Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 10 Ordinary Differential Equations Ex 10.1 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.1

Question 1.
For each of the following equations, determine its order, degree (if exists)
(i) \(\frac { dy }{ dx }\) + xy = cot x
(ii) (\(\frac { d^3y }{ dx^3 }\))^2/3 – 3 \(\frac { d^2y }{ dx^2 }\) + 5\(\frac { dy }{ dx }\) + 4 = 0
(iii) (\(\frac { d^2y }{ dx^2 }\))² + (\(\frac { dy }{ dx }\))² = x sin (\(\frac { d^2y }{ dx^2 }\))
(iv) \(\sqrt{\frac { dy }{ dx }}\) – 4 \(\frac { dy }{ dx }\) – 7x = 0
(v) y(\(\frac { dy }{ dx }\)) = \(\frac { x }{ (\frac { dy }{ dx })+(\frac { dy }{ dx })^3 }\)
(vi) x²\(\frac { d^2y }{ dx^2 }\) + [1 + (\(\frac { dy }{ dx }\))²]^1/2 = 0
(vii) (\(\frac { d^2y }{ dx^2 }\))³ = \(\sqrt{1+(\frac { dy }{ dx })}\)
(viii) \(\frac { d^2y }{ dx^2 }\) = xy + cos (\(\frac { dy }{ dx }\))
(ix) \(\frac { d^2y }{ dx^2 }\) + 5 \(\frac { dy }{ dx }\) + ∫ ydx = x³
(x) x = e^{xy(\(\frac { dy }{ dx }\))}
Solution:
(i) \(\frac { dy }{ dx }\) + xy = cot x
In the given equation, the highest order derivative is \(\frac { dy }{ dx }\) only its power is 1
∴ Its order = 1 & degree = 1

(ii) (\(\frac { d^3y }{ dx^3 }\))^2/3 – 3 \(\frac { d^2y }{ dx^2 }\) + 5\(\frac { dy }{ dx }\) + 4 = 0
Taking power 3 on both sides, we get
(\(\frac { d^3y }{ dx^3 }\))² = (3 \(\frac { d^2y }{ dx^2 }\) – 5\(\frac { dy }{ dx }\) – 4)³
In the equation (1), the highest order derivative is \(\frac { d^3y }{ dx^3 }\) and its power is 2.
∴ Its order = 3 & degree = 2

(iii) (\(\frac { d^2y }{ dx^2 }\))² + (\(\frac { dy }{ dx }\))² = x sin (\(\frac { d^2y }{ dx^2 }\))
In the equation, the highest order derivative is \(\frac { d^2y }{ dx^2 }\) and its order is 2.
It has a term sin (\(\frac { d^2y }{ dx^2 }\)), so its degree is not defined or degree does not exist.

on squaring both sides,
\(\frac { dy }{ dx }\) = 16 (\(\frac { dy }{ dx }\))² + 49 x² + 56x \(\frac { dy }{ dx }\)
clearly, it is a differential equation of order = 1 & degree = 2.

In this equation, the highest order derivative is \(\frac { dy }{ dx }\) & its power is 4.
∴ Its order = 1 & degree = 4

In this equation, the highest order derivative is \(\frac { d^2y }{ dx^2 }\) & its power is 2.
∴ Its order = 2 & degree = 2

In this equation, the highest order derivative is \(\frac { d^2y }{ dx^2 }\) & its power is 6.
∴ Its order = 2 & degree = 6

(viii) \(\frac { d^2y }{ dx^2 }\) = xy + cos (\(\frac { dy }{ dx }\))
In this equation, the highest order derivative is \(\frac { d^2y }{ dx^2 }\) & its power is 2.
It has a term cos(\(\frac { dy }{ dx }\)), so its degree is not defined or degree does not exist.

(ix) \(\frac { d^2y }{ dx^2 }\) + 5 \(\frac { dy }{ dx }\) + ∫ ydx = x³
differentiating with respect to x, we get
\(\frac { d^3y }{ dx^3 }\) + 5 \(\frac { d^2y }{ dx^2 }\) + y = 3x²
In this equation the highest order derivative is \(\frac { d^3y }{ dx^3 }\) & its power is 1
∴ Its order = 3 & degree = 1

(x) x = e^{xy(\(\frac { dy }{ dx }\))}
In this equation the highest order derivative is \(\frac { dy }{ dx }\) & its order is 1
It has the term e^{xy(\(\frac { dy }{ dx }\))}
So its degree is not defined or degree does not exist.

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• Chapter 1 Principles of Management
• Chapter 2 Functions of Management
• Chapter 3 Management By Objectives (MBO) and Management By Exception (MBE)
• Chapter 4 Introduction to Financial Markets
• Chapter 5 Capital Market
• Chapter 6 Money Market
• Chapter 7 Stock Exchange
• Chapter 8 Securities Exchange Board of India (SEBI)
• Chapter 9 Human Resource Management
• Chapter 10 Recruitment Methods
• Chapter 11 Employee Selection Process
• Chapter 12 Employee Training Method
• Chapter 13 Concept of Market and Marketer
• Chapter 14 Marketing and Marketing Mix
• Chapter 15 Recent Trends in Marketing
• Chapter 16 Consumerism
• Chapter 17 Rights, Duties & Responsibilities of Consumers
• Chapter 18 Grievance Redressal Mechanism
• Chapter 19 Environmental Factors
• Chapter 20 Liberalization, Privatization and Globalization
• Chapter 21 The Sale of Goods Act, 1930
• Chapter 22 The Negotiable Instruments Act, 1881
• Chapter 23 Elements of Entrepreneurship
• Chapter 24 Types of Entrepreneurs
• Chapter 25 Government Schemes for Entrepreneurial Development
• Chapter 26 Companies Act, 2013
• Chapter 27 Company Management
• Chapter 28 Company Secretary

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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Accountancy Guide Pdf Chapter 7 Company Accounts Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 7 Company Accounts

12th Accountancy Guide Company Accounts Text Book Back Questions and Answers

I Multiple Choice Questions

Choose the correct answer

Question 1.
A preference share is one
(i) which carries preferential right with respect to payment of dividend at fixed rate
(ii) which carries preferential right with respect to payment of capital on winding up
(a) Only (i) is correct
(b) Only (ii) is correct
(c) Both (i) and (ii) are correct
(d) Both (i) and (ii) are incorrect
Answer:
(c) Both (i) and (ii) are correct

Question 2.
That part of share capital which can be called up only on the winding up of a company is called:
(a) Authorised capital
(b) Called up capital
(c) Capital reserve
(d) Reserve capital
Answer:
(d) Reserve capital

Question 3.
At the time of forfeiture, share capital account is debited with
(a) Face value
(b) Nominal value
(c) Paid up amount
(d) Called up amount
Answer:
(d) Called up amount

Question 4.
After the forfeited shares are reissued, the balance in the forfeited shares account should be transferred to
(a) General reserve account
(b) Capital reserve account
(c) Securities premium account
(d) Surplus account
Answer:
(b) Capital reserve account

Question 5.
The amount received over and above the par value is credited to
(a) Securities premium account
(b) Calls in advance account
(c) Share capital account
(d) Forfeited shares account
Answer:
(a) Securities premium account

Question 6.
Which of the following statement is false?
(a) Issued capital can never be more than the authorised capital
(b) In case of under subscription, issued capital will be less than the subscription capital
(c) Reserve capital can be called at the time of.winding up
(d) Paid up capital is part of called up capital
Answer:
(b) In case of under subscription, issued capital will be less than the subscription capital

Question 7.
When shares are issued for purchase of assets, the amount should be credited to
(a) Vendors A/c
(b) Sundry assets A/c
(c) Share capital A/c
(d) Bank A/c
Answer:
(c) Share capital A/c

Question Match the pair and identify the correct option
(1) Under subscription – (i) Amount prepaid for calls
(2) Over subscription – (ii) Subscription above the offered shares
(3) Calls in arrear – (iii) Subscription below the offered shares
(4) Calls in advance -(iv) Amount unpaid on calls

Answer:
C

Question 9.
If a share ₹ 10 on the which ₹ 8 has been paid up is forfeited. Minimum reissue price is
(a) ₹ 10 per share
(b) ₹ 8 per share
(c) ₹ 5 per share
(d) ₹ 2 per share
Answer:
(d) ₹ 2 per share
Hint:
One share Rs. 10
(-) Paid up value Rs. 8 Each
Minimum Reissue Price 2 Each

Question 10.
Supreme Ltd. forfeited 100 shares of ₹ 10 each for non-payment of final call of ₹ 2 per share. All these shares were re-issued at ₹ 9 per share. What amount will be transferred to capital reserve account?
(a) ₹ 700
(b) ₹ 800
(c) ₹ 900
(d) ₹ 1,000
Answer:
(a) ₹ 700

II Very Short Answer Questions

Question 1.
What is a share?
Answer:
The capital of a company is divided into small units of a fixed amount. These units are called shares. There are two types

1. Preference shares and
2. Equity shares.

Question 2.
What is Over-Subscription?
Answer:
When the number of shares applied for is more than the number of shares offered for subscription, it is said to be oversubscription.

Question 3.
What is meant by calls in arrears?
Answer:
When a shareholder fails to pay the amount due on the allotment or on calls, the amount remaining unpaid is known as calls in arrears.

Question 4.
Write a short note on the securities premium account.
Answer:
When a company issues shares at a price more than the face value (nominal value), the shares are said to be issued at a premium. The excess is called a premium amount and is transferred to a securities premium account.
Securities premium account is shown under reserves and surplus as a separate head in the Note to Account to the balance sheet.

Question 5.
Why are the shares forfeited?
Answer:
When a shareholder defaults in making payment of allotment/call money, the shares may be forfeited. On forfeiture, the share allotment is cancelled and to that extent, paid-up share capital reduced.

III Short Answer Questions

Question 1.
State the differences between preference shares and equity shares.
Answer:
(i) Preference shares
Preference shares are the shares which have the following two preferential right over, the equity shares:

• Preference towards the payment of dividend at a fixed rate during the lifetime of the company and.
• Preference towards the repayment of capital on winding up of the company.

(ii) Equity shares
Equity shares are those shares which do not preference shares. These shares do not enjoy any preferential rights. The rate of dividend is not fixed on equity shares and it depends upon the profits earned by the company. In case of winding up of a company equity shareholders are paid after the payment are made to preference shareholders. Equity shares are also known as ordinary shares.

Question 2.
Write a brief note on calls in advance.
Answer:
The excess amount paid over the called up value of a share is known as calls in advance. It is the excess money paid on application or allotment or calls. Such excess amounts can be returned or adjusted towards future payments. If the company decides to adjust such amount towards future payment, the excess amount is transferred to a separate account called a calls-in advance account.

Question 3.
What is a reissue of forfeited shares?
Answer:
Shares forfeited can be reissued by the company. The shares can be reissued at any price.
But, the reissue price cannot be less than the amount unpaid on forfeited shares.
When forfeited shares are reissued at a loss, such loss is to be debited to the forfeited shares account.
when forfeited shares are reissued at a premium, the amount of such premium will be credited to the securities premium account.

Question 4.
Write a short note on

1. Authorized capital
2. Reserve capital

Answer:

1. Authorised Capital: According to Sec 2(8) of the Companies Act, “It means such capital as is authorized by the memorandum of a company to be the maximum amount of share capital of the company”.

2. Reserve Capital: Sec 65 of Companies Act 2013, only an unlimited company having share capital while converting into a limited company, may have a reserve capital. The company reserve a part of its subscribed capital to be called up only at the time of winding up. It is called reserve capital.

Question 5.
What is meant by the issue of shares for consideration other than cash?
Answer:
A company may issue shares for consideration other than cash when the company acquires fixed assets such as land and buildings, machinery, etc.

IV Exercises

Question 1.
Progress Ltd. issued 50,000 ordinary shares of ₹ 10 each, payable ₹ 2 on the application, ₹ 4 on allotment ₹ 2 on first call and ₹ 2 on final call. All the shares are subscribed and the amount was duly received. Pass journal entries.
Solution :

Under Subscription

Question 2.
Sampath company issued 25,000 shares at ₹ 10 per share payable ₹ 3 on the application, ₹ 4 on allotment; ₹ 3 on first and final call. The public subscribed for 24,000 shares. The directors allotted all the 24,000 shares and received the money duly. Pass necessary journal entries.
Solution:

Over Subscription

Question 3.
Saranya Ltd. issued 20,000 equity shares of ₹ 10 each to the public at par. The details of the amount payable on the shares are as follows:
On application – ₹ 3 per share
On allotment – ₹ 4 per share
On first and final call – ₹ 3 per share
Application money was received on 30,000 shares. Excess application money was refunded immediately. Pass journal entries are as follows:
Solution:

Question 4.
Gaja Lid issued 40,000 shares of ₹ 10 each of the public payable ₹ 2 on the application, ₹ 5 on the allotment, and ₹ 3 on the first and final call. The application was received for 50,000 shares. The Directors decided to allot 40,000 shares on a pro-rata basis and a surplus of application money was utilized for allotment. Pass journal entries assuming that the amount due was received.
Solution :

Question 5.
Lalitha Ltd. offered 30,000 equity shares ₹10 each to the public payable ₹ 2 per share on the application, ₹ 3 on share allotment, and the balance when required. Applications for 50,0 shares were received on which directors allotted as:
Applicants for 10,000 shares Full
Applicants for 35,000 shares 20,000 shares (excess money will be utilized for allotment
Applicants for 5,000 shares Nil
All the money due was received. Pass journal entries upto the receipt of allotment.
Solution :

Calls In advance

Question 6.
Das Ltd, offered 50,000 equity shares of ₹ 1 0 each to the public payable as follows: On applications were ₹ 4; on allotment ₹ 3; on first call 1 and on second and final call ₹ 2. Applications were received for 1,00,000 shares. All the applicants were allotted 1 share for every two shares applied. Excess application money was used for the amount due on allotment and call. Pass necessary journal entries.
Solution :

Question 7.
Anjali Flour Ltd. with a registered capital of ₹ 4,00,000 in equity shares of ₹ 10 each, issued 30,0 of such shares; payable ₹ 2 per share on the application, ₹ 5 per share on the allotment, and ₹ 3 shares on the first call. The issue was duly subscribed.
All the money payable was duly received but on the allotment, one shareholder paid the entire balance on his holding of 500 shares. Give journal entries to record the I transactions.
Solution :

Calls in Advance

Question 8.
Muthu Ltd. issued 50,000 shares of ₹ 10 each payable as follows; ₹ 2 on the application; ₹ 4 on allotment; ₹ 4 on first and final, call.
All money payable was duly received except one shareholder holding 1,000 shares failed to pay the call money. Pass the necessary journal entries for calls by using calls in the arear account.
Solution :

Answer:
Calls in arrear: ₹ 4,000

Forfeiture os Shares

Question 9.
Arjun was holding 1,000 shares ₹ 10 each of Vanavill Electronics Ltd, issued at par. He paid ₹ 3 on the application, ₹ 4 on the allotment but could not pay the first and final call of ₹ 3. The directors forfeited the shares for nonpayment of call money. Give Journal entry for forfeiture of shares.
Solution :

Answer:
Forfeited shares account: ₹ 7,000

Question 10.
Lakshmi was holding 50 hares of ₹ 10 each on which he paid ₹ 2 on application but could not pay ₹ 4 on the allotment and ₹ 2 on first call. Directors forfeited the shares after the first call. Give journal entry for recording the forfeiture of shares.
Solution :

Answer:
Forfeited shares account: ₹ 100

Reissue of shares

Question 11.
Goutham Ltd. forfeited 500 equity shares of ₹ 10 each issued at par held by Ragav for nonpayment of the final call of ₹ 2 per share. The shares were forfeited and reissued to Madhan at ₹ 8 per share. Show the journal entries for forfeiture and reissue.
Solution :

Answer:
Capital reserve account: ₹ 3,000

Question 12.
Nivetha Ltd. forfeited 1,000 equity shares of ₹ 10 each for non-payament of call of ₹ 4 per share. Of these 800 shares were reissued @ ₹ 7 per share. Pass journal entries for forfeiture and reissue?
Solution :

Answer:
Capital reserve account: ₹ 2,400

Question 13.
Nathiya Textiles Ltd. forfeited 100 shares of ₹ 10 each, ₹ 8 called up, on which Mayuri had paid application and allotment money of 6 per share. Of these 75 shares were reissued to soundayya by receiving ₹ 7 per share. Pass journal entries for forfeiture and reissue?
Solution :

Answer:
Capital reserve account ₹ 375.

Question 14.
Simon Ltd issued 50,000 equity shares of ₹ 10 each at par payable on-application ₹ 1 per share, on allotment ₹ 5 per share, on first call ₹ 2 per share, and on second and final call ₹ 2 per share. The issue was fully subscribed and all the amounts were duly received with exception of 2,000 shares held by chezhian, who failed to pay the second and final call. His shares were forfeited and reissued to Elango at ₹ 8 per share. Journalise the above transactions?
Solution

Answer:
Capital reserve account: ₹ 12,000

Question 15.
Kanchana Ltd. issued 50,000 shares ₹ 10 each payable as under?
On application ₹ 1
On allotment ₹ 5
On first call ₹ 2
On final call ₹ 2
Applications were received for 70,000 shares. Applications for 8,000 shares were rejected and allotment was made proportionately towards the remaining applications. The directories made both the calls and all the amounts were received except the final call on 1,500 shares which were subsequently forfeited. Later 1.200 forfeited shares were reissued by receiving ₹ 8 per share. Give journal entries.
Solution :

Answer:
Capital reserve account: ₹ 7,200

Shares issued premium

Question 16.
Viswanath Furniture Ltd. invited applications for 20,000 shares of ₹ 10 each at a premium of 2 per share payable?
₹ 2 On application
₹ 5 (including premium) on the allotment
₹ 5 On the first and final call
There were oversubscription and applications were received for 30,000 shares and the excess applications were rejected by the directors. Pass the journal entries.
Solution :

Question 17.
United Industries Ltd. issued shares of ₹ 10 each at 10% premium payable ₹ 3 on the application, ₹ 4 on the allotment (including premium), ₹ 2 on the first call, and ₹ 2 on the final call.
Journalise the transections relating to forfeiture of shares for the following situations: Manoj who holds 250 shares failed to pay the second and final call and his shares were forfeited.
Manoj who holds 250 shares field to pay the allotment money and first call and second and final call and his shares were forfeited.
Manoj who holds 250 shares failed to pay the allotment money and first call money and his shares were forfeited after the first call.
Solution:

Answer:
Forfeited shares account: (i) ₹ 2,000; (ii) ₹ 750; (iii) ₹ 750

Question 18.
Kasthuri Ltd. had allotted 20,000 shares to applicants of 30,000 shares on a pro-rata basis. The amount payable was ₹ 1 on the application, ₹ 5 on the allotment (including a premium of ₹ 2 each), and ₹ 2 on the first call and ₹ 2 on final calls, Subin, a shareholder failed to pay the first call and final call on his 500 shares. All the shares were forfeited and out of the 400 shares were re-issued @ ₹ 8 per share. Pass necessary journal entries.
Solution :

Answer:
Capital reserve account: ₹ 1,600

Question 19.
Vairam Ltd. issued 60,000 shares of ₹ 10 each at a premium of ₹ 2 per share payable as follows:
On application ₹ 6
On allotment ₹ 4 (including premium)
On the first and final call ₹ 2
The issue was fully subscribed and the amount due was received except Saritha to whom 1,000 shares were allotted who failed to pay the allotment money and first and final call money. Her shares were forfeited. All the forfeited shares were reissued to Parimala at ₹ 7 per share.
Solution :

Answer:
Capital reserve account: ₹ 3,000

Issue of shares for cash in lumpsum

Question 20.
Abdul Ltd. issue 50,000 shares of ₹ 10 each at a premium of ₹ 2 per share. Pass journal entry if the amount is fully received along with a premium amount of ₹ 2 per share.
Solution :

Answer:
Share capital account ₹ 5,00,000; Securities premium account ₹ 1,00,000

Issue of shares for consideration other than cash

Question 21.
Paradise Ltd. purchased assets of ₹ 4,40,000 from Suguna Furniture Ltd. It issued equity shares of ₹ 10 each fully paid in satisfaction of their claim. What entries will be made if such issue is: (a) at par and (b) a premium of 10%.’
Solution :
Calculation of the number of shares to be issued

Answer:
(a) Share capital account: ₹ 4,40,000;
(b) Share capital account ₹ 4,00,000; Securities premium account ₹ 40,000

12th Accountancy Guide Company Accounts Additional Important Questions and Answers

I Multiple Choice Questions

Question 1.
A company issued 3,00,000 shares of Rs.10 each to the public but only 1,50,000 shares were subscribed. It’s subscribed capital is
(a) Rs. 30,00,000
(b) Rs. 15,00,000
(c) Rs. 10,00,000
Hint:
issued No. of share = 3,00,000
Subscribed No. of share = 1,50,000
Each share Rs. 10
Now ask the Question
No. of Subscribed Capital × Each Share
= 1,50,000 × 10 = ₹ 15,00,000
Answer:
(b) Rs. 15,00,000

Question 2.
The directors of a company forfeited 100 shares of Rs.10 each on which the final call, money of Rs.3 was not paid Later these shares were reissued of Rs.800 capital reserve will be
(a) Rs. 700
(b) Rs.500
(c) Rs. 500

Answer:
(b) Rs.500

Question 3.
When more applications are received than that are offered to the public it is called.
(a) Under subscription
(b) Full subscription
(c) Oversubscription
Answer:
(c) Oversubscription

Question 4.
Captial reserve is shown on the ………….. ride of the B/s
(a) Liability
(b) Assets
(c) Both
Answer:
(a) Liability

Question 5.
The balance of forfeited shares A/c is ……………. is in the B/s
(a) Added to paid-up capital
(b) Added to the authorized capital
(c) deducted from paid to capital
Answer:
(a) Added to paid-up capital

Question 6.
Calls-in-arrears are shown in the B/s as
(a) Deduction from called up capital
(b) Addition to paid-up capital
(c) Addition to issued capital
Answer:
(a) Deduction from called up capital

Question 7.
A company issued 20,000 Es of Rs.100 each to the public, but only 18,000 shares were subscribed, its subscribed capital is Rs.
(a) Rs. 20,00,000
(b) Rs. 18,00,000
(c) Rs. 21,00,000
Hint:
issued No. of share = 20,000
Paid up No. of share = 18,000
Each share = ₹ 10
∴ Now
Paid up Capital is
= 18,000 x 100
= ₹ 18,00,000 .
Answer:
(b) Rs. 18,00,000

Question 8.
A company issued 50,000 shares of Rs. 10 each at a premium of Rs. 2 each …………… is the securities premium amount
(a) Rs. 50,000
(b) Rs. 1,00,000
(c) Rs. 1,50,000
Hint:
Premium – ₹ 2 Each
Securities premium = 50,000 × 2 = ₹ 1,00,000
Answer:
(b) Rs. 1,00,000

Question 9.
Securities premium will appear in the _
(a) Assets
(b) Liability
(c) Assets & Liability
Answer:
(b) Liability

Question 10.
A Company is an association of
(a) workers
(b) merchants
(c) persons
Answer:
(c) persons

III Short Answer Questions

Question 1.
Define Company.
Answer:
According to Lord Justice Lindsey, “A Company is an association of many persons who contribute money or money’s worth to common stock and employ it in some trade or business and who share the profit and loss arising therefrom. The common stock so contributed is denoted in money and is the capital of the company. The persons who contributed in it or from it, or to whom it belongs, are members. The proportion of capital to which each member is entitled is his share”.

Question 2.
What are the characteristic features of the company?
Answer:
Following are the characteristics of a company:

• Voluntary association: A Company is a voluntary association of persons. No law can compel persons to form a company.
• Separate legal entity: Company is an artificial person. It has a separate legal entity which is separate and distinct from its members.
• Common seal: A company may have a common seal which can be affixed on the documents.
• Perpetual succession: A company continues forever. Its continuity is not affected by the changes in its members. It can be wound up only by law.
• Limited liability: The liability of the shareholders of the company is limited to the extent of the face value of the shares held by the shareholders.
• Transferability of shares: The shares of a company are freely transferable except in case of a private company.

Question 3.
Write short notes on

1. Issued capital
2. Subscribed capital
3. Called up capital
4. Paid-up capital

Answer:

1. Issued capital: This represents that part of authorized capital which is offered for subscription.
2. Subscribed capital: It refers to that part of issued capital which has been applied for and also allotted by the company.
3. Called up capital: It refers to that part of subscription capital which has been called up by the company for payment.
4. Paid-up capital: It is that part of called up capital which has been actually paid by the shareholders.

Question 4.
What do you mean by under subscription?
Answer:
All the shares offered to the public may not be subscribed in full. When the number of shares subscribed is less than the number of shares offered, it is known as under subscription.

Question 5.
What are the journal entries passed when shares are issued for cash in lumpsum?
Answer:

IV Exercises

Question 1.
A company issued 20,000 shares of ₹ 10 each payable as follows:
₹ 3 on Application,
₹ 3 on Allotment,
₹ 4 on First and Final call.
All the shares were subscribed and duly paid for. Pass journal entries.
Solution :

Question 2.
Preeti Ltd. invited applications for 5,000 shares of ₹ 10 each payable as follows:
₹ 3 on Application,
₹ 2 on Allotment,
₹ 2 on First call and
₹ 3 Final call
All these shares were subscribed and paid. Pass journal entries.
Solution:
 

Question 3.
A Joint Stock Company had a Nominal Capital of 5,00,000 divided into 5,000 shares ₹ 100 each payable:
₹ 30 per share on Application,
₹ 20 per share on Allotment,
₹ 30 on First call and
₹ 20 Final call
All the shares were subscribed and fully paid for by the public. Pass journal entries.
Solution :

Question 4.
Mary Ltd. Issued 1,000 shares of ₹ 10 each at a premium of ₹ 2 per share payable as follows:
₹ 4 on Application,
₹ 4 on Allotment, (including premium
and the balance when required. All the shares were subscribed for and duly paid. Pass necessary Journal entries.
Solution :

Question 5.
Global Ltd. issued 6,000 shares on ₹ 100 each at a premium of ₹ 20 per share payable as follows:
₹ 30 on Application,
₹ 50 Oon Allotment, (including premium)
₹ 30 on First call and
₹ 10 Final call
All shares were duly subscribed and money due was fully received. Pass journal entries.
Solution :
Journal Entries in the Books of Global Ltd

Question 6.
On 1.1.98 Alpha Ltd issued 1,00,000 shares of ₹ 10 each payable ₹ 2 on application. The company received applications for 1,20,000 shares. The excess applications were rejected. and money refunded Pass necessary entries.
Solution :
Journal Entries in the Books of Alpha Ltd

Question 7.
Good Luck Co. Ltd issued 1,00,000 shares @ 10 each payable:
₹ on Application,
₹ on Allotment,
and the balance when required. 1,50,000 shares were, applied for. The directors rejected the excess applications and refunded the application money. All money was received. Pass entries to record the transactions.
Solution :

Question 8.
Bhagavathi Ltd. issued 10,000 shares of ₹ 10 each at a discount of 10% payable as follows:
on Application, ₹2.50
on Allotment, ₹ 3.00
On the First and Final call ₹ 3.50
All money due was received except the final call on 100 shares which were forfeited by the company after giving due notice. Pass the forfeited entry.
Solution :

Question 9.
Ganthimathi Ltd. was registered with a nominal capital of ₹ 1,00,000 in equity shares of ₹ 10 each. It offered to the public 6,000 shares for subscripted The application was, however, received for 8,000 shares. The directors had to reject the application for 1,000 shares and to return the money received by Theron. The application money received on the other 1,000 shares was adjusted to the allotment account. The amount payable on shares was
₹ 3 per share on Application,
₹ 3 per share on Allotment and the balance on the first and final call
On shareholder holding, 100 shares failed to pay the call money and as a result, his shares were forfeited. Pass the necessary journal entries.
Solution :
Journal Entries in the Books of Ghandhimathi Ltd

Question 10.
Gani Ltd. forfeited 20 shares of ₹ 10 each fully called up, held by Santha for nonpayment of final call of 4 per sháre. These shares were re-issued to Josephin for ₹ 8 per share as fully paid up. Give journal entries for the forfeited and re-issue of shares.
Solution :

Question 11.
A Company forfeited 100 equity shares of ₹ 100 each issued at a premium of 10% (to be paid at the time of allotment) on which first call money of ₹ 30 per share and final call of ₹ 20 were not received. These shares were forfeited and subsequently re-issued at 90 per share. Give necessary journal entries regarding forfeited and re-issue of shares.
Solution :

Question 12.
The Directors of a company forfeited 200 Equity shares of ₹ 10 each fully called upon which the final call of ₹ 2 has not been paid. The shares were re-issued upon payment of ₹ 1,500. Journalise the above transactions.
Solution :

Question 13.
The Director of a company forfeited 100 shares of ₹ 10.each fully called up for non-payment of the First call of ₹ 2 per share and Final call of ₹ 3 per share. 60 of these shares were subsequently re-issued at 6 per share fully paid up. Pass necessary Journal entries to record the above.
Solution :

Question 14.
The directors of Sheeia Ltd. forfeited 2,000 shares ₹ 10 each for non-payment of final call of ₹ 2.50. 1,800 of these shares were re-issued for ₹ 6 per share fully paid up. Give the necessary Journal entries.
Solution :

Amount to be transferred to Capital reserve A/c = Profile on reissue × No.of share reissued
= 3.50 × 1,800
= Rs.6,300

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Accountancy Guide Pdf Chapter 8 Financial Statement Analysis Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis

12th Accountancy Guide Financial Statement Analysis Text Book Back Questions and Answers

I Multiple Choice Questions

Choose the correct answer

Question 1.
Which of the following statements is not true?
(a) Notes and schedules also form part of financial statements.
(b) The tools of financial statement analysis include common -size statement
(c) Trend analysis refers to the study of movement of figures for one years
(d) The common-size statements show the relationship of various items with some common base, expressed as percentage of the common base.
Answer:
(c) Trend analysis refers to the study of movement of figures for one years

Question 2.
Balance sheet provides information about the financial position of a business concern
(a) Over a period of time
(b) As on a particular date
(c) For a period of time
(d) For the accounting period
Answer:
(b) As on a particular date

Question 3.
Which of the following tools of financial statement analysis is suitable when data relating to several years are to be analysed?
(a) Cash flow statement
(b) Common size statement
(c) Comparative statement
Answer:
(d) Trend analysis

Question 4.
The financial statements do not exhibit
(a) Non-monetary data
(b) Past data
(c) Comparative statement
(d) Standard costing
Answer:
(a) Non-monetary data

Question 5.
Which of the following is not a tool of financial statement analysis?
(a) Trend analysis
(b) Common size statement
(c) Comparative statement
Answer:
(d) Standard costing

Question 6.
The term fund’ refers to
(a) Current liabilities
(b) Working capital
(c) Fixed assets
(d) Non-current assets.
Answer:
(b) Working capital

Question 7.
Which of the following statement is not true?
(a) All the limitations of financial statements are applicable to financial statement analysis also.
(b) Financial statement analysis is only the means and not an end.
(c) Expert knowledge is not required in analyzing the financial statements
(d) Interpretation of the analysed data involves personal judgment.
Answer:
(c) Expert knowledge is not required in analyzing the financial statements

Question 8.
A limited company’s sales have increased from ‘ 1,25,000 to 1,50,000. How does this appear in the comparative income statement?
(a) +20%
(b) +120%
(c) -120%
(d) -20%
Hint:
Percentage increase or Decrease = \(\frac{\text { Absolute amount of increase or decrease }}{\text { Year } 1 \text { amount }}\) × 100
Sales increase = ₹ 1,25,000 – ₹ 1,50,000 = ₹ 25,000
\(\frac{25,000}{1,25,000}\) × 100 = +20% = 1,25,000
Answer:
(a) +20%

Question 9.
In a common-size balance sheet, if the percentage of non-current assets is 75, what would be the percentage of current assets?
(a) 175
(b) 125
(c) 25
(d) 100
Hint:
Let Assets = ₹ 100
Non current assets = ₹ 75
∴ Current assets = 100 – 75 = 25
Answer:
(c) 25

Question 10.
Expenses for a business for the first year were ₹ 80,000. In the second year, it was increased to ₹ 88,000. What is the trend percentage in the second year?
(a) 10%
(b) 110%
(c) 90%
(d) 11%
Hint:
Computation of trend percentage = \(\frac{\text { Other year }}{\text { Earliest base year }}\) × 100
Earliest base year x 100
For second year = \(\frac{88,000}{80,000}\) × 100 = 110%
Answer:
(b) 110%

II Very Short Answer Questions

Question 1.
What are the financial statements?
Answer:
Financial statements are the statements prepared by the business concerns at the end of the accounting period to ascertain the operating results and the financial position.

Question 2.
List the tools of financial statement analysis.
Answer:

• Comparative statement
• Common size statements
• Trend analysis
• Funds flow statement
• Cash flow analysis

Question 3.
What is working capital?
Answer:
The term ’fund’ refers to working capital. Working capital refers to the excess of current assets over current liabilities.

Question 4.
When is trend analysis preferred to other tools?
Answer:
Trend analysis discloses the changes in financial and operating data between specific periods when data for more than two years are to be analyzed. It may be difficult to use a comparative statement.

III Short Answer Questions

Question 1.
‘Financial statements are prepared based on past data’. Explain how this is a limitation.
Answer:
The nature of the financial statement is historical. Past cannot be the index of the future and cannot be cent percent basis for future estimation, forecasting, budgeting, and planning.

Question 2.
Write a short note on cash flow analysis?
Answer:
Cash flow analysis is concerned with the preparation of a cash flow statement which shows * the inflow and outflow of cash and cash equivalents in a given period of time. Cash includes cash in hand and demand deposits with banks. Cash equivalents denote short term investments which can be realized easily within a short period of time, without much loss in value. Cash flow analysis helps in assessing the liquidity and solvency of a business concern.

Question 3.
Briefly explain any three limitations of financial statements.
Answer:

1. Lack of qualitative information: Qualitative information, that is non – monetary information is also important for business decisions. For example Efficiency of the employees and efficiency of the management. But this is ignored in financial statements.
2. Record of historical data: Financial statement are prepared based on historical data. They may not reflect the current position.
3. Ignores price level changes: Adjustments for price level changes are not made in the financial statements. Hence financial statements may not reveal the current position.

Question 4.
Explain the steps involved in preparing comparative statements?
Answer:
Following are the steps to be followed in preparation of the comparative statement:

1. Column 1: In this column, particulars of items of income statement or balance sheet are written.
2. Column 2: Enter absolute amount of year 1.
3. Column 3: Enter absolute amount of year 2.
4. Column 4: Show the difference in amounts between year 1 and year 2. If there is an increase in year 2, put plus sign and if there is decrease put minus sign.
5. Column 5: Show percentage increase or decrease of the difference amount shown in column 4 by dividing the amount shown in column 4 (absolute amount of increase or decrease) by column 2 (year 1 amount).

That is,
Percentage increase or decrease = \(\frac{\text { Absolute amount of increase or decrease }}{\text { Year } 1 \text { amount }}\) × 100

Question 5.
Explain the procedure for preparing a common – Size statement.
Answer:
Common-size statements can be prepared with three columns. Following are the steps to be followed in the preparation of the common – size statement.

1. Column 1. In this column, the particulars of items of the income statement or balance sheet are written.
2. Column 2. Enter the absolute amount.
3. Column 3. Choose a common base as 100.

For example Revenue from operations can be taken as the base for income statement and total of the balance sheet can be taken as the base for the balance sheet. Work out the percentage for all the items of column 2 in terms of the common base and enter them in column 3.

IV Exercises

Comparative statement analysis

Question 1.
From the following particulars, prepare a comparative income statement of Arul Ltd.

Solution:
The comparative income statement of Arul Ltd for the year ended 31.3.16 & 31.3.17

Answer :
Revenue from operation: 20 %;
Other income: 200%;
Total revenue: 50%;
Expenses: 25%;
Profit before tax: 100%

Question 2.
From the following particulars, prepare a comparative income statement of Barani Ltd.

Solution:
Computation of % increase for revenue from operation
Percentage increase = \(\frac{15,000}{30,000}\) × 100 = 50%
Comparative Income statement of Barani Ltd. for the year ended 31.3.17 & 31.3.18


Answer :
Revenue from operations: 50%
Other income: 50%
Total revenue: 50%
Expenses: 50%;
Profit before tax: 50%;
Profit after tax: 50%

Question 3.
From the following Particulars, prepare comparative income ‘statement of Daniel Ltd.

Solution:
The comparative Income statement of Daniel Ltd. for the year ended 31.3.16 & 31.3.17

Answer :
Revenue from operations: 25%
Expenses: 10%
Tax: 50%;
Profit before tax: 50%;
Profit after tax: 50%

Question 4.
From the following particulars, prepare a comparative .statement of the financial position of Muthu Ltd.

Solution:
Comparative B/s of Muthu Ltd. As on 31.3.17 & 31.3.18

Answer :
Shareholder’s fund: 10%;
Non-current liabilities: 10%
current liabilities: 10%
Total equity and liabilities: 10%;
Non -current assets: 20%;
Current assets: 30%
Total Assets: 10%

Question 5.
From the following particulars, prepare a comparative statement of the financial position of Kala Ltd.


Solution:



Answer :
Share capital: 20%;
Reserves and surplus: Nil.
Non-Current liabilities: -20%;
Current liabilities: -40%;
Total equity and liabilities: 10%;
Fixed assets: 16%;
Non-current investments: -20%;
Inventories: 25%;
Cash and cash equivalents: 20%;
total assets: 10%

Question 6.
Prepare a common-size income statement for the following particulars Raja Ltd. for the year ended 31st March 2017.

Solution:

Answer :
2016-2017:
Other income: 15%
total revenue: 115%
Expenses: 30%
Profit before tax: 85%

Question 7.
From the following particulars of Maria ltd and Kala Ltd. Prepare a common-Size income statement for the year ended 31st March 2019.

Solution:


Answer :
Maria Ltd: Other Income: 10%; total revenue: 110%; Expenses: 70%; Profit before tax:40%
Kala Ltd: Other income: 15%; Total revenue: 115% Expenses: 60%; Profit before tax:55%

Question 8.
Prepare a common-size income statement for the following particulars of Sam Ltd. table

Solution:

Answer :
2015-16: Other income:20%; Total revenue: 120% Expenses:60%; Profit before tax:60%: Tax: 18%; Profit after tax:42%;
2016-17: Other income: 10%; total revenue: 110%; Expenses:50%; Profit before tax: 60%: Tax:18%;
Profit after tax:42%

Question 9.
Prepare Common-size balance sheet of Meena Ltd. as of 31st March 2018.

Solution:
Common size B/s Meena Ltd as on 31.3.18


Answer :
Shareholder’s fund:50%; Non-current liabilities: 40%; Current liabilities: 10%; Non-current assets:75%; current assets:25%

Question 10.
Prepare a common-size statement of financial position for the following particulars of Rani Ltd.

Solution:


Answer :
2015-16 Shareholder’s fund:60%; Non-current liabilities: 30%; Current liabilities: 10%; Non-current assets:80%; current assets:20%
2016-17 Shareholder’s fund:60%;Non-current liabilities: 25%; Current liabilities: 15%; Non-current assets:80%; Current assets:20%

Question 11.
Prepare a common-size statement of financial position for the following particulars of Yasmin Ltd. and Sakthi Ltd.


Solution:


Answer:
Yasmin Ltd: Share capital 40%; Reserves and surplus: 10%; Noncurrent liabilities:30%; current liabilities:20%. Fixed assets:40%; Non -Current investments: 10% Inventories:40%; Cash & cash equivalents: 10% Sakthi Ltd: Share „ capital: 50%; Reserves and surplus: 10%; Non-current liabilities:30%; Current liabilities: 10%; Fixed assets:50% Non-current investmesits:20%; Inventories: 15% cash & cash equivalents: 15%
Trend Analysis:

Question 12.
From the following particulars, calculate the trend percentages of Kala Ltd.

Solution:

Answer:
2016-17: Revenue from operations: 125%; Other income: 150%; Total revenue: 130% Expenses:145% Profit before tax: 120%
2017-18: Revenue from operations: 150%; Other income 200%; Total revenue: 160% Expenses:175%; Profit before tax: 150%

Question13.
From the following particulars, calculate the Trend percentages of Kavitha Ltd.

Solution:

Answer:
2016-17: Revenue from operations: 125%; Other income: 125%; total revenue:125%; Expenses: 120%; Profit before tax:150%; Tax :150%; Profit after tax: 150%;
2017-2018; Revenue from operations: 150%; Other income:150%; total revenue:150% Expenses:80%; Profit before tax: 500% Tax:500%; Profit after tax:500%

Question 14.
From the following particulars, calculate the trend percentage of Kumar Ltd.

Solution:

Answer :
2016-17: Revenue from operations:90%; Other income: 160%; total revenue: 100%; Expenses:80%; Profit before tax: 150%; Tax: 150%; Profit after tax:150%;
2017-18: Revenue from operations:50%; Other income: 120%; total revenue:60%; Expenses:50%; Profit before tax: 85%; Tax:85%; Profit after tax:85%

Question 15.
From the following particulars, calculate the trend percentages of Anu Ltd.

Solution:


Answer :
year 2: Shareholder’s Fund: 110%, Non-current liabilities: 125%; Current liabilities:80%; Total equity and liabilities: 110%; Non-Current assets: 120%, Current assets:80%; total assets: 10%;
year 3: Shareholder’s fund: 120%; Non- current liabilities: 120%; Current liabilities: 120% Total equity and liabilities: 120%; Non-current assets: 130% Current assets:90%; total assets: 120%

Question 16.
From the following particulars, calculate the trend percentages of Babu Ltd.

Solution:



Answer:
Year 2: Share capital: 127%; Reserves and surplus: 100%; Non-current liabilities: 110%; Current liabilities: 150% Total equity and liabilities: 120%; Fixed assets:118%; Non-current investments: 125%; Inventories:! 10%; Cash & cash equivalents: 150%; total assets:120%;
Year 3: Share capital: 106%; Reserves and surplus :150%; Non-current liabilities: 120%; Current liabilities: 200%; total equity and liabilities: 125%; Fixed assets:103%; Non – Current investments: 150%; Inventories: 120%; Cash & cash equivalents:200%; total assets: 125%

12th Accountancy Guide Financial Statement Analysis Additional Important Questions and Answers

Other Important questions & Answers

Question 1.
…………………………. are the tools of financial analaysis
(a) Comparative statements
(b) Trend analysis
(c) Common size statement
(d) all the above
Answer:
(d) all the above

Question 2.
Analysis of financial statements involves
(a) B/S
(b) Trading A/c
(c) All the above
Answer:

III Short Answer Questions

Question 1.
What are the features of a financial statement?
Answer:
Following are the features of financial statements:

• Financial statements are generally prepared at the end of an accounting period based on transactions recorded in the books of accounts.
• These statements are prepared for the organization as a whole.
• Information is presented in a meaningful way by grouping items of similar nature such as fixed assets.
• Financial statements are prepared based on historical cost.
• Financial statements are prepared based on accounting principle and Accounting Standards, which make financial statements comparable and realistic.
• Financial statements involve personal judgment in a certain case. For example, the selection of the method of depreciation, percentage of reserve, etc.

Question 2.
Explain the significance of financial statements.
Answer:
Financial statements reveal the operating results and financial position of the business concern. The significance of financial statements to various stakeholders is as follows:

• To Management: Financial statements provide information to the management to take decision and to have control over business activities, in various areas.
• To shareholders: Financial statements help the shareholders to know whether the business has potential for growth and to decide to continue their shareholding.
• To potential investors: Financial statements help to value the securities and compare it with those of other business concerns before making their investment decisions.
• To creditors: Creditors can get information about the ability of the business to repay the debts from financial statements.
• To bankers: Information given in the financial statements is significant to the bankers to assess whether there is adequate security to cover the amount of the loan or overdraft.
• To the government: Financial statements involve personal judgment in a certain case. For example, the selection of the method of depreciation, percentage of reserve, etc.
• To employees: Through the financial statements, the employees can assess the ability of the business to pay salaries and whether they have future growth in the concern.

Question 3.
Explain the provisions of the Indian Companies Act 2013 to be followed while preparing the financial statement of or company.
Answer:
Following provisions of the Indian Companies Act, 2013 have to be followed while preparing the financial statements of a company:

• As per Section 2(40), financial statements include balance sheet, profit and loss account/income and expenditure account, cash flow statement, statement of changes in equity and any explanatory note annexed to the above.
• Section 129 (1) of the Indian Companies Act, 2013 states that the financial statements shall give a true and fair view of the state of affairs of the company and shall comply with the Accounting Standards notified under section 133.
• Section 129 (1) also states that the financial statements shall be prepared in the form provided in schedule III of the Indian Companies Act, 2013.

Question 4.
What do you mean by financial statement analysis?
Answer:
Financial statement analysis is a comparison of the various items in the financial statement by establishing and evaluating relationships among them so that, it gives a better understanding of the performance and financial status of the business concern.

Question 5.
What are the objectives of financial statement analysis?
Answer:
Financial statement analysis may be done with any of the following objectives:

• To analyze profitability and earning capacity.
• To study the long term and short term solvency of the business.
• To determine the efficiency in operations and use of assets.
• To determine the efficiency of management and employees.
• To determine the trend in sales, production, etc.
• To forecast for the future and prepare budgets.
• To make inter-firm and intra-firm comparisons.

Question 6.
What are the limitations of fin. State analysis.
Answer:
Following are the limitations of financial statement analysis:
1. All the limitations of financial statements such as ignoring non-monetary information, ignoring price level changes, etc., are applicable to financial statement analysis also.

2. Financial statement analysis is only the means and not an end, that is, it is only a tool in the hands of management and other shareholders. Interpretation of the results has to be done only by the financial analysts with due regard to the internal and external environmental factors.

3. Expert knowledge is required in analyzing the financial statements.
4. Interpretation of the analyzed data involves personal judgments as different experts may give different views.

Question 7.
What do you mean by Horizontal analysis?
Answer:
Horizontal analysis
When figures relating to several years are considered for the purpose of analysis, the analysis is called horizontal analysis. Generally, one year is taken as the base year and the figures relating to the other years are compared with that of the base year. Comparative statements and trend percentages are examples of horizontal analysis.

Question 8.
What do you mean by vertical analysis?
Answer:
Vertical analysis
When figures relating to one accounting year alone are considered for the purpose of analysis, the analysis is called vertical analysis. Here, the relationship is established among items from various financial statements relating to the same accounting period. Preparation of common size statements and computation of ratios are examples of vertical analysis.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.10 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10

Choose the most suitable answer from the given four alternatives:

Question 1.
The value of \(\int_{0}^{2/3}\) \(\frac { dx }{ \sqrt{4-9x^2} }\)
(a) \(\frac { π }{ 6 }\)
(b) \(\frac { π }{ 2 }\)
(c) \(\frac { π }{ 4 }\)
(d) π
Solution:
(a) \(\frac { π }{ 6 }\)
Hint:

Question 2.
The value of \(\int_{-1}^{2}\) |x| dx
(a) \(\frac { 1 }{ 2 }\)
(b) \(\frac { 3 }{ 2 }\)
(c) \(\frac { 5 }{ 2 }\)
(d) \(\frac { 7 }{ 2 }\)
Solution:
(c) \(\frac { 5 }{ 2 }\)
Hint:
|x| = x if x > 0
= -x if x < 0

Question 3.
For any value of n ∈ Z \(\int_{0}^{π}\) e^cos²x cos3 [(2n + 1) x] dx
(a) \(\frac { π }{ 2 }\)
(b) π
(c) 0
(d) 2
Solution:
(c) 0
Hint:

Question 4.
The value of \(\int_{-π/2}^{π/2}\) sin² x cos x dx
(a) \(\frac { 3 }{ 2 }\)
(b) \(\frac { 1 }{ 2 }\)
(c) 0
(d) \(\frac { 2 }{ 3 }\)
Solution:
(d) \(\frac { 2 }{ 3 }\)
Hint:
f(x) = sin²x cos x
f(-x) = sin²(-x) cos(-x) = sin²x cos x
f(x) = f(-x)
f is an even function

Question 5.
The value of \(\int_{-4}^{4}\)[tan^-1(\(\frac { x^2 }{ x^4+1 }\)) + tan^-1(\(\frac { x^4+1 }{ x^2 }\))] dx
(a) π
(b) 2π
(c) 3π
(d) 4π
Solution:
(d) 4π
Hint:

Question 6.
The value of \(\int_{-π/4}^{π/4}\) (\(\frac { 2x^7-3x^5+7x^3-x+1 }{ cos^2x }\)) dx is
(a) 4
(b) 3
(c) 2
(d) 0
Solution:
(c) 2
Hint:

Odd function – 3 Odd function + 7 Odd function – Odd function + even function
= 0 + 2 \(\int_{0}^{π/4}\) sec² x dx
= 2 [tan x]\(_{0}^{π/4}\)
= 2(1 – 0)
= 2

Question 7.
If f(x) = \(\int_{0}^{x}\) cos t dt, then \(\frac { df }{ dx }\) =
(a) cos x – x sin x
(b) sin x + x cos x
(c) x cos x
(d) x sin x
Solution:
(c) x cos x
Hint:
f(x) = \(\int_{0}^{x}\) cos t dt
u = t, dv = cos t dt
u’ = 1, v = sin t
v₁ = -cos t
f(x) = [t sin t + cos t]\(_{0}^{x}\)
f(x) = x sin x + cos x – 1
\(\frac { df }{ dx }\) = x cos + sin x – sin x
\(\frac { df }{ dx }\) = x cos x

Question 8.
The area between y² = 4x and its latus rectum is
(a) \(\frac { 2 }{ 3 }\)
(b) \(\frac { 4 }{ 3 }\)
(c) \(\frac { 8 }{ 3 }\)
(d) \(\frac { 5 }{ 3 }\)
Solution:
(c) \(\frac { 8 }{ 3 }\)
Hint:
y² = 4x
y = 2√x

Question 9.
The value of \(\int_{0}^{1}\) x (1 – x)⁹⁹ dx is
(a) \(\frac { 1 }{ 11000 }\)
(b) \(\frac { 1 }{ 10100}\)
(c) \(\frac { 1 }{ 10010 }\)
(d) \(\frac { 1 }{ 10001 }\)
Solution:
(b) \(\frac { 1 }{ 10100}\)
Hint:

Question 10.
The value of \(\int_{0}^{π}\) \(\frac { dx }{ 1+5^{cosx} }\) is
(a) \(\frac { π }{ 2 }\)
(b) π
(c) \(\frac { 3π }{ 2 }\)
(d) 2π
Solution:
(a) \(\frac { π }{ 2 }\)
Hint:

Adding (1) and (2)

Question 11.
If \(\frac { Γ(n+2) }{ Γn }\) = 90 then n is
(a) 10
(b) 5
(c) 8
(d) 9
Solution:
(d) 9
Hint:

n² + n = 90
n² + n – 90 = 0
(n + 10) (n – 9) = 0
n = 9

Question 12.
The value of \(\int_{0}^{π/6}\) cos³ 3x dx is
(a) \(\frac { 2 }{ 3 }\)
(b) \(\frac { 2 }{ 9}\)
(c) \(\frac { 1 }{ 9 }\)
(d) \(\frac { 1 }{ 3 }\)
Solution:
(b) \(\frac { 1 }{ 10100}\)
Hint:
\(\int_{0}^{π/6}\) cos³ 3xdx = \(\frac { 1 }{ 3 }\) × \(\frac { 2 }{ 3 }\) × 1
= \(\frac { 2 }{ 9 }\)

Question 13.
The value of \(\int_{0}^{π}\) sin⁴ x dx
(a) \(\frac { 3π }{ 10 }\)
(b) \(\frac { 3π }{ 8}\)
(c) \(\frac { 3π }{ 4 }\)
(d) \(\frac { 3π }{ 2 }\)
Solution:
(b) \(\frac { 3π }{ 8}\)
Hint:
(x) = sin⁴x dx
f(π – x) = sin⁴ (π – x) = sin⁴x

Question 14.
The value of \(\int_{0}^{∞}\) e^-3x x² dx
(a) \(\frac { 7 }{ 27 }\)
(b) \(\frac { 5 }{ 27 }\)
(c) \(\frac { 4 }{ 27 }\)
(d) \(\frac { 2 }{ 27 }\)
Solution:
(d) \(\frac { 2 }{ 27 }\)
Hint:

Question 15.
If \(\int_{0}^{a}\) \(\frac { 1 }{ 4+x^2 }\) dx = \(\frac { π }{ 8 }\) then a is
(a) 4
(b) 1
(c) 3
(d) 2
Solution:
(d) 2
Hint:

Question 16.
The volume of solid of revolution of the region bounded by y² = x(a – x) about the x-axis is
(a) πa³
(b) \(\frac { πa^3 }{ 4 }\)
(c) \(\frac { πa^3 }{ 5 }\)
(d) \(\frac { πa^3 }{ 6 }\)
Solution:
(d) \(\frac { πa^3 }{ 6 }\)
Hint:
y² = x(a – x)
To find limit put y = 0
x (a – x) = 0
x = 0, x = a

Question 17.
If f(x) = \(\int_{1}^{x}\) \(\frac { e^{sinx} }{ u }\) du, x > 1 and \(\int_{1}^{3}\) \(\frac { e^{sinx^2} }{ x }\) dx = \(\frac { πa^3 }{ 6 }\) [f(a) – f(1)] then one of the possible value of a is
(a) 3
(b) 6
(c) 9
(d) 5
Solution:
(c) 9
Hint:

Question 18.
The value of \(\int_{0}^{1}\) (sin^-1 x)² dx is
(a) \(\frac { π^2 }{ 4 }\) – 1
(b) \(\frac { π^2 }{ 4 }\) + 2
(c) \(\frac { π^2 }{ 4 }\) + 1
(d) \(\frac { π^2 }{ 4 }\) – 2
Solution:
(d) \(\frac { π^2 }{ 4 }\) – 2
Hint:

Question 19.
The value of \(\int_{0}^{a}\) (\(\sqrt{a^2-x^2}\))³ dx is
(a) \(\frac { πa^3 }{ 16 }\)
(b) \(\frac { 3πa^4 }{ 16 }\)
(c) \(\frac { 3πa^2 }{ 8 }\)
(d) \(\frac { 3πa^4 }{ 8 }\)
Solution:
(b) \(\frac { 3πa^4 }{ 16 }\)
Hint:

Question 20.
If \(\int_{0}^{x}\) f(t) dt = x + \(\int_{x}^{1}\) f(t) dt, then the value of f(1) is
(a) \(\frac { 1 }{ 2 }\)
(b) 2
(c) 1
(d) \(\frac { 3 }{ 4 }\)
Solution:
(a) \(\frac { 1 }{ 2 }\)
Hint:

= 1 + 0 – x f(x)
f(x) + x f(x) = 1
f(x)[1 + x] = 1
when x = 1,
f(1)(2) = 1
f(1) = \(\frac { 1 }{ 2 }\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Accountancy Guide Pdf Chapter 6 Retirement and Death of a Partner Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner

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

I Multiple Choice Questions

Choose the correct answer

Question 1.
A partner retires from the partnership firm on 30th June. He is liable for all the acts of the firm up to the
(a) End of the current accounting period
(b) End of the previous accounting period
(c) Date of his retirement
(d) Date of his final settlement
Answer:
(c) Date of his retirement

Question 2.
On the retirement of a partner from a partnership firm, accumulated profits and losses are distributed to the partners on the basis of
(a) New profit sharing ratio
(b) Old profit sharing ratio
(c) Gaming ratio
(d) Sacrificing ratio
Answer:
(b) Old profit sharing ratio

Question 3.
On the retirement of a partner, general reserve will be transferred to the
(a) Capital account of all the partners
(b) Revaluation account
(c) Capital account of the continuing partners
(d) Memorandum revaluation account
Answer:
(a) Capital account of all the partners

Question 4.
On revaluation, the increase in liabilities leads to ……………….
(a) Gain
(b) Loss
(c) Profit
(d) None of these
Answer:
(b) Loss

Question 5.
At the time of retirement of a partner, determination of gaining ratio is required
(a) To transfer revaluation profit or loss
(b) To distribute accumulated profits and losses .
(c) To adjust goodwill
(d) None of these
Answer:
(c) To adjust goodwill

Question 6.
The final amount due to a retiring partner is not paid immediately,it is transferred to
(a) BankA/c
(b) Retiring partners capital A/c
(c) Retiring partner s loan A/c
(d) Other partner s capital A/c
Answer:
(c) Retiring partner s loan A/c

Question 7.
‘A’was a partner in a partnership firm. He died on 31st March 2019. The final amount due to him is ? 24,000 which is not paid immediately. It will be.transferred to
(a) As capital account
(b) A’s loan account
(c) As Executors account
(d) As Executors loan account
Answer:
(d) As Executors loan account

Question 8.
A, B and C are partners sharingprofits in the ratio of 2:2:1. On retirement of B, goodwill of the firm was valued as ₹ 30,000. Find the contribution of A and C to compensate B:
(a) ₹ 20,000 and ₹ 10,000
(b) ₹ 8,000 and ₹ 4,000
(c) ₹ 10,000 and ₹ 20,000
(d) ₹ 15,000 and ₹ 15,000
Hint:

∴ Retire partner share will be compensate of‘A’ and ‘Copartners
’A’ Goodwill value = 30,000 x \(\frac{2}{5}\) = 5 12,000
‘B’ Goodwill Value = 30,000 x \(\frac{2}{5}\) = 5 12,000
‘C’ Goodwill Value = 30,000 x \(\frac{1}{5}\) = 5 6,000
Now Compensate A& C
‘B’ Share good will – 12,000 x \(\frac{2}{3}\) = 8,000
12,000 x \(\frac{1}{3}\) = 4,000
Answer:
(b) ₹ 8,000 and ₹ 4,000

Question 9.
A, B and C are partners sharing profits in the ratio of 4:2:3. C retires. The new profit sharing ratio between A and B will be
(a) 4:3
(b) 3:4
(c) 2:1
(d) 1:2
Hint:
OldPartnersABC

Answer:
(c) 2:1

Question 10.
X, Y and Z were partners sharing profits and losses equally. X died on 1st April 2019. Find out the share of X in the profit of 2019 based on the profit of 2018 which showed ₹ 36,000.
(a) ₹ 1,000
(b) ₹ 3,000
(c) ₹ 12,000
(d) ₹ 36,000
Hint:
Partner’s X: Y: Z
Sharing ratio = equally =\(\frac{1}{3}: \frac{1}{3}: \frac{1}{3}\)
Profit = ₹ 36,000
Share of ‘X’
36,000 x \(\frac{1}{3}\) = ₹ 12,000
Answer:
(c) ₹ 12,000

II Very Short Answer Questions

Question 1.
What is meant by the retirement of a partner?
Answer:
When a partner leaves from partnership firm it is known as retirement. The reasons for the retirement of a partner may be illness, old age, and disagreement with other partners, etc.

Question 2.
What is the gaining ratio?
Answer:
The continuing partners may gain a portion of the share of profit of the retiring partner. The gain may be shared by all the partners or some of the partners. The gaining ratio is the proportion of the profit which is gained by the continuing partners.

Question 3.
What is the purpose of calculating the gaining ratio?
Answer:
The purpose of finding the gaining ratio is to bear the goodwill to be paid to the retiring partner.

Question 4.
What Is the journal entry to be passed to transfer the amount due to the deceased partner to the executor of the deceased partner?
Answer:
To transfer the amount due to the deceased partner to the executor or legal representative of the deceased partner.

III. Short Answer Questions

Question 1.
List out the adjustments made at the time of retirement.
Answer:
The Following adjustments are necessary at the time of retirement of a partner:

• Distribution of accumulated profits, reserves, and losses
• Revaluation of assets and liabilities
• Determination of new profit sharing ratio and gaining ratio
• Adjustment for goodwill
• Adjustment for current year’s profit or loss up to the date of retirement
• Settlement of the amount due to the retiring partner

Question 2.
Distinguish between sacrificing ratio and gaining ratio.
Answer:

Question 3.
What are the ways in which the final amount to an outgoing partner can be settled?
Answer:
The amount due to the retiring partner may be settled one of the following ways:

1. Paying the entire amount due immediately in cash
2. Transfer the entire amount due to the loan account of the partner
3. Paying part of the amount immediately in cash and transferring the balance to the loan account of the partner.

IV Exercise

Distribution of accumulated profits, reserves, and losses

Question 1.
Dheena, Surya, and Jankai are partners sharing profits and losses in the ratio of 5:3:2. on 31.3.2018, Dheena retired. On the date of retirement, the books of the firm showed a reserve fund of ₹ 50,000. The pass journal entry to transfer the reserve fund.
Solution:

Answer:
Reserves fund: dheena: ? 25,000(Cr); Surya: ₹ 15,000(Cr.); Janaki: ₹ 10,000(Cr.);

Question 2.
Rosi, Rathi, and Rani are partners of firms sharing profits and losses equally. Rathi retired from the partnership on 1.1.2018. On the date, their balance sheet showed an accumulated loss of ₹ 45,000 on the asset side of the balance sheet. Give the journal [ entry to distribute the accumulated loss.
Solution:

Answer:
Rosi: ₹ 15,000(Dr.); Rathi: ₹ 15,000(Dr.);Rani: ₹ 15,000(Dr.);

Question 3.
Akash, Mukesh, and Sanjay are partners in firm sharing profits and losses in the ratio of 3:2:1. Their balance sheet as on 31st March 2017 is as follows:

Pass journal entry to transfer the accumulated profit and prepare the capital account of the partners
Solution:


Answer:
Akash capital: ₹ 67,000(Cr.); Mugesh Capital: ₹ 78,000(Cr.); Sanjay’s Capital:₹ 39,000(Cr.);

Question 4.
Roja, Neeia ‘and Kanaga are partners, sharing profit and losses in the ratio of 4:3:3. On 1st April 2017, Roja retires and on retirement, the following adjustments are agreed upon.
(i) Increase the value of the Building by ₹ 30,000.
(ii) Depreciate stock by ₹ 5,000, and furniture by ₹ 12,000.
(iii) Provide for an outstanding liability of ₹ 1,000.
Pass journal entries and prepare revaluation account.
Solution:

Answer:
Revaluation profit : ₹ 12,000

Question 5.
Vinoth, Karthi, and Pranav are partners sharing profits and. losses’ in the ratio of 2:2:1 Pranav retires from the partnership on 1st April 2018. The following adjustments are to be made.
(i) increase the value of the land building by ₹ 18,000.
(ii) Reduce the value of machinery by ₹ 15,000.
(iii) A provision would also be made for outstanding expenses for ₹ 8,000.
Give journal entries and prepare a revaluation account.
Solution:

Question 6
Chandru, Vishal, and Ramanan are partners in firms sharing profits and losses equally. Their balance sheet as of 31st March 2018 is as follows.

Ramanan retired on 31st March 2019 subject to the following conditions:
(i) Machinery is valued at ₹ 1,50,000
(ii) Value of furniture brought down by ₹ 10,000
(iii) Provision for doubtful debts should be increased to ₹ 5,000. ,
(iv) investment of ₹ 30,000 not recorded in the books is to be recorded now.
Pass necessary journal entries and prepare revaluation account and capital account of partners.
Solution:

Answer:
Revaluation profit: ₹ 48,000;
Capital account:)
Chandru :₹ 76,000(Cr);
Vishal: ₹ 86,000(Cr);
Ramanan’s loan account: ₹ 86,000 (a))

Question 7.
Kayal, Mala and Neela are partners sharing profits in the ratio of 2:2:1. kayal retires and the new profit sharing ratio between Mala and Neela is 3:2. Calculate the gaining ratio.
Solution:
New Profit Sharing Ratio and Gaining Ratio
Gain Ratio = New Ratio – Old Ratio
Kayal –
Mala = \(\frac { 3 }{ 2 }\) – \(\frac { 2 }{ 5 }\) – \(\frac { 1 }{ 5 }\)
Neela = \(\frac { 2 }{ 5 }\) – \(\frac { 1 }{ 5}\) = \(\frac { 1 }{ 5 }\)
Gaining Ratio = 1:1

Question 8.
Sunil, Sumathi and Sundari are partners sharing profits in the ratio of 3:3:4. Sundari retires and her share is taken up entirely by Sunil. Calculate the new profit sharing ratio and gaining ratio.
Solution:
old ratio = 3:3:4 (Sunil:sumathi:sundar)
New Ratio = old ratio + gaining ratio

Answer:
Gaining ratio:4:0;
New ratio:7:3

Question 9.
Ramu, Somu and Gopu are partners sharing profits in the ratio of 3:5:7. Gopu retires and the share is purchased by Ramu and Somu in the ratio of 3:1. Find the new profit sharing ratio and gaining ratio.
Solution:

Question 10.
Navin, Ravi and Kumar are partners sharing profits in the ratio of 1/2, 1/4 and 1/4 respectively. Kumar retires and his share is taken up by Navin and Ravi equally. Calculate the new profit sharing ratio and gaining ratio.
Solution :

Answer:
New ratio:5:3;
Gaining ratio: 1:1

Question 11.
Mani, Gani and Soni are partners sharing the profits and losses in the ratio of 4:5:6. Mani retires from the firm. Calculate the new profit sharing ratio and gaining ratio.
Mani: Gani: Soni → 4:5:6
[If nothing is mentioned about the new ratio, old ratio of the continuing partners is equal to their new ratio.]
Solution:
Old ratio = 4:5:6
New ratio – 5:6

Answer:
New profit shaving ratio and gaining ratio is 5:6

Question 12.
Rajan, Suman and jegan were partners in firm sharing profits and losses in the ratio of 4:3:2 Suman retired from partnership. The goodwill of the firm on the date of retirement was valued at ₹ 45,000. Pass necessary journal entries for goodwill on the assumption that the fluctuating capital method is followed.
Solution :

Answer:
Suman’s share of goodwill: ₹ 15,000;
Rajan’s capital: ₹ 10,000(Dr);Jegan’s capital: ₹ 5,000(Dr);

Question 13.
Balu, Chandru and Nirmal are partners in a firm sharing profits and losses in the ratio of 5:3:2 on 31st March 2018, Nirmal retires from the firm. On the date of Nirmal’s retirement, goodwill appeared in the books of the firm at ₹ 60,000 By assuming thuctuating capital account, pass the necessary journal entry if the partners decide to
(a) Write off the entire amount of existing goodwill
(b) Write off half of the existing goodwill.
Solution :

Answer:
(a) Balu’s capital: ₹ 30,000(Dr.); Chandru’s capital: ₹ 18,000(Dr.)
Nirmala’s capital: ₹ 12,000(Dr);(b)balu’s capital: ₹ 15,000(Dr);
Chandru’s capital: ₹ 9,000(Dr);Nirmala’s capital: ₹ 6,000(Dr);

Question 14.
Rani, Jaya and Rathi are partners sharing profits and losses in the ratio of 2:2:1. On 31.3.2018, Rathi retired from the partnership. Profit of the preceding years is as follows: 2014: ₹ 10,000; 2015; ₹ 20,000; 2016; ₹ 18,000 and 2017; ₹ 32,000
Find out the share of profit of Rathi for the year 2018 till the date of retirement if
(a) profit is to be distributed on the basis of the previous year’s profit
(b) Profit is to be distributed on the basis of the average profit of the past 4 years.
Also pass necessary journal entries by assuming partners’ capitals are fluctuating.
Solution:
(a) on the basis of previous years profit
Profit of 2017 = Rs.32,000
Date of retirement = 31.3.2018
share of profit of Rathi for 3 months = \(32,000 \times \frac{3}{12} \times \frac{1}{5}\)
= Rs. 1600

Answer:
(a) Rathi’s share of profit: ₹ 1,600; (b) Rathi’s share of profit: ₹ 1,000;
Settlement of amount due to the retiring partner.

Question 15.
Kavin, Madhan, and Ranjith are partners sharing profits and losses in the ratio of 4:3:3 respectively. Kavin retires from the firm on 31st December 2018. On the date of retirement, his capital account shows a credit balance of ₹ 1,50,000. Pass journal entries if:
(a) The amount due is paid off immediately.
(b) The amount due is not paid immediately.
(c) ₹ 1,00,000 is paid and the balance in the future.
Solution:

Answer:
(a) Kavin’s loan:Nil;
(b) Kavin’s loan: ₹ 1,50,000;
(c) Kavin’s loan: ₹ 50,000;

Question 16.
Manju, Chara’and Lavanya are partners in firm sharing profits and losses in the ratio of 5:3:2. The balance sheet as of 31st March, 2018 was as follows:

Manju retired from the partnership firm on 31.3.2018 subject to the following adjustments:
(i) Stock to be depreciated by ₹ 10,000
(ii) Provision for doubtful debts to be created for ₹ 3,000
(iii) Buildings to be appreciated by ₹ 28,000.
Prepare revaluation account and capital accounts of partners after retirement.
Solution:

Answer:
Revaluation profit: ₹ 15,000; Manju’s loan account:
₹ 1,02,000(Cr).Capital account: Charu ₹ 89,OO0(Cr); Lavanya: ₹ 83,000(Cr))

Question 17.
Kannan, Rahim, and John are partners in a firm sharing profits and losses in the ratio of 5:3:2. The balance sheet as of 31st December 2017 was as follows:

John retires on 1st January 2018, subject to the following conditions :
(i) To appreciate building by 10%
(ii) Stock to be depreciated by 5%
(iii) To provide ₹ 1,000 for bad debts
(iv) An unrecorded liability of ₹ 8,000 has been noticed.
(v) The retiring partner shall be paid immediately.
prepare revaluation account, partner’s capital account, and the balance sheet of the firm after retirement.
Solution :

Answer:
Revaluation loss: ₹ 1,000; capital Account; Kannan: ₹ 1,04,500
Rahim: ₹ 82,700 Amount paid to John ₹ 41,800; Balance sheet total: ₹ 2,15,000

Question 18.
Saran, Arun, and Karan are partners in firms sharing profits and losses in the ratio of 4:3:3. The balance sheet as of 31.12.2016 was as follows:

Karan retires on 1.1.2017, subject to the following conditions:
(i) Goodwill of the firm is valued at ₹ 21,000 %
(ii) Machinery to be appreciated by 10%
(iii) Building to be valued at ₹ 80,000
(iv) provision for bad debts to be raised to ₹ 2,000
(v) Stock to be depreciated by ₹ 2,000
(vi) The final amount due to Karan is not paid immediately.
prepare the necessary ledger accounts and show the balance sheet of the firm after retirement. *
Solution:

Gaining ratio
Old ratio = 4:3:3
New ratio = 4:3
Gaining ratio = Saran = New ratio – old ratio

GR = 12:9 or 4:3
Karan’s share of G/w = 21,000 x \(\frac{3}{10}\) = Rs.6,300
K’s g/w In GR = Rs. 6,300
Saran’s share = 6300 x \(\frac{3}{10}\) = Rs.3,600
Aruns share = 6,300 x \(\frac{3}{7}\) = Rs.2,700
Answer:
Revaluation profit: ₹ 21,000; capital A/c: Saran: ₹ 70,800, Arun: ₹ 58,100
Karan’s loan Account ₹ 57,100; Balance sheet total: ₹ 2,21,000

Question 19.
Rajesh, Sathish and Mathan are partners sharing profits and losses in the ratio of 3:2:1. respectively. Their balance sheet as on 31.3.2017 is given below:

Mathan retires on 31st March, 2017 subject to the following conditions :
(i) Rajesh and Sathis will share profits and losses in the ratio of 3:2
(ii) Assets are to be revalued as follows;
Machinery ₹ 4,50,000, Stock ₹ 2,90,000 , Debtors ₹ 1,52,000
(iii) Goodwill of the firm is valued at ₹ 1,20,000
prepare the necessary ledger accounts and the balance sheet immediately after the retirement of Mathan.
Solution:

Gaining ratio = New ratio – old ratio
Rajesh = \(\frac{3}{5}-\frac{3}{6}=\frac{18-15}{30}=\frac{3}{30}\)
Sathish = \(\frac{2}{5}-\frac{2}{6}=\frac{12-10}{30}=\frac{2}{30}\)
GR = 3:2
Mathans share of G/w = 12,000 x \(\frac{1}{6}\) = Rs.20,000
In gaining ratio
Rajeshs share = 20,000 x \(\frac{3}{5}\) = Rs. 12,000
Sathishs share = 12,000 x \(\frac{2}{5}\) = Rs. 8,000

Answer:
Revaluation Profit: ₹ 12,000; Mathan loan: A/c: ₹ 2,92,000 Rajesh’s capital A/c ₹ 4,54,000; Sathish capital A/c: ₹ 3,36,000
Balance sheet total ₹ 13,10,000

Question 20.
Janani and Jamuna are partners sharing profits and losses in the ratio of 3:3:1. respectively. Janaki died on 31st December 2017. The final amount due to her showed a credit balance of ₹ 1,40,000. Pass Journal Entry if
(i) The amount due is paid off immediately.
(ii) The amount due is not paid immediately.
(iii) ₹ 75,000 is paid and the balance in the future.
Solution:

Question 21.
Varsha, Shanthi, and Madhuri are partners in a firm sharing profits in the ratio of 5:4:3. Their balance sheet as of 31st December 2017 is as under:

on 1st January 2018, Madhuri died and on her death, the following arrangements are made:
(i) Stock to be depreciated by ₹ 5,000
(ii) Premises is to be appreciated by 20%
(iii) To provide ₹ 4,000 for bad debts
(iv) The final amount due to Madhuri was not paid.
prepare revaluation account, partners capital account, and the balance sheet of the firm after death.
Solution:

Answer: Revaluation Profit ₹ 15,000;
Varsha’s capital A/c ₹ 1,01,250; Sathish capital A/c: ₹ 77,000
Madhuri’s executors account ₹ 32,750; Balance sheet total: ₹ 2,43,000

Question 22.
Vijayan, Sudhan, and Suman are partners who share profits and losses; in the capital ratio. Their balance sheet as of 31st December 2018 is as follows:

Suman died on 31.3.2019. On the death of Suman, the following adjustments are made:
(i) Building is to be valued at ₹ 1,00,000
(ii) Stock to be depreciated by ₹ 5,000
(iii) Goodwill of the firm is valued at ₹ 36,000
(iv) Share of profit from the closing of the last financial year to the date of death on the basis of the average of the three completed year’s profit before death.
profit for 2016, 2017 and 2018 were ₹ 40,000,₹ 50,000 and ₹ 30,000 respectively.
Prepare the necessary ledger accounts and the balance sheet immediately after the death of Suman.
Solution:

Suman share of G/w
Vijayan : Sudhan : suman
Old ratio = 7:5:3
New ratio = 7:5
Gaining ratio New ratio – old ratio

Answer:
Revaluation Profit: ₹ 15,000; Vijayan’s capital A/c: ₹ 81,200;
Sudhan’s capital A/c ₹ 58,000; Suman’s Executor’s A/c: ₹ 45,800
Balance sheet total: ₹ 2,02,000

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

I. Choose the best answer

Question 1.
In the absence of any specific agreement, between the partners, partners loan to the firms will carry an inters at the rate of …………… %
(a) 5%
(b) 6%
(c) 4%
Answer:
(b) 6%

Question 2.
A,B & C shares profit as 1/2 to A, 1/3 to B, and 1/6 to C. If B retries, then the new profit sharing ratio ………..
(a) 3:1
(b) 3:2
(c) 1:3
Hint:
Old Partner’s A, B, C
Old ratio = \(\frac{1}{2}: \frac{1}{3}: \frac{1}{6}\)
‘B’ Retire Partner
In order to equalise the denomination
\(\frac{3}{6}: \frac{2}{3}: \frac{1}{6}\)
Now profit sharing ration = \(\frac{3}{4}: \frac{1}{4}\). or 3 :1
Answer:
(a) 3:1

Question 3.
At the time of retirement, the revelation profits of the business will be shared by ……………… partners.
(a) all the partners
(b) Continuing partners
(c) Old partners
Answer:
(a) all the partners

Question 4.
At the time of retirement, of partners, the existing partners stand to
(a) gain
(b) loss
(c) no change
Answer:
(a) gain

Question 5.
A,B & C are sharing profits in the ratio of \(\frac{2}{5}\) , \(\frac{2}{5}\) \(\frac{1}{5}\) C retired from business and his share was purchased equally by A and B, Then new profit sharing ratio shall be

Hint:

Answer:
a

Question 6.
If the amount due to the outgoing partner is transferred to loan A/c then he is entitled to interest at ……………….. until it is paid out.
(a) 6%
(b) 5%
(c) 8%
Answer:
(a) 6%

Question 7.
At the time of retirement of a partner calculation of new profit, ratio is
(a) not necessary
(b) necessary
(c) optional
Answer:
(b) necessary

III Short Answer Questions

Question 1.
What are the journal entries to be passed to transfer the accumulated profits, losses & reserves?
Answer:

Question 2.
Write the format of Revaluation A/c.
Answer:
New profit sharing ratio is the agreed proportion in which future profit will be distributed to the continuing partners. If the new profit sharing ratio is not agreed, the continuing partners will share the profits and losses equally.

Question 3.
How will you adjust the share of profits loss of the retaining partner if he retires in between in an accounting year?
Answer:
When a partner retires in between in an accounting year, his share of the current year’s profit or loss upto the date of retirement has to be distributed to the retiring partner. It may be estimated based on the current year’s turnover. The previous year’s profit or the average of the past year’s profit may also be taken as the base to estimate the current year’s profit. The following journal entry is passed

Question 4.
What are the adjustments to be made on the death of a partner?
Answer:
The following adjustments are made on the death of a partner.

• Distribution of accumulated profits, reserves, and losses
• Revaluation of assets and liabilities
• Determination of new profit sharing ratio and gaining ratio
• Adjustment for goodwill
• Adjustment for current year’s profit or loss upto the date of death
• Settlement of the amount due to the deceased partner.

Question 5.
What are the journal entries to be passed for settlement of the amount due to the deceased partner
Answer:
To transfer the amount due to the deceased partner to the executor or legal representative of the deceased partner.

IV Additional Problems:

Question 1.
A, B, and C sharing profits in the ratio of 5:3:2. C retires. Find out the new profit sharing ratio and gaining ratio.
Solution:

Question 2.
G, H, and I are partners sharing profits in the ratio of 5:3:2.1 taken up by G and H equally. Find out the new profit sharing ratio of G and H.
Solution:

Question 3.
J, K and L are partners sharing profits in the ratio of 5:3:2. retires and his share was taken up entirely by K. Find out the new profit sharing ratio and gaining ratio of continuing partners
Solution:

Question 4.
X,Y and Z are partners sharing profits in- the ratio of 5:3:2 Z retires and the ratio between X and Y is 3:2. Find out the gaining ratio
Solution :

Question 5.
Prabha,Kavitha and Meena were partners of a firm sharing profit and loss in the ratio of 3:2:1 Meena Wanted to retire. They decided to revalue the assets and liabilities of the firm as indicated below:
To write down Machinery by Rs 10,000 and Stock by Rs. 4,000
To bring into books as unrecorded investments Rs. 5,000
To Write off Rs.3,000 from sundry creditors as it was no longer liable.
Pass entries to give effect to the above adjustments. Show also Revaluation account.
Solution:

Question 6.
Banumathi, Bharafhi and Shanthi are partners sharing, profits in the ratio of 5:3:2. On April 1, 2005 Shanthi decided to retire. On that date, there was a credit balance of Rs. 60,000 in their profit and loss account. Pass entry.
Solution:

Question 7.
Thangamuthu, Anaimuthu and Vairamuthu are partners sharing profit and loss in the ratio of 3:3:2. Thangamuthu wanted to retire on 1st June 2005, the firms books showed a general reserve of Rs. 40,000. Pass entry.
Solution:

Question 8.
Lalitha, Jothi and Kanaga were partners of a sharing profit and losses in the ratio of 3:2:3, Set out below was their balance sheet as on 31st March 2003.

Lalitha retired from the partnership on 1st April 2004 on the following terms:
Goodwill of the firm was to be valued at Rs.30,000
The assets are to be valued as under Stock Rs. 1,00,000; Furniture Rs.15,000; Plant and Machinery Rs.45,000; Building Rs.1,00,000.
A provision for doubtful debts is created at Rs.4,250.
Lalitha was to be paid off immediately.
Show the journal entries, prepare a revaluation account, capital account, Bank account, and balance sheet of the reconstituted partnership.
Solution:

Question 9.
‘9. A, B and C are partners sharing profits and losses in the ration of 5:3:2 respectively. Retries from the firm on 1st April 2005. After his retirement, his capital account shows a credit balance of Rs.1,35,000 after the necessary adjustment made. Give journal entries, if.
the amount due is paid off immediately.
When the amount due is not paid immediately.
Rs. 45,000 is paid and the balance In the future.
Solution :

Question 10.
Pallavan, Pandian and Chozhan were carrying on partnership business sharing profits in the ratio of 3:2:1. On March 31, 2005, the balance sheet of the firm stood as follows:

Chozhan retried on April 1, 2005 on the following terms:
Building to be appreciated by Rs. 15,000
Provision for doubtful debts to be made at 6% on debtors
Goodwill of the firm is valued at Rs. 18,000
Rs.50,000 to be paid to chozhan immediately and the balance transferred to his loan account.
Prepare Revaluation Account, Capital Accounts, Bank Account and the Balance Sheet after Chozhan’s retirement.
Solution :

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.9 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.9

Question 1.
Find by integration, the volume of the solid generated by revolving about the x axis, the region enclosed by y = 2x², y = 0 and x = 1
Solution:
The region to be revolved is sketched

Since revolution is made about the x axis, the volume of the solid generated is given by

Required volume = \(\frac { 4π }{ 5 }\) cubic units

Question 2.
Find, by integration, the volume of the solid generated by revolving about the x axis, the region enclosed by y = e^-2x, y = 0, x = 0 and x = 1.
Solution:
Since revolution is made about the x axis, the volume of the solid generated is given

Required volume = \(\frac { π }{ 4 }\) [1 – e^-4] cubic units

Question 3.
Find, by integration, the volume of the solid generated by revolving about the y axis, the region enclosed by x² = 1 + y and y = 3.
Solution:
The region to be revolved is sketched.

Since revolution is made about the y axis, the volume of the solid generated is given by

Required volume = 8π cubic units

Question 4.
The region enclosed between the graphs of y = x and y = x² is denoted by R. Find the volume generated when R is rotated through 360° about x axis.
Solution:
The region to be revolved is sketched.

Find the intersecting point of y = x and y = x²
x² = x
x² – x = 0
x (x – 1) = 0 x = 0, x = 1
If x = 0, y = 0, x = 1, y = 1
∴ Points of intersection are (0, 0), (1, 1)

Required volume = \(\frac { 2π }{ 15 }\) cubic units

Question 5.
Find, by integration, the volume of the container which is in the shape of a right circular conical frustum as shown to figure.

Solution:
By using integration we have to find the volume of the frustum. So first find the equation of the curve.
Let A(0, 1) and B(2, 2) be two points. Line joining these two points form a straight line. That straight line revolves around x axis.

Volume of the solid revolves around x axis

Volume of the frustum = \(\frac { 14 }{ 3 }\) π m³

Question 6.
A watermelon has an ellipsoid shape which can be obtained by revolving an ellipse with major axis 20 cm and minor axis 10 cm about its major axis. Find its volume using integration.
Solution:
A watermelon has an ellipsoid shape.
2a = 20
a = 10
2b = 10
b = 5

∴ Volume of the frustum = \(\frac { 1000π }{ 3 }\) cubic units.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.8 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.8

Question 1.
Find the area of the region bounded by 3x – 2y + 6 = 0, x = -3, x = 1 and x axis.
Solution:
3x – 2y + 6 = 0
2y = 3x + 6
y = \(\frac { 1 }{ 2 }\)(3x + 6)

Question 2.
Find the area of the region bounded by 2x – y + 1 = 0, y = -1, y = 3 and y axis.
Solution:
Given straight line is 2x – y + 1 = 0
y = 2x + 1, x = \(\frac { y-1 }{ 2 }\)

= 1 + 1
= 2 sq. units
Area required = 2 sq. units

Question 3.
Find the area of the region bounded by the curve 2 + x – x² + y = 0, x axis, x = -3 and x = 3
Solution:
Given curve is
2 + x – x² + y = 0
y = x² – x – 2
y = (x – 2)(x + 1)

Question 4.
Find the area of the region bounded by the line y = 2x + 5 and the parabola y = x² – 2x.
Solution:
First, we find the point of intersection of
y = 2x + 5 and y = x² – 2x
x² – 2x = 2x + 5
x² – 4x – 5 = 0
(x – 5) (x + 1) = 0
x = 5, x = – 1
when x = 5, y = 15
x = -1. y = 3
(5, 15) (-1, 3) are intersecting points.

54 – 18 = 36
Area required = 36 sq. units

Question 5.
Find the area of the region bounded between the curves y = sin x and y = cos x and the lines x = 0 and x = π
Solution:
First find the intersecting point of two curves
sin x = cos x
tan x = 1
x = \(\frac { π }{ 4 }\)

Area required = 2√2 sq. units

Question 6.
Find the area of the region bounded by y = tan x, y = cot x and the lines x = 0, x = \(\frac { π }{ 2}\), y = 0.
Solution:
First find the intersecting point of y = tan x and y = cot x
tan x = cot x
\(\frac { tan x }{ cot x }\) = 1
tan²x = 1
tan x = 1

= log √2 + log √2
= 2 log √2
= log(√2)²
= log 2 sq. units

Question 7.
Find the area of the region bounded by the parabola y² = x and the line y = x – 2.
Solution:
First find the intersecting point of y² = x and y = x – 2
y = y² – 2
y² – y – 2 = 0
y = 2, y = -1
Intersecting points are (4, 2), (1, -1)

Required Area = \(\frac { 9 }{ 2 }\) sq. units

Question 8.
Father of a family wishes to divide his square field bounded by x = 0, x = 4, y = 4 and y = 0 along the curve y² = 4x and x² = 4y into three equal parts for his wife, daughter and son. Is it possible to divide? If so, find the area to be divided among them.
Solution:
Given curve y² = 4x and x² = 4y
Draw these two curves
Also draw the square bounded by the lines
x = 0, x = 4, y = 4 and y = 0
To prove Area A₁ = Area A₂ = Area A₃
Now the point of intersection of the curves y² = 4x and x² = 4y is given by
(\(\frac { y^2 }{ 4 }\))² = 4y
y⁴ = 64y ⇒ y (y³ – 64) = 0
y = 0, y = 4
when y = 0 ⇒ x = 0
y = 4 ⇒ x = 4
Point of intersection are O (0, 0) and B (4, 4)
Now, the area of the region bounded by the curves y² = 4x and x² = 4y is
A₂ = \(\int_{0}^{4}\)(\(\sqrt { 4x }\) – \(\frac { x^2 }{ 4 }\)) dx

Now the area of the region bounded by the curves x² = 4y, x = 4 and x axis is

Similarly the area of the region bounded by the curve y² = 4x, y axis and y = 4 is

Hence we see that
A₁ = A₂ = A₃ = \(\frac { 16 }{ 3 }\) sq. units

Question 9.
The curve y = (x – 2)² + 1 has a minimum point at P. A point Q on the curve is such that the slope of PQ is 2. Find the area bounded by the curve and the chord PQ.
Solution:
Given curve is y = (x – 2)² + 1
(i.e) (y – 1) = (x – 2)²
Vertex of the parabola is (2, 1)
Minimum point P is (2, 1)
Slope of PQ is 2.

Equation of PQ is y – 1 = 2 (x – 2)
y – 1 = 2x – 4
y = 2x – 3
Intersecting point of y = 2x – 3 and y = (x – 2)² + 1
2x – 3 = (x – 2)² + 1
2x – 4 = (x – 2)²
2(x – 2) = (x – 2)²
x – 2 = 2
x = 4
when x = 4, y = 5

Required Area = \(\frac { 4 }{ 3 }\) sq. units

Question 10.
Find the area of the region common to the circle x² + y² = 16 and the parabola y² = 6x
Solution:
First find the intersecting point of the curves
x² + y² = 16 and y² = 6x
x² + 6x = 16
x² + 6x – 16 = 0
(x + 8) (x – 2) = 0
x = -8, x = 2
x = -8 is impossible
x = 2, y = 2√3
Radius of the circle x² + y² = 16 is 4

Area OABC = 2 (Area of OAB)
= 2 (Area of the curve y² = 6x in [0, 2] + Area of the curve x² + y² = 16 in [2, 4])

Required Area = \(\frac { 4 }{ 3 }\) [4π + √3] sq. units

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 4 Transition and Inner Transition Elements Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 4 Transition and Inner Transition Elements

12th Chemistry Guide Transition and Inner Transition Elements Text Book Questions and Answers

Part – I – Text Book Evaluation

I. Choose the correct answer

1. Sc (Z = 21) is a transition element but Zinc (Z = 30) is not because
a) both Sc³⁺ and Zn²⁺ ions are colourless and form white compounds
b) in case of Sc, 3d orbital are partially filled but in Zn these are completely filled
c) last electron as assumed to be added to 4s level in case of zinc
d) both Sc and Zn do not exhibit variable oxidation states
Answer:
b) in case of Sc, 3d orbital are partially filled but in Zn these are completely filled

2. Which of the following d block element has half-filled penultimate d sub shell as well as half-filled valence sub shell?
a) Cr
b) Pd
c) Pt
d) none of these
Answer:
a) Cr

3. Among the transition metals of 3d series, the one that has highest negative \(\left(\mathrm{M}^{2+} / \mathrm{M}\right)\) standard electrode potential is
a) Ti
b) Cu
c) Mn
d) Zn
Answer:
a) Ti

4. Which one of the following ions has the same number of unpaired electrons as present in V³⁺?
a) Ti³⁺
b) Fe³⁺
c) Ni²⁺
d) Cr³⁺
Answer:
c) Ni²⁺

5. The magnetic moment of Mn²⁺ ion is
a) 5.92BM
b) 2.80BM
c) 8.95BM
d) 3.90BM
Answer:
a) 5.92BM

6. The catalytic behaviour of transition metals and their compounds is ascribed mainly due to
a) their magnetic behaviour
b) their unfilled d orbitals
c) their ability to adopt variable oxidation states
d) their chemical reactivity
Answer:
c) their ability to adopt variable oxidation states

7. The correct order of increasing oxidizing power in the series

Answer:
a)VO₂⁺ < Cr₂O₇^2- < MnO₄^–

8. In acid medium, potassium permanganate oxidizes oxalic acid to
a) oxalate
b) Carbon dioxide
c) acetate
d) acetic acid
Answer:
b) Carbon dioxide

9. Which of the following statements is not true?
a) on passing H₂S, through acidified K₂Cr₂O₇ solution, a milky colour is observed
b) Na₂Cr₂O₇ is preferred over K₂Cr₂O₇ in volumetric analysis
c) K₂Cr₂O₇ solution in acidic medium is orange in colour
d) K₂Cr₂O₇ solution becomes yellow on increasing the pH beyond 7
Answer:
b) Na₂Cr₂O₇ is preferred over K₂Cr₂O₇ in volumetric analysis

10. Permanganate ion changes to ________ in acidic medium
a) MnO₄^2-
b) Mn²⁺
c) Mn³⁺
d) MnO₂
Answer:
b) Mn²⁺

11. How many moles of I₂ are liberated when 1 mole of potassium dichromate react with potassium iodide?
a) 1
b) 2
c) 3
d) 4
Answer:
c) 3

12. The number of moles of acidified KMnO₄ required to oxidize 1 mole of ferrous oxalate (FeC₂O₄) is
a) 5
b) 3
c) 0.6
d) 1.5
Answer:
c) 0.6

13. Which one of the following statements related to lanthanons is incorrect?
a) Europium shows +2 oxidation state
b) The basicity decreases as the ionic radius decreases from Pr to Lu.
c) All the lanthanons are much more reactive than aluminium.
d) Ce⁴⁺ solutions are widely used as oxidising agents in volumetric analysis.
Answer:
c) All the lanthanons are much more reactive than aluminium.

14. Which of the following lanthanoid ions is diamagnetic?
a) Eu²⁺
b) Yb²⁺
c) Ce²⁺
d) Sm²⁺
Answer:
b) Yb²⁺

15. Which of the following oxidation states is most common among the lanthanoids?
a) 4
b) 2
c) 5
d) 3
Answer:
d) 3

16. Assertion : Ce⁴⁺ is used as an oxidizing agent in volumetric analysis
Reason : Ce⁴⁺ has the tendency of attaining +3 oxidation state.
a) Both assertion and reason are true and reason is the correct explanation of assertion.
b) Both assertion and reason are true but reason is not the correct explanation of assertion.
c) Assertion is true but reason is false
d) Both assertion and reason are false.
Answer:
a) Both assertion and reason are true but reason is the correct explanation of assertion.

17. The most common oxidation state of actinoids is (PTA -2)
a) +2
b) +3
c) +4
d) +6
Answer:
b) +3

18. The actinoid elements which show the highest oxidation state of +7 are
a) Np, Pu, Am
b) Li, Fm, Th
c) U, Th, Md
d) Es, No, Lr
Ans : a) Np, Pu, Am

19. Which one of the following is not correct? (PTA -5)
a) La (OH)₃, is less basic than Lu (OH)₃
b) In lanthanoid series ionic radius of Ln³⁺ ions decreases
c) La is actually an element of transition metal series rather than lanthanide series
d) Atomic radii of Zr and Hf are same because of lanthanide contraction,
Answer:
a) La(OH)₃, is less basic than Lu(OH)₃

II. Answer the following questions

Question 1.
What are transition metals? Give four examples (PTA – 3)
Answer:

1. Transition metal is an element whose atom has an incomplete d sub shell or which can give rise to cations with an incomplete d sub shell.
2. They occupy the central position of the periodic table, between s-block and p-block elements.
3. Their properties are transitional between highly reactive metals of s-block and elements of p-block which are mostly non-metals.
4. Example – Iron, Copper, Tungsten, Titanium.

Question 2.
Explain the oxidation states of 4d series elements.
Answer:

• The oxidation states of 4d series elements vary from +3 for Y to +8 for Ru.
• The highest oxidation state of 4d elements are found in their compounds with higher electronegative elements like O, F and Cl.
• Ex: In RuO₄, the oxidation state of Ru is +8.

Question 3.
What are inner transition elements?
Answer:

1. The inner transition elements have two series of elements.

• Lanthanoids
• Actinoids

2. Lanthanoid series consists of 14 elements from Cerium (₅₈Ce) to Lutetium (₇₁Lu) following Lanthanum (₅₇La). These elements are characterised by the preferential filling of 4f orbitals. .

3. Actinoids consists of 14 elements from Thorium (₉₀Th) to Lawrencium (₁₀₃Lr) following Actinium (₈₉Ac). These elements are characterized by the preferential filling of 5f orbital.

The elements which in their elemental or ionic form have partly filled f orbitals are called f block elements.
As the f orbitals lie inner to the penultimate shell, therefore these elements having partially filled f orbitals, are also called inner transition elements.

Question 4.
Justify the position of lanthanides and actinides in the periodic table. (PTA – 1)
Answer:
Lanthanides:

• The actual position of lanthanides in the periodic table is at group number 3 and period number 6.
• In the sixth period, the electrons are preferentially filled in inner 4f-sub shell.
• The fourteen elements following lanthanum (Ce to Lu) show similar chemical properties.
• Hence they are grouped together and placed at the bottom of the periodic table.
• This position is justified as follows:
  (1) General electronic configuration:
  [ Xe] 4f^1-145d^0,16S²
  (2) Common Oxidation state: +3
  (3) They have similar physical and chemical properties.

Actinides:

• The actual position of actinides in the periodic table is at group number 3 and period number 7.
• In the seventh period, the electrons are perferentially filled in inner 5f-Sub shell.
• The fourteen elements following actinides (Th to Lr) show similar chemical properties.
• Hence they are grouped together and placed at the bottom of the periodic table.
• This position is justified as follows:
  (1) General electronic configuration:
  [Rn] 5f^2-146d^0-27S²
  (2) Common oxidation state: +3
  (3) They have similar physical and chemical properties.

Question 5.
What are actinides? Give three examples.
Answer:
The fourteen elements following actinium is from thorium to lawrencium are called actinoids.
Examples: Thorium, uranium, Neptunium.

Question 6.
Describe the preparation of potassium dichromate. (PTA – 2)
Answer:

1. The fourteen elements following actinium ,i.e., from thorium (Th) to lawrentium (Lr) are called actinoids.
2. All the actinoids are radioactive and most of them have short half lives.
3. The heavier members being extremely unstable and not of natural occurrence. They are produced synthetically by the artificial transformation of naturally occuring elements by nuclear reactions.
4. Example – Thorium, Uranium, Plutonium, Californium.

Question 7.
What is lanthanide contraction and what are the effects of lanthanide contraction? (PTA – 3)
Answer:
As we move across 4f series, the atomic and ionic radii of lanthanoids show gradual decrease with increase in atomic number. This decrease in ionic size is called lanthanide contraction.

Consequences:
Basicity differences:
As we from Ce³⁺ to Lu³⁺, basic character of LU³⁺ ions decrease. Due to the decrease in the size of LU³⁺ ions, the ionic character of Ln-OH bond decreases (covalent character increases)

Similarities among lanthanoids:
In the complete f-series very small change in radii of Lanthanoids are observed and their chemical properties a requite similar.

Second and third transition series resemble each other due to lanthanide contraction.

Question 8.
Complete the following.
Answer:

Question 9.
What are interstitial compounds?
Answer:
An interstitial is a compound that is formed when small atoms like hydrogen, boron, carbon or nitrogen are trapped in the interstitial holes in a metal lattice. They are non-stoichiometric compounds.
Ex: Tic, ZrH_1.92

Question 10.
Calculate the number of unpaired electrons in Ti³⁺, Mn^2+, and calculate the spin-only magnetic moment.
Answer:
Electronic configuration of Ti³⁺ = [Ar]3d¹
It has only one unpaired electron, ie. n = 1
Spin only magnetic moment
\(=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{1(1+2)}=\sqrt{3}\) = 1.732BM

Electronic configuration of Mn²⁺ = [Ar] 3d⁵
It has five unpaired electrons ie. n = 5
Spin only magnetic moment
\(=\sqrt{n(n+2)}=\sqrt{5(5+2)}=\sqrt{35}\) = 5.92BM

Question 11.
Write the electronic configuration of Ce⁴⁺ and Co²⁺
Answer:
Electronic cofiguration of Ce⁴⁺ = [Xe] 4f⁰5d⁰6s⁰
Electronic configuration of Co²⁺ = [Ar] 3d⁷.

Question 12.
Explain briefly +2 states become more and more stable in the first half of the first-row transition elements with increasing atomic number.
Answer:
In the elements of the first half of the first row, with the removal of valance 4s electrons (+2 oxidation state) the 3d orbital get gradually occupied. Since the number of empty d orbital decreases, the stability of the cations (M²⁺) increases from Sc²⁺ to Mn²⁺.

Question 13.
Which is more stable? Fe³⁺ or Fe²⁺ – explain.
Answer:

• Electronic configuration of Fe³⁺ = [Ar] 3d⁵. It consists of 5 unpaired electrons. Half filled and stable.
• Electronic configuration of Fe²⁺ = [Ar]3d⁶. It consists of 4 unpaired electrons, partially filled d subshell.
• Hence Fe³⁺ is more stable than Fe²⁺.

Question 14.
Explain the variation in E°_{M³⁺/M²⁺} 3d series.
Answer:
Transition metals in their high oxidation states tend to be oxidising. The standard reduction potential for the M³⁺/M²⁺ half cell gives the relative stability between M³⁺ and M²⁺.

The negative values for titanium, vanadium and chromium indicate that the higher oxidation state is preferred.

The high reduction potential of M³⁺/M²⁺ indicates Mn²⁺ is more stable than Mn³⁺

Question 15.
Compare lanthanides and actinides.
Answer:

Question 16.
Explain why Cr²⁺ is slrongly reducing while Mn³⁺ is strongly oxidizing. (PTA – 5)
Answer:
E° value for Cr³⁺/Cr²⁺ is negative (-0.41 V), where as E° Value for Mn³⁺/Mn²⁺ is positive (+1.57V).

Hence Cr²⁺ ions can easily undergo oxidation to give Cr³⁺ ions and therefore, acts as strong reducing agent. On the otherhand, Mn³⁺ can easily undergo reduction to give Mn²⁺ and hence acts as oxidising agent.

Question 17.
Compare the ionization enthalpies of first series of the transition elements.
Answer:

• Ionisation energy of transition element is intermediate between those of s and p block elements.
• As we move from left to right in a transition metal series, the ionisation enthalpy increase.
• This is due to the increase in nuclear charge corresponding to the filling of d electrons.
• The following figure shows the trends in ionisation enthaply of first series of transiton elements.

Question 18.
Actinoid contraction is greater from element to element than the lanthanoid contraction why?
Answer:
This is due to poor shielding by 5f electrons in actinoids as compared to that by 4f electrons in lanthanoids.

Question 19.
Out of Lu(OH)₃ and La(OH)₃ which is more basic and why?
Answer:

Question 20.
Why europium (II) is more stable than cerium (II)
Answer:

Question 21.
Why do zirconium and Hafnium exhibit similar properties?
Answer:
This is because Zr and Hf have similar atomic sizes which is due to lanthanoid contraction.

Question 22.
Which is stronger reducing agent Cr²⁺ or Fe²⁺?
Answer:
Cr²⁺ is a stronger reducing agent than Fe.
Reason :
E°(Cr³⁺/Cr²⁺) is negative (-0.41 V) where as E°(Fe³⁺/Fe²⁺) is positive (+0.77V). Thus Cr³⁺ is easily oxidised to Cr³⁺ but Fe²⁺ cannot be easily oxidised to Fe³⁺. Hence Cr²⁺ is a stronger reducing agent than Fe²⁺.

Question 23.
The E°_M⁺/M value for copper is positive. Suggest a possible reason for this.
Answer:
E°_M⁺/M for any metal depends upon the sum of the enthalpy changes taking place in the following steps.
M_(s) + ∆H_(a) → M_(g) (∆H_(a) = enthalpy of atomisation)
M_(g) + ∆H_(i) → M²⁺_(g) (∆H_(i) = ionisation enthalpy)
M²⁺_(g) + ∆_aq → M²⁺_(aq) (∆H_(Hyd) = hydration enthalpy)

Coppes possesses a high enthalpy of atomisation and low enthalpy of hydration. Hence E° Cu²⁺/ Cu is positive.

Question 24.
Describe the variable oxidation on state of 3d series elements.
Answer:

• First transition metal Sc exhibits only +3 oxidation state
• All other transition elements exhibit variable oxidation states by losing electrons from (n-l)d orbital and ns orbital
• Because the energy difference between these two orbitals is very small.

(ex)

• At the beginning of the series, +3 oxidation is stable.
• Number of oxidation states increases with the number of electrons available.
• Number of oxidation states decreases as the number of paired electrons increases.
• Hence the first and last elements show less number of oxidation states and the middle elements show more number of oxidation states.
• (Ex): First element Sc has only one oxidation state +3. Middle elements Mn has Six different oxidation states from +2 to +7. Last element Cu has +1 and +2 oxidation states.
• Relative stability of different oxidation states of 3d-metals is correlated with the extra stability of half filled and fully filled electronic configurations.
• (Ex): Mn²⁺(3d⁵) is more stable than Mn⁴⁺(3d³)

Question 25.
Which metal in the 3d series exhibits +1 oxidation state most frequently and why?
Answer:
Copper has electronic configuration 3d¹⁰4s¹. It can easily lose 4s¹ electron to give the stable 3d¹⁰ configuration. Hence, it shows +1 oxidation state.

Question 26.
Why first ionization enthalpy of chromium is lower than that of Zinc.
Answer:

Question 27.
Transition metals show high melting points, why? (PTA -6)
Answer:
Transition metals show high melting points due to the presence of unpaired d-electrons available for metallic bonding.

III. Evaluate Yourself

Question 1.
Compare the stability of Ni⁴⁺ and Pt⁴⁺ from their ions enthalpy value.
Answer:

• The value of the ionisation enthalpies can be used is estimating the relative stability.
• Pl⁴⁺ relative to Ni⁴⁺ can be explained as follows:
  Ni⁴⁺ – IE₃ +IE₄ = (3395 +5297) = 8692 KJmol^-1
  Pt⁴⁺ – IE₃ +IE₄ = (2800 +4150) = 6950 KJmol^-1
• Pt⁴⁺ requires lesser energy as compared to the formation of Ni⁴⁺
• Therefore Pt⁴⁺ compounds are more stable than Ni⁴⁺ compounds.

Question 2.
Why ion is more stable in +3 oxidation state than in +2 and the reverse is true for manganese?
Answer:
(i) Atomic number of Fe=26
(ii) Electric configuration of Fe = [Ar] 3d⁶4S²
(iii) After removal of two electrons [Fe²⁺]=[Ar]3d⁶
(iv) Further, with removal of one more electron it becomes Fe³⁺ and d⁵ (half filled) that is most stable.
(v) Atomic number of Mn=25
(vi) Electronic configuration of Mn= [Ar] 3d⁵4S²
(vii) After removal of two outer electrons it will become Mn²⁺ and remaining d will be half filled d⁵ and that is most stable.
(viii) Removal of one electron Mn³⁺ having d⁴ configuration which is incompletely filled.

12th Chemistry Guide Transition and Inner Transition Elements Additional Questions and Answers

Part – II – Additional Questions

I. Choose the correct answer

1. Transition elements are good conductors because
a) They are metals
b) They are all solids
c) They have free electrons in outer energy orbits
d) All of these
Answer:
d) All of these

2. Transition elements are
a) All metals
b) All non metals
c) Metals and non metals
d) Gases
Answer:
a) All metals

3. Transition elements form complexes very readily because of
a) Small cation size
b) Vacant – d – orbitals
c) Large ionic charge
d) All are correct
Answer:
d) All are correct

4. The transition metal present in vitamin B12 is
a) Fe
b) Co
c) Ni
d) Na
Answer:
b) Co

5. Transition elements are frequently used as catalysts, because of
a) Large ionic charge
b) Large surface area for the reactions to be absorbed
c) Unpaired d electrons
d) Both (b) and (c) are correct
Answer:
d) Both (b) and (c) are correct

6. d – Block elements are arranged in
a) Three series
b) Six series
c) Two series
d) Four series
Ans : d) Four series

7. d – Block element generally for
a) Covalent hydrides
b) Metallic hydrides
c) Interstitial hydrides
d) Salt like hydrides
Answer:
c) Interstitial hydrides

8. Metals which are hard and lustrous substances with high melting points form highly coloured compounds are known as
a) Alkali metals
b) Alkaline earth metals
c) Transition metals
d) None of these
Answer:
c) Transition metals

9. In the first transition series, the incoming electrons enters
a) 5 d orbitals
b) 4 d orbitals
c) 3 d orbitals
d) 2 d orbitals
Answer:
c) 3 d orbitals

10. Transition elements form alloys easily because they have
a) Same atomic number
b) Same electronic configuration
c) Nearly same atomic size
d) None
Answer:
c) Nearly same atomic size

11. Transition elements that show anaomalous electronic configuration in first series are.
a) Cr and Ni
b) Cu and Co
c) Fe and Ni
d) Cr and Cu
Answer:
d) Cr and Cu

12. Lightest transition element is
a) Fe
b) Sc
c) Os
d) None
Answer:
b) Sc

13. Densest transition element is
a) Fe
b) Sc
c) Os
d) Mn
Answer:
b) Sc

14. Variable valencies of transition elements is due to
a) Different energies of (n -1) d electrons
b) Different energies of ns electrons
c) Similar energies of (n – 1)d electrons
d) Similar energies of (n – 1)d and ns electrons
Answer:
d) Similar energies of (n-l)d and ns electrons

15. Which of the following ions has minimum ionic radius
a) Ni²⁺
b) Co²⁺
c) Cr²⁺
d) V²⁺
Answer:
a) Ni²⁺

16. Ionic radii of zirconium and hofminium become almost identical because
a) They are d block elements
b) They belong to the same group
c) Of increased nuclear charge
d) Of lanthanoids contraction
Answer:
d) Of lanthanoids contraction

17. The colour of Fe ions is
a) Blue
b) Light green
c) Very dark green
d) Yellow
Answer:
b) Light green

18. Magnetic property of transition metal is due to
a) Spin of electron
b) Orbital moment
c) Both
d) Neither of the two
Answer:
a) Spin of electron

19. All ferromagnetic substances
a) Can be magnetised
b) Can be electrolysed
c) Have completely filled d – orbitals
d) None
Answer:
a) Can be magnetised

20. Maxmium paramagnetism in 3d series shown by
a) Mn
b) Co
c) Ni
d) Fe
Answer:
a) Mn

21. Which of the following has the minimum magnetic moment
a) Mn²⁺
b) Fe²⁺
c) Cr²⁺
d) V³⁺
Answer:
d) V³⁺

22. Paramagnetism in the substance increases as
a) The number of paired electrons increases
b) The number of unpaired electrons increases
c) The number of unpaired electrons decreases
d) The number of paired electrons decreases
Answer:
b) The number of unpaired electrons increases

23. The number of unpaired electrons present in Cr³⁺ ion is
a) 1
b) 2
c) 3
d) 4
Answer:
c) 3

24. Which of the following has more unpaired d electrons
a) Zn²⁺
b) Fe²⁺
c) N^3-
d) Cu⁺
Answer:
b) Fe²⁺

25. The lanthanoids contraction is responsible for the fact that
a) Zr and Y have the same radius
b) Zr and Nb have similar oxidation state
c) Zr and Hr have almost the same radius
d) Zr and Zn have same oxidation state
Answer:
c) Zr and Hr have almost the same radius

26. Which of the following is coloured
a) Cu⁺
b) Cu²⁺
c) Ti⁴⁺
d) V⁵⁺
Answer:
Cu⁺

27. In the four successive transition elements (Cr, Mn, Fe & Co) the stability of +2 oxidation state will be are in which of the following order? (PTA – 5)
a) Fe>Mn>Co>Cr
b) Co>Mn>Fe>Cr
c) Cr>Mn>Co>Fe
d) Mn>Fe>Cr>Co
Answer:
d) Mn>Fe>Cr>Co

28. Chromyl chloride when dissolved in NaOH solution gives yellow solution. The yellow solution contains (PTA – 3)
a) Cr₂O₇^2-
b) CrO₄^2-
c) CrO₅
d) Cr₂O₃
Answer:
b) CrO₄^2-

29. In the dichromate anion (Cr₂O₇)^2- (PTA – 1)
a) 4Cr – O bonds are equivalent
b) 6 Cr – O bonds are equivalent
c) All Cr – O bonds are equivalent
d) All Cr-O bonds are non-equivalent
Answer:
b) 6 Cr – O bonds are equivalent

30. Which of the following forms colourless compound?
a) Sc³⁺
b) V³⁺
c) Ti³⁺
d) Cr³⁺
Answer:
a) Sc³⁺

31. The basic character of the transition metal monoxides follows the order (Atomic number, Ti = 22; V = 23; Cr = 24)
a) VO > CrO > TiO > FeO
b) Cro > VO > FeO > TiO
c) TiO > FeO > VO > CrO
d) TiO > VO > CrO > FeO
Answer:
d) TiO > VO > CrO > FeO

32. The correct order of ionic radii of Y³⁺, La³⁺, EU³⁺, and Lu³⁺ is (Atomic number Y = 39; La = 57; Eu = 60; Lu = 71)
a) Y³⁺ < La³⁺ < Eu³⁺ < Lu³⁺
b) Y³⁺ < Lu³⁺ < Eu³⁺ < La³⁺
c) Lu³⁺ < Eu³⁺ < La³⁺ < Y³⁺
d) La³⁺ < Eu³⁺ < Lu³⁺ < Y³⁺
Answer:
b) Y³⁺ < Lu³⁺ < Eu³⁺ < La³⁺

33. During the smelting process silica is added to roasted copper to remove
a) Copper sulphide
b) Ferrous sulphide
c) Ferrous oxide
d) Ferrous Chloride
Answer:
c) Ferrous oxide

34. The ore which contains copper and iron both
a) Malachite
b) Chalcopyrite
c) Chalocite
d) Azurite
Answer:
b) Chalcopyrite

35. According to Ellingham diagram, the oxidation reaction of carbon to carbon monoxide may be used to reduce which one of the following oxides at lowest temperature?
a) Al₂O₃
b) Cu₂O
c) MgO
d) ZnO
Answer:
a) Al₂O₃

36. In electro chemical process (Electrolysis of fused salt) is used to extract
a) Iron
b) Lead
c) Sodium
d) Silver
Answer:
c) Sodium

37. In metallurgy, flux is a substance used to convert
a) Mineral into silicate
b) Fusible impurities to soluble impurities
c) Infusible impurities to soluble impurities
d) none of these
Answer:
c) Influsible impurities to soluble impurities

38. Heating of Iron pyrites in air to remove sulphur is called
a) Fusion
b) Calcination
c) Smelting
d) Roasting
Answer:
d) Roasting

39. Which of the following metal is leached by cyanide process?
a) Silver
b) Sodium
c) Aluminium
d) Copper
Answer:
a) Silver

40. The elements in which extra electrons enter into (n – 2) orbitals are called
a) f – block elements
b) p – block elements
c) d – block elements
d) s – block elements
Answer:
a) f – block elements

41. The most common oxidation state of lanthanoides
a) +4
b) +2
c) +6
d) +3
Answer:
d) +3

42. ………………. form complexes.
a) lanthanoides
b) actinides
c) Thorium
d) Cerium
Answer:
b) Actnides

43. ……………. form oxocations.
a) Actnides
b) lanthanoides
c) s – block
d) p – block
Answer:
a) Actnides

44. The correct electronic configuration of Gd³⁺ is
a) [Xe]4f¹⁴
b) [Xe]4f⁷
c) [Xe]4f⁰
d) [Xe]4f⁷
Answer:
b) [Xe]4f⁷

45. The general electronic configuration of lanthanoides
a) [Xe]4f^0-14
b) [Xe]5d^0-1
c) [Xe]4f^2-14 5d^0-1 bs²
d) [Xe]4f^0-14 5d^1-10 6s²
Answer:
c) [Xe]4f^2-14 5d^0-1 bs²

46. As we move from lanthanum to Lutetium, basic character of hydroxides.
a) decreases
b) increases
c) decrease and then increases
d) none of these
Answer:
a) decreases

47. The colour of U³⁺ is
a) red
b) green
c) yellow
d) pink
Answer:
a) red

II. Assertion and Reason

Question 1.
Assertion : For chromium, the ground state electronic configuration is 3d⁵4s¹ rather than 3d⁴4s².
Reason : The energy of the Cr atom is lower, when the six valence electrons are in different atomic orbitals with parallel spins
A) Both Assertion and Reason are true and Reason is the correct explanation of A.
B) Both Assertion and Reason are true but Reason not is the correct explanation of Assertion.
C) Assertion is true but Reason is false
D) Assertion is false but Reason is true
Answer:
A) Both Assertion and Reason are true and Reason is the correct explanation of A.

Question 2.
Assertion : The energies of the 5s and 4d orbitals are very close.
Reason: The relative energies of the 4d and 5s orbitals vary with the nuclear charge and the electronic distribution.
A) Both Assertion and Reason are true and Reason is the correct explanation of A.
B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
C) Assertion is true but Reason is false
D) Assertion is false but Reason is true
Answer:
B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 3.
Assertion : The densities, melting and boiling points of the transition elements are generally very high
Reason: Zn, Hg and Cd have low melting and boiling points as the d – block is complete.
A) Both Assertion and Reason are true and Reason is the correct explanation of A.
B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
C) Assertion is true but Reason is false
D) Assertion is false bu t Reason is true
Answer:
B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 4.
Assertion: The transition metals more similar to one another than are representative metals of group 1 and group 2
Reason: Inner d orbitals are being filled
A) Both Assertion and Reason are true and Reason is the correct explanation of A.
B) Both Assertion and Reason are true but Reason is the correct explanation of Assertion.
C) Assertion is true but Reason is false
D) Assertion is false but Reason is true
Answer:
A) Both Assertion and Reason are true and Reason is the correct explanation of A.

Question 5.
Assertion : All the transition elements are metals and good conductors of heat and electricity.
Reason : The penultimate shell of electrons of all these elements is expanding and they are, therefore, expected to have physical and chemical properties in common.
A) Both Assertion and Reason are true and Reason is the correct explanation of A.
B) Both Assertion and Reason are true but Reason is the correct explanation of Assertion.
C) Assertion is true but Reason is false
D) Assertion is false but Reason is true
Answer:
A) Both Assertion and Reason are true and Reason is the correct explanation of A.

III. Match the following

1. Match the catalysts given in column I with the processes given in column II

Answer:

2. Match the properties given in column I with the metals in colum II

Answer:

3. Match the statements given in column I with oxidation states given in column II

Answer:

4. Match the following column I with column II

Answer:

5. Match the following

Answer:

IV. Choose the correct statements

Question 1.
Consider the following statements which is/are correct.
I. Sc ion has no unpaired electrons and the expected magnetic moemnt value is zero B.M
II. The oxidation state of Cr in Cr₂O₇^2- is +4
III. The spin only magnetic moment is given by \(\sqrt{n(n+2)} B M\)
a) I, III
b) II, I
c) III, II
d) II
Answer:
a) I, III

Question 2.
Consider the following statements which is / are correct
I. All Lanthanoids displaces hydrogen from acidified water.
II. The colour of tripositive Lanthanides becomes darker as one goes from Cr³⁺ to LU³⁺
III. The ionic size of tripositive Lanthanoid ions increases with atomic number
a) I
b) II
c) II, III
d) I, III
Answer:
a) I

Question 3.
Consider the following statement which is / are correct.
I. The maximum oxidation state of Osmium is +8
II. The highest oxidation state of a transition metal is given by outermost ‘s’ plus ‘d’ electrons.
III. The maximum magnetic moment is shown by the ion having d orbital electronic configuration is 3d⁵
a) I
b) II
c) I, III
d) I, II, III
Answer:
d) I, II, III

Question 4.
Consider the following statements which is / are correct.
I. Compared to Cu²⁺ salts Cu⁺ salts are less stable.
II. Iron does not form interstitial compounds
III. Ag⁺ is isoelectronic with Cd²⁺
a) I, III
b) II, III
c) III
d) II
Answer:
a) I, III

V. Choose the incorrect statements

Question 1.
Consider the following statements which is / are incorrect.
I. The common oxidation state of lanthanides is +3
II. All lanthanides form oxocations
III. All lanthanides are non – radioactive
a) I
b) II
c) II, III
d) II, III
Answer:
d) II, III

Question 2.
Consider the following statements which is / are incorrect.
I. The common oxidation state of Lanthanum is +4
II. UO₂²⁺ is colourless.
III. Actindes forms oxocations.
a) I, II
b) II, III
c) III, I
d) I, II, III
Answer:
a) I, II

Question 3.
Consider the following statements which is incorrect regarding potassium dichromate.
a) It oxidises ferric salt to ferrous salts.
b) It oxidises KI to I₂
c) It oxidises H₂S to S
d) None of the above
Answer:
a) It oxidises ferric salt to ferrous salts

Question 4.
Consider the following statements which is / are correct.
I. Potassium permanganate is extracted from chromite iron ore.
II. Potassium dichromate is extracted from Pyrolusite ore.
III. Potassium permangate is mainly used in Tanning of leather.
a) III
b) II
c) I, II
d) I, III
Answer:
c) I, II

VI. Fill in the blanks

1. The electronic configuration of scandium is ……………….. .
Answer:
[Ar] 3d¹4s²

2. The metal with highest electrical conductivity at room temperature is ……………….. .
Answer:
Silver

3. ……………….. values are used to predict the thermodynamic stability of the compounds
Answer:
Ionisation energy

4. The common oxidation state of scandium is ………………… .(MARCH 2020)
Answer:
+3

5. Mn²⁺ is ………………… than Mn⁴⁺
Answer:
More stable than

6. The oxidation state of W in Wcl₆ is
Answer:
+6

7. ………………… metal is unique in 3d series
Answer:
Copper

8. The metal shows maximum ferromagnetic character is
Answer:
Iron

9. Paramagnetism is the property of ………………… .
Answer:
Unpaired electrons

1o. Paramagnetism is common in
Answer:
Transition elements

11. Hydroformylation catalyst is
Answer:
(Co₂(CO)₈)

12. Potassium dichromate is in colour
Answer:
Orange red

13. The oxidation state of chromium in potassium dichromate is
Answer:
+6

14. The colour of potassium permanganate is
Answer:
dark purple colour

15. The oxidation state uranium in UP6 is
Answer:
+6

VII. Choose the correct Pair

1. a) Lead – Lead pencil
b) Cerusite – Zinc
c) Zinc sulphate – antiseptic
d) Lead pipes – softwater
Answer:
a) Lead – Lead pencil

2. a) Ag + hot NaOH – Products
b) Zn + hot NaOH – Products
c) Au + NaOH – Products
d) Cr + NaOH – Products
Answer:
b) Zn + hot NaOH – Products

3. a) Cu²⁺ – Diamagnetic
b) Cu²⁺ – Colourless
c) Cu²⁺ – Zero magnetic moment
d) Cu²⁺ – One unpaired electron
Answer:
d) Cu²⁺ – One unpaired electron

4. a) TiCl₄ – Polymerisation catalyst
b) Ni – Hydrogenation catalyst
c) Fe – Haber’s process
d) All the above pairs are correct
Answer:
All the above pairs are correct

VIII. Choose the incorrect Pair

1. a) Sc – 3d series
b) Zn – 3d series
c) Cr – 3d series
d) Cu – 4d series
Answer:
d) Cu – 4d series

2. a) Mn⁴⁺ – 3d³
b) Mn³⁺ – 3d⁴
c) Mn⁵⁺ – 3d⁵
d) Mn⁵⁺ – 3d²
Answer:
c) Mn⁵⁺ – 3d⁵

3. a) Cu⁺, Zn²⁺ – Diamagnetic
b) Sc³⁺, Ti⁴⁺, V⁵⁺ – Paramagnetic
c) Co⁵⁺, Fe²⁺ – Paramagnetic
d) Cu²⁺ – Paramagnetic
Answer:
b) Sc³⁺, Ti⁴⁺, V⁵⁺ – Paramagnetic

4.a) Potassium dichromate – +6
b) Potassium dichromate – oxidising
c) Potassium dichromate – Chrome tanning
d) Potassium dichromate – Bayer’s reagent
Answer:
d) Potassium dichromate – Bayer’s reagent

IX. Choose the odd man outer

1. Scandium, Titanium, Vanadium, Cadmium
Answer:
Cadmium
Hint : Cadmium belongs to 4d series. All others are 3d series.

2. Zirconium, Niobium, Technicium,
Ruthenium
Answer:
Technicium
Hint: Technicium is radio active. All others are non radio active.

3. U³⁺, UO₂²⁺, U⁴⁺, Ce³⁺
Answer:
Ce⁴⁺
Hint: Ce⁴⁺ is colourless. All others are coloured.

X. Additional Questions – 2 Mark

Question 1.
Write the classfication of transition elements.
Answer:

• d – Block elements composed of 3d series (4th period) Scandium to Zinc (10 elements),
• 4d series (5th period) Yttrium to Cadmium (10 elements)
• 5d series (6th period) Lanthanum, Hafinium to mercury (10 elements).
• 6d series (7th period) Actinium, Ruther Fordium to copernicium (10 mercury)

Question 2.
Give the electronic configuration of copper and chromium.
Answer:
The electronic configurations of
Chromium Cr [Ar] 3d⁵ 4S¹
Copper, Cu [Ar] 3d¹⁰ 4S¹

Question 3.
What are the type of packing possible in transition metals?
Answer:

• Most of the transition elements are hexagonal close packed,
• cubic close packed
• body centrered cubic which are the characteristics of true metals.

Question 4.
Why transition elements show variable oxidation states? (PTA – 4)
Answer:
Transition elements exhibit variable oxidation states due to

• loosing electrons from (n-1) d orbital
• ns orbital as the energy difference between them is very small.

Question 5.
Explain why Mn²⁺ is more stable than Mn³⁺?
Answer:

• Mn²⁺ has outer electronic configuration as 3d⁵
• Mn³⁺ has outer electronic configuration as 3d⁴
• Half filled orbitals and completely filled orbital are more stable than the partially filled orbitals so Mn²⁺ with half filled d orbital is more stable than Mn³⁺

Question 6.
Explain about ferromagnetic materials.
Answer:

• Ferromagnetic materials have domain structure.
• In each domain the magnetic dipoles are arranged.
• But the spin dipoles of the adjacent domains are randomly oriented.
• Some transition elements or ions with unpaired d electrons show ferromagnetism.

Question 7.
Explain why the melting and boiling points of Cd, Hg and Zn are low?
Answer:
The elements of group 12 [Zn, Hg, Cd] are quite soft with low melting points. Mercury is a liquid and melts at – 38°C. These elements has no unpaired electrons available for metallic bonding and therefore their melting and boiling points are low.

Question 8.
Which of the following ions would from colourless complexes?
Answer:
Cu²⁺, Zn²⁺, Ti , Ti⁴⁺, Cd²⁺
Zn²⁺ – 3d¹⁰ > Ti⁴⁺ – 3d° > Cd²⁺ – 4d¹⁰

These elements from colourless complexes because they have fully filled d – orbitals and do not possess impaired electrons and so no d – d transition is possible.

Question 9.
State Hume – Rothery rule for alloy formation.
Answer:
According to Hume – Rothery rule to form a substitute alloy the difference between the atomic radii of solvent and solute is less than 15%.

Both the solvent and solute must have the same crystal structure and valence and their electronegativity difference must be dose to zero.

Question 10.
Explain why transition metals form alloys? Transition metals form more alloys themselves because of their
Answer:
a tomic sizes are similar

one metal atom can be easily replaced by another metal atom from its crystal lattice to form an alloy.

Question 11.
Silver atom has completely filled d-orbitals (4d¹⁰) in its ground state, How can you say that it is a transition element?
Answer:
The outer electronic configuration of Ag (Z = 47) is 4d¹⁰ 5S¹, In addition to +1, it shows an oxidation state of +2 also. In +2 oxidation state, the configuration is d⁹
ie. the d – subshell is incompletely filled. Hence, it is transition element.

Question 12.
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Answer:
Oxygen and fluorine have small size high electro negativity. Hence, they can oxidise the metal to the highest oxidation state by prompting all the valance electrons to participate in bonding.

Question 13.
Calculate the ‘Spin only’ magnetic moment of M²⁺ (aq) ion (z=27)
Answer:
Electronic configuration of M with z = 27 is [Ar] 3d⁷4S²
Thus, electronic configuation of M²⁺ will be [Ar] 3d⁷4S°

Question 14.
In what way is the electronic configuration of the transition elements different from that of the non – transition elements?
Answer:
Transition elements contain incompletely filled d-sub shell i.e their electronic configuration is (n-l)d^”1( ns0’1, , whereas non – transition elements have either no d- subshell or their

d-subshell is completely filled and have ns^1,2 or n² np^1-6 in their outermost shell.

Question 15.
Name the oxometa! anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
Answer:
Cr²O₇^2- and Cr²O₄^2-
(Group number = Oxidation state of Cr = 6 MnO₄^–
(Group = oxidation stale of Mn=7)

Question 16.
What is the effect of increasing pH of a solution of Potassium dichromate?
Answer:
On increasing pH of potassium dichromate solution (i.e. on adding alkali) it changes to potassium chromate.

Question 17.
Write chemical equations for the reactions involved in the manufacture of potassium permanganate from Pyrolusite. (PTA- 5)
Answer:
Relevant equations for the manufacture of KMn04 from pyrolusite ore are given below:

Question 18.
Classify the following elementsintod-block and f-block elements. (MARCH 2020)
Answer:
(i) Tungsten
(ii) Ruthenium
(iii) Promethium
(iv) Einstenium
d-block elements – Tungsten, Ruthenium
f-block elements- promethium, Einstenium

Question 19.
The halides of transition elements become more covalent with increasing oxidation state of the metal why?
Answer:
As the oxidation slate of the metal increases, its charge increases. According to Eajan’s Rules, as the charge of the metal ion increases covalent character increases because the positively charged cation attracts the electron could on the anion towards itself.

Question 20.
Although Cr³⁺ and Co²⁺ ions have the same number of unpaired electrons but the magnetic moment of Cr³⁺ is 3.87 B.M. and that of Co²⁺ is 4.87 B.M. why?
Answer:
Cr³⁺ ion has symmetrical electronic configuration in the outermost orbit i.e. 3d³. In such ions there is no orbital configuration to magnetic moment. However appreciable orbital contribution takes place in Co²⁺ with 3d⁷ configuration.

Question 21.
Why E° value for Mn, Ni and Zn are more megative than expected?
Answer:
Negative values for Mn²⁺ and Zn²⁺ are related to the stabilities of half – filled and fully filled configurations respectively.

Question 22.
Calculate the spin only magnetic moment of Mn²⁺
Answer:
E.C of Mn²⁺ = [Ar] 3d⁵

Question 23.
Why Ni^II complexes are thermodynamically more stable than Pt^II complexes?
Answer:
The ionisation energy of Ni²⁺ less than that of Pt²⁺. Hence Ni²⁺ complexes are more stable than Pt²⁺ complexes.

Question 24.
Why do transition elements and its compounds act as catalyst? (PTA – 5)
Answer:
Transition metals has energetically available d orbitals that can accept electrons from reactant molecule or metal can form bond with reactant molecule using it’s d electrons.

Question 25.
WhatisZeigler-Natta Polymerisation
Answer:
A mixture of TiCl₄ and Tri alkyl aluminium is used for polymerisation.

Question 26.
Actinoid atoms are generally coloured? Justify your answer.
Answer:
The actinoid ions are generally coloured. This can be explained interms of unpaired
electrons u nde rgoi ng f – f Transi t ion.

Question 27.
How many unpaired electrons are present in Mn²⁺ ion. (Z=25) How does it influences magnetic behaviour of Mn²⁺ ion.
Answer:
Mn²⁺ :3d⁵ has 5 unpaired electorns. It is highly paramagnetic and it is attracted by magnet.

Question 28.
Transition metal atoms/ions are usually coloured. Justify.
Answer:
Transition metal ions have unpaired electrons. They can undergo d-d transition by absorbing light from visible region and radiating complementary colour.

Question 29.
Give the disproportionation of manganese (VI) in acidic medium.
Answer:
3MnO₄^2- + 4H⁺ → 2MnO^– + MnO₂ + 2H₂O

Question 30.
Oxoanions of a metal show higher oxidation state. Give reason.
Answer:
Oxo anions of a metal show higher oxidation state due to the ability of oxygen to form multiple bonds and its high electronegativity.

XI. Additional Questions – 3 Mark

Question 1.
In the series Sc (Z = 21) to Zn (Z = 30) the enthalpy of atomisation of Zinc is the lowest i.e. 126 KJ mol¹. why?
Answer:
In the series Sc to Zn, all elements have one or more unpaired electrons except zinc which has no unpaired electrons, its outer electronic configuration being 3d¹⁰ 4s². Lower the number of unpaired electrons, lower is the metal – metal bonding. Hence, metal – metal bonding is weakest in zinc. Therefore, enthalpy of atomisation is lowest.

Question 2.
Explain why Cu⁺ ion is not stable in aqueous solutions.
Answer:
Cu²⁺ (aq) is much more stable than Cu⁺ (aq). Although, second ionisation enthalpy of copper is large but ∆H_hyd for Cu²⁺ (aq) is much more negative than thatfor Cu⁺ (aq) and hence it more than compensates for second ionisation enthalpy of copper. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproport¬ionation.

Question 3.
Write down the electronic configuration of :
Answer:
i) Cr³⁺
ii) Cu⁺
iii) Co²⁺
iv) Mn²⁺
v) Pm³⁺
vi) Ce⁴⁺
vii) Lu²⁺
viii) Th⁴⁺
Answer:

Question 4.
What are the different oxidation states exhibited by lanthanoids?
Answer:
The most common oxidation state of lanthanoids is +3. However, some lanthanoids also show an oxidation state of +2 and +4. For example, Eu shows an oxidation state of +2 and Ce shows on oxidation state of +4.

Question 5.
Describe the oxidising action of potassium dichromate and write the ionic equations for its reactions with (i) iodide (ii) iron II solution and (iii) H₂S (PTA – 5)
Answer:
i) Cr₂O₇^2- + 14H⁺ + 6I^– → 2Cr³⁺ + 7H₂O + 3I₂
ii) Cr₂O₇^2- + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺+ 7H₂O
iii) Cr₂O₇^2- + 8H⁺ + 3H₂S → 2Cr³⁺ + 7H₂O + 3S

Question 6.
What is meant by disproportionations’? Give two examples of disporportionation reaction in aqueous solution.
Answer:
When the oxidation state of an element in a reactant increases in one of the products and decreases in other product, the phenomenon is called disproportionation. For example, Mn(VI) in MnO₄² changes to Mn (VII) in the product MnO₄ and Mn(IV) in the product MnO₂ as shown by the reaction.
3MnO₄^2- + 4H⁺ → 2MnO_4-+ MnO₂ + 2H₂O
similarly, Cr (V) undergoes disproportionation in acidic medium as follows.
CrQ₄^3-+ 8H⁺ → 2CrO₄^2- + Cr³⁺ + 4H₂O

Question 7.
Calculate the no. of unpaired electrons in the following gaseous ions :
Answer:
Mn³⁺, Cr³⁺, V³⁺ and Ti³⁺ which one of these is the most stable in aqueous solution?
Mn³⁺ : 3d⁴ has 4 unpaired electrons,
Cr³⁺ : 3d³ has 3 unpaired electrons,
V³⁺ : 3d² has 2 unpaired electrons,
Ti³⁺ : 3d¹ has 1 unpaired electrons.
Cr³⁺ is most stable out of these in aqueous solution because it has half filled t₂g level (i.e.t³_2g)

Question 8.
Why is the +2 oxidation state of manganese quite stable, while the same is not true for iron?
Answer:
Half filled configuration is more stable than others. Mn²⁺ is more stable due to half-filled d – orbitals.
Fe²⁺ is not stable because it does not have half filled d – orbitals. Configurations of Mn²⁺ and Fe²⁺ are as follows :
Mn²⁺ : 3d⁵ 4s⁰
Fe²⁺ : 3d⁶ 4s⁰

Question 9.
Write two characteristics of the transition elements. (PTA – 3)
Answer:
All the transition elements are metals.

All the transition metals are good conductors of heat and electricity.

Question 10.
Explain about diamagnetic materials.
Answer:

• Materials with no elementary magnetic dipoles are diamagnetic or a species with all paired electrons exhibits diamagnetism.
• This kind of materials are repelled by the magnetic field.
• Because the presence of external magnetic field, a magnetic induction is introduced to the material which generates weak magnetic field that oppose the applied field.

Question 11.
Explain about paramagnetic materials.
Answer:

• Paramagnetic solids having unpaired electrons.
• In the absence of external magnetic field, the dipoles are arranged at random and hence the solid shows no net magnetism.
• In the presence of magnetic field, the dipoles are aligned parallel to the direction of the applied field and therefore, they are attracted by an external magnetic field.

Question 12.
A substance is found to have a magnetic moment of 3.9 BM. How many number of unpaired electrons does it certain?
Answer:
Magnetic moment µ = 3.9
\(\sqrt{\mathrm{n}(\mathrm{n}+2)}\) = 3.9B.M
n(n + 2) = (3.9)² = 15
n (n + 2 ) = 3(3+ 2)
n = 3
The number of unpaired electrons is 3.

Question 13.
Why Zn²⁺ salts are white while Ni²⁺ salts are coloured? (PTA – 3)
Answer:

• Zn²⁺ has the configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰
• It has completely filled d orbital and there is no unpaired electrons.
• Hence Zn²⁺ is colourless.
• Ni²⁺ has the configurations 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸
• It has two unpaired electrons and d – d transitions are possible.
• Hence Ni²⁺ is coloured.

Question 14.
[Ti(H₂O)₆]³⁺ is coloured while
[Sc(H₂O)₆]³⁺ is colourless. Explain.
Answer:

• In [Ti(H₂O)₅]¹⁺ ,Tr³⁺ has outer electronic configuration as 3d¹
• In this case d-d transition is possible by the absorption of energy from the visble light and produce purple colour.
• But is [Sc(H₂O)₆]³⁺, Sc³⁺ has outer electronic configuration as 3d°.
• In this case d-d transition is not possible and it is colourless.

Question 15.
What are the characteristics of interstitial compounds?
Answer:

• They are hard and show electrical and thermal conductivity
• They have high melting points higher than those of pure metals
• Transition metal hydrides are used as powerful reducing agents
• Metallic carbides are chemically inert.

Question 16.
Explain why d block elements form more complexes?
Answer:

• Transition metal ions are small and highly charged
• They have vacant low energy orbitals to accept an electron pair donated by other groups.
• Examples : [Fe(CN)₆]^4-, [Co(NH₃)₆]³⁺

Question 17.
Explain chromyl chloride test. (MARCH 2020)
Answer:
When potassium dichromate is heated with any chloride salt in the presence of Cone H₂SO₄, orange red vapours of chromyl chloride (CrO₂Cl₂) is evolved.

This reaction is used in the detection of chloride ions in qualitative analysis.

Question 18.
Write the uses of potassium dichromate.
Answer:

• It is used as a strong oxidizing agent.
• It is used in dyeing and printing.
• It used in leather tanneries for chrome tanning.
• It is used in quantitiative analysis for the estimation of iron compounds and iodides.

Question 19.
Write the uses of potassium permanganate.
Answer:

• It is used as a strong oxidizing agent.
• It is used for the treatment of various skin infections and fungal infections of the foot.
• It used in water treatment industries to remove iron and hydrogen sulphide from well water.
• It is used as a Bayer’s reagent for detecting unsaturation in an organic compound.
• it is used in quantitative analysis for the estimation of ferroussalts, oxalates, hydrogen peroxide and iodides.

Question 20.
What is the action of heat on K₂Cr₂O₇ ?
Answer:
Potassium dichromate on heating gives potassium chromate.

XII. Additional Questions – 5 Mark

Question 1.
For M²⁺ /M and M³⁺ / M systems, the E° values for some metals are as follows.
Cr²⁺ /Cr = 0.9 V ; Cr³⁺ /Cr2+ = – 0.4 V
Mn²⁺ /Mn = -1.2 V ; Mn³⁺/Mn²⁺= +1.5 V
Fe²⁺ /Fe = +0.4 V ;Fe³⁺/Fe²⁺ = +0.8 V
Use this data to comment upon
(a) the stability of Fe³⁺ in acid solution as composed to that of Cr³⁺ or Mn²⁺ and
(b) the ease with which iron can be oxidised as compared to the similar process for either chromium or manganese metals.
Answer:
(a) Cr³⁺ / Cr²⁺ has a negative reduction potential Hence, Cr³⁺ cannot be reduced to Cr²⁺ i.e. Cr³⁺ is most stable. Mn³⁺/ Mn²⁺ has large positive E° value. Hence, Mn³⁺ can be easily reduced to Mn²⁺, i.e. Mn³⁺ is least stable. Eo value for Fe³⁺ / Fe²⁺ is positive but small. Hence, Fe³⁺ is more stable than Mn³⁺ but less stable than Cr³⁺. Thus, the stability follows the order.
Cr³⁺>Fe³⁺>Mn³⁺

(b) Oxidation potentials for the given pairs will be +0.9V, +1.2V, and +0.4V. Thus the order of getting oxidised will be
Mn > Cr > Fe

Question 2.
(i) Why do transition elements show variable oxidation states?
(ii) Why do most transition metal ions exhibit para magnetism?
(iii) How is the magnetic moment of a species related to the number of unpaired electrons?
Answer:
(i) This is because ns and (n-1) d orbitals do not differ much in energy. Electrons from both may take part in bonding. Hence, they show variable valency.
(ii) This is because of presence of unpaired electrons in most of the transition metal ions.
(iii) \(\mu_{\mathrm{s}}=\sqrt{\mathrm{n}(\mathrm{n}+2)}\) BM, where n stands for the number of unpaired electrons.

Question 3.
(i) Why is copper (z=29) considered as transition element?
(ii) K₂PtCl₆ is well – known compound while corresponding Ni compound is not known.
(iii) Why is radius of Fe²⁺ less than that of Mn²⁺?
(iv) Why is electronic configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁴ not correct for the ground state of Cr (Z-24)?
Answer:
(i) This is because Cu²⁺ ion has incomplete d – orbital
(ii) This is because Pt⁴⁺ is more stable than Ni⁴⁺ Energy required to remove 4 electrons from pt is less and that in Ni than.
(iii) This is because effective nuclear charge in Fe²⁺ is more compared to that in Mn²⁺
(iv) Correct electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵ because half – filled configuration

Question 4.
(i) Which trivalent cation is the largest in lanthanoid series?
(ii) One unpaired electron in atom contributes a magnetic moment of 1.1 B.M. Calculate the magnetic moment of Cr. (Atomic number = 24)
(iii) In a paramagnet ion, all the bonds formed between Mn and O are covalent. Give reasons.
Answer:
(i) La³⁺ is the largest ion
(ii) As Cr has 6 unpaired electrons, its magnetic moment is 6 x 1.1 = 6.6 B.M.
(iii) Oxidation state of Mn is MnO₄ is +7. It is energitacally not possible to lose 7 electrons to give ionic species. It forms bonds by sharing of elctrons. Hence, covalent bonds are formed.

Question 5.
There is only a marginal difference in decrease in ionisation elthalpy from Aluminium to Thallium. Explain why? (MARCH 2020)
Answer:
This is due to the presence of inner d-electrons and f-electrons which has poor shielding effect compared to s and p electrons.

Question 6.
(i) Name the metal with tripositive charge represented by the following electronic configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³
(ii) Why is K₂Cr₂O₇ generally preferred over Na₂Cr₂O₇ in volumetric analysis though both are oxidising agents?
(iii) Why does V₂O₅, act as catalyst?
Answer:
(i) Cr³⁺ is represented by Ihe configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³
(ii) Na₂Cr₂O₇ absorbs moisture from, the atmosphere.
(iii) It can form unstable intermediates with the reactants which readilv change into products

Question 7.
Explain the following observations.
(a) The elements of the d – series exhibit a large number of oxidation states than the elements of f – series.
(b) The Cu⁺ salts are colourless while Cu²⁺ salts are coloured (Atomic number of [Cu=29]
Answer:
(a) In d – series there are large number of unpaired electrons which take part in bond formation due to less effective nuclear charge, therefore, number of oxidation states are more. In f – block, there is more effective nuclear charge due to poor shielding effect of f – orbitals. Therefore, less number of electrons take part in bond formation.

(b) Cu⁺ does not have unpaired electtron, therefore the electrons cannot undergo d-d transitions. That is why, Cu⁺ salts are colourless. Cu²⁺ salts are coloured due to presence of one unpaired electron, it can undergo d-d transition by absorbing light from visible region and radiating blue colour.

Question 8.
How do you account for the following?
(i) All Scandium salts are white [Atomic number of Sc=21]
(ii) The first ionisation energies of the 5d transition elements are higher than those of the 3d and 4d transition elements in respective groups.
Answer:
(i) In Scandium Salts, Scandium has +3 oxidation state. Sc³⁺ does not have unpaired electrons and has empty d – orbitals. Therefore, there is no d-d transition, Hence its salts are white.

(ii) Due to poor shielding effect of 5d and 4f – electrons, effective nuclear charge increases, Hence, ionisation energy of 5d transition elements is more than that of 3d and 4d transition elements in respective groups.

Question 9.
How would you account for the following Situations?
(i) The transition metals generally form coloured compounds.
(ii) With 3d⁴ configuration, Cr²⁺ acts as a reducing agent, actinoids exhibit a larger number of oxidation agent.
(iii) The actinoids exhibit a larger number of oxidation states than the corresponding lanthanoids.
Answer:
(i) Transition metals contain unpaired d- electrons. The d – electrons absorb light from the visible range and are excited to higher energy d – orbitals (from t₂g to e_g orbitals). Transmitted light is the colour shown by the transition metals.

(ii) Cr²⁺ has d⁴ configuration while Cr³⁺ has more stable d⁵ configuration. Thus Cr, has a tendency to change from Cr²⁺ to Cr³⁺ or Cr²⁺ acts as reducing agent. Mn³⁺ has a d⁴ configuration. It has a tendency to change into Mn²⁺ with more stable d⁵ configuation. Thus, Mn³⁺ acts as an oxidising agent.

(iii) Actinoids exhibit a larger number of oxidation states than the corresponding lanthanoids because 5f, 6d and 7s levels are comparable energies.

Question 10.
A Violet compound of manganese (A) decomposes on heating to liberate oxygen and compound (B) and (C) of manganese are formed. Compound (C) reacts with KOH in presence of potassium nitrate to give compound.
(B) On heating compound (C) with Cone. H₂SO₄ and NaCl, Chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds (A) to (D) and also explain the reactions involved.
Answer:
The compounds A,B,C and D are given as under : A = KMnO₄, B = K₂MnO₄, C = MnO₂, D = MnCl₂
The reactions are explained as under:

2MnO₂ + 4KOH+ O₂ → 2K₂MnO₄ + 2H₂O
MnO₂ + 4NaCl + 4H₂SO₄ → MnCl₂ + 4NaHSO₄ + Cl₂ + 2H₂O

Question 11.
When a chromite ore (A) is fused with Sodium Carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid, compound (C) can be crystallized from the solution. When compound (C) is treated with KC1, orange crystals of compound (D) crystallises out. Identify (A) to (D) and also explain the reactions.
Answer:
The compounds A,B,C and Dare given as under
A = FeCr₂O₄
B = Na₂CrO₄
C = Na₂Cr₂O₇-2H₂O
D = K₂Cr₂O₇
The reactions are explained as under :

Question 12.
Explain the following facts:
(a) Transition metals acts as catalysts.
(b) Chromium group elements have the highest melting points in their respective series.
(c) Transition metals form coloured complexes.
Answer:
(a) Transition metals have incomplete d-orbitals. They combine with the reactants to form intermediate products which change into the final products.
(b) Chromium group elements have the greatest number of unpaired electrons in d-orbitals. They ae capable of forming maximum interatomic metallic bonds. This raises the melting point of the chromium group elements.
c) Transition metals have incompletely filled d-orbitals and have unpaired electrons which are excited to higher energy d-orbitals in the same sub-shell (from t2g to eg). These are called d-d transitions. During this process, the molecules absorb energy from the visible region. The transmitted light is the colour shown by the substance.

Question 13.
Discuss the structure of dichromate ion:
Answer:
(i) Both dichromate and chromate ions are oxo anions of chromium and they are oxidising agents.
(ii) In these ions Cr is in +6 oxidation state.
(iii) In aqueous solution these ions are interconvertible.
(iv) In alkaline solution chromate ion is predominant.
(v) In acidic solution dichromate ion is predominant.

Question 14.
Discuss the oxidising power of KMnO₄ in
a) Acidic medium
b) neutral medium
c) alkaline medium
Answer:
a) Acidic medium
It oxidises ferrous salt to ferric salt
2MnO₄^– + 10Fe²⁺ + 16H⁺ → 2Mn²⁺ + 10Fe³⁺ + 8H₂O

b) Neutral medium:
It oxidises H2S to sulphur
2MnO₄^– + 3H₂S → 2MnO + 3S + 2OH^– + 2H₂O

c) Alkaline medium:
In the presence alkali the permanganate ion is converted into manganate ion.
MnO₄^– + 2H₂O + 3e^– → MnO₂ + 40H^–

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Accountancy Guide Pdf Chapter 9 Ratio Analysis Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 9 Ratio Analysis

12th Accountancy Guide Ratio Analysis Text Book Back Questions and Answers

I Multiple Choice Questions

Choose the correct answer

Question 1.
The mathematical expression that provides a measure of the relationship between two figures is called
(a) Conclusion
(b) Ratio
(c) Model
(d) Decision
Answer:
(b) Ratio

Question 2.
Current ratio indicates
(a) Ability to meet short term obligations
(b) Efficiency of management
(c) Profitability
(d) Long term solvency
Answer:
(a) Ability to meet short term obligations

Question 3.
Current assets excluding inventory and prepaid expenses is called
(a) Reserves
(b) Tangible assets
(c) Funds
(d) Quick assets
Answer:
(d) Quick assets

Question 4.
Debt equity ratio is measure of
(a) Short term solvency
(b) Long term solvency
(c) Profitability
(d) Efficiency
Answer:
(b) Long term solvency

Question 5.
Which of the following is not a tool of financial statement analysis?

Answer:
(a) (i) – 1,(ii) – 4,(iii) – 3,(iv) – 2

Question 6.
To test the liquidity of a concern, which of the following ratios are useful?
(i) Quick ratio
(ii) Net Profit ratio
(iii) Debt – equity ratio
(d) Current ratio
Select the correct answer using the codes given below:
(a) (i) and (ii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and iv)
Answer:
(b) (i) and (iv)

Question 7.
Proportion of share holders’ funds to total assets is called
(a) Proprietary ratio
(b) Capital gearing ratio
(c) Debt equity ratio
(d) Current ratio
Answer:
(a) Proprietary ratio

Question 8.
Which one of the following is not correctly matched?
(a) Liquid ratio – Proportion
(b) Gross profit ratio – Percentage
(c) Fixed assets turnover ratio – Percentage
(d) Debt – equity ratio – Proportion
Answer:
(c) Fixed assets turnover ratio – Percentage

Question 9.
Current liabilities ₹ 40,000; Current assets ₹ 1,00,000; Inventory ₹ 20,000. Quick ratio is
(a) 1:1
(b) 2,5:1
(c) 2:1
(d) 1:2
Hint:
Quick ratio or Liquid ratio = \(\frac{\text { Liquid Assets }}{\text { Current liabilities }}\)
Liquid assets = Current Assets – Inventory
= 1,00,000 – 20,000
= 80,000
= \(\frac{80,000}{40,000}\)
= 2:1
= 110%
Answer:
(c) 2:1

Question 10.
Cost of revenue from operation 3,00,000; Inventory at the beginning of the year 60,000; Inventory at the close of the year’ 40,000. Inventory turnover ratio is.
(a) 2 times
(b) 3 times
(c) 6 times
(d) 8 times
Hint:

Answer:
(c) 6 times

II Very Short Answer Questions

Question 1.
What is meant by accounting ratios?
Answer:
The ratio is a mathematical expression of the relationship between two related or interdependent items. It is the numerical or quantitative relationship between two items. It is calculated by dividing one item by the other related item. When ratios are calculated on the basis of accounting information, these are called ‘accounting ratios’.

Question 2.
What is the quick ratio?
Answer:
The quick ratio gives the proportion of quick assets to current liabilities. It indicates whether the business concern is in a position to pay its current liabilities and when they become due, out of its quick assets.

Question 3.
What is meant by debt-equity ratio?
Answer:
It is calculated to assess the long-term solvency position of a business concern. The debt equity ratio expresses the relationship between long term debt and shareholder’s funds.
Debt equity ratio = \(\frac{\text { Long term debt }}{\text { Shareholders funds }}\)
Capital employed = Shareholder’s funds + Noncurrent liabilities
Greater the return on investment better is than the profitability of a business and vice versa.

Question 4.
What does the return on investment ratio indicate?
Answer:
Return on investment shows the proportion of net profit before interest and tax to capital employed (shareholders’ funds and long term debts). This ratio measures how efficiently the capital employed is used in the business. It is an overall measure of the profitability of a business concern.

Question 5.
Statement any two limitations of ratio analysis.
Answer:
Ratios are only means: Ratios are not ended in themselves but they are only means to achieve a particular purpose. Analysis of related items must be done by the management or experts with the help of ratios. Change in price level: Ratio analysis may not reflect price level changes and current values as they are calculated based on historical data given in the financial statement.

III Short Answer Questions

Question 1.
Explain the objectives of ratio analysis.
Answer:
Following are the objectives of ratio analysis:

• To simplify accounting figures
• To facilitate analysis of financial statements
• To analysis the operational efficiency of a business
• To help in budgeting and forecasting
• To facilitate intra firm and inter-firm comparison of performance

Question 2.
What is the inventory conversion period? How is it calculated?
Answer:
The inventory conversion period is the time taken to sell the inventory. A shorter inventory conversion period indicates more efficiency in the management of inventory. It is computed as follows:

Question 3.
How is operating profit ascertained?
Answer:
Operating profit = Revenue from operations – Operating cost
Cost of revenue from operations = Purchases of stock – in – trade + Change in inventories of stock in trade + Direct expenses.
Operating expenses = Administrative expenses + Selling and distribution expenses.
Operating cost = Cost of revenue from operations + Operating expenses.

Question 4.
State any three advantages of ratio analysis.
Answer:
Following are the advantage of ratio analysis:

• Measuring operational efficiency: Ratio analysis helps to know the operational efficiency of a business by finding the relationship between operating cost and revenues and also by comparison of present ratios with those of the past ratios.
• Intra firm comparison: Comparison of the efficiency of different divisions of an organization is possible by comparing the relevant ratios.
• Inter-firm comparison: Ratio analysis helps the firm to compare its performance with other firms.

Question 5.
Bring out the limitations of ratio analysis:
Answer:

• Consistency in preparation of financial statements: Inter firm comparisons with the help of ratio analysis will be meaningful only if the firms follow uniform accounting procedures consistently.
• Non-availability of standards or norms: Ratios will be meaningful only if they are compared with accepted standards or norms. Only few financial ratios have universally recognized standards. For other ratios, comparison with standards is not possible.
• Change in price level: Ratio analysis may not reflect price level changes and current values as they are calculated based on historical data given in financial statements.

IV Exercises

Liquidity ratios

Question 1.
Calculate the current ratio from the following information.

Solution:
Current ratio = \(\frac{\text { Current Assets }}{\text { Current liabilities }}\)
Current Assets = Current Investments + Inventories + Trade Dr’s + B/R+Cash & Cash equivalents
= 40,000 + 2,00,000 + 1,20,000 + 80,000 + 10,000
= Rs. 4,50,000

Current Liabilities
= Trade Cr’s + B/P + Exps. Payable.
= 80,000 + 50,000 + 20,000
= Rs. 1,50,000
Cur. Ratio = \(\frac{4,50,000}{1,50,000}\)
= 3:1
Answer:
Current ratio : 3:1

Question 2.
Calculate quick ratio: Total current liabilities ₹ 2,40,000; total current assets ₹ 4,50,000; Inventories ₹ 70,000; Prepaid Expenses ₹ 20,000
Solution:
Quick Ratio = \(\frac{\text { Quick assets }}{\text { Current liabilities }}\)
Quick assets = Current Assets – Inventories & Prepaid exps.
= 4,50,000 – (70,000 + 2000)
= Rs.3,60,000
Quick Ratio = \(\frac{3,60,000}{2,40,000}\)
=1:5:1
Answer:
Quick ration: 1:5:1

Question 3.
Following is the balance sheet of Lakshmi Ltd. as of 31st March 2019.

Calculate: (i) Current ratio (ii) Quick ratio
Solution:
Current ratio = \(\frac{\text { Current Assets }}{\text { Current liabilities }}\)
Current Assets = Inventories + Trade Dr’s + Cash & Cash equivalents + Prepaid Exps
= 1,60,000 +3,20,000 + 80,000 + 40,000
= Rs. 6,00,000
Current Liabilities = Short term borrowings + Trade Payable + Expenses payable + Short term provisions.
= 50,000 + 3,10,000 + 15,000 + 25,000
= Rs. 4,00,000

Answer:
(i) Current ratio: 1.5:1;
(ii) Quick ratio: 1:1

Question 4.
From the following information calculate debt equity ratio.

Solution:
Debt Equity Ratio = \(\frac{\text { Long term debt }}{\text { Shareholder’s Funds }}\)
Long term debt = Debenture = Rs. 6,00,000
Shareholder’s Fund = Equity share capital + Reserves & Surplus
= 6,00,000 + 2,00,00 = Rs. 8,00,000
Debt Equity Ratio = \(\frac{6,00,000}{8,00,000}\)
= 0.75:1
Answer:
Debt equity ratio: 0.75:1

Question 5.
From the following Balance Sheet of Sundaram Ltd. Calculate proprietary ratio:

Answer:
Proprietary ratio: 0.75:1

Question 6.
From the following information calculate the capital gearing ratio:

Answer:
Capital gearing ratio: 0.5:1

Question 7.
From the following Balance Sheet of James Ltd. as of 31.03.2019 calculate
(i) Debt- Equity ratio
(ii) Proprietary ratio
(iii) Capital gearing ratio

Solution:
Debt Equity Ratio =  \(\frac{\text { Long Term Debt }}{\text { Shareholder’s Fund }}\)|
Shareholder’s Fund
Long term debt = Debentures = Rs. 3,00,000
Shareholder’s Fund = Eq. share capital +Pref. Shares capital + Reserves & surplus
= 2,50,000 ÷ 2,00,000 + 1,50,000
= Rs. 6,00,000

(3) Capital Gearing ratio

= \(\frac{\text { Funds bearing fixed interest (or) Fixed divided }}{\text { Equity shareholder’s Fund }}\)
Funds bearing fixed interest (or) fixed dividend
= Pref. Share Cap + Debentures
= 2,00,000 + 3,00,000 = Rs. 5,00,000
Equity share holder’s Fund .
= Equity Share cap + Reserves & Surplus
= 2,50,000 + 1,50,000
= Rs. 4,00,000
Capital gearing ratio = \(\frac{5,00,000}{4,00,000}\)
= 1.25:1

Answer:

• Debt-equity ration; 0.5:1;
• Proprietary ration; 0.5:1;
• Capital gearing ratio:1.25:1

Question 8.
From the given information calculate the inventory turnover ratio and inventory conversion period (in months) of Devi Ltd.

Solution:
Inventory Turnover Ratio = \(\frac{\text { cost of revenue from operations }}{\text { Average Inventory }}\)
Cost of revenue from operations = Purchase of stock + change in inventories of finished goods operations + Direct Exps.
= Rs. 6,90,000
AverageInventory = \(\frac{\text { Opening Inventory + Closing inventory }}{2}\)
= \(\frac{1,70,000+1,30,000}{2}\)
= Rs. 1,50,000
Change in inventory = Opening inventory – Closing inventory
= 1,70,000 – 1,30,000
= Rs. 40,000
Cost of revenue from operation
= 6,90,000 + 40,000 + 20,000
= Rs. 7,50,000

Answer:
Inventory turnover ratio 5times; Inventory conversion period 2.4 months

Question 9.
The credit revenue from operations of Velavan Ltd, amounted to ₹ 10,00,000. Its debtors and bills receivables at the end of the accounting period amounted to ₹ 1,10,000 and ₹ 1,40,000 respectively. Calculate trade receivables turnover ratio and also.collection period in months.
Solution:
Trade receivable Turnover ratio = \(\frac{\text { Credit revenue from Operations }}{\text { Average trade receivables }}\)
Average trade receivables = \(\frac{\text { Opening trade receivables + Closing trade receivables }}{2}\)
Trade receivable = Trade Drs + B/R
Inventory Turnovers Ratio = \(\frac{10,00,000}{2,50,000}\)
= 4 times
Average Trade receivable
= 1,10,000 + 1,40,000
= Rs. 2,50,000
Debt collection period = \(\frac{\text { Number of months in a year }}{\text { Trade receivable turnover ratio }}\)
= \(\frac{12}{4}\)
= 3 months
Answer:
Trade receivables turnover ratio: 4 time; Debt collection period: 3 months

Question 10.
From the following figures obtained from Arjun Ltd, calculate the trade payable turnover ratio and credit payment period (in days)

Solution:


Answer:
Trade payable turnover ratio: 10 times; Credit payment period: 36.5 days

Question 11.
From the following information of Geetha Ltd., Calculate fixed assets turnover ratio
(i) Revenue from operations during the year was ₹ 55,00,000.
(ii) Fixed assets at the end of the year ₹ 5,00,000
Solution:

Answer:
Fixed assets turnover ratio: 11 times

Question 12.
Calculate

• Inventory turnover ratio
• Trade receivable turnover ratio
• Trade payables turnover ratio and
• Fixed assets turnover ratio from the following obtained from Aruna Ltd.
  

  Particulars	As of 31st March 2018 ₹	As of 31st March 2019 ₹
  Inventory	3,60,000	4,40,000
  Trade receivables	7,40,000	6,60,000
  Trade Payable	1,90,000	2,30,000
  Fixed assets	6,00,000	8,00,000

   

Additional information:

• Revenue from operations for the year ₹ 35,00,000
• Purchases for the year ₹ 21,00,000
• Cost of revenue from operation ₹ 16,00,000
  Assume that sales and purchases are for credit.

Solution




Answer:

• Inventory turnover ratio; 4 times;
• Trade receivable turnover ratio; 5 times;
• Trade payables turnover ratio: 10 times;
• Fixed assets turnover ratio: 5 times

Question 13.
Calculate gross profit ratio form the following: Revenue from operations ₹ 2,50,000, Cost of revenue from operation ₹ 2,10,000 and Purchases ₹ 1,80,000.
Solution:

Answer:
Gross Profit ratio 16%

Question 14.
Following is the statement of profit and loss of Padma Ltd. for the year ended 31st March, 2018. Calculate the operating cost ratio.


Notes to Accounts

Solution:
Operating cost Ratio = \(\frac{\text { Operating cost }}{\text { Revenue from operation }}\) × 100
Operating cost = Cost of revenue from operations + Operating expenses.
Cost of revenue from operations = Purchase + Change in inventory + Direct Expenses
= 8,60,000 + 40,000 + Nil
= Rs. 9,00,000
Operating Exps = Salaries + Office & Administration Exps + Selling+Distribution Exps
= 1,60,000 + 50,000 + 90,000
= Rs. 3,00,000
Operating cost = 9,00,000 + 3,00,000
= Rs, 12,00,000
Operating cost Ratio = \(\frac {12,00000}{15,00000}\) × 100
= 80%
Answer:
Operating cost ratio 80%

Question 15.
Calculate operating profit ratio under the following cases.
Case 1 : Revenue from operations ₹ 8,00,000 Operating Profit ₹ 2,00,000.
Case 2 : Revenue from operations ₹ 20,00,000 Operating Cost ₹ 14,00,000.
Case 3 : Revenue from operations ₹ 10,00,000 Gross profit 25% on revenue from operations, operating expenses ₹ 1,00,000.
Solution:


Answer:
Operating profit ratio – Case 1: 25%; Case 2:30%; Case 3:15%

Question 16.
From the following details of a business, concern calculates net profit ratio.

Solution
Net Profit Ratio = \(\frac{\text { Net Profit }}{\text { Revenue from operations }}\)× 100
Gross profit = Revenue from operations – Cost of revenue from operation
= 9,60,000-5,50,000 Rs. 4,10,000
Operating Profit = Gross Profit – Operating Exps
Operating Exps = Office & Administrative Exps + Selling & Distribution Exps
= 1,45,000 + 25,000 = Rs. 1,70,000
Operating Profit = 4,10,000 – 1,70,000
Operating Profit = Rs. 2,40,000
Net Profit ratio = \(\frac{2,40,000}{9,60,000}\)× 100 = 25%
Answer :
Net Profit ratio 25%

Question 17.
From the following statement of profit. and loss of Dericston Ltd. Calculate
(i) Gross Profit ratio
(ii) Net Profit ratio.

Solution
Gross Profit Ratio = \(\frac{\text { Gross Profit }}{\text { Revenue from operations }}\)×100
Gross profit = Revenue from operations – Cost of revenue from operation
Cost of revenue from operations = Purchase + Change in inventories
= 18,80,000 – 80,000 = Rs. 18,00,000
Gross Profit = 24,00,000 – 18,00, 000 = Rs. 6,00,000

Answer :

• Gross profit ratio of 25%
• net Profit ratio 10%

Question 18.
From the following trading activities of Jones Ltd. Calculate

• Gross profit ratio
• Net Profit ratio
• Operating cost ratio
• Operating profit ratio

Solution
(1) Gross Profit Ratio = \(\frac{\text { Gross Profit }}{\text { Revenue from operations }}\)× 100
cost of Revenue from operations = Purchase + Change in inventories
= 2,10,000 + 30,000
= Rs. 2,40,000
Gross profit = Revenue from operations – cost of revenue from operations
= 4,00,000 – 2,40,000
= Rs. 1,60,000
Gross Profit ratio = \(\frac{1,60,000}{4,00,000}\)× 100 = 40%


Answer :

• Gross profit ratio 40%
• Net Profit ratio 20%
• Operating cost ratio 75%
• Operating profit ratio 25%

Question 19.
Following is the extract of the balance sheet of Abdul Ltd., as of 31st March 2019

Net Profit before interest and tax for the year was ₹ 60,000. Calculate the return on capital employed for the year.
Solution:

Answer :
Return on capital employed: 15%

12th Accountancy Guide Ratio Analysis Additional Important Questions and Answers

Other Important question & Answers
Question 1.
All solvency ratios are express in terms of
(a) Proportion
(b) Times
(c) Percentage
Answer:
(b) Times

Question 2.
All activity ratios, (or) Turnover ratios in terms of
(a) Proportion
(b) Times
(c) Percentage
Answer:
(b) Times

Question 3.
All profitability ratios are expressed in terms of ………………
(a) Proportion
(b) Times
(c) Percentage
Answer:
(c) Percentage

Question 4.
Shareholders funds include
(a) Equity share capital, preference share capital reserves & Surplus
(b) Loans from banks & financial institutions.
(c) Equity share capital, preference share capital, reserves & surplus, and loans from banks & financial institutions.
Answer:
(a) Equity share capital, preference share capital reserves & Surplus

Question 5.
The current ratio is a
(a) Solvency ratio
(b) Profitability ratio
(c) Liquidity ratio
Answer:
(a) Solvency ratio

Question 6.
The ratio is expressed in ……………… way
(a) 2
(b) 4
(c) 3
Answer:
(c) 3

Question 7.
The cost of revenue from the operation is Rs. 4,00,000. Average inventories Rs. 8,00,000 Inventory turnover ratio is
(a) 5 times
(b) 4 times
(c) 7 times
Answer:
(a) 5 times

Question 8.
The operating ratio is equal to
(a) 100-operating profit ratio
(b) 100 +operating profit ratio
(c) Operating profit ratio
Answer:
(a) 100-operating profit ratio

Question 9.
Operating expenses include
(a) Selling & administration expenses
(b) Selling & administration expens
(c) a & b
Answer:
(c) a & b

Question 10.
Equity share capital Rs. 2,00,000 Reserves & Surplus Rs. 30,000 Debenture Rs. 40,000 and shareholders fund will be.
(a) Rs. 200,000
(b) 2,70,000
(c) Rs. 2,30,000
Hint:
Share holder fund = Equity share capital + Reserve and surplus
= 2,00,000 + 30,000
= ₹ 2,30,000
Answer:
(c) Rs. 2,30,000

III Short Answer Questions

Question 1.
Define Ratio Analysis
Answer:
According to Myers, “Ratio analysis is a study of the relationship among various financial factors in a business”.

Question 2.
What do you mean by ratio Analysis?
Answer:
Ratio analysis is a tool which involves analyzing the financial statements by calculating various. It is a tool of financial statement analysis, in which, inferences are drawn based on the computation and analysis of different ratios.

Question 3.
What are the two ways of classifying the ratios?
Answer:
Ratios may be classified in the following two ways:

• Traditional classification
• Functional classification

Question 4.
What do you mean by the traditional classification of ratio? Explain
Answer:
The traditional classification of ratios is done on the basis of the financial statements from which the ratios are calculated. Under the traditional classification, the ratio is classified as:

• Balance sheet ratio: If both items in a ratio are from the balance sheet, it is classified as a balance sheet ratio.
• Income statement ratio: If the two items in a ratio are from the income statement, it is classified as an income statement ratio.
• Inter – Statements ratio: If a ratio is computed with a tone item from the income statement and another item from the balance sheet, it is called an inter-statement ratio.

Question 5.
What is the functional classification of ratios?
Answer:
Under the functional classification, the rations are classified as follows:

• Liquidity ratios
• Long term solvency ratios
• Turnover ratios.
• Profitability ratios.

Question 6.
What do you mean by Liquidity ratios?
Answer:
Liquidity means the capability of being converted into cash with ease. Liquidity ratios help to assess the ability of a business concern to meet its short term financial obligations. Short term assets (current assets) are more liquid as compared to long term assets (fixed assets). Liquidity ratios are also called as short term solvency ratios.

Question 7.
What do you mean by the current ratio?
Answer:
Current ratio gives the proportion of current assets to current liabilities of a business concern. It is computed by dividing current assets by current liabilities. The current ratio indicates the ability of an entity to meet its current liabilities as and when they are due for payment.

Question 8.
What do you mean by Long-term solves ratios? what are its types?
Answer:
Long term solvency means the firm’s ability to meet its liabilities in the long run. Long term solvency ratios help to determine the ability of the business to repay its debts in the long run. The following ratios are normally computed for evaluating long term solvency of the business:

• Debt equity ratio
• proprietary ratio
• Capital gearing ratio

Question 9.
What do you mean by Turnover Ratios? What are its types?
Answer:
Turnover ratios show how efficiently assets or other items have been used to generate revenue from operations. They are also called as activity ratios or efficiency ratios. They how the speed of movement of various items. They are expressed as number of times in relation to the item compared.
The important turnover ratios are:

• Inventory turnover ratio
• Trade receivable turnover ratio
• Trade payable turnover ratio
• Fixed assets turnover ratio.

Question 10.
What do you mean by Trade Receivable Turnover ratio?
Answer:
Trade receivable turnover ratio is the comparison of credit revenue from operations with average trade receivables during an accounting period. It gives the velocity of the collection of cash from trade receivables.

Question 11.
What do you mean by Debt collection Period?
Answer:
Debt collection period is the average time taken to collect the amount due from trade receivables. Lesser the debt collection period, grater is the efficiency of management in the collection of cash from trade receivables. It is calculated as follows.

What do you mean by Trade payable Turnover ratio?
Trade payable turnover ratio is the comparison of net credit purchases with average trade payables during an accounting period. It gives the velocity to payment of cash towards trade payables.

Question 12.
What do you mean by credit payment period?
Answer:
It is the average time taken by the business for payment of accounts payable. Lesser the credit payment period, greater is the efficiency of the management in managing accounts payable as it indicated quicker settlement of trade payables. It is calculated as follows:

Question 13.
What do you mean by Fixed Assets Turnover
Answer:
The fixed assets turnover ratio gives the number of times the fixed assets are turned over during the year in relation to the revenue from operations. This ratio indicates the efficiency of utilization of fixed assets.

Question 14.
What do you mean by profitability ratios? What are its types?
Answer:
Profitability ratios help to assess the profitability of a business concern. These rations also help to analyze the earning capacity of the business in terms of utilization of resources employed in the business. Generally, these rations are expressed as a percentage.
The profitability ratios commonly used are:

1. Gross profit ratio
2. Operating cost ratio
3. Operating profit ratio
4. Net profit ratio
5. Return on investment.