of the normal to the curve x2 = 4y which passes through the point (-1, 4). 

Solution: Suppose the normal at P(x1, y1) on the parabola x 1 = 4y passes through (-1, 4)

Since, P(x1, y1) lies on x1 = 4y

= 4y1

The equation of curve is x2 = 4y.

Differentiating with respect to x, we have

2x = 4

 

Q.19. Find

Solution:

 

Q.20. Prove that  and hence evaluate .

Solution:

Q.21. Solve the differential equation: 

Solution:

 

Q.22. The scalar product of the vector    with a unit vector along the sum of the vectors  and  is equal to 1. Find the value of λ and hence find the unit vector along  .

Solution:

 

Q.23. If the lines  and  are perpendicular, find the value of λ. Hence find whether the lines are intersecting or not.

Solution: 

Section – D

Q.24. If A =, Find A-1.

Hence solve the system of equations

x + 3y + 4z = 8

2x + y + 2z = 5

and 5x + y + z = 7

Solution :

OR

Q.24. Find the inverse of the following matrix, using elementary transformation:

A =  

Solution:

Q.25. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is . Also find the maximum volume.

Solution: Let, ‘x’ be the diameter of the base of the cylinder and let ‘h’ be the height of the cylinder.

Q.26. Using method of integration, find the area of the triangle whose vertices are (1, 0), (2, 2) and (3, 1).

Solution: A(1, 0), B(2, 2) and C(3, 1)

OR

Q.26. Using method of integration, find the area of the region enclosed between two circles x2 + y2 = 4 and (x – 2)2 + y2 = 4.

Solution: Equations of the given circles are,

x2 + y2 = 4 …(i)

(x – 2)2 + y2 = 4 …(ii)

Equation (i) is a circle with centre O at the origin and radius 2. Equation (ii) is a circle with centre C (2, 0) and radius 2.

Solving equation (i) and (ii) we have

(x – 2)2 + y2 = x2 + y2

or x2 – 4x + 4 + y2 = x2 + y2 or x = 1 which gives y = ±

Thus, the points of intersection of the given circles are A (1, 3) and A (1, 3)

Q.27. Find the vector and cartesian equations of the plane passing through the points having position vectors and . Write the equation of a plane passing through a point (2, 3, 7) and parallel to the plane obtained above. Hence, find the distance between the two parallel planes.

Solution: Let A, B, C be the points with position

OR

Q.27. Find the equation of the line passing through (2, -1, 2) and (5, 3, 4) and of the plane passing through (2, 0, 3), (1, 1, 5) and (3, 2, 4). Also, find their point of intersection.

Solution: We know that the equation of a line passing through points (x1, y1, z1,) and (x2, y2, z2) ) is given by

 

Q.28. There are three coins. One is a two-headed coin, another is a biased coin that comes up heads 75% of the time and the third is an unbiased coin. One of the three coins is chosen at random and tossed. If it shows heads, what is the probability that it is the two-headed coin?

Solution: Given, there are three coins.

Let,

E1 = coin is two headed

E2 = biased coin

E3 = unbiased coin

A = shows only head

 

Q.29. A company produces two types of goods, A and B, that require gold and silver. Each unit of type A requires 3g of silver and 1 g of gold while that of type B requires 1 g of silver and 2g of gold. The company can use at most 9 g of silver and 8 of gold. If each unit of type A brings a profit of ₹ 40 and that of type B ₹ 50, find the number of units of each type that the company should produce to maximize profit. Formulate the above LPP and solve it graphically and also find the maximum profit.

Solution: There are two types of goods, A and B and let units of type A be x and units of type B be y. 

Next Paper SET-II and SET-III Click Here!

Physics 12th Previous Year Question Paper 2018 (CBSE)

Physics

Section – A

Q.1. A proton and an electron travelling along parallel paths enters a region of uniform magnetic field, acting perpendicular to their paths. Which of them will move in a circular path with higher frequency ? 

Answer: Frequency of revolution of a particle 

OR

Answer: Since mass of electron is less than that of proton, therefore, its frequency of revolution will be higher than that of proton.

 

Q.2. Name the electromagnetic radiations used for

(a) water purification, and

(b) eye surgery

Answer: (a) Water purification : Ultraviolet radiation.

(b) Eye surgery : Ultraviolet radiation/laser

 

Q.3. Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity. 

Answer : Graph for photoelectric current (I) versus applied potential for radiation of the same frequency and varying intensity.

 

Q.4. Four nuclei of an element undergo fusion to form a heavier nucleus, with release of energy. Which of the two parent or the daughter nucleus would have higher binding energy per nucleon ? 

Answer: When lighter nuclei combine to form a heavier nucleus, binding energy per nucleon increases and energy is released. Thus, the daughter nucleus would have higher binding energy per nucleon.

 

Q.5. Which mode of propagation is used by short wave broadcast services.

Answer: Short wave broadcast services use sky wave propagation.

 

Section – B

Q.6. Two electric bulbs P and Q have their resistances in the ratio of 1 : 2. They are connected in series across a battery. Find the ratio of the power dissipation in these bulbs. 

Answer :

 

Q.7. A 10V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38 Ω as shown in the figure. Find the value of current in the circuit.

OR

Q.7. In a potentiometer arrangement for deter-mining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9 Ω is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell.

 

Q.8. (a) Why are infra-red waves often called heat waves ? Explain.

(b) What do you understand by the statement, “Electromagnetic waves trans-port momentum” ? 

Answer: (a) Infra-red waves are called heat waves because they raise the temperature of the object on which they fall and hence increase their thermal motion. They also affect the photographic plate and are readily absorbed by most materials.

(b) Electromagnetic waves transport momentum. This means that when an electromagnetic wave travels through space with energy U and speed c, then it transports linear momentum p = UC. If a surface absorbs the waves completely, then momentum ‘p’ is delivered to the surface. If the surface reflects the wave, then momentum delivered by both incident and reflected wave adds on to give ‘2P’ momentum.

 

Q.9. If light of wavelength 412.5 nm is incident on each of the metals given below, which ones will show photoelectric emission and why? 

Since, the energy of incident radiation is greater than the work function of sodium and potassium, but less than that of calcium and molybdenum, therefore, photoelectric emission will take place in sodium and potassium.

 

Q.10. A carrier wave of peak voltage 15 V is used to transmit a message signal. Find the peak voltage of the modulating signal in order to have a modulation index of 60%.

Answer: Given  AC= 15v  and modulation intex μ= 60% = 60⁄100

                                           Now, we know that 

                                               μ = AmAc 

                                               60⁄100 = Am15

                                                                             Am= 15 × 0.6

                                              Am= 9v

Hence, the peak voltage of modulating signal should be 9v in order to have a modulation index of 60%.

 

Section – C

Q.11. Four point charges Q, q, Q and q are placed at the corners of a square of side ‘a’ as shown in the figure.

Find the

(a) resultant electric force on a charge Q and

(b) potential energy of this system.

OR

Q.11. (a) Three point charges q – 4q and 2q are placed at the vertices of an equilateral triangle ABC of side ‘l’ as shown in the figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge q.

(b) Find out the amount of the work done to separate the charges at infinite distance.

Answer : (a) Force on charge Q at B due to charge q at A

Answer: 

 

Q.12. (a) Define the term ‘conductivity’ of a metallic wire. Write its SI unit.

(b) Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field E.

Answer: (a) Conductivity of a metallic wire is defined as its ability to allow electric charges or heat to pass through it. Numerically, conductivity of a material is reciprocal of its resistivity.

SI unit : ohm-1 m-1 or mho m-1 or Siemen m-1.

(b) Consider a potential difference V be applied across a conductor of length l and cross section A.

Electric field inside the conductor, E = v/l.

Due to the external field the free electrons inside the conductor drift with velocity Vd.

Let, number of electrons per unit volume = n,

charge on an electron = e

Total electrons in length, l = nAl And,

total charge, q = neAl

 

Q.13. A bar magnet of magnetic moment 6 J/T is aligned at 60° with a uniform external magnetic field of 0.44 T. Calculate

(a) the work done in turning the magnet to align its magnetic moment

(i) normal to the magnetic field,

(ii) opposite to the magnetic field, and

(b) the torque on the magnet in the final orientation in case (ii).

 

Q.14. (a) An iron ring of relative permeability μ has winding’s of insulated copper wire of n turns per meter. When the current in the winding is I, find the expression for the magnetic field in the ring.

(b) The susceptibility of a magnetic material is 0.9853. Identify the type of magnetic material. Draw the modification of the field pattern on keeping a piece of this material in a uniform magnetic field.

 

Q.15. (a) Show using a proper diagram how unpolarised light can be linearly polarised by reflection from a transparent glass surface.

(b) The figure shows a ray of light falling normally on the face AB of an equilateral glass prism having refractive index 3/2 placed in water of refractive index 4/3 Will this ray suffer total internal reflection on striking the face AC ?

 

Q.16. (a) If one of two identical slits producing interference in Young’s experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringes in the interference pattern.

(b) What kind of fringes do you expect to observe if white light is used instead of monochromatic light ?

Answer : (a) The resultant intensity in Young’s experiment is given by

When slit is not covered, then I0 is the intensity from each slit.

Maximum intensity (Imax) occurs when Φ = 0°.

Minimum intensity (Imin) occurs when (Φ ) = 180°.

If one slit is covered with glass to reduce its intensity by 50%, then

(b) If instead of monochromatic light, white light is used, then the central fringe will be white and the fringes on either side will be coloured. Blue colour will be nearer to central fringe and red will be farther away. The path difference at the centre on perpendicular bisector of slits will be zero for all colours and each colour produces a bright fringe thus resulting in white fringe. Further, the shortest visible wave, blue, produces a bright fringe first.

 

Q.17. A symmetric biconvex lens of radius of curvature R and made of glass of refractive index 1.5, is placed on a layer of liquid placed on top of a plane mirror as shown in the figure. An optical needle with its tip on the principal axis of the lens is moved along the axis until its real, inverted image coincides with the needle itself. The distance of the needle from the lens is measured to be x. On removing the liquid layer and repeating the experiment, the distance is found to be y. Obtain the expression for the refractive index of the liquid in terms of x and y.

Answer: Given, refractive index of lens, μg = 1.5. The distance of the needle from the lens in the first case = The focal length of the combination of convex lens and plano concave lens formed by the liquid, f= x And, the distance measured in the second case = Focal length of the convex lens, f1 = y If the focal length of plano concave lens formed by the liquid be f2, then

 

Q.18. (a) State Bohr’s postulate to define stable- orbits in hydrogen atom. How does de Broglie’s hypothesis explains the stability of these orbits ?

(b) A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 level. Estimate the frequency of the photon.

Answer : (a) Bohr’s postulate for stable orbits in hydrogen atom : An electron can revolve only in those circular orbits in which its angular momentum is an integral multiple of h/2π, where h is Planck’s constant.

If n is the principal quantum number of orbits, then an electron can revolve only in . certain orbits or definite radii. These are called stable orbits.

de Broglie explanation of stability of orbits:

According to de Broglie, orbiting electrons around the nucleus is associated with a stationary wave. Electron wave is a circular standing wave. Since destructive interference will occur if a standing wave does not close upon itself, only those de Broglie waves exist for which the circumference of . circular orbit contains a whole number of wavelengths i.e., for orbit circumference of nth orbit as 2nπrn.

OR

 

Q.19. (a) Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon (BE/A) versus the mass number A.

(b) A radioactive isotope has a half-life of 10 years. How long will it take for the activity to reduce to 3.125% ?

Answer: (a) Plot of binding energy per nucleon mass number :

1. When we move from the heavy nuclei region to the middle region of the plot, we find that there will be a gain in the overall binding energy and hence results in release of energy. This indicates that energy can be released when a heavy nucleus (A ~ 240) break into two roughly equal fragments. This process is called nuclear fission.

2. Similarly, when we move from lighter nuclei to heavier nuclei, we again find that there will be gain in the overall binding energy and hence the release of energy takes place. This indicates that energy can be released when two or more lighter nuclei fuse together to form a heavy nucleus. This process is called nuclear fusion.

 

Q.20. (a) A student wants to use two p-n junction diodes to convert alternating current into direct current. Draw a labelled circuit diagram she would use and explain how it works.

(b) Give the truth table and circuit symbol for NAND gate.

Answer : (a) Full wave rectifier :

Explanation : In positive half cycle of AC, end A becomes positive and D1 becomes forward biased and D2 is reverse biased, so conducts and D2 doesn’t. So conventional current flows through D1, RL and upper half of secondary winding. Similarly, during negative half cycle of AC, diode D2 becomes forward biased and D1 is reverse biased, current flows through D2, RL and lower half of secondary winding. Thus, current flows in the same direction in both half cycles of input AC voltage.

 

Q.21. Draw the typical input and output characteristics of an n-p-n transistor in CE configuration. Show how these characteristics can be used to determine (a) the input resistance (r ), and (b) current amplification factor (β). 

Answer: The input and output characteristics of n-p-n transistor in CE configuration is given below:

Input resistance is defined as the ratio of change in base-emitter voltage (ΔVBE) to the change in base current (ΔIB) at a constant collector-emitter voltage.

Current amplification factor (β) is defined as the ratio of the change in the collector current to the change in the base current at a constant collector-emitter voltage when the transistor is in active state.

 

 

Q.22. (a) Give three reasons why modulation of a message signal is necessary for long distance transmission.

(b) Show graphically an audio signal, a carrier wave and an amplitude modulated wave. 

Answer: 

(a) The three reasons necessary for long-distance transmission are:

i) A reasonable length of the transmission antenna.

ii) Increase in effective power radiated by the antenna.

iii) Reduction in the possibility of “mix-up‟ of different signals.

(b) In Amplitude modulation, the transmission of a wave signal is done by modulating the amplitude. At the source, the information signal is modulated and sent to the receiver, once it is received, it is then demodulated and filtered to remove any disturbance in the original signal that may have occurred during the transmission. The amplitude modulation is the process of superimposing the information signal with a carrier generated signal. 

The information signal is the data message signal stored on a network that has to be transmitted from a source to a receiver.A carrier wave is an electromagnetic wave of fixed amplitude and of a constant frequency that is modulated with an input message signal for the purpose of transmission. 

 

Section – D

Q.23. The teachers of Geeta’s school took the students on a study trip to a power generating station, located nearly 200 km away from the city. The teacher explained that electrical energy is transmitted over such a long distance to their city, in the form of alternating current (ac) raised to a high voltage. At the receiving end in the city, the voltage is reduced to operate the devices. As a result, the power loss is reduced. Geeta listened to the teacher and asked questions about how the ac is converted to a higher or lower voltage. 

(a) Name the device used to change the alternating voltage to a higher or lower value. State one cause for power dissipation in this device.

(b) Explain with an example, how power loss is reduced if the energy is transmitted over long distances as an alternating current rather than a direct current.

(c) Write two values each shown by the teachers and Geeta.

Answer : (a) The device used to change the alternating voltage to a higher or lower value is step up or step-down transformer. It works on the principle of Faraday’s law of induction by converting one value of electrical energy into another.

One of the reasons for power dissipation in this device is the production of eddy currents in core. The changing magnetic field induces an EMF in the secondary winding. However, due to this changing magnetic field through the core, an EMF is induced in the core itself as the core is made of conducting material and thus current is generated in the core. These circulating currents are in the form of swirling eddies, called eddy currents. 

(b) The relation between the voltage in the secondary coil with a voltage in the primary coil is given as

Vs = (Ns/Np) Vp

And the relation between current in the secondary coil with a current in the primary coil is given as

Is =(Np/Ns) Ip

In a step-up transformer, if the voltage is getting stepped up, then current in the secondary coil is getting reduced.

Hence the power loss (P = I2R) is reduced considerably while such stepping up is not possible for direct current.

(c) The teacher is well informed and willing to share her knowledge with Geeta that reflects her concerns for her. She wants her students to learn the practical aspects of the theoretical learning that is done in the class. 

Geeta, on the other hand, manifest the value of a good listener who is interested in learning the practical application of the concepts that has been taught in the class. It reflects her curious nature and the sheer willingness to learn something new. 

 

Section – E

Q.24. (a) Define electric flux. Is it a scalar or a vector quantity ? A point charge q is at a distance of dl2 directly above the center of a square of side d, as shown in the figure. Use Gauss’ law to obtain an expression for the electric flux through the square.

(b) If the point charge is now moved to a distance ‘d ‘ from the center of the square and the side of the square is doubled, explain how the electric flux will be affected. 

OR

Q.24. (a) Use Gauss’ law to derive the expression for the electric field (E) due to a straight uniformly charged infinite line of charge density λ C/m.

(b) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.

(c) Find the work done in bringing a charge q from perpendicular distance r1to r2( r2 > r1).

Answer: (a) Electric flux : Electric flux through an area is defined as the product of electric field strength E and area dS perpendicular to the field. It represents the field lines crossing the area. It is a scalar quantity. Imagine a cube of edge d, enclosing the charge. The square surface is one of the six faces of this cube. According to Gauss’ theorem in electrostatics,

Total electric flux through the cube =

This is the total flux through all six surface

∴ Electric flux through the square surface =  

(b) On moving the charge to distance d from the center of the square and making side of square 2d, does not change the flux at all because flux is independent of side of square or distance of charge in this case.

OR

Answer: (a) Electric field  due to a straight uniformly charged infinite line of charge density λ : Consider a cylindrical Gaussian surface of radius r and length l coaxial with line charge. The cylindrical Gaussian surface may be divided into three parts :

(i) curved surface S1 

(ii) flat surface S2 and

(iii) flat surface S3.

By symmetry, the electric field has the same magnitude E at each point of curved surface Sj and is directed radially outward. We consider small elements of surfaces S1, S2 and S3.

 

b) Graph showing variation of E with perpendicular distance from the line of charge : The electric field is inversely proportional to distance V from line of charge.

Q.25. (a) State the principle of an ac generator and explain its working with the help of a labelled diagram. Obtain the expression for the emf induced in a coil having N turns each of cross-sectional area A, rotating with a constant angular speed ω in a magnetic field , directed perpendicular to the axis of rotation. 

(b) An aeroplane is flying horizontally from west to east with a velocity of 900 km/ hour. Calculate the potential difference developed between the ends of its wings having a span of 20 m. The horizontal component of the Earth’s magnetic field is 5 × 10-4T and the angle of dip is 30°. 

OR

Q.25. A device X is connected across an ac source of voltage V = V0sin ωt . The current through X is given as I = I0sin

(a) Identify the device X and write the expression for its reactance.

(b) Draw graphs showing variation of voltage and current with time over one cycle of ac, for X.

(c) How does the reactance of the device X varies with frequency of the ac ? Show this variation graphically.

(d) Draw the phasor diagram for the device X.

Answer : (a) Principle of ac generator :

The ac generator is based on the principle of electromagnetic induction. When closed coil is rotated in a uniform field with its axis perpendicular to the field, then the magnetic flux changes and emf is induced.

Working : When the armature coil rotates, the magnetic flux linked with it changes and produces induced current. If initially, coil PQRS is in vertical position and rotated clockwise, then PQ moves down and SR moves up. By Fleming’s right hand rule, induced current flows from Q to P and S to R which is the first half rotation of coil. Brush B1is positive terminal and B2 is negative. In second half rotation, PQ moves up and SR moves down. So induced current reverses and the alternating current is produced in this manner by the generator.

 

Q.26. (a) Draw a ray diagram to show image formation when the concave mirror produces a real, inverted and magnified image of the object.

(b) Obtain the mirror formula and write an expression for the linear magnification.

(c) Explain two advantages of a reflecting telescope over a refracting telescope. 

OR

Q.26. (a) Define a wave front. Using Huygens’ principle, verify the laws of reflection at a plane surface.

(b) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band ? Explain.

(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the center of the obstacle. Explain why ?

Answer : (a) Concave mirror produces real, inverted and magnified image for object placed between F and C :

(b) Derivation for mirror formula and magnification: Consider an object AB be placed in front of a concave mirror beyond center of curvature C.

(c) Advantages of reflecting telescope over a refracting telescope are :

1. Reflecting telescope is free from chromatic and spherical aberrations unlike refracting telescope. Thus image formed is sharp and bright.

2. It has a larger light gathering power so that a bright image of even far off object is obtained.

3. Resolving power of reflecting telescope is large.

OR

Answer: (a) Wave front is defined as the continuous locus of all the particles of a medium which are vibrating in the same phase. Verification of laws of reflection using Huygens’s principle : Let XY be a reflecting surface at which a wave front is being incident obliquely. Let v be the speed of the wave front and at time t = 0, the wave front touches the surface XY at A. After time t, point B of wave front reaches the point B’ of the surface.

According to Huygens’s principle each point of wave front acts as a source of secondary waves. When the point A of wave front strikes the reflecting surface, then due to the presence of reflecting surface, it cannot advance further; but the secondary wavelet originating from point A begins to spread in all directions in the first medium with speed v. As the wave front AB advances further, its points A1, A2, A3 etc. strike the reflecting surface successively and send spherical secondary wavelets in the first medium.

 

Physics 12th Previous Year Question Paper 2019 SET-II,III (CBSE)

Previous Paper click Here !

Physics

SET-II

Q.1. Write the relation for the force acting on a charged particle q moving with velocity  in the presence of a magnetic field  .

Answer:

 

Q.2. Draw the pattern of electric field lines due to an electric dipole. 

Answer:

 

Q.3. Identify the semiconductor diode whose I-V characteristics are as shown. 

Answer: Photodiode/Solar cell.

 

Section – B

Q.4. How is the equation for Ampere’s circuital law modified in the presence of displacement current? Explain. 

Answer:

Case-1

Let the case-1 where a point P is considered outside capacitor charging from Ampere’s law. Magnetic field at point will be

Case-2

Now take case-2 where shape surface under consideration covers capacitor’s plate as we consider there is no current through capacitor then this value of B will be zero.

Hence, there is a contradiction. Therefore this Ampere’s law was modified with addition of displacement current inside capacitor.

Where, Id displacement current.

During charging of capacitor, outside the capacitor, Ic (conduction current) flows and inside Id displacement current flows.

 

Q.5. Under what conditions does the phenomenon of total internal reflection take place? Draw a ray diagram showing how a ray of light deviates by 90° after passing through a right-angled isosceles prism. 

Answer: The phenomenon of total internal reflection occurs when,

  • Angle of incidence is equal or greater than the critical angle. i ≥ C
  • When light travels from a denser medium to less denser medium.In case of right angle isosceles triangle if light rays fall normally on AB then light incident face AC with angle of incidence &gt; critical angle.

Hence, total internal reflection will occur with normal to the surface of BC.

 

Q.6. A beam of light converges at a point P. Draw ray diagrams to show where the beam will converge if

(i) a convex lens, and

(ii) a concave lens is kept in the path of the beam. 

Answer:

 

Section – C

Q.7. (a) How is the stability of hydrogen atom in Bohr model explained by de-Broglie’s hypothesis?

(b) A hydrogen atom initially in the ground state absorbs a photon which excites it to n = 4 level. When it gets de-excited, find the maximum number of lines which are emitted by the atom. Identify the series to which these lines belong. Which of them has the shortest wavelength? 

Answer: (a) From Bohr’s model-An atom has a number of stable orbits in which an electron can reside without the emission of radiant energy. Each orbit corresponds to a certain energy level. Electron revolves is circular orbit

The motion of an electron in circular orbits is restricted in such a manner that its angular momentum is an integral multiple h/2Π

 

Q.8. What is the reason to operate photo-diodes in reverse bias? A p-n photodiode is fabricated from a semiconductor with a band gap of range of 2.5 to 2.8 eV. Calculate the range of wavelengths of the radiation which can be detected by the photodiode. 

Answer : Photo diode are reverse biased for working in photo-conductive mode. This reduces the response time because the additional reverse bias increases the width of the depletion layer, which decreases the junction capacitance. The reverse bias also increases the dark current without much change in the photo current.

The range of wavelengths of radiation which can be detected by the photodiode is 443.97 nm to 497.25 nm.

 

Q.9. A ray of light incident on the face AB of an . isosceles triangular prism makes an angle of incidence (i) and deviates by angle P as shown in the figure. Show that in the position of minimum deviation ∠β = ∠α . Also find out the condition when the refracted ray QR suffers total internal reflection. 

Answer:

 

Q.10. A 100 μF parallel plate capacitor having plate separation of 4 mm is charged by 200 V dc. The source is now disconnected. When the distance between the plates is doubled and a dielectric slab of thickness 4 mm and dielectric constant 5 is introduced between the plates, how will

(i) its capacitance,

(ii) the electric field between the plates, and

(iii) energy density of the capacitor get affected? Justify your answer in each case 

Answer:

As dielectric of 4 mm is inserted between the plates of capacitor and the spacing between the plates is doubled then it will acts as following Fig-A and Fig-B.

Physics

SET-III

Q.1. Draw a pattern of electric field lines due to two positive charges placed a distance d apart. 

Answer : Electric field lines due to two positive charge placed at a distance d apart:

Q.2. When a charge q is moving in the presence of electric (E) and magnetic (B) fields which are perpendicular to each other and also perpendicular to the velocity v of the particle, write the relation expressing v in terms of E and B. 

Answer : v = E/B

 

Q.3. Draw the I-V characteristics of a Zener diode. 

Answer:

 

Section – B

Q.3. State, with the help of a ray diagram, the working principle of optical fibers. Write one important use of optical fibers. 

Answer:

Optical fiber works on the principle of total internal reflection. When the angle of incidence is greater than Critical angle then incident rays are totally reflected back in the same media.

When, θi < θc , Total internal reflection occurs and if θi < θc , refraction occurs.

Application: Optical fiber are used for communication due to very high bandwidth of media.

 

Q.4. How are electromagnetic waves produced by oscillating charges ? What is the source of the energy associated with the em waves.

Answer : Oscillating charges are responsible for the generation of periodically varying electric field in the space. The oscillating charges generate varying electric current which in turn is responsible for the generation of periodical varying magnetic field. This way the electromagnetic waves are generated.

 

Q.5. The wavelength of light from the spectral emission line of sodium is 590 nm. Find the kinetic energy at which the electron would have the same deBroglie wavelength.

 

Section – C

Q.6. (a) Draw the energy level diagram for the line spectra representing Lyman series and Balmer series in the spectrum of hydrogen atom.

(b) Using the Rydberg formula for the spectrum of hydrogen atom, calculate the largest and shortest wavelengths of the emission lines of the Balmer series in the spectrum of hydrogen atom. (Use the value of Rydberg constant R = 1.1 x 107 m-1 

Answer: a) Energy level diagram showing lyman and balmer series :

Spectrum wavelengths of both series for hydrogen atom

 

Q.7. In a network, four capacitors C1, C2, C3 and C4 are connected as shown in the figure.

(a) Calculate the net capacitance in the circuit.

(b) If the charge on the capacitor C1 is 6 μC,

(i) calculate the charge on the capacitors C3 and C4 and,

(ii) net energy stored in the capacitors C3 and C4 connected in series.

Answer: (a)

 

Q.8.Draw the circuit diagram of a full wave rectifier. Explain its working principle. Show the input waveforms given to the diodes D1 and D2 and the corresponding output waveforms obtained at the load connected to the circuit.

Answer:

During the first half of input sinusoidal ac signal diode D1 conducts as it is forward bias and during the second half of input ac signal diode D2 conducts as it is forward bias now. D2 and D1 are inverse bias conditions during first and second half respectively and doesn’t conduct. Due to this output appears as waveform (b).

 

Q.9. (a) When a convex lens of focal length 30 cm is in contact with a concave lens of focal length 20 cm, find out if the system is converging or diverging.

(b) Obtain the expression for the angle of incidence of a ray of light which is incident on the face of a prism of refracting angle A so that it suffers total internal reflection at the other face. (Given the refractive index of the glass of the prism is μ). 

Answer:

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