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Andhra Pradesh BIEAP AP Inter 1st Year Accountancy Study Material 7th Lesson Journal Proper Textbook Questions and Answers.

AP Inter 1st Year Accountancy Study Material 7th Lesson Journal Proper

Questions

Question 1.
What is Journal Proper ?
Answer:
There are some transactions which cannot be recorded in any of the remaining seven subsidiary books, are entered in special book known as “Journal proper”.

Question 2.
Explain various kinds of transactions that are recorded in Journal Proper.
Answer:
The following are the transactions recorded in journal proper.
1) Opening entries : Opening entries are passed at the commencement of the new year, to record the balances of assets and liabilities brought forward from the previous year.
The rule to be applied as
Assets A/C
To Liabilities a/c
To Capital a/c
(Being last year balance brought forward).

2) Purchase and sale of fixed assets: Business organisations either purchase or sell the assets for cash and sometimes on credit. If it is purchased on credit, then the entry should be recorded in the journal proper.
Example : Purchased machinery from Godrej Company 40,000.

3) Rectification entries : Sometimes errors may occur while recording transactions, posting them into the ledger or while balancing the ledger accounts. In such cases certain entries should be passed in order to rectify the errors. Such entries are called “Rectification entries”.
Example : Rama paid Rs. 1000. This was credited to Bheema’s A/c.

4) Adjustment entries: At the time of preparation of final accounts of the business firm some adjustments are to be made. The journal entries relating to these adjustments are known as “Adjustment entries”.
Example: Outstanding expenses, prepaid expenses, incomes receivable, income received in advance, depreciation, etc.

5) Closing entries: At the end of every financial year the balances of all the Nominal accounts are transferred to the Trading and Profit & Loss accounts. These transferred entries are known as Closing entries. All expenses and losses are debited and incomes and gains are credited to Trading, Profit & Loss account.

6) Transfer entries: Sometimes the trader transfers the amount from one account to another account. This type of journal entries are known as ‘Transferred journal entries’.
Example : Transfer of Profit to Reserve Fund.

7) Other entries : It is not possible to enter same transactions in journal and hence they are recorded in Journal Proper.
Example: Interest on drawings, interest on capital, goods lost due to fire & theft, goods sent on consignment, etc.

Question 3.
Explain the following in not more than 5 lines.
a) Opening Entries
b) Rectification Entries
c) Adjustment Entries
d) Closing Entries
Answer:
a) Opening Entries : Opening entries are passed at the commencement of the new year, to record the balances of assets and liabilities brought forward from the previous year.
The rule to be applied as
Assets a/c Dr
To Liabilities a/c
To Capital a/c
(Being the last year balance brought forward).

b) Rectification Entries : Sometimes errors may occur while recording transactions, posting them into the ledger or while balancing the ledger accounts. In such cases certain entries should be passed in order to rectify the errors. Such entries are called “Rectification entries”.
Example : Rama paid Rs. 100. This was credited to Somu’s A/c.

c) Adjustment Entries: At the time of preparation of final accounts of the business firm some adjustments are to be made. The journal entries relating to these adjustments are known as “Adjustment entries”.
Example: Outstanding expenses, prepaid expenses.

d) Closing Entries: At the end of every financial year the balances of all the Nominal accounts are transferred to the Trading and Profit & Loss accounts. These transferred entries are known as Closing entries. All expenses and losses are debited and incomes and gains are credited to Trading, Profit & Loss account.

Problems

Question 1.
Record the Opening entry from the following particulars on 1st April 2013: (Mar. 2019 – T.S.) (Mar. 2018 : May 17 – A.P.)

Answer:

Question 2.
Write the Opening entry on 1st Jan. 2013 in the books of Ram from the following:

Answer:

Question 3.
Record the Opening entry from the following assets and liabilities as on Jan. 1st, 2014:

Answer:

Question 4.
Pass Opening Journal Entry from the following particulars on 1^st April, 2013.

Question 5.
Record Opening entry on 1^st Jan. 2014 from the following assets and liabilities:


Answer:

Question 6.
Rectify the following errors dated on Dec. 31^st, 2013:
1. Machinery purchased Rs. 20,000, wrongly debited to Purchases a/c,
2. Commission received Rs. 3,000, wrongly credited to Interest received a/c,
3. Salaries paid to Accountant Prakash Rs. 10,000, debited to his personal a/c,
4. Purchased goods for Rs. 8,000, recorded as Rs. 80,000,
5. Paid cash to Rohit Rs. 5000, wrongly debited to Mohit a/c.
Answer:

Question 7.
Pass adjustment journal entries for the following:
1. Salaries Outstanding Rs. 2,000,
2. Insurance Paid in Advance Rs. 500,
3. Credit 5% Reserve for Bad and Doubtful Debts on Debtors amounted to Rs. 10,000,
4. Provide 10% Depreciation on Machinery. Machinery Value Rs. 20,000,
5. Closing Stock Rs. 15,000.
Answer:

Question 8.
Record the Closing Entries from the following Ledger Balances ofAshrith :

Answer:

Question 9.
Write Opening Entry as on Jan. 1^st 2014 from the following balance sheet of Venkat: Balance Sheet of Venkat as on 31^st Dec. 2013

Answer:

Question 10.
Write Opening Journal Entry from the following as on 1^st April 2013 : (Mar. 2018-T.S.)

Answer:

Student Activity

Visit any organisation/firm and record opening, closing and other entries of Journal proper.

Andhra Pradesh BIEAP AP Inter 1st Year Accountancy Study Material 6th Lesson Subsidiary Books Textbook Questions and Answers.

AP Inter 1st Year Accountancy Study Material 6th Lesson Subsidiary Books

Questions

Question 1.
Explain the various Subsidiary Books with suitable examples.
Answer:
Separate Books or special books which are maintained to record a particular category of trans¬actions are known as “Subsidiary books”.
The journal is divided into eight subsidiary books. They are :
a) Purchase book : This book records all the goods purchased on credit. Cash purchases and purchase of assets are not recorded in this book.
Whenever goods are bought on credit we receive an invoice from the seller. It is called “Inward Invoice”. It is evidence for recording in purchase book.

b) Sales book : The goods sold on credit are recorded in this book. Cash sales and assets sold for cash or credit are not recorded in this book.
Whenever goods are sold on credit we send an Invoice to the customer. It is called “Outward Invoice”. It is evidence for recording in sales book.

c) Purchase returns book: This book is also known as returns outward book. This book records the goods returned to the suppliers. When goods are returned to their sellers a Debit Note’ is prepared and sent to the suppliers along with goods.

d) Sales returns book: This book is also known as returns inward book. This book records the goods returned by the customers. When the sellers receive goods back from the customer along with a debit note, then he prepares a credit note and sends to the customer.

e) Cash book: Cash book is used for recording all cash transactions i.e. cash receipts and cash payments either in cash or by cheque are recorded in this book.

f) Bills receivable book: This book is used to record all the bills received from the customers for the amount due. It contains details of acceptor of the bill, its due date, date bill, the amount due, etc.

g) Bills payable book: If the goods are purchased on credit and bills are accepted for the due amounts, all these bills payable are recorded in this book. It contains the amount due, date of bill, place of payment, due date, etc .

h) Journal proper : This book is used to record only those transactions, which cannot be recorded in any one of the seven subsidiary books side above.

Question 2.
Give the advantages of subsidiary books.
Answer:
Subsidiary Book Advantages:
a) Saving of Time : No need of writing the Journal Entries. Transactions are directly entered into their respective journals.
b) Division of Work: By entrusting different Subsidiary Books to different persons, division of principle can be implemented.
c) Easy Recording: Transactions can be recorded very fast and easy.
d) Improves Efficiency: Accounting work will be done efficiently by allotting work to different experts who prepare the Special Books.
e) Detection of Errors : Since Separate Books are maintained to record a particular set of transactions, errors can be easily noticed.

Question 3.
Explain about Purchases book and draw the pro forma of it.
Answer:
This book records all the goods purchased on credit. Cash purchases and purchase of assets are not recorded in this book.

Whenever goods are bought on credit we receive an invoice from the seller. It is called “Inward Invoice”. It is evidence for recording in purchase book. It is numbered serially and filed in a separate file.
Proforma of purchase book:

Purchase book contains five columns, such as Date, Particulars, Invoice No., L.F.No. and Amount.

Question 4.
Explain about Sales book and draw the pro forma of it.
Answer:
The goods sold on credit are recorded in this book. Cash sales and assets sold for cash or credit are not recorded in this book.
Whenever goods are sold on credit we send an invoice to the customer. It is called “Outward invoice”. It is evidence for recording in sales book. It is serially numbered and kept in a separate file.

Proforma of sales book:

Sales book contains five columns, such as Date, Particulars, Invoice No., L.F.No. and Amount.

Question 5.
Answer the following in not more than 5 lines. (Mar. ’17 – A.P.)
a) Invoice
b) Debit Note
c) Credit Note
d) Trade Discount
e) Journal Proper
Answer:
a) Invoice:
It is the document prepared by the supplier of goods with all the details like quantity, price discount offered and other terms and conditions. This is also known as “Inward Invoice”.

b) Debit Note: (Mar. 2019 – A.P.) (Mar. 2018 – T.S.)
If goods are returned to the supplier, a debit note or letter is sent along with goods, informing the supplier about the debit given to his account for the value of the goods returned. This note is known as debit note.

c) Credit Note:
If the customer returned the goods, a credit note is sent to him by the traders informing the customer that his account has been credited with the value of the goods returned.

d) Trade Discount:
It is a reduction in the catalogue price of an article. This is given by the wholesaler to the retailer to enable him to sell at a catalogue price and make a profit. It is not entered in books.

e) Journal Proper: (May ’17 – A.P.)
This book is used for recording only those transactions which cannot be recorded in any one of the seven subsidiary books said above.
Examples: Opening entries, closing entries, adjustment entries, transfer entries.

Problems

Question 1.
Record the following transactions in the purchase book.

Hint 1: For transaction dated on 11^th Calculate trade discount @ 10% on Rs. 5,000/- i.e. 5000 × \(\frac{10}{100}\) = 500, and take net purchases as 5000 – 500 = 4500.
Hint 2: Transaction dated 17^th March is a cash transaction. So, it is not to be taken in purchase book.
Answer:

Question 2.
Prepare Purchase Book from the following:

Hint 1: For transaction dated on 10^th calculate discount @ 10% on 10,000/-
i.e. 1,000 and takes net purchases as 10,000 – 1,000 = 9,000.
Hint 2: Transaction dated on 12^th is a cash transaction. So, it is not entered in purchases book.
Answer:

Question 3.
Record the following in Purchases Book:
2012
March 1^st Purchase goods on credit from Chand traders Invoice No. 301.
120 reams of white paper @15 per ream
60 dozen ink pots @ 25 per dozen

March 8^th Purchased from Gupta traders Invoice No. 403
10 Drawing boards @ 30 each
40 Notebooks @ 20 each

March 10^th Purchased goods from Goyal and Company for cash 3,000.

March 15^th Purchased goods from Ankit Rs. 10,000 as per Invoice No. 450.

Hint 1 : Transaction dated on March 1^st Net Purchases amount is to be calculated as, (120 × 15) + (60 × 25) = 1800 + 1500 = 3300
Hint 2 : Transaction dated on March 8^th Net Purchases amount is,
(10 × 30) + (40 × 20) = 300 + 800 = Rs. 1100
Answer:

Question 4.
Enter the following in Purchase Book and Purchase Returns Book: (Mar. 2019 – A.P. & T.S.)

Answer:

Question 5.
Prepare Sales Book (May 17 – T.S.)

Hint 1: Trade discount is 4000 × 5/100. Net sales = 3,800.
Hint 2: Transaction dated on 14th is a Cash transaction.
Answer:

Question 6.
Enter the following transactions in Sales Book and prepare Ledger.

Hint 1: For transaction dated on 18^th calculate trade discount i.e. 8000 × 10/100 = 800, net sales = 7200.
Hint 2: Transaction dated on 20th is a cash transaction.
Answer:

Ledgers:

Question 7.
Prepare Sales Book and Sales Returns Book from the following. (Mar. 2018 – A.P. & T.S.)

Answer:

Question 8.
Record the following transactions in Proper Subsidiary Books.

Answer:

Question 9.
Enter the following transactions in the related Subsidiary Books.

Hint 1: For transaction dated on 10^th Trade discount is 15000 × 10/100 = 1500
Net purchases = 13500
Hint 2: Transaction dated on 18^th is a cash transaction.
Hint 3: For transaction dated on 20^th Trade discount is 3000 × 5/100 = 150
Net sales = 2850
Hint 4: Transaction dated on 26^th is to be entered in journal proper.
Hint 5: For transaction dated on 27^th Trade discount is 8000 × 15/100 = 1200
Net sales = 6,800
Answer:

Question 10.
Prepare Purchases Book.

Answer:

Question 11.
Enter the following transactions in Purchases Book and post them in Ledger.

Answer:

Ledgers:

Question 12.
Record the following in Sales Book and Sales Returns Book.


Answer:

Question 13.
Enter the following transactions in Proper Subsidiary Books: (May ’17 – A.P.)

Hint : Transaction dated Nov. 19^th trade discount 15% of 10000 i.e.,
10000 × 15/100 = 1500 is to be deducted from purchase amount Rs. 10000.
Net purchases 10000 – 1500 = 8500

Question 14.
From the following transactions prepare Sales Book and Sales Returns Book. (IPE. Mar. 14)

Answer:

Question 15.
Enter the. following transactions in proper Subsidiary Books. (IPE. Mar. 14)

Answer:

Student Activity

Collect the invoices, debit notes, credit notes from any business organization.

Andhra Pradesh BIEAP AP Inter 1st Year Economics Study Material 2nd Lesson Theory of Consumers Behaviour Textbook Questions and Answers.

AP Inter 1st Year Economics Study Material 2nd Lesson Theory of Consumers Behaviour

Essay Questions

Question 1.
Explain the law of diminishing marginal utility and are its limitations. [March 17]
Answer:
Hermann Heinrich Gossen was the first economist to explain the law of diminishing marginal utility in 1854. It is also known as Gossen’s ‘first law’. In 1890 Marshall in his principles of economics developed and popularised this analysis. This law explains the functional relationship between the stock of commodity the marginal utility of commodity.

According to Marshall “The additional benefit which a person derives from a given increase of his stock thing diminishes with every increase in stock that he already has”.

“A consumer increases the consumption of any one commodity keeping constant the consumption of all other commodities the marginal utility of the variable commodity must eventually decline”. Kenneth E.Boulding.

The law says that as we gone consuming a commodity satisfaction derives from its additional units goes on diminishes.
Assumptions :

1. Rationality : Consumer is a rational man which means he always tries to get maximum satisfaction.
2. Cardinal measurement of utility : Utility is a cardinal concept, i.e., utility can be measured and compared numerically.
3. Utilities are independent: It implies that utility of any commodity depends as its own quantity
4. Homogeneous : Units of the commodity are similar in quantity, size, taste and colour etc.
5. No time gap : There should not be any time gap between the consumption of one unit and other it.
6. Constant marginal utility : It is assumed that the marginal utility of money remains constant. ’
7. Total & marginal utility : Total utility : Total satisfaction obtained by the consumer from the consumption of a given quantity of commodity.
   TU_n = f(Q_n)
   Where TU_n = Total utility of n commodity,
   f = functional relationship,
   Q_n = Quantity of n commodity.

Marginal utility: Marginal utility is the addition made to the total utility by consuming one more unit of the commodity.
It can be explained as
MU_n = TU_n – TU_n-1
MU_n = Marginal utility of n^th unit
TU_n = Total utility of n^th unit
TU_n-1 = Total utility of n – 1 units.
MU may also be expressed as follows.
Marginal utility is the additional utility derived from the consumption of an extra unit of commodity.
MU = \(\frac{\Delta \mathrm{TU}}{\Delta \mathrm{C}}\)
Where ∆TU = Change in total utility
∆C = Change in no. of units consumed.

Explanation of the law : The law of diminishing marginal utility explains the relation between the quantity of good and its marginal utility. If a person goes on increasing his stock of a thing, the marginal utility derived from an additional unit declines. We show this tendency with an imaginary table given below.

In the table let us suppose that one is fond of apples. As he consumes one apple after another he derives less and less satisfaction. The first unit consumed with atmost pleasure. For the second, the intensity of his desire diminishes. The third will be still less and so on. The total utility increasing until the consumption of fourth unit of good but diminishing rate. Fifth unit of apple gives him maximum total utility. But marginal utility becomes zero. Further consumption of sixth unit TU diminishes and MU becomes negative.

The relationship between total utility and marginal utility is explained in the following three ways.

1. When total utility increases at diminishing rate, marginal utility falls.
2. When total utility is maximum, marginal utility becomes zero.
3. When total utility decreases, marginal utility becomes negative.

This can be shown in the following diagram.

In the diagram on ‘X’ axis measures units of apples and OY axis measures total utility and marginal utility. TU curve represents total utility and MU curve represents marginal utility. TU curve is maximum at 5^th unit where MU curve will become zero. TU curve slopes downwards from 6^th unit, while MU will become negative.

Limitations or Exceptions :

1. Hobbies: This law does not operate in the case of hobbies like collection of stamps, old paintings, coins etc. Greater the collections of a person, greater is his satisfaction. Marginal utility will not diminish.
2. Drunkers : It is pointed out that the consumption of liquor is not subject to the law of diminishing marginal utility. The more a person drinks liquor, the more he likes it.
3. Miser : This law does not apply to money. The more money a person has the greater is the desire to acquire still more of it.
4. Further, this law does not hold good if any change in income of the consumer, tastes and preferences.

Question 2.
Critically examine the law of equi-marginal utility.
Answer:
Law of equi – marginal utility is an important law of consumption. It is called as “Gossen’s second law”, as its formulations is associated with the name of H.H.Gossen.

According to Marshall “If person has a thing which can be put to several uses, he will distribute it among these uses in such a way that it has the same marginal utility in all uses. If it had a greater marginal utility in one use than in another, he would gain by taking away some of it from the second and applying it to the first”.

According to this law the consumer has to distribute his money income on different uses in such a manner that the last rupee spent on each commodity gives him the same marginal utility. Equalisation of marginal utility in different uses will maximise his total satisfaction. Hence this law is known as the “Law of equi-marginal utility”.

The fundamental condition for consumer’s equilibrium can be explained in the following way.
\(\frac{\mathrm{MU}_{\mathrm{x}}}{\mathrm{P}_{\mathrm{x}}}=\frac{\mathrm{MU}_{\mathrm{y}}}{\mathrm{P}_{\mathrm{y}}}=\frac{\mathrm{MU}_{\mathrm{z}}}{\mathrm{P}_{\mathrm{z}}}\) = MU_m
Where MU_x, MU_y, MU_z, MU_m = Marginal utilities of commodities x, y, z, m;
P_x, p_y, p_z = Prices of X, y, z.
This law can be explained with the help of a table. Suppose the consumer is prepared to spend his money income is ₹ 26/- on two goods X and Y. Market prices of two goods are ₹ 4/- & ₹ 5/- respectively. Now the marginal utilities of good x & good y are shown below.

For explaining consumer’s maximum satisfaction and consequent equilibrum position we need to reconstruct the above table by dividing marginal utilities of x by its price ₹ 4/- and marginal utility of y by ₹ 5/-. This is shown in the following table.

In the table it is clear that when consumer purchase 4 units of goods x & 2 units of good y. Therefore, consumer will.be in equilibrium when he is spending (4 × 4 = 16 + 2 × 5 = 10) ₹ 26 on them.
Limitations of the law : The law of equi marginal utility has been subject to certain limitations which are as given below.

1. The law assumes that consumer is a rational man and always tries to get maximum satisfaction. But it is not possible always to compare the utilities derived from different commodities.
2. This law not applicable when goods are indivisible.
3. The law is based on unrealistic assumptions like cardinal measurement of utility and marginal utility of money remains constant. In real world MU of money does not remain constant.
4. This law will not be applicable to complementary goods.
5. Another limitations of this law is that there is no fixed accounting period for the consumer in which he can buy and consume goods.

Question 3.
Explain consumer’s equilibrium with law of equi-marginal utility. [March 18]
Answer:
Law of equi-marginal utility is an important law of consumption. It is called as “Gossen’s with the name of H.H.Gossen. According to Marshall “If person has a thing which can be put to several uses, he will distribute it among these uses in such a way that it has the same marginal utility in all uses. If it had a greater marginal utility in one use than in another, he would gain by taking away some of it from the second and applying it to the first”.

According to this law the consumer has to distribute his money income on different uses in such a manner that the last rupee spent on each commodity gives him the same marginal utility. Equalisation of marginal utility in different uses will maximise his total satisfaction. Hence this law is known as “law of equi-marginal utility”.

Assumptions of the law : The law of equi-marginal utility depends on the following assumptions.

1. This law is based on cardinal measurement of utility.
2. Consumer is a national man always aiming at maximum satisfaction.
3. The marginal utility of money remains constant.
4. Consumer’s income is limited and he is proposed to spent the entire amount on different goods.
5. The price of goods are unchanged.
6. Utility derived from one commodity is independent of the utility of the other commodity.

The fundamental condition for consumer’s equilibrium can be explained in the following way.
\(\frac{\mathrm{MU}_{\mathrm{x}}}{\mathrm{P}_{\mathrm{x}}}=\frac{\mathrm{MU}_{\mathrm{y}}}{\mathrm{P}_{\mathrm{y}}}=\frac{\mathrm{MU}_{\mathrm{z}}}{\mathrm{P}_{\mathrm{z}}}\) = MU_m
Where MU_x, MU_y, MU_x, MU_m = marginal utility of commodities X, Y, Z, m;
P_x, P_y, P_z = prices of x, y, z.
This law can be explained with the help of a table. Suppose the consumer is prepared to spend his money income is ₹ 26/- on two goods x and y. Market prices of two goods are ₹ 4/- & ₹ 5/- respectively. Now the marginal utilities of goods x & y are shown below.

For explaining consumer’s maximum satisfaction and consequent equilibrum position we need to reconstruct the above table by dividing marginal utilities of x its price ₹ 4/- and marginal utility of y by ₹ 5/-. This is shown in the following table.

In the table it is clear that when consumer purchase 4 units of goods x & 2 units of good y. Therefore, consumer will be in equilibrium when he is spending (4 × 4 = 16 + 2 × 5 = 10) ₹ 26 on them.

Consumer’s equilibrium may be shown in the diagram.

Consumer’s equilibrium by using principle of equi-marginal utility

In the diagram marginal utility curves of goods slope downwards i.e. AB & CD taking of the income of the consumer as given, suppose his M>_U of money constant at OE. \(\frac{\mathrm{MU}_{\mathrm{x}}}{\mathrm{P}_{\mathrm{x}}}\) is equal to OE when OG quantity of good X is brought. \(\frac{\mathrm{MU}_{\mathrm{y}}}{\mathrm{P}_{\mathrm{y}}}\) is equal to OE, when OF quantity of good y is purchased. Thus consumer purchasing OG of X and OF of Y. \(\frac{\mathrm{MU}_{\mathrm{x}}}{\mathrm{P}_{\mathrm{x}}}=\frac{\mathrm{MU}_{\mathrm{y}}}{\mathrm{P}_{\mathrm{y}}}\) = MU_m. This is consumer’s equilibrium.

Question 4.
Explain the consumer’s equilibrium using indifference curve.
Answer:
The point where the consumer gets maximum possible satisfaction, where the budget line is tangent to the indifference curve and the MRS is equal to the price ratio of the two goods will be defined as equilibrium of the consumer.
Assumptions :

1. Consumer scale of preferences must remain constant.
2. Money income of thfe consumer must remain constant.
3. The price of two goods must remain unchanged.
4. There should be no change in the tastes and habits of the consumer.
5. The consumer is rational and thus maximises his satisfaction.

Conditions of equilibrium : There are two conditions that must be satisifed for the consumer to be in equilibrium. They are :

1. At the point of equilibrium, the budget / price line must be tangent to the indifference curve.
2. At the point of equilibrium, the consumer’s MRS_xy and the price ratio must be equal.
   i.e„ MRS_xy = \(\frac{P_x}{P_y}\)

This can be shown in the following diagram.

In the diagram ‘AB’ is consumer’s budget or price line. IC, IC₁, IC₂ are indifference curves. In the diagram the consumer is equilibrium OM of x and ON of y. At point E the price line touches to an IC₁. At point ‘S’ consumer will be on ‘O’ lower indifference curve IC and will be getting less satisfaction than at E on IC. IC₂ is beyond the capacity of consumer. So it is outside to the budget line.

Question 5.
Define price line / budget line and explain shifts in the budget line.
Answer:
The budget line or price line shows all possible combinations of two goods that a consumer can buy with the given income of the consumer and prices of the two goods.

The concept of budget / price line will be shown in the following example. Suppose that a consumer has ₹ 150 (income) to buy two goods namely X and Y. Whose prices are ₹ 15 and ₹ 30 each. With the given information now we can draw the budget or price line as shown in the diagram.

In the above diagram ‘AB’ is the ‘budget or price line’. The slope of the line AB represents the ratio of the prices of X and Y in such a manner that 10 of x will be equal to 5 of y.

Shifts in the Budget line : The position of the budget line depends upon size of money income of the consumer. If his income increase and the price of the two commodities remaining the same, the consumer can buy more of both the commodities. On the other hand, if his income decreases, the prices of the two commodities remaining the same the consumer now to reduce his purchase. As a result of changes in the consumer income, there will be shifts in budget line also. The same is shown in the following diagram.

1. When the income of the consumer increase, the budget line moves towards right from the original to AB to A’B’.
2. When the income of the consumer decrease, the budget line moves to the left from the original / initial AB to A”B”.

Change in price line : The slope of the price line depends on the prices of both the commodities there will be a change in the slope of the price line when there is a change in the price or either of the two commodities.
i) Suppose that the price of ‘X’ falls, while the price of Y and money income of the consumer remaining the same.

In the above diagram, the initial price line of AB, before change in the price of X’. Suppose that the price of X’ has fallen and the price of ‘Y’ remaining the same.

ii) Suppose that the price Y falls, while the price of X and money income of the consumer re-maining the same.

In the above diagram, the initial line is AB that is before a change in price of Y.

Short Answer Questions

Question 1.
Difference between cardinal and ordinal utility.
Answer:
The concept of utility was introduce by Benham in 1789. Utility means want satisfying power of a commodity. It is a psychological phenomenon. The measurement of utility, there are two different approaches.

1. Cardinal utility
2. Ordinal utility

1) Cardinal utility : This approach was developed by Alfred Marshall. According to him utility is psychological concept. So it can be measured ‘util’. The numbers 1, 2, 3, 4 etc are cardinal numbers. According to this analysis the utilities derived from consumption of different commodities can be measured in terms of arbitary units, such as 1, 2, 3 … and so on.

2) Ordinal utility : This approach was developed by R.J.D. Allen and J.R.Hicks. According to them utility is psychological concept. So we cannot measure in numerically much less compared. The numbers 1^st, 2^nd, 3^rd, 4^th etc., are ordinal numbers. The ordinal numbers are ranked. It means the utilities obtained by the consumer from different goods can be arranged in a serial order such as 1^st, 2^nd, 3^rd, 4^th etc.

Question 2.
Properties of IC (Indifference curves). [March 18, 17, 16]
Answer:
The important properties of indifference curves are :

1. Indifference curves slopes downwards from left to right there exists negative slope.
2. Indifference curves are convex to the origin because of diminishing marginal rate of substitution.
3. Indifference curves can never intersect each other.
4. Higher indifference curve represent higher level of satisfaction. Indifference curve to the right represents higher satisfaction.

Question 3.
Marginal Rate of Substitution
Answer:
The concept of MRS (Marginal Rate of Substitution) is the basis of indifference curves. The MRS may be defined as the rate at which an individual will exchange successive unity of one commodity for another. This can be explained with the help of following example.

From the table it is clear that consumer has 15 units of good Y and 1 of good X, he is willing to forego 4Y for IX. Here marginal rate of substitution of X for Y is 4 :1 and for 3^rd combination it is 3 : 1 and so on. The MRS xy diminishes as the consumer goes on substituting X for Y.
MRS_xy = \(\frac{\Delta Y}{\Delta X}\)

Question 4.
Indifference Map
Answer:
A set of indifference curves drawn for different income levels is called as “indifference map”. In other words indifference map is the locus of various indifference curves various levels of satisfaction to the consumer.

From the above diagram it is clear that an indifference map of IC₁, IC₂, IC₂. Each curve shows a certain level of satisfaction to the consumer: The different indifference curves are always arranged and numbered in ascending order.

As are moves from IC₁ to IC₃ on an indifference map, the IC labelled higher number (IC₃) is preferred to the IC labelled lower number (IC₁).

Question 5.
Price Line.
Answer:
The budget / price line shows all possible combinations of two goods that a consumer can buy, with given income of the consumer and prices of the two goods. But which particular combination of two goods on IC he can get depends on two factors.

1. Consumer’s money income
2. Prices of two goods.

The concept of budget / price line will be shown in the following diagram.

In the above diagram PL is Budget line. If consumer spend his total income on good X, he could get L. Any point outside the given price line H will beyond the capacity of consumer, K is under spending Capacity.

The concept of price line or budget line can be known the following example. Suppose the consumer has ₹ 5/- to buy two goods say X and Y prices of X and Y are ₹ 1/- and 0.50 paisa. Then the following are the opportunities available before him.

1. If consumer spends his total amount of ₹ 5 on X only he gets 5 units of X and none of Y.
2. If he spends whole of his money i.e., ₹ 5 on Y only he gets 10 units of Y and none of X.
3. If consumer wants to have both X & Y. Therefore he can move within OAB of price line.

Additional Questions

Question 6.
Explain the importance of law of equi-marginal utility.
Answer:
The law of equi-marginal utility states that a consumer will be in equilibrium when the marginal utility of the various commodity are equal.
Importance :

1. Basis of consumer expenditure : It is basis for consumers expenditure and guide the consumers while allocating resources.
2. In the field of production : This law is useful to the producer and it explains how a producer maximises his profits and reduces cost of production.
3. Exchange : In all our exchanges, this law works, for exchange is nothing but substitution of one thing for another.
4. Public Finance : This law helps the government in the allocation of scarce resources and also the government levied taxes on the basis of this principle.
5. Price determination : This principle has an important bearing on the determination of value.

Question 7.
Explain superiority of the indifference curve technique.
Answer:

1. It is an analysis of multi-goods model.
2. It does not assume marginal utility of money remains constant.
3. It analyses income effect, price effect and substitution effect.
4. The assumption of ordinal measurement of utility made by indifference curve is less restrictive and more realistic.

Question 8.
What are the important assumptions of indifference techniques ?
Answer:
An indifference curve represents satisfaction of a consumer from two commodities. An IC curve can defined as the locus of points each representing a different combination of two goods yielding the same level of satisfaction.
Assumptions :

1. Rationality: It is assumed that the consumer tries to obtain maximum satisfaction from his expenditure.
2. Scale of preference : Consumer is able to arrange the available combinations of goods according to scale of preference.
3. Ordinal utility : It assumes ordinal utility approach. So utility measurable only ordinal terms i.e., 1^st, 2^nd, 3^rd ………… etc.
4. Diminishing marginal rate of substitution : It is the rate at which a consumer is willing to substitute commodity to another. So that this satisfaction remains the same.
5. Consistency : Consumer’s choices have to consistent. It means if consumer prefers A to B and B to C his choice reflects his rationality.
6. Completeness : The consumer’s scale of preferences is to complete that he is able to choose any one of the two combinations of commodities presented to him.

Very Short Answer Questions

Question 1.
Utility
Answer:
The want satisfying power or capacity of a commodity or service is known as utility. It is the basis of consumer’s demand for a commodity.

Question 2.
Cardinal utility [March 17]
Answer:
Alfred Marshall developed cardinal utility analysis. According to this analysis the utilities derived from consumption of different commodities can be measured in terms of arbitary units called utils. 1, 2, 3, 4 are called cardinal numbers.

Question 3.
Ordinal utility
Answer:
This was developed by J.R.Hicks, Allen. Utility is subjective and measurement of utility in numerical terms is not possible. We can observe the preference one for a good more than for another. Ordinal numbers such as 1^st, 2^nd, 3^rd ………….. etc. The ordinal numbers are ranked.

Question 4.
Scale of preferences
Answer:
Guides the consumer in his purchases. It reflects his tastes and preferences.

Question 5.
Price Line [March 16]
Answer:
It shows all possible combinations of two goods that a consumer can buy, with the given income of the consumer’s and prices of the two goods.

Question 6.
MRS
Answer:
The additional amount of one product required to compensate a consumer for a small decrease in the quantity of another, per unit of the decrease.

Additional Questions

Question 7.
Total utility
Answer:
Total utility is the total amount of utility which a consumer derives from a given stock of a commodity.
TU_n = f(Q_n)

Question 8.
Marginal utility
Answer:
Marginal utility is the additional utility obtained from the consumption of additional unit of the commodity.
MU_n = TU_n – TU_(n-1)
(or)
MU = \(\frac{\Delta \mathrm{TU}}{\Delta \mathrm{Q}}\)

Question 9.
Consumers equilibrium
Answer:
The term equilibrium implies a position of rest or changelessness. A consumer attains equilibrium only when he secures maximum satisfaction out of his expenditure. In distributing a commodity various uses, the consumer will secure maximum satisfaction if the marginal utility of the commodity equalised in all its uses.
\(\frac{\mathrm{MU}_{\mathrm{x}}}{\mathrm{P}_{\mathrm{x}}}=\frac{\mathrm{MU}_{\mathrm{y}}}{\mathrm{P}_{\mathrm{y}}}\) ………… and ao on.

Question 10.
Indifference curve
Answer:
It represents the satisfaction of a consumer from two goods of various combinations. It is drawn on the assumption that for all possible combinations of the two goods on an indifference curve, the satisfaction level remains the same.

Question 11.
Iso-utility curve
Answer:
Iso-utility curve is also known as indifference curve or the curve of equal utility. The situation where consumers yield the same level of total satisfaction at various combinations of the commodities called Iso-utility curve.

Question 12.
Indifference schedule
Answer:
It is a table representing the various combinations of goods which give equal satisfaction to the consumer. An indifference curve is drawn on the basis of an “indifference schedule”.

Question 13.
Indifference map
Answer:
A set of indifference curves drawn for different income levels is called indifference map.

From the above diagram it is clear that an indifference map of IC₁, IC₂, IC₃. Each curve shows a certain level of satisfaction to the consumer.

Andhra Pradesh BIEAP AP Inter 1st Year Economics Study Material 1st Lesson Introduction Textbook Questions and Answers.

AP Inter 1st Year Economics Study Material 1st Lesson Introduction

Essay Questions

Question 1.
Discuss Wealth definition.
Answer:
Adam Smith was the first person to give a precise definition of Economics and separate this study from other social sciences. Adam Smith is considered as ‘Father of Economics’. He defined it in his famous book Wealth of Nations’, as “An enquiry into the nature and causes of wealth of nations”. Most of the economists in the 19th century held this view.

J.B. Say states that “The aim of political economy is to show the way in which wealth is produced, distributed and consumed”. The other economists who supported this definition are J.B. Say, J.S.Mill, Walker and others.
The main features of Wealth definition :

1. Acquisition of wealth is considered as the main objective of human activity.
2. Wealth means material things.
3. Human beings are guided by self-interest, whose objective is to accumulate more and more wealth.

Criticism : The wealth definition was severely criticised by many writers due to its defects.

1. Economists like Carlyled and Ruskin pointed out that economics must discuss ordinary man’s activities. So they called it as a ‘Dismal Science’.
2. Adam Smith’s definition, wealth was considered to consist of only material things and services are not included. Due to this the scope of economics is limited.
3. Marshall pointed out wealth is only a means to an end but not an end in itself.
4. This definition concentrated mainly on the production side and neglected distributed side.

Question 2.
Explain Welfare definition.
Answer:
Alfred Marshall tried to remedy the defects of wealth definition in 1890. He shifted emphasis from production of wealth to distribution of wealth.

According to Marshall “Political economy or Economics is a study of mankind in the : ordinary business of life. It examines that part of individual and social action which is most closely connected with the attainment and with the use of material requisites of well-being. Thus Economics is on one side, a study of wealth and on the other and more important side, a part of study of man”.

Edwin Cannan defined it as “The aim of political economy is the explanation of the general causes on which the material welfare of human beings depends”.

In the words of Pigou “The range of enquiry becomes restricted to that part of social welfare that can be brought directly or indirectly into relation with the measuring rod of money”.

The main features of Welfare definition :

1. Economics as a social science is concerned with man’s ordinary business of life.
2. Economics studies only economic aspects of human life and it has no concern with the social, political and religious aspects of human life. It examines that part of individual and social action which is closely connected with acquisition and use of material wealth for promotion of human welfare.
3. According to Marshall, the activities which contribute to material welfare are considered as economic activities.
4. He gave primary importance to man and his welfare and to wealth as means for the promotion of human welfare.

Criticism:

1. Robbins criticised Marshall’s economics is a ‘social science’ rather than a human science, which includes the study of actions of every human being.
2. Marshall’s definition mainly concentrated on the welfare derived from material things only. But non – materialistic goods which are also’ very important for the well being of the people. Hence, it is incomplete.
3. Critics pointed out that quantitative measurement of welfare is not possible. Welfare is a subjective concept and changes according to time, place and persons.
4. According to Marshall, economics deals with those activities of men which will promote human welfare. But production of alcohol and drugs do not promote human welfare. Hence the scope of economics is limited.
5. Another important criticism is that it is not concerned with the fundamental problem of scarcity of resources.
6. According to Robbins the economic problem arises due to unlimited wants and limited resources. These factors are ignored in this definition.

Question 3.
Explain how Robbings definition is superior to the welfare definition.
Answer:
Lionel Robbin’s of London School of Economics introduced the ‘Scarcity’ definition of Economics, in his book.

‘An Essay on the nature and significance of Economic Science’.

According to Robbin’s “Economics is the science which studies human behaviour as a relationship between ends and scarce means which have alternative uses”. Scarcity of resources is the central idea in Robbin’s definition.
Main features of Robbin’s definition :

1. Unlimited wants or ends
2. Means are scarce or limited
3. Means have alternative uses
4. Problem of choice.

Welfare definition: According to Marshall “Economics is on one side a study of wealth and on the other and more important side a part of study of man.” Marshall in his definition gave more importance to man than wealth.

Marshall defined “Political economy or Economics is a study of mankind in the ordinary business of life, it examines that part of individual and social action which is most closely connected with the attainment and with the use of the material requisites of well-being.

Main features of welfare definition :

1. He assumed that Economics must be a science which is study of man kind in the ordinary business of life.
2. Economics is concerned with real man influenced by human considerations and it has no concern with the political, social and religious aspects of life.
3. Wealth is a means for promoting human welfare, i 4) The main emphasis of Marshall is on material welfare and the immaterial aspects are ignored.

Superiority of Robbin’s definition over Marshall’s definition :

1. According to Marshall “Economics Studies the activities of those people who live in society”. But Robbin’s says that Economics studies, all human activities whether they promote human welfare or not.
2. According to Marshall “The scope of Economics is limited. But Robbin’s the scope of Economics is wide”.
3. Robbin’s definition has universal applicability. Because it is applicable to all types of societies.
4. Robbin’s definition of Economics is neutral between ends. He made economics a positive science. It does not pass value judgements.

Question 4.
Define Prof. Samuelson’s growth definition.
Answer:
Robbin’s definition does not take into consideration the dynamic problem of economic growth. As the time passes the scarcity of means ends, targets choices undergo a change. The inherent defect of Robbins definition has been rectified by Paul Samuelson in his definition of Economics.

Prof. Paul Samuelson, a Nobel Prize winner of 1970 provided a new definition of economics in which he introduced time element and it is dynamic in nature. Therefore his definition is known as growth oriented definition.

According to Samuelson “Economics is the study of how people and society choosing with or without the use of money, to employ scarce productive resources that could have alternative use to produce various commodities and distribute them for consumption. Now or in the future among various persons and groups in society.
Important features of the definition :

1. Scarcity : Like Robbins, Samuelson emphasises the scarcity of resources, unlimited wants and the alternative uses for the means.
2. Dynamism : Samuelson’s definition is dynamic. He talks about production, distribution and consumption in the present and also in the future.
3. Wide Scope : This definition widen the scope of Economics. It deals with problems of choice in a dynamic society.
4. Economic growth : He gave importance to economic growth the future consumption is safeguard by productive investment which leads to economic growth.

Thus Samuelson definition of economics is considered to be the most satisfactory definition of economics as it clearly states.

Question 5.
Distinguish between “Micro” and “Macro Economics”.
Answer:
Modem economic theory divided it into two branches, namely (i) Micro Economics (ii) Macro Economics. Ragnar Frisch was the first economist to use the words “Micro and Macro” in economic theory in 1930.

Micro Economics : The term “Micro Economics” is derived from the Greek word MIKROS’ which means small. Thus micro economics is the theory of small. It was developed by classical economists like Adam Smith, J.B.Say, J.S.Mill, Ricardo, Marshall etc. It studies about individual units or behaviour of that particular units like individual income, price, demand etc. Micro Economics is also known as partial analysis. If main, concentrates on the determination of prices of commodities and factors of production. It is also known as “Price theory”. According to K.E. Boulding Micro Economics is the study of particular firms, particular households, individual prices, wages, incomes individual industries and particular commodities.

Shapiro says “Micro Economics has got relation with small segments of the society.
Macro Economics : The term Macro Economics is derived from the Greek word ‘MAKROS’ which means large. Thus Macro Economics is the study of economic system as a whole. It was developed by J.M. Keynes. It studies aggregates in the economy like national income, total consumption, total saving and total employment etc. It is also known as Income and Employment theory.

According to Boulding “Macro Economics studies National Income not Individual income, general price level instead of individual prices and national output instead of individual output. Macro Economics also studies the economic problems like poverty, unemployment, economic growth, development etc. It is also deals with the theory of distribution.

The difference between Micro Econoinics and Macro Economics : Micro and Macro Economics are interrelated to each other. Inspite of close relationship between the two branches of economics, fundamentally they differ from each other.

Micro Economics

1. The word micro derived from the greek word ‘ Mikros’ means “small”.
2. Micro Economics is the study of individual units of the economy.
3. It is known as Price theory.
4. Micro Economics explains price determination both commodity and factor markets.
5. Micro Economics is based on price mechanism which depends on demand and supply.

Macro Economics

1. The word macro derived from the greek word ‘Makros’ which means large”.
2. Macro Economics is the study of economy as a whole.
3. It is known as Income and Employment theory.
4. Macro Economics deals with national income, total employment, general price level and economic growth.
5. Macro Economics based on aggregate demand. and aggregate supply.

Short Answer Questions

Question 1.
Free goods and Economic goods.
Answer:
Free goods

1. Free goods are nature’s gift.
   Ex : Air, Sunshine etc.
2. Their supply is abundant.
3. They do not have price.
4. These goods don’t have cost of production.
5. Free goods have only value in use.
6. These goods are not included in National Income.

Economic goods

1. Economic goods are man made.
   Ex: Book, Pen etc.
2. Supply is always less than their demand.
3. These goods command price.
4. Economic goods have cost of production.
5. Economic goods have both use value and exchange value.
6. Economic goods are included in National Income.

Question 2.
Characteristics of Wants. [March 18, 17, 16]
Answer:
Human wants are starting point of all economic activities. They depend on social and economic conditions of individuals.
Characteristic features of wants :
1) Unlimited wants : Human wants are unlimited. There is no end to human wants. When one want is satisfied another want takes its place. Wants differ from person to person, time to time and place to place.

2) A particular want is satiable : Although a man cannot satisfy all his wants, a particular want can be satisfied completely in a period of time.
Ex: If a person is thirsty he can satisfy it by drinking a glass of water.

3) Competition : Human wants unlimited. But the means to satisfy them are limited of scarce. Therefore they complete with each other in priority of satisfaction.

4) Complementary: To satisfy a particular want we need a group of commodities at the same time.
Ex: Writing need is satisfied only when we have pen, ink and paper together.

5) Substitution : Most of our wants can be satisfied by different ways.
Ex : If we feel hungry, we take some food and satisfy this want.

6) Recurring : Many wants appear again and again thought they are satisfied at one point of time.

7) Habits : Wants change into habits, which cannot be given up easily.
Ex : Smoking cigarettes for joke results into a habit if it is not controlled.

8) Wants vary with time, place and person : Wants go on changing with the passage of time. They are changing from time to time, place to place and person to person. Human wants are divided into 1. Necessities, 2. Comforts and 3. Luxuries.

Question 3.
Various types of utility.
Answer:
The want satisfying capacity of a commodity at a point of time is known as utility.
Types of utility:
1) Form utility: Form utilities are created by changing the shape, size and colour etc., of a commodity so as to increase its want satisfying power.
Ex : Conversion of a wooden log into a chair.

2) Place utility : By changing the place some goods acquire utility.
Ex : Sand on the sea shore has no utility. If it is brought out and transported to market, it gains utility. This is place utility.

3) Time utility : Time utilities are created by storage facility.
Ex : Business men store food grains in the stock points in the off season and releases them to markets to meet high demand and obtained super normal profits.

4) Service utility : Services also have the capacity to satisfy human wants.
Ex: Services of Lawyer, Teacher, Doctor etc. These services directly satisfy human wants.
Hence they are called as service utilities.

Question 4.
Jacob Viner’s definition. [March 18, 17, 16]
Answer:
Jacob Viner’s-definition of Economics is considered as modem definition of Economics. He is an American economist known for his short run and longrun cost curve analysis. According to Jacob Yiner “Economics is what economists do”.
The problems of the economy are :

1. What to produce and in what quantities : The economy has to decide whether to produce consumer goods and capital goods. These decisions are influence by individuals as well as government.
2. How to produce these goods : A decision has to be made whether to use labour intensive or capital intensive techniques.
3. For whom to produce these goods and services: It is concerned with the distribution of income and wealth among different sections of the society.
4. How efficient the productive resources are in use : This refers to the efficiency of economic system.
5. Whether the available resources are fully utilised : If resources are fully utilised that it can provide more employment opportunities.
6. Is the economy growing or static over a period of time.

Question 5.
Various economic investigations.
Answer:
According to Peterson “The term method refers to the techniques and producers used by economists for both construction and verification economic principles. There are plainly two methods used by the economists for conducting economic investigations. They are :

1. Deductive method
2. Inductive method.

1. Deductive method : This method is also known as the analytical and abstract method. The method of studying phenomenon by taking same assumptions and deducing conclusions from those assumptions is known as the deductive method. It proceeds from general to the particular or from universal to the individual. This was advocated by economists belonging to the classical school. There are four steps involved in drawing inference through deductive method. They are :

1. Selecting the problem
2. Formulating assumptions
3. Formulating hypothesis
4. Verifying the hypothesis.

The law of diminishing marginal utility is one law derived using this deductive method.
Merits of deduction:

1. It is less expensive and less time consuming.
2. It analyses complex economic phenomena and bring exactness to economic generalizations.
3. It helps in laying down basic principles of human behaviour.

Inspite of the above stated advantages, it is not free from limitations. It is based on unrealistic assumptions with little empirical content.

2. Inductive method : This method is also known as Historical, Empirical, Concrete, Ethical or Realistic. This method was strongly advocated and made use of by economists belonging to the historical school. This method proceeds from a part to the whole from particular to general or from the individual to the universal.
The following are the important steps involved in deriving economic generalisations through inductive method.

1. Selection of the problem
2. Collection of data
3. Observation
4. Generalisation

Merits of induction method :

1. It is nearer to reality and therefore expected to depict reality.
2. This method involves less chances of mistakes.

Inspite of several advantages it has its own defects. This method is expensive and consuming. It can be used by those who possess skill and competance in handling ex data.

Additional Questions

Question 6.
What is meant by Micro Economics ? Discuss its importance.
Answer:
The tehn ‘Micro Economics’ is derived from the greek word ‘MIKROS’ which means ‘small’. Thus Micro Economics deals with individual units like individual demand, price, supply etc. It was popularised by Marshall. It is also called as ‘Price Theory’ because it explains pricing in product market as well as factor market.
Importance :

1. Micro Economics provides the basis for understanding the working of the economy as a whole.
2. This study is useful to the government to frame suitable policies to active economic growth and stability.
3. This study is applicable to the field of international trade in the determination of exchange rates.
4. Micro Economics provides an analytical tool for evaluating the economic policies of the government.
5. It can be used to examine the condition of economic welfare and it suggests ways and means to bring about maximum social welfare.

Question 7.
What is meant by Macro Economics ? Discuss its importance.
Answer:
The term Macro Economics is derived from the greek word ‘MAKROS’ which means large. It was developed by j.M.Keynes. Macro Economics deals with economic system as a while like national income, aggregate demand, aggregate supply, general price level etc. It is also known as ‘Income and Employment’ theory.
Importance :

1. Macro Economics study is more useful to the government for formulation and execution of policies for achievement of maximum social benefit.
2. It helps in understanding the problems of unemployment poverty, inflation etc, and suggests has to solve them.
3. It gives us a picture of the working of the economy as a whole;
4. The study of Macro Economics is helpful in analysing the causes of business cycles and in providing-remedies.
5. Macro Economics includes economic growth and suggests how developing countries can use their resources to maximise their growth.
6. Macro Economic study is useful for making international comparisons in terms of average national income.

Question 8.
Explain the circular flow of income with suitable diagram.
Answer:
Income is a flow over a period of time. Income flow is of circular in character. Where beginning and end cannot be traced. National output originates in private and public sectors. It moves to the households. The household is the basic consuming unit in economic life. In every economy income flows from households to firms and vice versa. Thus the factor market and the product market are closely related to each other.

The circular flow of income can be explained with the help of the following diagram.

According to the above diagram, it is clear that the factor market and commodity market are closely related to one another.

The households supply the resource services and receives in returm payment interms of money. Thus money flows from producing units to households. The household exchange that money for goods and services they want. As a result the money flows from households to firms. Thus there is a circular flow of income and output.

Question 9.
Explain the differences between Consumer goods and Capital goods.
Answer:
Anything that can satisfy a human want is called a good. Goods can be classified into two types namely,

1. Free goods
2. Economic goods.

Further the economic goods divided into two types namely :

1. Consumer goods
2. Capital goods or Producer goods.

1) Consumer goods: A consumer good is an economic good or commodity purchased by households for final consumption. Thus these goods satisfy human wants directly.
Ex : Foods, books etc.
Consumer goods further divided into two types.
a) Perishable goods
b) Durable goods
a) Perishable goods : They lose their value in single use.
Ex : Milk, fruits etc.
b) Durable goods: These goods which yields service over period of time. Hence utility from these goods can be derive’d for a long time.
Ex : T.Vs & Computers.

2) Producer or Capital goods: Goods which are used in the production of other goods -lied producer or capital goods. They satisfy human wants indirectly.
Ex : Machines, buildings etc.
Differences between Consumer goods and Capital goods.
Consumer goods

1. These goods satisfy human wants directly. Ex: Milk, fruits etc.
2. They have direct demand.
3. These are the goods of first order.
4. They are net used in the production process of other goods.
5. They yield utility to the owners of these goods.

Capital goods

1. These goods satisfy human wants indirectly. Ex: Machines, raw-materials.
2. They have indirect or derived demand.
3. These are the goods of second order.
4. These are used in the production process of other goods.
5. They yield income to the owners of these goods.

Very Short Answer Questions

Question 1.
Economic goods
Answer:
Economic goods are man-made, they have cost of production and price. They are limited in supply. They have both value in use and value in exchange.
Ex : Pen, Book etc.

Question 2.
Capital goods
Answer:
Goods which are used in the production of other goods are called producer or capital goods. They satisfy human wants indirectly.
Ex : Machines, tools, buildings etc.

Question 3.
Intermediary goods [March 17]
Answer:
Goods which are under the process of production and semi finished goods are known as intermediary goods.
Ex : Cotton and fibre etc.

Question 4.
Wealth [March 18, 16]
Answer:
Wealth means stock of assets held by an individual or institution that yields has the potential for yielding income in some form. Wealth includes money, shares of companies, land etc. Wealth has three properties. 1. Utility 2. Scarcity 3. Transferability

Question 5.
Income
Ans. Income is a flow of satisfaction from wealth per unit of time. In every economy income’ flows from households to firms and vice versa. Income can be expressed in two types.

1. Money income which is in terms of money.
2. Real income which is in terms of goods and services.

Question 6.
Utility [March 16]
Answer:
Utility is the wants satisfying capacity of goods and services. It is a subjective concept. No one can measure it in mathematical terms.

Question 7.
Exchange value
Answer:
Exchange value is the purchasing power of one commodity to another. All economic goods have exchange value.

Question 8.
Price [March 18, 17]
Answer:
The price of anything is its value measured in terms of money.
Ex: A commodity is exchanged for 50 rupees then the price of commodity is 50 rupees.

Question 9.
Choice problem
Answer:
The choice problem is the central problem of Economics. The problems of the economy “What to produce ? How to produce” ? The problem of choice between commodities and the problem of choice of production techniques respectively.

Question 10.
Deductive Method
Answer:
Deductive method is the process from general to particular or from the universal to the individual.

Question 11.
Inductive method
Answer:
It is the process from particular to general or from the individual to the universal.
Economic statistics : It refers to the functional relationship between the two variables whose values are related to the same point. This concept was introduced by J.S. Mill.

Question 13.
Economic dynamics
Answer:
J.S. Mill introduced this concept in Economics. It refers to the analysis where the functional relationship is established between relevant variables whose values belong to different point of time.

Question 14.
Partial equilibrium
Answer:
It was popularised by Marshall. It exists when an equilibrium relates to a single variable.

Question 15.
General equilibrium
Answer:
This concept was popularised by Leon Walras. General equilibrium exists when an equilibrium relates to number of variables or even the economy as a whole.

Question 16.
Micro Economics
Ans. The word ‘Micro’ derived from Greek word ‘Mikros’ which means ‘small’. It was developed by Marshall. It is the study of the individual units like individual demand, price, supply etc.

Question 17.
Macro Economics
Answer:
The word ‘Macro’ derived from Greek word ‘Makros’ which means large. It was developed by J.M. Keynes. It studies aggregates or economy as a whole like national income, employment, general price level etc. It is also called “Income and Employment” theory.

Andhra Pradesh BIEAP AP Inter 1st Year Accountancy Study Material 5th Lesson Ledger Textbook Questions and Answers.

AP Inter 1st Year Accountancy Study Material 5th Lesson Ledger

Short Answer Questions

Question 1.
Define ledger and explain the advantages of ledger.
Answer:
Ledger is a book which facilitates recording of all types of transactions related to Personal, Real and Nominal. All the debit and credit aspects which are recorded in the journal are transferred to the respective accounts in the ledger.
According to L.C.Cropper, “the book which contains a classified and permanent record of all transactions of a business is called the Ledger”.

Advantages:

The following are the advantages of ledger.

1. Complete information at a glance : All the transactions pertaining to an account are collected at one place in the ledger. By looking at the balance of that account, one can understand the collective effect of all such transactions at a glance.
2. Arithmetical accuracy: With the help of ledger balances, Trial balance is prepared to know the arithmetical accuracy of accounts.
3. Result of business operations: It facilitates the preparation of final accounts for ascertaining the operating result and the financial position of the business concern.
4. Accounting information : The data supplied by various ledger accounts are summarized, analyzed and interpreted for obtaining various accounting information.

Question 2.
What do you mean by posting ? Explain the rules relating to posting.
Answer:
The process of transferring the entries recorded in the journal or subsidiary books to the respective accounts opened in the ledger is called ‘Posting’. In other words’, posting means process of grouping of all the transactions relating to a particular account at one place.

RULES:

Following steps should be taken into account while making posting.

A) Opening of separate accounts: Each transaction affects minimum two accounts for which separate accounts are to be opened in the ledger. All transactions relating to an account, debit as well as credit, are to be posted to know the net position of the account.

B) Posting journal entry to concerned side: If an account is debited in the journal, posting will be made on the debit side of the account in the ledger. Similarly, if an account is credited in the journal, that account would be credited in the ledger.

C) Use of word ‘To” and “By” : While writing the debit side, commence with word “To” and write the name of the account, which is credited in the journal. Write the word “By” on the credit side before writing the name of the account that is debited in the journal.

D) Balance in account: The difference between debit and credit totals of an account is the net position of the account, known as balance of the account.

Very Short Answer Questions

Question 1.
What is ledger ? (Mar. ’15 – T.S.)
Answer:
Ledger is a book which facilitates recording of all types of transactions related to Personal, Real and Nominal accounts separately in related accounts. In other words, the group of accounts recorded in a book is called Ledger’. Ledger is also called ‘Book of Final Entry’.

Question 2.
What is posting ? (Mar. ’17 – T.S.)
Answer:
The process of transferring the entries recorded in the journal or subsidiary books to the respective accounts opened in the ledger is called Posting’. It is also process of grouping of all the transactions relating to a particular account at one place.

Question 3.
What do you mean by balancing of an account ?
Answer:
Balancing means the writing of the difference between the amount columns of the two sides in the ledger (smaller total) side, so that the totals of the two sides become equal.

Question 4.
What is debit balance ?
Answer:
The excess of debit total over the credit total is called the debit balance.

Question 5.
What is credit balance ?
Answer:
The excess of credit total over the debit total is called the credit balance.

Problems

Question 1.
Journalise the following transactions of Mr. Prahlad and post them in the ledger and balance the same.
2013
February 01 Prahlad invested Rs. 5,00,000 cash in the business
February 03 Paid into Bank Rs. 60,000
February 05 Purchased building for Rs. 2,00,000
February 07 Purchased goods for Rs. 50,000
February 10 Sold goods for Rs. 80,000
February 15 Withdrew cash from bank Rs. 12,000
February 25 Paid internet Rs. 2,000
February 28 Paid Salary Rs. 18,000
Answer:
Journal Entries in the books of Mr. Prahlad


Ledger Accounts :

Question 2.
Prepare Pavan account from the following: (May ’17 – T.S.)

Answer:

Question 3.
Prepare Sudha account from the following: (Mar. 2019 – T.S.) (May 17 – A.P.)

Answer:

Question 4.
Prepare Swami’s account from the following: (Mar. 2018 – T.S.)

Answer:

Question 5.
Prepare Machinery account from the following:

Answer:

Question 6.
Prepare the Ledger account of Bhavya from the following particulars:

Answer:
Journal Entries in the books of Bhavya

Question 7.
From the following information prepare Praveen’s Account as on 31-3-2014. (Mar. 2018 – A.P.)

Answer:

Question 8.
Prepare Vamsi’s Account from the following.

Answer:

Question 9.
Prepare Anurudh ‘s Account from the following. (Mar. 2019 – A.P.)

Answer:

Student Activity

Visit any bank or any organisation and find the process of preparation of accounts.

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 7th Lesson Systems of Particles and Rotational Motion Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 7th Lesson Systems of Particles and Rotational Motion

Very Short Answer Questions

Question 1.
Is it necessary that a mass should be present at the centre of mass of any system ?
Answer:
No. Any mass need not be present at the centre of mass of a system.
Ex : a hollow sphere, centre of mass lies at its centre.

Question 2.
What is the difference in the positions of a girl carrying a bag in one of her hands and another girt carrying a bag in each of her two hands ?
Answer:
When the girl carries a bag in one hand (left) her centre of mass shifts towards other hand (right). In order to bring it in the middle, the girl has to lean towards her other side. When the girl carries a bag in each of her two hands (left and right), the centre of mass does not shift. The girl does not bend any side beacuse of same hags’ .w in her two hands.

Question 3.
Two rigid bodies have same moment of inertia about their axes of symmetry. Or the two, which body will have greater kinetic energy ?
Answer:
E = \(\frac{1}{2}\) I ω² = \(\frac{1}{2} \frac{\mathrm{L}^2}{1}\) , E oc \(\frac{1}{l}\) (∵ L = constant)
The rigid body having less moment of inertia will have greater kinetic energy.

Question 4.
Why are spokes provided in a bicycle wheel ?
Answer:
By connecting to the rim of wheel to the axle through the spokes the mass of the wheel gets concentrated at its rim. This increases its moment of inertia. This ensures its uniform speed.

Question 5.
We cannot open or close the door by applying force at the hinges, why ?
Answer:
When the force is applied at the hinges, the line of action of the force passes through the axis of rotation i.e, r = 0, so we can not open or close the door by pushing or pulling it at the hinges.

Question 6.
Why do we prefer a spanner of longer arm as compared to the spanner of shorter arm ?
Answer:
The torque applied on the nut by the spanner is equal to the force multiplied by the perpendicular distance from the axis of rotaion.
A spanner with longer arm provides more torque compared to a spanner with shorter arm. Hence longer arm spanner is preferred.

Question 7.
By spinning eggs on a table top, how will you distinguish a hard boiled egg from a raw egg ? [Mar. 13]
Answer:
A raw egg has some fluid in it and a hard boiled egg is solid form inside. Both eggs are spinning on a table top, the fluid is thrown outwards. Therefore (I_r > I_b) That means M. I of raw egg is greater than boiled egg. As I × ω = constant; ∴ ω_r < ω_b. That means Angular Velocity of raw egg is less than angular velocity of boiled egg.

Question 8.
Why should a helicopter necessarily have two propellers ?
Answer:
If there were only one propeller in the hellicopter then, due tio conservation of angular momentum, the helicopter itself would have turned in the opposite direction. Hence, if should have two propellers.

Question 9.
It the polar ice caps of the earth were to melt, what would the effect of the length of the day be ?
Answer:
Earth rotates about its polar axis. When ice of polar caps of earth melts, mass concentrated near the axis or rotation spreads out. Therefore, moment of Inertia I increases.

As no external torgue acts, L = I × ω = \(I\left(\frac{2 \pi}{T}\right)\)= constant. With increase of I, T will increase i.e. length of the day will increase.

Question 10.
Why is it easier to balance a bicycle in motion?
Answer:
When a bicycle is in motion, it is easy to balance because the principle of conservation of angular momentum is involved.

Short Answer Questions

Question 1.
Distinguish between centre of mass and centre of gravity. [A.P. – Mar. ‘18, ‘16, ‘15, ‘14, ‘13; TS – Mar. ‘16, ‘15, ‘14, ‘13]
Answer:
Centre of mass

1. Point at which entire mass of the body is supposed to be concentrated, and the motion of the point represents motion of the body.
2. It refers mass of to body.
3. In a uniform gravitational field centre of mass and centre of gravity coincide
4. Centre of mass of the body is defined to describe the nature of motion of a body as a whole.

Centre of gravity

1. Fixed point through which the weight of the body act.
2. It refers to the weight acting on all particles of the body
3. In a non-uniform gravitational field, centre of gravity and centre of mass do not coincide.
4. Centre of gravity of body is defined to know the amount of stability of the body when supported.

Question 2.
Show that a system of particles moving under the influence of an external force, moves as if the force is applied at its centre of mass.
Answer:
Consider \(\overrightarrow{r_1}, \overrightarrow{r_2}, \overrightarrow{r_3} \ldots \ldots \overrightarrow{r_n}\) be the position vectors of masses m₁, m₂, m₃ …………. m_n of n particle system.
According to Defination of centre of mass.

Differentiating the above equation w.r.t. time, we obtain

Where F_ext represents the sum of all external forces acting on the particles of the system. This equation states that the centre of mass of a system of particles moves as if all the mass of the system was concentrated at the centre of mass and all the external forces were applied at that point.

Question 3.
Explain about the centre of mass of earth-moon system and its rotation around the sun.
Answer:
In the solar system the planets have different velocities and have complex two dimensional motion. But the motion of the centre of mass of the planet is simple and translational. consider the earth and moon system. We consider that the earth is moving around the sun in an elliptical path. But actually the centre of mass of earth and moon moves in an elliptical path round the sun. But the motion of either earth or moon is complicated when considered separately, more over we say that moon goes round the earth.

But actually earth and moon are revolving round their centre of mass such that they are always on opposite sides of the centre of mass. Here the forces of attraction between earth and moon are internal forces.

Question 4.
Define vector product. Explain the properties of a vector product with two examples. [T.S. AP – Mar. 15]
Answer:
The cross product of two vectors is given by \(\vec{C}=\vec{A} \times \vec{B}\). The magnitude of the vector defined from cross product of two vectors is equal to product of magnitudes of the vectors and sine of angle between the vectors.

Direction of the vectors is given by right hand corkscrew rule and is perpendicular to the plane containing the vectors.
∴ \(|\vec{C}|\) = AB sin θ. and \(\vec{C}\) = AB sin θ \(\hat{n}\). Where, \(\hat{n}\) is the unit vector perpendicular to the plane containing the vectors \(\vec{A}\) and \(\vec{B}\)
Example: 1) Torque is cross product of position vector and Force. i.e., \(\overrightarrow{\mathrm{T}}=\vec{r} \times \vec{F}\)
2) Angular momentum is cross product of position vector and momentum
i.e., \(\vec{L}=\vec{r} \times \vec{p}\)

Properties:

1. Cross product does not obey commutative law. But its magnitude obeys commutative low.
   \(\vec{A} \times \vec{B} \neq \vec{B} \times \vec{A} \Rightarrow(\vec{A} \times \vec{B})=-(\vec{B} \times \vec{A}),|\vec{A} \times \vec{B}|=|\vec{B} \times \vec{A}|\)
2. It obeys distributive law \(\vec{A} \times(\vec{B} \times \vec{C})=\vec{A} \times \vec{B}+\vec{A} \times \vec{C}\)
3. The magnitude of cross product of two vectors which are parallel is zero.
   Since θ = 0; \(|\vec{A} \times \vec{B}|\) = AB sin 0° = 0
4. For perpendicular vectors, θ = 90°, \(|\vec{A} \times \vec{B}|\) = AB sin 90° \(|\hat{n}|\)= AB

Question 5.
Define angular velocity(u). Derive v = r ω. [T.S. Mar. 16]
Answer:
Angular velocity (ω):
The rate of change of angular displacement of a body is called angular velocity, i.e., ω = \(\frac{\mathrm{d} \theta}{\mathrm{dt}}\)
Derivation of v = rω
consider a rigid body be moving with uniform speed (v)along the circumference of a circle of radius r. Let the body be displaced from A to B in a small interval of time At making an angle ∆θ at the cantre. Let the linear displacement be ∆x from A to B.
From the property of circle, length of arc = radius × angle
∆x = r ∆θ
This equation is divided by ∆t, and taking

Question 6.
Define angular acceleration and torque. Establish the relation between angular acceleration and torque. [T.S. Mar. 18]
Answer:
Angular acceleration :
The rate of change of angular velocity is called angular acceleration
i.e., α = \(\frac{\mathrm{d} \omega}{\mathrm{dt}}\)
Torque : The rate of change of angular momentum is called torque or The moment of Force is called Torque.

Relation between angular acceleration and Torque: Consider a rigid body of mass ‘M’ rotating in a circular path of radius ‘R’ with angular velocity ‘ωω’ about fixed axis.

By definition, τ = \(\frac{\mathrm{dL}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{I} \omega)}{\mathrm{dt}}\)
Where I = MR² = moment of inertia of a body
τ = I \(\frac{\mathrm{d} \omega}{\mathrm{dt}}\) [∵ I = constant]
But \(\frac{\mathrm{d} \omega}{\mathrm{dt}}\) α
∴ τ = Iα

Question 7.
Write the equations of motion for a particle rotating about a fixed axis?
Answer:
Equations of motion for a particle rotating about a fixed axis:
1. ω_f = ω_f + αt [∵ like v = u + at]
2. θ = \(\left(\frac{\omega_f+\omega_t}{2}\right) t\) [∵ like v = \(\left(\frac{v_1+v_2}{2}\right) t\)]
3. θ = ω_it + \(\frac{1}{2}\) α t² [∵ like s = ut + \(\frac{1}{2}\) at²]
4. ω_f² – ω_i² = 2 α θ [∵ like v² – u² = 2as]
Where ω_i = Initial angular velocity
ω_f = Final angular velocity
α = Angular acceleration
θ = Angujar displacement
t = time

Question 8.
Derive expressions for the final velocity and total energy of a body rolling without slipping.
Answer:
Expression of velocity of a body Rolling down an inclined plane:
Consider a rigid body of mass M and radius R rolling down an inclined plane from a height h . Let v the linear speed acquired by the body when it reaches the bottom of the plane and k is its radius of gyration.

According to law of conservation of Energy, we have P.E of body on top of inclined plane = K.E of a body at the bottom of inclimed plane
When a body rolls down an incline of height h, we apply the principle of conservation of energy.
P.E at the top – K.E of translation + K.E of rotation

Expression of Total energy of a body Rolling down on an inclined plane:
Suppose a body (Sphere) is rolling on a surface. Its motion can be treated as a combination of the translation of the centre of mass and rotation about an axis passing through the centre of mass. The total kinetic energy E can written as
E = E_T + E_R
Where E_T = Translational kinetic energy
and E_R = Rotational kinetic energy
E = \(\frac{1}{2}\) Mv² + \(\frac{1}{2}\) Iω²
= \(\frac{1}{2}\) MR² ω² + \(\frac{1}{2}\) Mk²ω²
E = \(\frac{1}{2}\) M ω² (R² + K²)
where k is radius of gyration
E = \(\frac{1}{2}\) Mv² (1 + \(\frac{K^2}{R^2}\)) [∵ ω = \(\frac{V}{R}\)]

Long Answer Questions

Question 1.
a) State and prove parallel axis theorem.
Answer:
Statement: The moment of inertia of a rigid body about any axis is equals to the sum of moment of inertia about a parallel axis passing through centre of mass (lg) and product of mass of body and square of perpendicular distance between two parallel axis.
∴ I = I_G + mr²
Proof : Consider a rigid body of mass M. Let I and I_G are the moment of inertia of a body about parallel axes X and Y respectively.

Let ‘r’ be the perpendicular distance between two axes.
Moment of inertia about ‘X’ axes is
I = Σm(OP)² ………………. (1)
Moment of inertia about Y axes is
I_G = Σm(GP)² …………….. (2)
from ∆⁴ OQP, OP² = (OQ)² + (PQ)²
= (OG + GQ)² + (PQ)²
= OG² + (GQ)² + 2.OG.GQ + (PQ)²
= OG² + (GP)² + 2(OG) (GQ) (∴ (GP)² = (GQ)² + (PQ)²)
Multiplying both sides by Σm,
⇒ Σm(OP)² = Σm(OG)² + Σm(GP)² + 2Σm(OG)(GQ)
from (1) and (2), I = mr² + IG + 2Σm (OG) (GQ)
But sum of moments of particles about centre of mass is zero.
i.e., Σm (OG) (GQ) = 0
∴ I = I_G + mr²
Hence proves.

b) For a thin flat circular disk, the radius of gyration about a diameter as axis is k. If the disk is cut along a diameter AB as shown in to two equal pieces, then find the radius of gyration of each piece about AB.

Answer:
For thin circular disk, the radius of gyration about a diameter
AB is, K = \(\sqrt{\frac{I}{M}}\)
Where M = mass of disk
I = M.I of disk
The disk is cut into two halves about AB, when each,
Mass M’ = \(\frac{\mathrm{M}}{\mathrm{2}}\) and each M.I, I = \(\frac{1}{2}\)
Therefore Radius of gyration of each piece is
K’ = \(\sqrt{\frac{I^{\prime}}{M^{\prime}}}=\sqrt{\frac{\left(\frac{I}{2}\right)}{\left(\frac{M}{2}\right)}}=\sqrt{\frac{I}{M}}\) = k

Question 2.
a) State and prove perpendicular axes theorem.
Answer:
Statement: The moment of inertia of a plane lamina about an axes perpendicular to its plane is equal to sum of moment of inertia of lamina about the perpendicular axes in its plane intersecting each other at a point, where the perpendicular axes passes.
i.e., I_z = I_x + I_y
Proof’ Consider a particle of mass’m’ at p. Let it be at a distance Y from Z-axis. Here ‘X’ and Y axes are in plane Ramina and Z-axes perpendicular to plane lamina.

Now,
Moment of inertia about X-axix, I_x = Σmx²
Moment of inertia about Y-axis, I_y = Σmy²
Moment of inertia about Z-axis, I_z = Σmr²
From ∆⁴ OQP, r² = x² + y²
multiplying both side with Σm,
⇒ Σmr² = Σmx² + Σmy²
⇒ I_z = I_x + I_y
Hence proved.

b) If a thin circular ring and a thin flat circular disk of same mass have same moment of inertia about their respective diameters as axes. Then find the ratio of their radii.
Answer:
For a thin flat circular dist, M.I., I_r = \(\frac{\mathrm{MR}_{\mathrm{r}}^2}{2}\)
For a thin flat circular dist, M.I., I_d = \(\frac{\mathrm{MR}_{\mathrm{d}}^2}{2}\)
Given that I_r = I_d
\(\frac{M R_r^2}{2}=\frac{M R_d^2}{4} \Rightarrow \frac{R_r^2}{R_d^2}=\frac{2}{4}=\frac{1}{2}\)
∴ \(\frac{R_r}{R_d}=\frac{1}{\sqrt{2}} \text { or } R_r: R_d=1: \sqrt{2}\)

Question 3.
State and prove the principle of conservation of angular momentum. Explain the principle of conservation of angular momentum with examples. [A.P. Mar 16]
Answer:
Statement: Angular momentum of a body remains constant when the external torque is zero.
L = I ω = constant K.
or I₁ ω₁ = I₂ ω₂.
If the moment of Inertia of a body is lowered, the angular velocity of the body co increases.
Proof:
By derfination, the rate of change of angular momentum is called Torque.
i.e., τ = \(\frac{\mathrm{dL}}{\mathrm{dt}}\)
If τ = 0 ⇒ \(\frac{\mathrm{dL}}{\mathrm{dt}}\) = 0
or L = constant k
⇒ L₁ = L₂
∴ I₁ ω₁ = I₂ ω₂
Example 1): When a man with stretched out arms stands on a turn table which is revolving then his moment of inertia is high. If he folded his hands, the moment of inertia decreases and hence the angular velocity, linear velocity increase, but the period decreases. In both cases angular momentum remains constant.

Example 2): An acrobat from a swing in a circus, leaves the swing with certain angular momentum, with his arms and legs stretched. As soon as he leaves the swing he pulls his hands and legs together thus lowering his M.I. and increasing his angular velocity. He then quickly makes somersaults in air and finally lands on a net or ground.

Problems

Question 1.
Show that a(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors a, b and c.
Solution:
Let a parallele piped be formed on three vectors.

Which is equal in magnitude to the volume of the parallel piped.

Question 2.
A rope of negligible mass is wound 4. round a hollow cylinder of mass 3kg and redius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30N ? What is the linear acceleration of the rope ? Assume that there is no slipping.
Solution:
Here M = 3kg, R = 40 cm = 0.4 m
Moment of intertia of the hollow cylinder about its axis
I = MR² = 3(0.4)² = 0.48 kgm²
Force applied F = 30 N
∴ Torque, τ = F × R= 30 × 0.4 = 12 N – m.
If α is the angular acceleration, produced, then from τ = Iα
α = \(\frac{\tau}{\mathrm{l}}=\frac{12}{0.48}\) = 25 rads^-2
Linear acceleration, a = Rα = 0.4 × 25 = 10 m/s²

Question 3.
A coin is kept a distance of 10 cm from the centre of a circular turntable. If the coefficient of static friction between the table and the coin is 0.8 find the frequency of rotation of the disc at which the coin will just begin to slip.
Solution:
Here, radius, r = 10cm = 0.1 m; μ_s = 0.8
F = μmg
mrω² = μmg
rω² = μg
ω = \(\frac{\mu \mathrm{g}}{\mathrm{r}}=\sqrt{\frac{0.8 \times 9.8}{0.1}}=\sqrt{8 \times 9.8}\)
= 8.854 rad/s
w = 2πn
∴ frequency n = \(\frac{\omega}{2 \pi}=\frac{8.854}{2 \times 3.14}\) = 1 .409rps
n = 1.409 × 60 = 84.54 rpm

Question 4.
Particles of masses 1g, 2g, 3g…. 100g are kept at the marks 1 cm. 2cm, 3cm ….100cm respectively on 2 meter scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the meter scale. –
Solution:
By adding all the masses, we get M = 5050g
= 5.050 kg
= 5.1 kg
and L = 1 m
M.I. of the meter scale = \(\frac{\mathrm{mL}^2}{12}=\frac{5.1 \times 1^2}{12}\)
= 0.425kg m²
= 0.43kg -m²

Question 5.
Three particles each of mass 100g are placed at the vertices of an equilateral triangle of side length 10 cm. Find the moment of inertia of the system about an axis passing through the centroid Of the triangle and perpendicular to its plane.
Solution:
m = 100 g = 100 × 10^-3kg
Side a = 10 cm

Question 6.
Four particles each of mass 100g are placed at the corners of a square of side 10 cm. Find the moment of inertia of the system about an axis passing through the centre of the square and perpendicular to its plane. Find also the radius of gyration of the system.
Solution:
mass m = 100g
= 100 × 10^-3kg
M = 4m = 400 × 10^-3kg
radius r = 10cm
= 10 × 10^-2m
Moment of Inertia I = Mr²

Question 7.
Two uniform circular discs, each of mass 1 kg and radius 20 cm, are kept in contact about the tangent passing through the point of contact. Find the moment of inertia of the system about the tangent passing through the point of contact.
Solution:
Mass m = 1kg; r = 20cm = 20 × 10^-2m
I = I₁ + I<sub2

I₁ = \(\frac{\mathrm{MR}^2}{4}\) + MR²
I₁ = \(\frac{5\mathrm{MR}^2}{4}\)
Similarly I₂ = \(\frac{5\mathrm{MR}^2}{4}\)
∴ I = \(\frac{10 \mathrm{MR}^2}{4}=\frac{10 \times 1 \times\left(20 \times 10^{-2}\right)^2}{4}\) = 0.1 kg-m²

Question 8.
Four spheres, each diameter 2a and mass ‘m’ are placed with their centres on the four corners of a spuare of the side b. Calculate the moment of inertia of the system about any side of the square.
Solution:
I₁ = mb²
I₂ = \(\frac{2}{5}\) ma²
I₃ = \(\frac{2}{5}\) ma²
I₄ = mb²
∴ Moment of inertia of the system
I = I₁ + I₂ + I₃ + I₄
= mb² + \(\frac{2}{5}\) ma² + \(\frac{2}{5}\) ma² + mb²
I = \(\frac{4}{5}\) ma² + 2mb²

Question 9.
To maintain a rotor at a uniform angular speed or 200 rads^-1, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine ? (Note : uniform angular velocity in the absence of friction implies zero torque. In practice applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.
Solution:
Here, ω = 200 rad/s, torque τ = 180 N – m, power p = ?
As p = τ ω
∴ p = 180 × 200 = 3600.watt = 36 kw.

Question 10.
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick ?
Solution:
Let M be the mass of stick concentrated at L, the 50 cm, mark.
For equilibrium about G’ the 45 cm mark,
10g (45 – 12) = mg(50 – 45)
10 g × 33 = mg × 5
M = \(\frac{10 \times 33}{5}\) = 66 gram.

Question 11.
Determine the kinetic energy of a circular disc rotating with a speed of 60 rpm about an axis passing through a point on its circumference and perpendicular to its plane. The circular disc has a mass of 5kg and radius 1m.
Solution:
Here M = 5kg; R = 1 m;
ω = 2π × \(\frac{\mathrm{N}}{\mathrm{t}}\) = 2n × \(\frac{60}{60}\) rad /s = 2π rad/s
The M.I of disc about parallel axis passing through a point on its circumferance
I = \(\frac{\mathrm{MR}^2}{2}\) + MR² = \(\frac{3}{2}\) MR²
∴ Kinetic energy = \(\frac{1}{2}\) I ω²
= \(\frac{1}{2}\) × \(\frac{3}{2}\) MR² ω² = \(\frac{3}{4}\) × 5 × (1)² × (2π)²
= \(\frac{3}{4}\) × 5 × 4π² = 15 × (\(\frac{22}{7}\)²
∴ K.E = 148.16 J.

Question 12.
Two particles each of mass m and speed v Travel in opposite directions along parallel lines separated by a distance. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.
Solution:
Given that m₁ = m₂ = m;
\(\vec{V}_1=\vec{V}_2=\vec{V}\)
momentum of 1^st particle p₁ = \(\mathrm{m} \vec{\mathrm{V}}\)
∴ p₁ = \(\mathrm{m} \vec{\mathrm{V}}\)
momentum of 2^nd particle \(\vec{\mathrm{P}}_2=\mathrm{mV}\)

If two particles moves oppositely on
Circumference of circle, distance d = 2r
Angular momentum of 1^st particle w.r.t
Centre ‘O’ is \(\vec{L_1}=\vec{r} \times \vec{P}_1=\vec{r} \times m \vec{V}\)
Angular momentum of 2^nd particle w.r.t
centre ‘O’ is \(\vec{\mathrm{L}_2}=\vec{\mathrm{r}} \times \vec{\mathrm{P}_2}=\vec{\mathrm{r}} \times \mathrm{m} \vec{\mathrm{V}}\)
∴ \(\vec{L_1}=\vec{L_2}\)

Question 13.
The moment of inertia of a fly wheel making 300 revolutions per minute is 0.3 kgm ² Find the torque required to bring it to rest in 20s.
Solution:

Question 14.
When 100 J of work is done on a fly wheel, its angular velocity is increased from 60 rpm to 180 rpm. What is the moment of inertia of the wheel ?
Solution:
Here, Initial frequency
n₁ = \(\frac{60}{60}\) = 1Hz
Initial angular velocity
ω₁ = 2π n₁ = 2π rad /sec
Final frequency n₂ = \(\frac{180}{60}\) = 3Hz
Final angular velocity
ω₂ = 2π n₂ = 2π × 3 = 6π rad/sec
Work done = 100 J from work – energy therorem,
Workdone = change in K.E.
W = \(\frac{1}{2}\) I ω₂² – \(\frac{1}{2}\) I ω₁²
100 = \(\frac{1}{2}\) I [(6π)² – (2π)²]
100 = \(\frac{1}{2}\) I (32π)²
I = \(\frac{200}{32 \pi^2}\) = 0.634 kg – m²
∴ I = 0.634 kg- m²

Additional Problems

Question 1.
Give the location of the centre of mass of a (i) sphere (ii) cylinder (iii) ring and (iv) cube. each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?
Solution:
In all the four cases, as the mass density is uniform, centre of mass is located at their respective geometrical centres.
No, it is not necessary that the center of mass of a body should lie on the body for example in caSe of a circular ring. Center of mass is at the centre of the ring, where there is no mass.

Question 2.
In the HCl molecule the separation between the nuclei of the two atoms Is about 1.27 Å(1 Å = 10^-10m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom Is concentrated in its nucleus.
Solution:
Let the mass of the H atom = m unit, mass of the Cl atom = 35.5 m units
Let cm be at the distance xÅ from Htom
∴Distance of cm from CL atom = (1.27 – x) Å
It cm is taken at the origin,
then mx + (1.27 – x) 55.5m = 0
mx = (1.27 – x) 35.5 m.
Negative sign indicates that if chlorine atom is on the right side of cm (+), the hydrogen atom is on the left side of cm, so leavning negative sign, we get
x + 35.5 x = 1.27 × 35.5
36.5 x = 45.085
x = \(\frac{45.085}{36.5}\)
= 1.235
x = 1.235 Å
Hence cm is located on the line joining centres of H and d atoms at a distance 1.235 Å from H.

Question 3.
A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system ?
Solution:
The speed of the centre of mass of the system , (trolley + child) shall remain unchanged, when the child gets up and runs about on the trolley in any manner. This is because forces involved in the exercise are purely internal i.e., from within the system. No external force acts on the system and hence there is no change in velocity of the system.

Question 4.
Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b
Solution:
Let \(\vec{\mathrm{a}}\) represented by \(\vec{\mathrm{OP}}\) and \(\vec{\mathrm{b}}\) be represented by \(\vec{\mathrm{OQ}}\).
Let ∠POQ = θ

∴ area of ∆ OPQ = \(\frac{1}{2}|\vec{a} \times \vec{b}|\), which was to be proved.

Question 5.
Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors a, b and c.
Solution:
Let a parallel piped be formed on three vectors.

Where \(\hat{n}\) is unit vector along \(\overrightarrow{\mathrm{OA}}\) perpendicular to the plane containing \(\) and \(\vec{\mathrm{b}}\). Now \(\vec{\mathrm{b}}\) = (a) (be) cos 0°
= abc
Which is equal in magnitude to the volume of the parallelepiped.

Question 6.
Find the components along the x. y. z axes of the angular momentum 1 of a particle whose position vector is r with components x, y, z and momentum is p with components p_x, p_y, and p_z,. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.
Solution:
For motion in 3D, the position vector \(\vec{\mathrm{r}}\) and linear momentum vector \(\vec{\mathrm{p}}\) can be written in terms of their rectangular components as follows.

Comparing the coefficient of k on both sides, we have L_z = xp_y – yp_x.
Therefore the particle moves only in x-y plane. The angular momentum has only z component.

Question 7.
Two particles, each of mass m and speed v. travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same what ever be the point about which the angular momentum is taken.
Solution:
From fig, vector angular momentum of the two particle system any point A on x₁ y₁ is.

Similary, vector angular momentum of the tw.o particle system, a bout any pt. B on x₂y₂ is
\(\vec{L_B}=m \vec{v} \times d+m \vec{v} \times 0=m \vec{v} d\)
Let us consider any other point (on AB, where AC = x)
∴ Vector angular momentum of the two particle system about c is
\(\vec{L_c}=m \vec{v}(x)+m \vec{v}(d-x)=m \vec{v} d\)
Clearly, \(\vec{\mathrm{L}_{\mathrm{A}}}=\vec{\mathrm{L}_{\mathrm{B}}}=\vec{\mathrm{L}_C}\)
Which has to be proved.

Question 8.
A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in figure. The angles made by the strings with the vertical are 36.9 and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.
Solution:

From fig (b)
θ₁ = 36.9°, θ₂ = 53.1°
If T₁, T₂ are the tensions in two strings, then for equilibrium along the horizontal,
T₁ sin θ₁ = T₂ sin θ₂
(or) \(\frac{T_1}{T_2}=\frac{\sin \theta_2}{\sin \theta_1}=\frac{\sin 53.1^{\circ}}{\sin 36.9^{\circ}}\)
= \(\frac{0.7407}{0.5477}\) = 1.3523
Let d be the distance of center of gravity c of the bar from the left end.
For rotational equilibrium about c,
T₁ cos θ₁ Xd = T₂ cos θ₂ (2 – d)
T₁ cos 36.9° × d = T₂ cos 53.1° (2 – d)
T₁ × 0.8366 d = T₂ × 0.6718 (2 – d)
Put T₁ = 1.3523 T₂ and solve to get
d = 0.745 m.

Question 9.
A car weight 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
Solution:
Here, m = 1800 kg
Distance between front and .back axles = 1.8 m
Distance of center of gravity (c) behind the front axle = 1.05 m
Let R₁ and R₂ be the forces exerted by the level ground on each front wheel and each back wheel. As it is dear from fig.
R₁ + R₂ = mg = 1800 × 9.8

For rotational equilibrium about C,
R₁ × 1.05 = R₂ (1.8 – 1.05) = R₂ × 0.75 ………….. (i)
\(\frac{R_1}{R_2}=\frac{0.75}{1.05}=\frac{5}{7}\)
Putting in (i)
\(\frac{5}{7}\) R₂ + R₂ = 1800 × 9.8
R₂ = \(\frac{7 \times 1800 \times 9.8}{12}\)
= 10290 N
R₁ = \(\frac{5}{7}\) R₂
= \(\frac{5}{7}\) × 10290
= 7350 N.

Question 10.
(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2 MR²/5. where M is the mass of the sphere and R is the radius of the sphere.
Solution:
a) Moment of inertia of sphere about any diameter = \(\frac{2}{5}\)MR²
Applying theorem of parallel axes,
Moment of inertia of sphere about a tangent to the sphere = \(\frac{2}{5}\)MR² + M(R)²
= \(\frac{7}{5}\)MR².

(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR²/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.
Solution:
We are given, moment of inertia of the disc about any of its diameters = \(\frac{1}{4}\)MR².
i) Using theorem of perpindicular axes, moment of inertia of the disc about an axis passing through its center and normal to the disc = 2 × \(\frac{1}{4}\)MR² = \(\frac{1}{2}\)MR²
ii) Using theorem of parallel axes, moment of inertia of the disc passing through a point on its edge and normal to the disc
= \(\frac{1}{2}\)R² + MR² = \(\frac{3}{2}\) MR².

Question 11.
Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time.
Solution:
If M is mass and R is radius of the hollow cylinder and the solid sphere, then
M.I of hollow cylinder about its axis of symmetry I₁ = MR² and
M.I of solid sphere about an axis through its 2 , centre, I₁ = \(\frac{2}{5}\) MR²
Torque applied, I = I₁ α₁ = I₂ α₂
\(\frac{\alpha_2}{\alpha_1}=\frac{l_1}{l_2}=\frac{M^2}{\frac{2}{5} M R^2}=\frac{5}{2}\)
α₂ > α₁
From ω = ω₀ + αt, we find that for given ω₀ and t, ω₂ > ω₁ to, i.e. angular speed of solid sphere will be greater than angular speed of hallow sphere.

Question 12.
A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s^-1. The radius of the cylinder is 0.25m. What is the kinetic energy associated with the rotation of the cylinder ? What is the magnitude of angular momentum of the cylinder about its axis ?
Solution:
Here, M = 20 kg, R = 0.25 m, w – 100 g^-1
Moment of inertia of solid cylinder
= \(\frac{\mathrm{MR}^2}{2}=\frac{20 \times(0.25)^2}{2}\) = 0.625 kg/m²
K.E of rotation = \(\frac{1}{2}\) Iω² = \(\frac{1}{2}\) × 0.625 × (100)² = 3125 J
Angular momentum, L = Iω
= 0.625 × 100
= 62.5 Js.

Question 13.
a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/miri. How much is the angular spped of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value ? Assume that the turntable rotates without friction.
Solution:
Here, intial angular speed ω₁ = 40 rev/min, ω₂ = ?
Final moment of inertia, I₂ = \(\frac{2}{5}\)I, Intial moment of Inertia
As no external torque acts in the process, therefore
Iω = constant
i.e. I₂ω₂ = I₁ω₁
ω₂ = \(\frac{I_1}{I_2}\) ω₁ = \(\frac{5}{2}\) × 40
= 100 rpm

b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy ?
Solution:
Final K.E of rotation, E₂ = \(\frac{1}{2}\)I₂ω₂²
Intial K.E of rotation, E₁ = \(\frac{1}{2}\)I₂ω₁²
\(\frac{E_2}{E_1}=\frac{\frac{1}{2} l_2 \omega_2^2}{\frac{1}{2} l_1 \omega_1^2}=\left(\frac{l_2}{l_1}\right)\left(\frac{\omega_2}{\omega_1}\right)^2\)
= \(\frac{2}{5} \times\left(\frac{100}{40}\right)^2=\frac{5}{2}\) = 2.5
∴ K.E of rotation increase. This is because child spends internal energy folding back his hands.

Question 14.
A rope of negligible mass is wound roung a hollow cylinder of mass 3kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ? what is the linear acceleration of the rope ? Assume that there is no slipping.
Solution:
Here M = 3kg, R = 40 cm = 0.4 m
Moment of Inertia of the hollow cylinder about its axis
I = MR² = 3(0.4)² = 0.48 kgm²
Force applied F = 30 N
∴ Torque, τ = F × R = 30 × 0.4 = 12 N – m.
If α is the angular acceleration, produced, then from τ = Iα
α = \(\frac{\tau}{\mathrm{I}}=\frac{12}{0.48}\) = 25 rads^-2
Linear acceleration, a = Rα = 0.4 × 25 = 10 m/s².

Question 15.
To maintain a rotor at a uniform angular speed of 200 rad s^-1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine ? (Note : uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.
Solution:
Here, ω = 200 rad/s, torque τ = 180 N – m, power p = ?
As p = τω
∴ p = 180 × 200 = 3600 watt = 36 kw.

Question 16.
From a uniform disc of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is a R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
Solution:
Suppose mass per unit area of the disc = M
∴ Mass of original disc

M = πR² × m
Mass of portion removed from the disc
M’ = π\(\left(\frac{R}{2}\right)^2 \times m=\frac{\pi R^2}{4} m=\frac{M}{4}\)
In fig, mass M is connected at O and mass
M’ is concentrated at O’, where OO’ = \(\frac{\mathrm{R}}{\mathrm{2}}\).
After the circular disc of mass M’ is removed, the remaining portion can be considered as a system of two masses M at 0 and – M’ = \(\frac{\mathrm{-M}}{\mathrm{4}}\) at O’. If x is the distance of centre of 4 mass (p) of the remaining part, then
x = \(\frac{M \times O-M^{\prime} \times \frac{R}{2}}{M-M^{\prime}}\)
= \(\frac{\frac{-M}{4} \times \frac{R}{2}}{M-\frac{M}{4}}=\frac{-M R}{8} \times \frac{4}{3 M}=\frac{-R}{6}\)
Negative sign shows that p is the left O.

Question 17.
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick ?

Solution:
Let m be the mass of stick concentrated at c, the 50 cm, mark,
For equilibrium about c’ the 45 cm mark,
10g (45 – 12) = mg(50 – 45)
10 g × 33 = mg × 5
m = \(\frac{10 \times 33}{5}\)
= 66 gram.

Question 18.
A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination, (a) Will it reach the bottom with the same speed in each case ? (b) Will jt take longer to roll down one plane than the other ? (c) If so, which one and why ?
Solution:
Let v be the speed of the solid sphere at the bottom of the incine. Applying principle of conservation of energy, we get
\(\frac{1}{2}\) mv² + \(\frac{1}{2}\) Iω² = mgh
As I = \(\frac{2}{5}\) mr²; \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) (\(\frac{2}{5}\) mr²)ω² = mgh
as rω = v, \(\frac{1}{2}\) mv² + \(\frac{1}{5}\) mv² = mgh
v = \(\sqrt{\frac{10}{7} \mathrm{gh}} .\)
As h is same in two cases, v must be same i.e., It will reach the bottom with the same speed. Time taken to roll down the two planes will also be the same, as their height is the same.

Question 19.
A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it ?
Solution:
Here, R = 2m, M = 100 kg v = 20 cm/s = 0.2 m/s.
Total energy of the hoop = \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) Iω²
= \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) (MR)²ω² = \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) mv²
= mv².
Work required to stop the hoop = total energy of the hoop
w = mv² = 100(0.2)² = 4 joule.

Question 20.
The oxygen molecule has a mass of 5.30 × 10^-26 kg and a moment of inertia of 1.94 × 10^-46 kg m² about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of trans-lation. Find the average angular velocity of the molecule.
Solution:
Here, m = 5.30 × 10^-26 kg
I = 1.94 × 10-^-46 . kgm²
v = 500 m/s
If \(\frac{\mathrm{m}}{\mathrm{2}}\) is mass of each atom of oxygen and 2r is the distance between the two atoms as shown in fig. then

= 6.7 × 10-¹² rod/s.

Question 21.
A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.
a) How far will the cylinder go up the plane ?
b) How long will is take to return to the bottom ?
Solution:
Here θ = 30°, v = θm/s
Let the cylinder go up the plane upto a height h
From \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) Iω² = mgh
\(\frac{1}{2}\) mv² + \(\frac{1}{2}\) (\(\frac{1}{2}\) mr²) ω² = mgh
\(\frac{3}{4}\) mv² = mgh
h = \(\frac{3 v^2}{4 g}=\frac{3 \times 5^2}{4 \times 9.8}\) = 1.913 m
If s is the distance up the inclined plane, then as sin θ = \(\frac{\mathrm{h}}{\mathrm{s}^2}\)
s = \(\frac{\mathrm{h}}{\sin \theta}=\frac{1.913}{\sin 30^{\circ}}\) = 3.826 m
Time taken to return to the bottom
t = \(\sqrt{\frac{2 s\left(1+\frac{k^2}{r^2}\right)}{g \sin \theta}}=t \sqrt{\frac{2 \times 3.826\left(1+\frac{1}{2}\right)}{9.8 \sin 30^{\circ}}}\) = 1.53 s.

Question 22.
As shown in Fig. the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F.1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder.
(Take g = 9.8 m/s²)

(Hint: Consider the equilibrium of each side of the ladder separately.)
Solution:
Data seems to be insufficient

Question 23.
A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand . The angular speed of the platform is 30 revolutions per minute. The man then brings his arms dose to his body with the distance of each weight from the axis changing from 90cm to 20 cm. Thd moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m²
(a) What is his new angular speed ? (Neglect friction.)
b) Is kinetic energy conserved in the process ? If not, form where does the change come about ?
Solution:
Here l₁ = 7.6 × 2 × 5(0.9)² = 15.7 kgm²
ω₁ = 30 rpm
l₂ = 7.6 + 2 × 5(0.2)² = 8.0 kgm²
ω₂ = ?
According to the principle of conservation of angular momentum
l₂ω₂ = l₁ω₁
ω₂ = \(\frac{l_1}{l_2} \omega_1=\frac{15.7 \times 30}{8.0}\) = 58.88 rpm

Question 24.
A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.
(Hint : The moment of inertia of the door about the vertical axis at one end is ML²/3.)
Solution:
Angular momentum imparted by the bullet
I_ = mv × r = (10 × 10^-3) × 500 × \(\frac{1}{2}\) = 2.5
Also, L = \(\frac{M L^2}{3}=\frac{12 \times 1.0^2}{3}\) = 4kgm
As L = Lω
∴ ω = \(\frac{\mathrm{L}}{1}=\frac{2.5}{4}\) = 0.625 rad/sec.

Question 25.
Two discs of moments of inertia I, and I2 about their respective axes (normal to the disc and passing through the centre), and rotawing with angular speeds ω₁, and ω₂ are brought into contact face to face with their axes of rotation coincident,
(a) .What is the angular speed of the two-disc system ?
(b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy ? Take ω₁ ≠ ω₁.
Solution:
Here, total intial angular momentum of the two discs L₁ = I₁ω₁ + I₂ω₂
Under, the given conditions, moment of intertia of the two disc system = (I₁ + I₂)
If ω is angular speed of the combined system, the final angular momentum of the system
L₂ = (I₁ + I₂)ω
As no external torque is involved in this excercise, therefore, L₂ = L₁
(I₁ + I₂)ω = I₁ω₁ + I₂ω₂
ω = \(\frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2}\)

b) Initial K.E of two disc E₁ = \(\frac{1}{2}\) I₁ω₁² + \(\frac{1}{2}\) I₂ω₂²
Final K.E of the system E₂ = \(\frac{1}{2}\) (I₁ + I₂)ω²

∴ E₁ – E₂ > 0 or E₁ > E₂ or E₂ < E₁
Hence there occurs a loss of K.E in the process. Loss of energy = E₁ – E₂. This loss must be due to friction in the contact of the two discs.

Question 26.
a) Prove the theorem of perpendicular axes.
(Hint: Square of the distance of a point (x, y) in the x – y plane from an axis through the origin and perpendicular to the plane is x² + y²).
Answer:
Statement: The sum of moments of inertia of a plane lamina about any two perpendicular axes in its plane is equal to its moment of inertia about an axis perpendicular to the plane and passing through the point of intersection of the first two axes.

Proof : Consider a plane lamina revolving about the Z axis. Let ‘O’ be the origin of the axis. Imagine a particle of mass ‘m’ lying at a distance ‘r’ from point ‘o’ on the plane. Let x, y be the coordinates of the point P.
Thus r² = x² + y²
Then the moment of the body about x-aixs
I_x = Σm y²
The moment of inertia of the body about y-axis.
I_y = Σm x²
Then the moment of inertia of the body about Z-axis

I_z = Σm r²
I_z = Σm(x² + y²)
I_z = Σm x² + Σm y² = I_y + I_x
∴ I_z = I_x + I_y
Hence perpendicular axes theorem is proved.

b) Prove the theorem of parallel axes.
(Hint: If the centre of mass is chosen to be the origin Σm_ir_i = 0).
Answer:
Statement: The moment of inertia of a plane lamina about an-axis is equal to the sum of the moment of inertia about a parallel axis passing through the centre of mass and the product of its mass and square of the distance between the two axes i.e., I₀ = I_G + Mr²
Let I_G is the moment of inertia of the plane lamina about the axis Z₂ passing through the centre of mass.
I₀ is the moment of inertia of the plane lamina about an axis Z₁
Let M be the mass of the lamina and r be the distance between the two axes . Then
I₀ = I_G + mr².
Proof : Let a particle of mass m is situated at P. Moment of inertia about the axis passing through 0 is
dl = m op² or I = Σm op².
Join the lines PO and PG and draw the line PQ and Join with the line extending from OG.
From the trainagle POQ, OP² = OQ² + PQ²
OP² – (OG + GQ)² + PQ² + OQ = OG + GQ
OP² = OG² + 2OG . GQ + (GQ² + PQ²)
OP² – OG² + 2OG. GQ + GP²
OP² = OG² + GP² + 2OG.GQ
[∵ From the ∆le PGQ, GP² = PQ² + GQ²]

Multiplying with Σm on both side
Σm OP² = Σm OG² + Σm GP² + Σm OG. GQ
But Σm OG² = Mr²
(∵ OG is constant and Σm = M, total mass of the body)
Σm GP² = I_G
Σm OP² = I₀
∴ I₀ = Mr² – I_G + 2r ΣmGQ
Σm.GQ = 0
[∵ The moment of all the particles about the centre of mass is always zero]
I₀ = I_G + Mr²
Thus the theorem is proved.

Question 27.
Prove the result that the velocity u of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by υ² = \(\frac{2 g h}{\left(1+k^2 / R^2\right.}\) using dynamical consideration (i.e. by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.
Solution:
When a body rolls down an incline of height h, we apply the principle of conservation of energy.

K.E of translation + K.E of rotation = P.E at the top
i.e . \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) Iω² = mgh
\(\frac{1}{2}\) mv² + \(\frac{1}{2}\) (mk²) ω² = mgh
As w = \(\frac{\mathrm{V}}{\mathrm{R}}\)
∴ \(\frac{1}{2}\)mv² + \(\frac{1}{2}\)m \(\frac{k^2}{R^2}\)v² = mgh
or mv² (1 + \(\frac{k^2}{R^2}\)) = mgh
v² = \(\frac{2 g h}{\left(1+\frac{K^2}{R^2}\right)}\)

Question 28.
A disc rotating about its axis with angular speed ω₀ is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in Fig. ? Will the disc roll in the direction indicated ?

Solution:
Using the reaction v = rω,
we get
for point A, V_A = Rω₀, along AX
for point B, V_B = Rω₀, along BX
for point C, V_c = (\(\frac{\mathrm{R}}{\mathrm{2}}\) ω₀ parallel to AX,
The disc will not rotate, because it is placed on a perfectly frictionless table, without, rolling is not possible.

Question 29.
Explain why friction is necessary to make the disc in Figure roll in the direction indicated.

(a) Give the direction of frictional force at B and the sense of frictional torque before perfect rolling begins.
(b) What is the force of friction after perfect rolling begins ?
Solution:
To roll a disc, we require a torque, which can be proved only by a tangential force. As force of friction is the only tangential force in this case, it is necessary.
(a) As frictional force at B opposes the velocity of point B, which is to the left, the frictional force must be to the right. The sense of frictional torque will be perpendicular to the plane of the disc and outwards.

(b) As frictional force at B decreases the velocity of the point of contact B with the surface, the perfect rolling begins only when velocity of point B becomes zero. Also, force of friction would become zero at this stage.

Question 30.
A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 π rad s^-1. Which of the two will start to roll earlier ? The co-efficient of kinetic friction is μ_k = 0.2.
Solution:
Here, intial velocity of centre of mass is zero i.e, u = 0.
Frictional force causes the GM to accelerate
μ_k mg = ma ∴ a = μ_k g
As v = u + at ∴ v = 0 + μ_k gt
Torque due to friction causes retardation in the intial angular speed ω₀.
i.e. μ_k mg × R = – Iα
α = \(\frac{\mu_k \mathrm{mgR}}{\mathrm{I}}\)
ω = ω₀ + αt
∴ ω = ω₀ – \(\frac{\mu_k \mathrm{mgRt}}{\mathrm{I}}\)
Rolling begins, when v = Rω from (ii) and (iv)

Comparing (vi) and (vii) we find that the disc would begin to roll earlier than the ring We can calculate the values of t from (vi) and (vii) using known values of μ_k, g, R and ω₀.

Question 31.
A cylinder of mass 10kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction μ_s = 0.25.
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination θ of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly?
Solution:
Here, m = 10 kg, r = 15 cm = 0.15 m
θ = 30°, μ_s = 0.25
Accelaration of the cylinder down the incline,
α = \(\frac{2}{3}\)g sin θ = \(\frac{2}{3}\) × 9.8 sin 30° = \(\frac{9.8}{3}\) m/s²
a) Force of friction, f = mg sin θ – ma = m(g sin θ – α) = 10(9.8 sin 30° – \(\frac{9.8}{3}\)) = 16.4 N
b) During rolling the point of contact is at rest. Therefore work done against friction is zero
c) For rolling without slipping/skidding μ = \(\frac{1}{3}\) tan θ
tan θ = 3μ
= 3 × 0.25 = 0.75
θ = 37°.

Question 32.
Read each statement below carefully, and state, with reasons, if it is true or false;
(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.
(b) The instantaneous speed of the point of contact during rolling is zero.
(c) The instantaneous acceleration of the point of contact during rolling is zero.
(d) For perfect rolling motion, work done against friction is zero.
(e) A wheel moving down a perfectly italic inclined plane will undergo slipping (not roiling) motion.
Solution:
a) The statement is false.

b) True. This is because rolling body can be imagined to be rotating about an axis passing through the point of contact of the body with the ground.
Hence its instantaneous speed is zero.

c) This is not true. This is because when the body is rotating its instantaneous acc is not zero.

d) It is true. This is because once the perfect rolling begins, force of friction becomes zero. Hence work done against friction is zero.

e) The statement is true. This is because rolling occurs only on account of friction which is tangential force capable of providing torque. When the inclined plane is perfectly smooth it will simply slip under the effect of its own weights.

Question 33.
Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass :
(a) Show P = \(\mathrm{P}_{\mathrm{i}}^{-}+\mathrm{m}_{\mathrm{i}} \mathrm{V}\)
Where P_i is the momentum of the i_i the particle (of mass m_i) and p’_i + m_iv’_i. Note relative velocity of the ith particle relative to the centre of mass. Also, prove using the definition of the centre of mass Σp’_i = 0
(b) Show K = K’ + 1/2MV²
Where K is the total kinetic energy of the system of particles. K’ is the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass and MV²/2 is the kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the system). The result has been used in Sec. 7.14.
(c) Show L = L’ + R × MV
Where L’ = Σr_i P_i is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember r’_i = r_i – R rest of the notation is the standard notation used in the chapter. Note L1 and MR × V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.
(d) Show \(\frac{d l^{\prime}}{d t}=\sum r_i^{\prime} \frac{d p^{\prime}}{d t}\)
Further, show that \(\frac{\mathrm{dL}^{\mathrm{l}}}{\mathrm{dt}}\) = τ’_ext
Whereτ’_ext is the sum of all external torques acting on the system about the centre of mass.
(Hint: Use the definition of centre of mass and Newton’s Third Law. Assume the internal forces between any two particles act along the line joining the particles.)
Solution:
a) Let m₁, m₂, … m_i mass points have the position vectors \(\overrightarrow{r_1}, \overrightarrow{r_2}, \overrightarrow{r_i}\) w.r.t origin ‘O’. The position vector of C.M. say \(\overrightarrow{\mathrm{OP}}\)
i.e., \(\overrightarrow{\mathrm{OP}}=\frac{m_1 \overrightarrow{r_1}+m_2 \overrightarrow{r_2}+\ldots}{m_1+m_2+\ldots}=\frac{m_i r_i}{M}\)
where i = 1, 2, 3 .
Now, let us change the origin to O’ and assume that the C.M is now at p’ with

Multiplying eqn. (1) by m_i and differentiating w.r.t to time, we get

even if we change the origin, the position of
centre of mass will not change i.e., Σp’_i = 0

b) In rotational kinematics, K.E_T of the system of particles = K.E_T of the system when the particle velocities are taken w.r.t to CM + K.E of the translation of the system as a whole (i.e. of the C.M motion of the system)
i.e. \(\frac{1}{2}\) m₁v₁² + \(\frac{1}{2}\) m₂v₂² + …..
= [\(\frac{1}{2}\) m₁v₁² + \(\frac{1}{2}\)m₂v₂ +……] + \(\frac{1}{2}\)Mv²
\(\frac{1}{2}\) m_iv_i² = \(\frac{1}{2}\)m_iv_i² + \(\frac{1}{2}\) mv²
∴ k = k’ + \(\frac{1}{2}\)
where M = total mass of particles
and y = velocity of C.M motion of the system.

c) From eq. (1), we have

Textual Examples

Question 1.
Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are 100g, 150 g, and 200 g respectively. Each side of the equilateral triangle is 0.5 long. [A.P. & T.S. Mar.18]
Solution:

With the x – and y-axes chosen as shown in Fig. 7.9, the coordinates of points O, A and B forming the equilateral triangle are
respectively (0, 0), (0, 5, 0), (0.25. \(\sqrt{3}\) ). Let the masses 100 g, 150g and 200g be located at O, A and B be respectively. Then.

Question 2.
Find the centre of mass of a triangular lamina.
Solution:
The lamina (∆LMN) may be subdivided into narrow strips each parallel to the base (MN) as shown in Fig.

By symmetry each strip has its centre of mass at its midpoint. If we joint the midpoint of all the strips we get the median LP. The centre of mass of the triangle as a whole therefore, has to lie on the median LP. Similarly, we can argue that it lies on the median MQ and NR. This means the centre of mass lies on the point of concurrence of the medians, i.e. on the centroid G of the triangle.

Question 3.
The density of the atmosphere at sea level is 1.29 kg/m³. Assume that it does not change with altitude. Then how high would the atmosphere extend ?
Solution:
The centres of mass C₁, C₂ and C₃ of the squares are, by symmetry, their geometric centres and have coordinates (1/2,1/2), (3/ 2,1/2), (1/2,3/2) respectively.

Question 4.
A 3m long ladder weighing 20 kg leans on a frictionless wall. Its feet reston the floor 1 m from the wall as shown in Fig. Find the reaction forcesof the wall and the floor.
Solution:

Let AB is 3 m long, A is at distance AC = 1 m from the wall. Pythagoras theorem,
BC = 2\(\sqrt{2}\) m.
For translational equilibrium, taking the forces in the vertical direction, N – W = 0 ……….. (i)
Taking the forces in the horizontal direction,
F – F₁ = 0 ……………. (ii)
taking the moments of the forces about A,
2\(\sqrt{2}\) F1 – (1/2) W = 0 ………….. (iii)
Now W = 20 g = 20 × 9.8 N = 1960.0 N
From (i) N = 196.0
From (iii) = F₁ = W/4 \(\sqrt{2}\) = 196.0/4 \(\sqrt{2}\)
= 34.6 N
From (ii) F = F₁ = 34.6 N
F₂ = \(\sqrt{F^2+N^2}\) = 199.0 N

Question 5.
Find the scalar and vector products of two vectors
a = \((3 \vec{i}-4 \vec{j}+5 \vec{k})\) and b = \((-2 \vec{i}+\vec{j}-3 \vec{k})\)
Solution:
a.b = (\((3 \vec{i}-4 \vec{j}+5 \vec{k})\)) . (\((-2 \vec{i}+\vec{j}-3 \vec{k})\))
= -6 – 4 – 15 = -25
a × b = \(\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
3 & -4 & 5 \\
-2 & 1 & -3
\end{array}\right|=7 \vec{i}-\vec{j}-5 \vec{k}\)
Note b × a = \(7 \vec{i}-\vec{j}-5 \vec{k}\)

Question 6.
Obtain Equation from first principles.
Solution:
The angular acceleration is uniform,
ω = αt + c (as α is comstant)
At t = 0, ω = ω₀ (given)
From (i) we get at t = 0, ω = c = ω₀
Thus, ω = αt + ω₀ as required

Question 7.
The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm is 16 seconds, (i) What is its angular acceleration, assuming the acceleration to be uniform ? (ii) How many revolutions does the engine make during this time ?
Solution:
(i) We shall use ω = ω₀ + αt
ω₀ = initial angular speed in rad/s
= 2π × angular speed in rev/s
= \(\frac{2 \pi \times \text { angular speed in rev } / \mathrm{min}}{60 \mathrm{~s} / \mathrm{min}}\)
= \(\frac{2 \pi \times 1200}{60} \mathrm{rad} / \mathrm{s}\)
= 40π rad/ s
Similarly ω = final angular speed in rad /s
= \(\frac{2 \pi \times 3120}{60} \mathrm{rad} / \mathrm{s}\)
= 2π × 52 rad/s = 104 π rad/s
∴ Angular acceleration
α = \(\frac{\omega-\omega_0}{t}\) = 4π rad/s²
The angular acceleration of the engine = 4π rad/s²

(ii) The angular displacement in time t is given by
θ = ω₀t + \(\frac{1}{2}\) αt²
= (40π × 16 + \(\frac{1}{2}\) × 4π × 16²) rad
= (640 π+ 512 π )rad
= 1150π rad
Number of revolutions = \(\frac{1152 \pi}{2 \pi}\) = 576

Question 8.
Find the torque of a force \((\bar{i}-\bar{j}+\bar{k})\) about the origin. The force acts on a particle whose position vector is \((\bar{i}-\bar{j}+\bar{k})\) (Mar.’14, 13)
Answer:
Here r = \(\vec{i}-\vec{j}+\vec{k}\)
and F = \(7 \vec{i}-3 \vec{j}+5 \vec{k}\)
We shall use the determinant rule to find the torque τ = r × F
τ = \(\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
1 & -1 & 1 \\
7 & 3 & -5
\end{array}\right|=(5-3) \vec{i}-(-5-7) \vec{j}\) + \((3-(-7)) \vec{k}\)
or τ = \(2 \vec{i}+12 \vec{j}+10 \vec{k}\)

Question 9.
Show that the angular momentum about any point of a single particle moving with constant velocity remains constant throughout the motion.
Solution:
Let the particle with velocity v be at point P at some instant t. We want to calculate the angular momentum of the particle about an arbitrary point O.

The angular momentum is 1 = r × mv. Its magnitude is mvr. sine, where θ is the angle between r and v as shown in Fig. Although the particle changes position with time, the line of direction of v remains the same and hence OM = r sin θ ; a constant.

Question 10.
Show that moment of a couple does not depend on the point about which you take the moments.
Solution:

The moment of the couple = sum of the moments of the two forces making the couple
= r₁ × (-F) + r₂ × F
= r₂ × F + r₁ × F
= (r₂ – r₁) × F
But r₂ + AB = r₂, and hence AB = r₂ – r₁.
The moment of the couple, therefore, is AB × F.

Question 11.
What is the moment of intertia of a disc about one of its diameters ?
Solution:

x and y-axes lie in the plane of the disc and z is perpendicular I_z = I_x = I_y
Now, x and y axes are along two diameters of the disc, and by symmetry the moment of inertia of the disc is the same about any diameter. Hence
I_x = I_y
and I_z = 2I_y
But I_x = I_y
So finally, I_x = I_z/2 = MR²/4
Thus the moment of inertia of a disc about any of its diameter is MR²/4.
Find similarly the moment of inertia of a ring about any of its diameter. Will the theorem be applicable to a solid cylinder?

Question 12.
What is the moment of inertia of a rod of mass M, length l about an axis perpendicular to it through one end ?
Solution:
For the rod of mass M and length l, I = Ml²/ 12. Using the parallel axes theorem, I’ = I + Ma² with a = l/2 we get.
I’ = \(\mathrm{M} \frac{l^2}{12}+\mathrm{M}\left(\frac{l}{2}\right)^2=\frac{\mathrm{M} l^2}{3}\)
We can check this independently since I is half the moment of inertia of a rod of mass 2M and length 21 about its midpoint,
I’ = \(2 \mathrm{M} \frac{4 l^2}{12} \times \frac{1}{2}=\frac{\mathrm{M} l^2}{3}\)

Question 13.
What is the moment of inertia of a ring about a tangent to the circle of the ring ?
Solution:
The tangent to the ring in the plane of the ring is parallel to one of the diameters of the ring. The distance between these two parallel axes is R, the radius of the ring. Using the parallel axes theorem.

= 0.4kg m²
α = angular acceleration
= 5.0 Nm/0.4 kg m² = 12.5 S^-2

b)Work done by the pull unwinding 2m of the cord
= 25 N × 2 m = 50 J

(c) Let ω be the final angular velocity. The kinetic energy gained = \(\frac{1}{2}\) Iω²
since the wheel starts from rest. Now,
ω₂ = ω²₀ +2αθ, ω₀ = 0
The angular displacement θ = length of unwound string / radius of wheel = 2m / 0.2 m = 10 rad
ω² = 2 × \(\frac{1}{2}\) × 12.5 × 10.0 = 250 (rad/s)²
∴ K.E. gained = \(\frac{1}{2}\) × 0.4 × 250 = 50 J

(d) The answers are the same, i.e. the kinetic energy gained by the wheel work done by the force. There is no loss of energy due to friction.

Question 14.
A cord of negligible mass is wound round the rim of a fly wheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord as shown in Fig. The flywheel is mounted on a horizontal axle -with frictionless bearings.
(a) Compute the angular acceleration of the wheel.
(b) Find the work done by the pull, when 2m of the cord is unwound
(c) Find also the kinetic energy of the wheel at this point. Assume that the wheel starts from rest.
(d) Compare answers to parts (b) and (c).
Solution:

a) We use Iα = τ
the toque τ = FR
= 25 × 0.20 Nm (as R = 0.20m)
= 5.0 Nm

Question 15.
Three bodies, a ring, a solid cylinder and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity ?
Solution:
We assume conservation of energy of the rolling body, i.e. there is no loss of energy due to friction etc. The protential energy lost by the body in rolling down the inclined plane (= mgh) must, therefore, be equal to kinetic energy gained. (See fig.) Since the bodies start from rest the kinetic energy gained is equal tot he final kinetic energy of the odies. From K = \(\frac{1}{2} m v_{\mathrm{cm}}^2\left[1+\frac{\mathrm{k}^2}{\mathrm{R}^2}\right]\)
K = \(\frac{1}{2} m v^2 1+\left(\frac{K^2}{R^2}\right)\) Where v is the final velocity of (the centre of mass of) the body.
Equating K and mgh

Note v² is independent of the mass of the rolling body:
For a ring, k² = R²
v_ring = \(\sqrt{\frac{2 g h}{1+1}}=\sqrt{g h}\)
For a solid cylinder K² = R²/0.2
v_disc = \(\sqrt{\frac{2 g h}{1+1 / 2}}=\sqrt{\frac{4 g h}{3}}\)
For a solid sphere K² = 2R²/5
v_sphere = \(\sqrt{\frac{2 g h}{1+1 / 52}}=\sqrt{\frac{10 g h}{7}}\)

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 5th Lesson Laws of Motion Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 5th Lesson Laws of Motion

Very Short Answer Questions

Question 1.
What is inertia ? What gives the measure of inertia ? [T.S. Mar. 17]
Answer:
The resistance of the body to change its state of rest of state of uniform motion in a straight line is called inertia of the body.
Acceleration = \(\frac{\text { Force }}{\text { Mass }}\)
∴ The more is the mass, less is the acceleration and more is the inertia. The mass of a body is a quantitative measure of its inertia.

Question 2.
According to Newton’s third law, every force is accompanied by an equal and opposite force. How can a movement ever take place ?
Answer:
Because both action and reaction are taking place on different bodies.

Question 3.
When a bullet is fired from a gun, the gun gives a kick in the backward direction. Explain. [A.P. Mar. 15]
Answer:
Due to law of conservation of momentum, Recoil of the gun V = \(\frac{\mathrm{mu}}{\mathrm{M}}\)
Where M – mass of the gun; m = mass of the bullet; u = velocity of the bullet

Question 4.
Why does a heavy rifle not recoil as strongly as a light rifle using the same cartridges ?
Answer:
Recoil of the Gun V = \(\frac{\mathrm{mu}}{\mathrm{M}}\)
Due to heavy mass of rifle the recoil is less.

Question 5.
If a bomb at . _st explodes into two pieces, the pieces must travel in opposite directions. Explain. [T.S. Mar. 16, 15]
Answer:
According to law of conservation of momentum,
Mu = m₁v₁ + m₂v₂
Initially the bomb is at rest u = 0
m₁v₁ + m₂v₂ = 0
or m₁v₁ – m₂v₂
(Negative sign indicates that the pieces must travel in opposite direction)

Question 6.
Define force. What are the basic forces in nature ?
Answer:
The force is on which changes or tends to change the state of rest or motion of a body. Basic forces :

1. Gravitational force
2. Electromagnetic force
3. Nuclear force
4. Weak interaction force

Question 7.
Can the coefficient of friction be greater than one ?
Answer:
Yes, coefficient of friction may be greater than one. In some particular cases it is possible. They are

1. Due to increase the inner molecular attractive forces between surfaces when the contact surfaces are highly polished.
2. When the contact surfaces of the bodies are inter locking the coefficient friction may be greater than one.

Question 8.
Why does the car with a flattened tyre stop sooner than the one with inflated tyres ?
Answer:
Flattened deforms more than the inflated tyre. Due to greater deformation of the type rolling friction is large nence it stops soon.

Question 9.
A horse has to pull harder during the start of the motion than later. Explain. [Mar. 13]
Answer:
We know the limiting frictional force is greater than kinetic frictional force. For starting motion of the cart, the limiting friction is to be overcome. Once motion is set, frictional force reduces. Therefore, the horse has to pull harder during starting of the cart.

Question 10.
What happens to the coefficient of friction if the weight of the body is doubled? [A.P. – Mar. 16]
Answer:
If weight of a body is doubled, coefficient of friction does not change. Coefficient of friction is independent of normal reaction. If weight is doubled, normal reaction doubled and correspondingly frictional force doubled. So, coefficient of friction does not change.

Short Answer Questions

Question 1.
A stone of mass 0.1 kg is thrown vertically upwards. Give the magnitude and direction of the net force on the stone
(a) during its upward motion.
(b) during the downward motion.
(c) at the highest point, where it momentarily comes to rest.
Answer:
Given that, mass of stone, m = 0.1 kg, g = 9.8 ms^-2.
a) During upward motion: Magnitude of net force on the stone,
F = |-mg|; F = 0.1 × 9.8 = 0.98N.
Direction of net force is in upward direction.

b) During downward motion: Magnitude of net force on the stone,
F = ma = 0.1 × 9.8 = 0.98N.
Direction of net force is in downward direction.

c) At the heighest point : Magnitude of net force, F = mg = 0.1 × 9.8 = 0.98N..
At highest point of stone, direction is indeterminate.

Question 2.
Define the terms momentum and impulse. State and explain the law of conservation of linear momentum. Give examples.
Answer:
Momentum : The product of mass and velocity of a body is called momentum momentum p = mv .
Impulse : The product of force and time that produces finite change in momentum of the body is called impulse.
Impulse (I) = Force × time duration = mat = \(m \frac{(v-u)}{t} t\)
= (mv – mu)
Law of conservation of linear momentum : The total momentum of an isolated system of interacting particles remains constant it there is no resultant external force acting on it.

Explanation : Consider two smooth, non-rotating spheres of masses m₁ and m₂ (m₁ > m₂). Let u₁ and u₂ be their initial velocities. Let v₁ and v₂ be final velocities after head on collision. According to law of conservation of linear momentum, we have
Momentum of the system before collision = Momentum of the system after collision.
t.e., m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Examples :

1. Motion of a Rocket
2. Bullet-Gun motion

Question 3.
Why are shock absorbers used in motor cycles and cars ?
Answer:
When a scooter or a car moves on a rough road it receives an impulse due to the Jerkey motion. In case the shockers are used in the vehicle, the time of impact increases, and decreases the impulsive force, due to increased value of the time of impact the force of impact is reduced. So it saves the vehicle and its occupants from experiencing reverse jerks.

Question 4.
Explain the terms limiting friction, dynamic friction and rolling friction.
Answer:
Limiting friction : The maximum value of static friction is called “Limiting friction”.
It is denoted by F_L = F_s(max) [∵ F_s ≤ μ_sN]
Dynamic friction (Kinetic friction) : The resistance encountered by a sliding body on a surface is called kinetic or dynamic friction F_k.
If the applied force overcomes the limiting friction and sets the body into motion. Then motion of the body is resisted by another friction called “Dynamic friction” or “Kinetic friction”.

Rolling friction : ‘The resistance encountered by a rolling body on a surface is called “Rolling friction”.
If a wheel or a cylinder or a spherical body like a marble rolls on horizontal surface, the speed of rolling gradually decreases and it finally stops.

Question 5.
Explain advantages and disadvantages of friction. [T.S., A.P. Mar. 15]
Answer:
Advantages of friction :

1. Safe walking on the floor, motion of vehicles etc., are possible only due to friction.
2. Nails, screws are driven into walls (or) wooden surfaces due to friction.
3. Friction helps the fingures hold the things (or) objects like pen, pencil and water tumbler etc.
4. Speed running vehicles etc. can be stopped suddenly when friction is present, otherwise accidents become large. Due to friction vehicles move on the roads without slipping and they can be stopped.
5. The mechanical power transmission of belt drive is possible due to friction.

Disadvantages of friction:

1. Due to friction there is large amount of power loss in machines and engines.
2. Due to friction wear and tear of the machines increases and reducing their life.
3. Due to friction some energy gets converted into heat which goes as waste.

Question 6.
Mention the methods used in decrease friction [A.P. Mar. 18; T.S. Mar. 16, Mar. 14]
Answer:

1. Polishing : By polishing the surfaces of contact, friction can be reduced.
2. Bearings : The rolling friction.is less than the sliding friction hence free wheels of a cycle, motor
   car, dynamos etc., are provided with ball bearings to reduce friction. Bearings convert sliding motion into rolling motion.
3. Lubricants: The lubricant forms a thin layer between surfaces of contact. It reduces the friction. In light vehicles or machines, oils like “three in one” are used as lubricants. In heavy machines greasure is used. In addition to this they guard the mechanical parts from over heating.
4. Streamlining : Automobiles and aeroplanes are streamlined to reduce the friction due to air.

Question 7.
State the laws of rolling friction.
Answer:
Laws of friction – rolling friction :

1. The smaller the area of contact, the lesser will be the rolling friction.
2. The larger the radius of the rolling body, the lesser will be the rolling friction.
3. The rolling friction is directly proportional to the normal reaction.
   If F_R is the rolling friction and ‘N’ is the normal reaction at the contact, then F_R ∝ N
   F_R = μ_RN; where μ_R is the coefficient of rolling friction.

Question 8.
Why is pulling the lawn roller preferred in pushing it ?
Answer:
Pulling of lawn roller : Let a lawn roller be pulled on a horizontal road by a force ‘F, which makes an angle θ with the horizontal, to the right as shown in the figure. The weight of the body “mg” acts vertically downwards.

Let the force ‘F’ be resolved into two mutually perpendicular components F sin θ, vertically upwards and F cos θ horizontally along the road.
∴ The normal reaction N = mg – F sin θ
Then the frictional force acting towards left is F_R = μ_RN, where μ_R is the coefficient of rolling friction between he roller and the road, or F_R= μ_R (mg – F sin θ)
∴ The net pulling force on roller is P = F cos θ – F_R = F cos θ – (μ_R mg – F sin θ)
or P = F(cos θ + μ_R sin θ) – μ_R mg ………………(1)

Pushing of lawn roller : When a lawn roller is pushed by a force ‘F1, which makes an angle ‘θ’ with the horizontal, the component of force acting vertically downwards is F sin θ. The horizontal component F cos θ pushes the roller to the right as shown in figure.
The weight ‘mg’ of the lawn roller acts vertically downwards. Therefore the normal reaction N of the surface on the roller
N = mg + F sin θ
Then the frictional force acting towards left.
F_R = μ_RN = μ_R (mg + F sin θ)
The net pushing force on roller is
P’ = F cos θ – F_R = F cos θ – μ_R (mg + F sin θ)
or P’ = F(cos θ – μ_R sin θ) – μ_R mg ……………… (2)
From equations (1) and (2) that it is easier to pull than push a lawn roller.

Long Answer Questions

Question 1.
a) State Newton’s second law of motion. Hence derive the equation of motion F = ma from it. [T.S. Mar. 18; A.P. Mar. 16, Mar. 13]
b) A body is moving along a circular path such that its speed always remains constant. Should there be a force acting on the body ?
Answer:
a) Newton’s Second of motion : “The rate of change of momentum of a body is directly proportional to the external force applied and takes place in the same direction”.
To show F = ma : Let a body of mass ‘m’ moving with a velocity ‘v’ under the action of an external force F in the direction of velocity.
Momentum ‘P’ of a body is the product of the mass and velocity V.
∴ P = mv ……………… (1)
According to Newton’s second law of motion, we have
\(\frac{\mathrm{dp}}{\mathrm{dx}}\) ∝ F, where F = external force
(or)F = K\(\frac{\mathrm{dp}}{\mathrm{dx}}\) ………………. (2)
From equations (1) and (2) we have
F = \(K \frac{d(m v)}{d t}=K \cdot m \frac{d v}{d t}\)
= Kma …………………. (3)
Since the rate of change of velocity \(\frac{\mathrm{dv}}{\mathrm{dx}}\) is the acceleration ‘a’ of the body.
In SI system the unit of force is Newton and is defined as that force which when acting on a body of mass 1 kg produces in it an acceleration of 1 ms^-2.
i.e., from equation (3),
If F = 1, m = 1 and a = 1
we get K = 1
Hence F = \(\frac{\mathrm{dp}}{\mathrm{dx}}\) = ma .
∴ F = ma

b) Suppose a body is moving along a circular part though its speed always remains constant its velocity changes at every point and resultant force acts on the body.

Question 2.
a) Define angle of friction and angle of repose. Show that angle of friction is equal to angle of repose for a rough inclined place.
Answer:
Angle of friction : The angle of friction is defined as the angle made by the resultant of the normal reaction and the limiting friction with the normal reaction is called angle of friction.

Angle of repose : The angle of repose is defined as the angle of inclination of a plane with respect to horizontal for which the body will be in equilibrium on the inclined plane is called angle of repose.

Angle of friction is equal to angle of repose for a rough inclined plane : Let us consider a body of mass’m1 on a rough inclined plane. The angle of inclination of the rough surface is ‘θ’. By increasing the angle of inclination at one end, the body tends to slide on the surface. Then the angle of inclination ’0′ is called angle of repose.

The weight (mg) of the body resolved into two components, the component mg cos 0 acts perpendicular to the inclined surface, which is equal to normal reaction ‘N‘.
i.e., N = mg cos θ ……………. (1)
The other component mg sin θ acts parallel to inclined plane and opposite to the frictional force ‘f’.
f_s = my sin θ ……………… (2)
From \(\frac{(2)}{(1)} \Rightarrow \frac{f_s}{N}=\frac{m g \sin \theta}{m g{con} \theta}\)
= \(\frac{f_s}{N}\) = tan θ = µ_s
∴ µ_s = tan θ …………………. (3)
when ‘α’ is angle of friction then from the definition of co-efficient of static friction,
µ_s = tan α ………………… (4)
from (3) and (4)
⇒ tan θ = tan α
θ = α
Hence angle of friction is equal to angle of repose.

b) A block of mass 4 kg is resting on a rough horizontal plane and is about to move when a horizontal force of 30 N is applied on it. If g = 10 m/s². Find the total contact force exerted by the plane on the block.
Solution:
Given m = 4kg
F = 30 N
g = 10 ms^-2
a = \(\frac{F}{m}=\frac{30}{4}\) = 7.5 ms^-2
µ = \(\frac{\mathrm{a}}{\mathrm{g}}=\frac{7.5}{10}=\frac{3}{4}\)
∴ Contact force = Frictional force
= µ mg
= \(\frac{3}{4}\) × 4 × 10 = 30 N.

Problems

Question 1.
The linear momentum of a particle as a function of time t is given by p = a + bt, where a and b are positive constants. What is the force acting on the particle ?
Answer:
Linear momentum of a particle p = a + bt
Force F = \(\frac{\mathrm{dp}}{\mathrm{dt}}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\)(a + bt) = 0 + b
∴ F = b

Question 2.
Calculate the time needed for a net force of 5 N to change the velocity of a 10 kg mass by 2 m/s.
Answer:
F = 5N, m = 10kg; (v-u) = 2m s^-1, t = ?
F = \(m \frac{(v-u)}{t} \Rightarrow 5=\frac{10 \times 2}{t}\)
∴ t = 4s.

Question 3.
A ball of mass m is thrown vertically upward from the ground and reaches a height h before momentarily coming to rest. If g is acceleration due to gravity. What is the impulse received by the ball due to gravity force during its flight ? (neglect air resistance).
Answer:
mass = m
Initial velocity of a ball to reach h is,
u = \(\sqrt{2 g h}\)

On return journey, velocity of a ball to reach the ground, v = –\(\sqrt{2 g h}\)
Impluse I = m (u – v) = m [\(\sqrt{2 g h}\) – (-\(\sqrt{2 g h}\))]
= m 2 \(\sqrt{2 g h}\) = 2m \(\sqrt{2 g h}\)
∴ I = \(\sqrt{8 m^2 g h}\)

Question 4.
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 in m s^-1 to 3.5 m s^-1 in 25 s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force ?
Answer:
m = 3.0 kg; u = 2.0 ms^-1,
v = 3.5ms^-1, t = 25 s;
F = m \(\left(\frac{v-u}{t}\right)\) = 3\(\left(\frac{3.5-2}{25}\right)\)
= \(\frac{3 \times 15}{25}=\frac{0.9}{5}\) = 0.18 N
This force acts in the direction of change in velocity.

Question 5.
A man in a lift feels an apparent weight W when the lift is moving up with a uniform acceleration of 1/3^rd of the acceleration due to gravity. If the same man were in the same lift now moving down with a uniform acceleration that is 1/2 of the acceleration due to gravity, then what is his apparent weight ?
Answer:
When the lift is moving up, a = \(\frac{g}{3}\)
Apparent weight W = m(g + a)
= m(g + \(\frac{g}{3}\)) = \(\frac{4mg}{3}\)
⇒ mg = \(\frac{3 W}{4}\) …………….. (1)
When the lift is moving down, a = \(\frac{g}{2}\)
Apparent weight W¹ = m(g – a)
= m(g – \(\frac{g}{2}\)) = \(\frac{mg}{2}\)
∴ |W¹| = \(\frac{\left[\frac{3 W}{4}\right]}{2}=\frac{3 W}{8}\)

Question 6.
A container of mass 200 kg rests on the back of an open truck, if the truck accelerates at 1.5 m/s², what is the minimum coefficient of static friction between the container and the bed of the truck required to prevent the container from sliding off the back of the truck ?
Answer:
m = 200kg, a = 1.5 ms^-2, g = 9.8 ms^-2
ma = μ_smg
μ_s = \(\frac{a}{g}=\frac{1.5}{9.8}\) = 0.153

Question 7.
A bomb initially at rest at a height of 40 m above the ground suddenly explodes into two identical fragments. One of them starts moving vertically down- wards with an initial speed of 10 m/s. If acceleration due to gravity is 10 m/s². What is the separa-tion between the fragments 2s after the explosion ?
Answer:
Bomb explodes into two fragments say 1 and 2.
For 1^st fragment, u₁ = 10m/s, t = 2 sec; g = 10 m/s^-2, s₁ = ?
Now displacement of 1^st fragment,
s₁ = u₁t + \(\frac{1}{2}\) gt²
= 10 × 2 × \(\frac{1}{2}\) × 10 × 2² = 40m

2^nd fragment is move opposite to 1^st fragment like an object from a tower.
For 2^nd fragment u₂ = – u₁ = 10m/s
t = 2 sec; g = 10 m/s²
Now displacement of 2^nd fragment
s₂ = u₂t + \(\frac{1}{2}\) gt²
= 10 × 2 × \(\frac{1}{2}\) × 10 × 2² = 0
∴ The seperation of two fragments
= s₁ + s₂ = 40 + 0 = 40 m

Question 8.
A fixed pulley with a smooth grove has a light string passing over it with a 4 kg attached on one side and a 3 kg on the other side. Another 3 kg is hung from the other 3 kg as shown with another light string. If the system is released from rest, find the common acceleration ? (g = 10 m/s²).

Answer:
From fig,
m₁ = 3 + 3
= 6 kg
m₂ = 4 kg
g = 10ms^-2
Acceleration of the system,
a = \(\left(\frac{\dot{m}_1-m_2}{m_1+m_2}\right) g=\left(\frac{6-4}{6+4}\right) \times 10\) = 2 ms^-2

Question 9.
A block of mass of 2 kg slides on an inclined plane that makes an angle of 30% with the horizontal. The coefficient of friction between the block and the surface is \(\frac{\sqrt{3}}{2}\).
(a) What force should be applied to the block so that it moves up without any acceleration ?
(b) What force should be applied to the block so that it moves up without any acceleration ?
Answer:
m = 2kg; θ = 30°; µ = \(\frac{\sqrt{3}}{2}\)
i) F_down = mg (sinθ – µ cosθ)
= 2 × 9.8 (sin30° – \(\frac{\sqrt{3}}{2}\)cos30°)
= 2 × 98 [\(\frac{1}{2}\) – \(\frac{\sqrt{3}}{2}\) × \(\frac{\sqrt{3}}{2}\)] = 49 N

ii) F_up = mg (sinθ + µ cosθ)
= 2 × 9.8 [sin30° + \(\frac{\sqrt{3}}{2}\) × cos30°]
= 2 × 9.8 [\(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\) × \(\frac{\sqrt{3}}{2}\)] = 24.5 N

Question 10.
A block is placed on a ramp of parabolic shape given by the equation g = \(\frac{x^2}{20}\), see If i_s = 0.5, what is the maximum height above the ground at which the block can be placed without slipping ? (tan θ = µ_s = \(\frac{\mathrm{d} y}{\mathrm{~d} x}\))

Answer:

Question 11.
A block of metal of mass 2 kg on a horizontal table is attached to a mass of 0.45 kg by a light string passing over a frictionless pulley at the edge of the table. The block is subjected to a horizontal force by allowing the 0.45 kg mass to fall. The coefficient of sliding friction between the block and table is 0.2.

Calculate (a) the initial acceleration, (b) the tension in the string, (c) the distance the block would continue to move if. after 2 s of motion, the string should break.
Answer:
Here, m₁ = 0.45kg
m₂ = 2kg
µ = 0.2
a) Initial acceleration.
a = \(\left(\frac{m_1-\mu m_2}{m_1+m_2}\right) g=\left[\frac{0.45-0.2 \times 2}{0.45+2}\right] \times 9.8\)
a = 0.2ms^-2

b) From fig, we have
T – f = m₂ a
T – 3.92 = 2 × 0.2
[∵ f = µm₂g = 0.2 × 2 × 9.8] = 3.92 N
⇒ N = 0.4 + 3.92 = 4.32 N

c) Velocity of string after 2 sec
= u in this case: µ’ = 0
Stoping distance s = \(\frac{\mu^2}{2 \mu g}=\frac{0.4 \times 0.4}{2 \times 0.2 \times 9.8}\) = 0.0408 m

Question 12.
On a smooth horizontal . surface a block A of mass 10 kg is kept. On this block a second block B of mass 5 kg is kept. The coefficient of friction between the two blocks is 0.4. A horizontal force of 30 N is applied on the lower block as shown.

The force of friction between the blocks is (take g = 10 m/s²).
Answer:
Here m_A = 10kg; m_B = 5kg;
F = 30N; µ = 0.4
F = (m_A + m_B)a
⇒ a = \(\frac{F}{\left(m_A+m_B\right)}\)
= \(\frac{30}{10+5}\) = 2ms^-2
f = m_Ba = 5 × 2 = 10 N

Additional Problems

(For simplicity in numerical calculations, take g = 10 ms^-2)

Question 1.
Give the magnitude and direction of the net force acting on
a) a drop of rain falling down with a constant speed,
b) a cork of mass 10 g floating on water,
c) a kite skillfully held stationary in the sky,
d) a car moving with a constant velocity of 30 km/h on a rough road,
e) a high-speed electron in space far from all material objects and free of electric and magnetic fields.
Answer:
a) As the rain drop is falling with a constant speed, its acceleration a = 0. Hence net force F = ma = 0.
b) As the cork is floating on water, its weight is being balanced by the upthrust (equal-to weight of water displaced). Hence net force on the cork is zero.
c) As the kite is held stationary, net force on the kite is zero, in accordance with Newton’s first law.
d) Force is being applied to overcome the force of friction. But as velocity of the car is constant, its acceleration, a = 0. Hence net force on the car F = ma = 0.
e) As no field (gravitational / electric / magnetic) is acting on the electron, net force on it is zero.

Question 2.
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble.
a) during its upward motion,
b) during its downward motion
c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at ah angle of 45° with the horizontal direction ?
Ignore are resistance.
Answer:
When a body is thrown vertically upwards (or) it moves vertically downwards, gravitational pull of earth gives it a uniform acceleration a = + g = + 9.8 ms^-2 in the downward direction. Therefore, the net force on the pebble in all the three cases is vertically downwards.
As m = 0.05 kg and a = + 9.8 m/s²
In all the three cases,
∴ F = ma
= 0.05 × 9.8 = 0.49 N, vertically downwards.
If the pebble were thrown at an angle of 45° with the horizontal direction, it will have horizontal and vertical components of velocity. These components do not affect the force on the pebble. Hence our answers do not alter in any case. However in each case (C), the pebble will not be at rest. It will have horizontal component of velocity at highest point.

Question 3.
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg.
a) just after it is dropped from the window of stationary train,
b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h,
c) just after it is dropped from the window of a train accelerating with 1 ms^-2,
d) lying on the floor of a train which is accelerating with 1 ms-2, the stone being at rest relative to the train.
Answer:
a) Here, m – 0.1 kg, a = +g = 9.8 m/s²
Net force, F = ma = 0.1 × 9.8 = 0.98 N This force acts vertically downwards.

b) When the train is running at a constant velocity its acc. = 0. No force acts on the stone due to this motion.
Therefore, force on the stone F = weight of stone mg = 0.1 × 9.8 = 0.98 N
This force also acts vertically downwards.

c) When the train is accelerating with 1m/s², an additional force F¹ = ma = 0.1 × 1 = 0.1 N acts on the stone in the horizontal direction. But once the stone is dropped from the train, F¹ becomes zero and the net force on the stone is F = mg = 0.1 × 9.8 = 0.98 N, acting vertically downwards.

d) As the stone is lying on the horizontal direction of motion of the train. Note the weight of the stone in this case is being balanced by the normal reaction.

Question 4.
One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is ;
i) T,
ii) T – \(\frac{\mathrm{m} v^2}{l}\),
iii) T + \(\frac{\mathrm{m} v^2}{l}\),
iv) 0
T is the tension in the string. (Choose the correct Answer)
Answer:
The net force on the particle directed towards the center is T. This provides the necessary centripetal force to the particle moving in the circle.

Question 5.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^-1. How long does the body take to stop ?
Answer:
Here, F = -50 N, m = 20 kg
μ = 15 m/s, v = 0, t = ?
From F = ma,
a = \(\frac{F}{m}=\frac{-50}{20}\) = -2.5 m/s²
From v = u + at
0 = 15 – 2.5t
t = \(\frac{15}{2.5}\) = 6s.

Question 6.
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 ms^-1 to 3.5 ms^-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force ?
Answer:
Here m = 3.0 kg
μ = 2.0 m/s
v = 3.5 m/s,
t = 25s, F = ?
F = ma = \(\frac{m(v-u)}{t}=\frac{3.0(3.5-2.0)}{25}\)
= 0.18 N.
The force is along the direction of motion.

Question 7.
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body ?
Answer:

θ = 36° 52
This the direction of resultant force and hence the direction of acceleration of the body, fig.
Also a = \(\frac{F}{m}=\frac{10}{5}\) = 2ms^-2

Question 8.
The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle ? The mass of the three wheeler is 400 kg and the mass of the driver is 65 kg.
Answer:
Here, u = 36km/h = 10 m/s, v = 0, t = s
m = 400 + 65 = 465 kg
Retarding force
F = ma = \(\frac{m(v-u)}{t}=\frac{465(0-10)}{4}\) = -1162 N.

Question 9.
A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms^-2. Calculate the initial thrust (force) of the blast.
Answer:
Here m = 20,000 kg = 2 × 10⁴ kg
Initial ace = 5 m/s²
Thrust, F = ?
Clearly, the thrust should be such that it overcomes the force of gravity besides giving it an upward acceleration of 5 m/s². Thus the force should produce a net acceleration of 9.8 + 5.0 = 14.8 m/s² As thrust = force = mass × acceleration
∴ F = 2 × 10⁴ × 14.8 = 2.96 × 10⁵N

Question 10.
A body of mass 0.40 kg moving initially with a constant speed of 10 ms-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at the time to be x = 0 and predict its position at t = -5 s, 25 s, 100 s.
Answer:
Here, m = 0.40 kg, µ = 10m/s due N
F = – 8.0 N
a = \(\frac{F}{m}=\frac{-8.0}{0.40}\) = -20 m/s²
for 0 ≤ t ≤ 30s.
i) At t = -5s, x = Ut = 10 × (-5) = -50 m

ii) At t = 25s, x = Ut + \(\frac{1}{2}\) at²
= 10 × 25 + \(\frac{1}{2}\) (-20) (25)² = – 6000m

iii) At t = 100s, The problem is divided into two parts, upto 30s, there is force/acc.
∴ from x₁ = Ut + \(\frac{1}{2}\) at²
= 10 × 30 + \(\frac{1}{2}\) (-20) (30)²
= -8700
At t = 30s, v = U + at = 10 – 20 × 30 = – 590 m/s,
∴ for motion from 30s to 100s
x₂ = vt = – 590 × 70 = – 41300 m
x = x₁ + x₂ = -8700 – 41300
= -50,000 m = – 50km.

Question 11.
A truck starts from rest and accelerates uniformly at 2.0 ms^-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity and (b) acceleration of the stone at t = 11 s ? (Neglect air resistance).
Answer:
Here, u = 0, a = 2 m/s², t = 10s

Let v be the velocity of the truck when the stone is dropped.
From v = u + at
v = 0 + 2 × 10 = 20m/s
a) Horizontal velocity of stone, when it is dropped, v_x = v = 20 m/s.
As air Resistance is neglected, v_x remains constant.
In the vertical direction, initial velocity of stone, µ = 0, a = g = 9.8 m/s²,
time t = 11 – 10 = 1s
From v = u + at
v_y = 0 + 9.8 × 1 = 9.8 ms^-1
Resultant velocity of stone, OC is given by
v = \(\sqrt{v_x^2+v_y^2}=\sqrt{20^2+(9.8)^2}\)
v = 22.3 m/s.
Let θ is the angle with the resultant velocity OC of stone makes with the horizontal direction OA, then from fig.
tan θ = \(\frac{v_y}{v_x}=\frac{9.8}{20}\) = 0.49
∴ θ = 29°

b) The moment the stone is dropped from the car, horizontal force on the stone = 0. The only acceleration the path followed by the stone is, however parabolic.

Question 12.
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 ms^-1. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.
Answer:
a) We shall study in unit x that at each extreme position, velocity of the bob is zero. If the string is cut at the extreme position, it is only under the action of ‘g’. Hence the bob will fall vertically downwards.

b) At the mean position, velocity of the bob is 1m/s. along the tangent to the arc, which is in the horizontal direction. If the string is let at mean position, the bob will be have as a horizontal projectile. Hence it will follow a parabolic path.

Question 13.
A man of mass 70 kg stands on a weighing scale in a lift which is moving
a) upwards with a uniform speed of 10 ms^-1
b) downwards with a uniform acceleration of 5 ms^-2
c) upwards with a uniform acceleration of 5 ms^-2
What would be the readings on the scale in each case ?
d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity ?
Answer:
Here, m = 70 kg, g = 9.8 m/s²
The weighing machine in each case measures the reaction R i.e. the apparent weight.
a) When lift moves upwards with a uniform speed, its accelerations is zero.
R = mg = 70 × 9.8 = 686N

b) When the lift moves downwards with a = 5 m/s²
R = m(g – a) = 70 (9.8 – 5) = 336 N

c) When the lift moves upwards with a = 5 m/s²
R = m(g + a) = 70 (9.8 + 5) = 1036 N
If the lift were to come down freely under gravity, downward acc. a = g
∴ R = m (g – a) = m(g – g) = zero

Question 14.
Figure shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t < 4 s, 0 < t < 4s ? (b) impulse at t = 0 and t = 4 s ?
(Consider one-dimensional motion only)

Answer:
i) For t < 0, the position time graph is OA which means displacement of the particle is zero.
i.e. particle is at rest at the origin. Hence force on the particle must be zero.

ii) For 0 < t < 4s, s, the position time graph OB has a constant slope. Therefore, velocity of the particle is constant in this interval i.e. particle has zero acceleration. Hence force on the particle must be zero. iii) For t > 4s, the position time graph BC is parallel to time axis. Therefore, the particle remains at a distance 3m from the origin, i.e. it is at rest. Hence force on the particle is zero.

iv) Impulse at t = 0 :
We know. Impulse = change in linear momentum, Before t = 0 particle is at rest i.e. u = 0. After t = 0, particle has a constant velocity v = \(\frac{3}{4}\)
= 0.75 m/s

∴ Impulse = m(v – u)
= u (0.75-0)
= 3kg m/s
∴ Impulse at t = 4s
Before t = 4s, particle has a constant velocity u = 0.75 m/s
After t = 4s, particle is at rest i.e. v = 0
Impulse = m(v – u) = 4 (0 – 0.75) = 3kg m/s.

Question 15.
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string, a horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of String. What is the tension in the string in each case ?
Answer:
Here, F = 500 N
m₁ = 10kg, m₂ = 20kg
Let T be the tension in the string and a be the acceleration of the system, in the direction of the force applied.
a = \(\frac{F}{m_1+m_2}=\frac{500}{10+20}=\frac{50}{3}\) m/s²
a) When force is applied on heavier block,
T = m₁ a = 10 × \(\frac{50}{3}\) N
T = 166.66 N

b) When force is applied on lighter block,
T = m₂a = 20 × \(\frac{50}{3}\) N
= 333.33 N
Which is different from value T in case (a) Hence our answer depends on which mass end, the force is applied.

Question 16.
Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
Answer:
Here, m₂ = 8 kg, m₁ = 12kg
as a = \(\frac{\left(m_1-m_2\right) g}{m_1+m_2}\)
a = \(\frac{(12-8) 9.8}{12+8}=\frac{39.2}{20}\)
= 1.96 m/s²
Again
T = \(\frac{2 m_1 m_2 g}{m_1+m_2}=\frac{2 \times 12 \times 8 \times 9.8}{(12+8)}\) = 94.1 N

Question 17.
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
Answer:
Let m₁m₂ be the masses of products and \(\vec{v}_1, \vec{v}_2\) be their respective velocities. Therefore total linear momentum after disintegration = \(m_1 \overrightarrow{v_1}+m_2 \overrightarrow{v_2}\). Before disintegration, the nucleus is at rest. There, its linear momentum before dis-integration is zero. According to the principle of conservation of linear momentum.
\(m_1 \overrightarrow{v_1}+m_2 \overrightarrow{v_2}\) (Or) \(\vec{v}_2=\frac{-m_1 \vec{v}_1}{m_2}\)
Negative sign shows that \(\overrightarrow{v_1}\) and \(\overrightarrow{v_2}\) are in opposite direction.

Question 18.
Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 ms^-1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other ?
Answer:
Here, initial momentum of the ball,
A = 0.05 (6) = 0.3 kg m/s
As the speed is reversed on collision, final momentum of the ball A = 0.05 (-6)
= – 0.3 kg ms^-1
Impulse imparted to ball A = Change in momentum of ball A = Final momentum – Intial momentum = – 0.3 – 0.3 = – 0.6 kg m/s

Question 19.
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms^-1, what is the recoil speed of the gun ?
Answer:
Here, mass of shell m = 0.02 kg
Mass of gun M = 100 kg
Muzzle speed of shell v = 80 m/s
Recoil speed of gun V = ?
According to the principle of conservation of linear momentum mv + MV = 0
(Or) V = \(\frac{-\mathrm{mv}}{\mathrm{M}}=\frac{-0.02 \times 80}{100}\) = 0.016 m/s

Question 20.
A batsman deflects a ball by the angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball ? (Mass of the balls is 0.15 kg).
Answer:
In fig. the ball hits the bat KL along AO and is deflected by the bat along OB. Where LAOB = 45°. ON is normal to the portion of the bat KL deflecting the ball.

∴ θ = LNOA
= 45°/2 = 22.5°
Intial vel along AO = u = 54 km/h
= 15 m/S₁ and mass of ball m = 0.15 kg Intial velocity along AO has the two rectangular components : u cos θ along NO produced and u sin 6 along the horizontal OL.
Final velocity along OB has the same magnitude = u
It is resolved into two rectangular components u cos 6 along ON and u sin θ along OL. We observe that there is no change in velocity along the horizontal, but velocity along vertical is just reserved.
∴ Impulse imparted to the ball
= Change in linear momentum of the ball
= m u cos θ – (- m u cos θ)
= 2 m u cos θ
= 2 × 0.15 × 15 cos 22.5°
= 4.5 × 0.9239
= 4.16 kg m/s

Question 21.
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./ min in a horizontal plane. What is the tension in the string ? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Answer:
Here, m = 0.25 kg, r = 15m
n = 40 rpm = \(\frac{40}{60}\), rps = \(\frac{2}{3}\), T = ?
T = mrw² = mr(2 π)² = 4 π²m²
T = 4 × \(\left(\frac{22}{7}\right)^2\) × 0.25 × 1.5 × \(\left(\frac{2}{3}\right)^2\) = 6.6N
If T_max = 200 N. then from
T_max = \(\frac{\mathrm{mv}_{\max }^2}{r}\)
\(v_{\max }^2=\frac{T_{\max } \times r}{m}=\frac{200 \times 1.5}{0.25}\) = 1200
v_max = \(\sqrt{1200}\) = 34.6 m/s

Question 22.
If, in Exercise 5.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks :
a) the stone moves radially outwards,
b) the stone flies off tangentially from the instant the string breaks,
c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ?
Answer:
The instant the string, breaks, the stone flies off tangentialy, as per Newton’s first law of motion.

Question 23.
Explain why
a) a horse cannot pull a cart and run in empty space,
b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
c) it is easier to pull a lawn mower than to push it.
d) a cricketer moves his hands back-wards while holding a catch.
Answer:
a) While trying to pull a cart, a horses pushes the ground backwards with a certain force at an angle. The ground offers an equal reaction in the opposite direction on the feet of the horse. The forward component of this reaction is responsible for the motion of the cart. In empty space, there is no reaction and hence a horse cannot pull the cart and run.

b) This is due to inertia of motion. When the speeding bus stops suddenly, lower part of the body in contact with the seats stops. The upper part of the body of the passengers tend to maintain its uniform motion. Hence the passengers are thrown forward.

c) While pulling a lawn roller, force is applied upwards along the handle. The vertical component of this force is upwards and reduces the effective weight of the roller, fig. (a) while pushing a lawn roller. Face is applied downwards along the handle. The vertical component of this force is downwards and increases the effective weight of the roller, fig. (b). As the effective weight is lesser in case of pulling than in a case of pushing, therefore pulling is easier than pushing.

d) While holding a catch, the impulse receives by the hands F × t = Change in linear momentum of the ball is constant. By moving his hands, backwards, the cricketer increases the time of impact (t) to complete the catch. As t increases, F decreases and as a reaction, his hands are not hurt severly.

Question 24.
Figure shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body ? What is the magnitude of. each impulse ?

Answer:
Here, m = 0.04 kg. position time graph shows that the particle moves from x = 0 at 0 to x = 2 cm at A in 2 sec.
As x – t graph is a st. line, the motion is with a constant velocity.
μ = \(\frac{(2-0) \mathrm{cm}}{(2-0) \mathrm{s}}\) = 1 cm s^-1
= 10^-2 ms^-1

From x = 2 cm at A, particle goes to x = 0 at B in 2 sec.
As AB is a stline, motion is with constant velocity v = 1 cm/s = 10^-2 m/s
Negative sign indicates the reversal of direction of motion. This is being repeated. We can visualise a ball moving between two walls located at x = 0 and x = 2 cm, getting rebounded repeatedly on striking against each wall on every collision with a wall, linear momentum of the ball changes. Therefore, the ball receives impulse after every two seconds. Magnitude of impulse = Total change in linear momentum.
= mu -(mv) = mu – mv = m(u – v)
= 0.04(10^-2 + 10^-2)
= 0.08 × 10^-2
= 8 × 10^-4 kg m/s

Question 25.
Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 ms-2. What is the net force on the man ? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt ? (Mass of the man = 65 kg).

Answer:
Here, acceleration of conveyer belt, a = 1 m/s²
As the man is standing stationary w.r.t the belt, acceleration of the man = Acceleration of belt = a = 1 m/s²
As m = 65 kg
∴ Net force on the man, F = ma = 65 × 1 = 65 N
Now, µ = 0.2
Force of limiting friction- F = µR = µmg
It the man remains stationary upto max.acc. a of the belt, then F = ma¹ = g mg
a¹ = mg = 0.2 × 9.8 = 1.96 ms^-2

Question 26.
A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are : (Choose the correct alternative)
Lowest Point – Highest Point
a) mg – T₁ – mg + T₂
b) mg + T₁ – mg – T₂
c) mg + T₁ – (mυ₁²)/R – mg – T₂ + (mυ₁²)/R
d) mg – T₁ – (mυ₁²/R – mg + T₂ + (mυ₁²)/R
T₁ and υ₁ denote the tension and speed at the lowest point T₂ and υ₂ denote corresponding values at the highest point.
Answer:
The net force at the lowest point α is
F_L = (mg – T₁) and the net force at the highest point H is F_H = mg + T₂. Therefore, alternative (a) is correct.

Question 27.
A helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms^-2. The crew and the passengers weight 300 kg. Give the magnitude and direction of the
a) force on the floor by the crew and passengers,
b) action of the rotor of the helicopter on the surrounding air,
c) force on the helicopter due to the surrounding air.
Answer:
Here, mass of the helicopter, m₁ = 100 kg
Mass of the crew and passengers m₂ = 300 kg
Upward acceleration a = 15 ms^-2 and g = 10 ms^-2
a) Force on the floor of helicopter by the crew and passengers = appeard weight of crew and passengers = m₂(g + a)
= 300(10 + 15) = 7500 N

b) Action of rotor of helicopter on surrounding air is obviously vertically downwards because helicopter rises on account reaction to this force. Thus, force of action F = (m₁ + m₂) (g + a)
= (1000 + 300) (10 + 15)
= 1300 × 25 = 32500 N

c) Force on the helicopter due to surrounding air is the reaction. As action and reaction are equal and opposite, therefore, force of reaction F¹ = 32500 N, vertically upwards.

Question 28.
A stream of water flowing horizontally with a speed of 15 ms^-1 gushes out of a tube of cross-sectional area 10^-2m² and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound ?
Answer:
Here v = 15 ms^-1
Area of cross section a = 10² m^-2, F = ?
Volume of water pushing out/sec
= a × v = 10^-12 × m³ s^-1
As density of water is 10³ kg/m², therefore, mass of water striking the wall per sec.
m = (15 × 10^-2) × 10³ = 150 kg/s
Change in linear momentum
As F = \(\frac{\text { Change in linear momentum }}{\text { Time }}\)
∴ F = \(\frac{m \times v}{t}=\frac{150 \times 15}{1}\) = 2250 N

Question 29.
Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of
a) the force on the 7^th coin (counted from the bottom) due to all the coins on its top
b) the force on the 7^th coin by the eighth coin,
c) the reaction of the 6^th coin on the 7^th coin.
Answer:
a) The force on 7^th coin is due to weight of the three coins lying above it. Therefore,
F (3m) kgf = (3mg)N
Where g is acceleration due to gravity. This force acts vertically downwards.

b) The eighth coin is already under the weight of two coins above it and it has its own weight too. Hence force on 7th coin due to 8th coin is sum of the two forces i.e.,
F = 2m + m = 3(m) kgf = (3mg)N
The force acts vertically downwards.

c) The sixth coin is under the weight of four coins above it.
Reaction, R = – F = -4m(kgf) = -(4 mg)N
Minor sign indicates that the reaction acts vertically upwards, opposite to the weight.

Question 30.
An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop ?
Answer:
Here θ = 15°
v = 720 km/h =\(\frac{720 \times 1000}{60 \times 60}\) = 200 ms^-1
g = 9.8 ms^-2
From tan θ = \(\frac{\mathrm{v}^2}{\mathrm{rg}}\)
v² = rg tan θ
r = \(\frac{v^2}{g \tan \theta}=\frac{(200)^2}{9.8 \times \tan 15^{\circ}}\)
= 15232 m = 15.232 km

Question 31.
A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 10⁶ kg. What provides the centripetal force required for this purpose – The engine or the rails ? What is the angle of banking required to prevent wearing out of the rail ?
Answer:
The centripetal force is provided by the lateral thrust exerted by the rails on the wheels. By Newton’s 3^rd law, the train exerts an equal and opposite thrust on the rails causing its wear and tear.
Obviously, the outer rail will wear out faster due to the larger force exerted by the train on it.
Here v = 54 km/h =\(\frac{54 \times 1000}{60 \times 60}\) = 15 m/s
g = 9.8 ms^-2
As tan θ = \(\frac{v^2}{r g}=\frac{15 \times 15}{30 \times 9.8}\) = 0.76
∴ θ = tan^-1 0.76 = 37.4°

Question 32.
A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. What is the action on the floor by the man in the two cases ? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding ?

Answer:
Here, mass of block m = 25 kg
Mass of man M = 50 kg
Force applied to lift the block
F = mg = 25 × 9.8 = 245 N
Weight of Man W = mg = 50 × 9.8 = 490 N
a) When block is raised by man as shown in Fig. force is applied by the man in the upward direction. This increases the apparent weight of the man. Hence action on the floor.
W¹ = W + F
= 490 + 245
= 735 N

b) When block is raised by man as shown in Fig. force is app ed by the man in the downward direction. This decreases the apparent weight of the man. Hence action on the floor in this case would be
W¹ = W – F = 490 – 245 = 245 N
As the floor yeilds to a normal force of 700 N, the mode (b) has to be adopted by the man to lift the block

Question 33.
A monkey of mass 40 kg climbs on a rope (Fig.) which can stand a maximum tension of 600 N. In which of the following cases will the rope break : the money
a) climbs up with an acceleration of 6 m s^-2
b) climbs down with an acceleration of 4 ms^-2
c) climbs up with a uniform speed of 5 ms^-1
d) falls down the rope nearly freely under gravity ?
(ignore the mass of the rope)

Answer:
Here, mass of monkey m = 40 kg
Maximum tension the rope can stand T = 600 N
In each case, actual tension in the rope will be equal to apparent weight of money (R). The rope will break when R exceeds T.
a) When monkey climbs up with a = 6 ms^-2
R = m(g + a)
= 40(10 + 6)
= 640 N (Which is greater than T)
Hence the rope will break.

b) When monkey climbs down with a = 4 ms^-2,
R = m(g – a) = 40(10 – 4) = 240 N
Which is less than T.
∴ The rope will not break.

c) When monkey climbs up with a uniform speed v = 5 ms^-1.
Its acceleration, a = 0
∴ R = mg = 40 × 10 = 400 N,
which is less than T.
∴ The role will not break

d) When monkey falls down the rope nearly freely under gravity
a = g
∴ R = m(g – a) = m(g – g) = zero
Hence the rope will not break.

Question 34.
Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Fig.). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are (a) the reaction of the partition (b) the action-reaction forces between A and B ? What happens when the wall is removed ? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between μ_s and μ_k.

Answer:
Here, mass of body A, m₁ = 5 kg
Mass of body B, m₂ = 10 kg

Coefficient of friction between the bodies and the table, μ = 0.15
Horizontal force applied on A,
F = 200 N

a) Force of limiting friction acting to the left
f = μ (m₁ + m₂)g
= 0.15(5 + 10) × 9.8 = 22.05 N
∴ Net force to the right exerted on the partition
F’ = 200 – 22.05 = 177.95 N
Reaction of partition = 177.95 N to the left.

b) Force of limiting friction acting on body A
f₁ = μm₁g = 0.15 × 5 × 9.8
= 7.35 N
∴ Net force exerted by body A on body B.
F” = F -f₁ = 200 – 7.35
= 19265 N
This is to the right
Reaction of body B on body A = 192.65
N to the left when the portion is removed, the system of two bodies will move under the action of net force.
F₁ = 177.95 IM
Acceralation produced in the system
a = \(\frac{F^1}{m_1+m_2}=\frac{177.95}{5+10}\)
= 11.86 ms^-2
Force producing motion in body A
F₁ = m₁ a = 5 × 11.86
= 59.3 N
∴ Net force exerted by A on body B, when partition is removed
= F” – F₁ = 192.65 – 59.3
= 133.35 N.
Hence the reaction of body B on A, when partition is removed = 133.35 N to the left.
Thus answers to (b) do change.

Question 35.
A block of mass 15 kg is placed on a long trolley. The cofficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 ms^-2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.
Answer:
a) Flere, m = 15 kg;
μ = 0.18,
a = 0.5 ms^-2
t = 20 s

Force on the block due to motion of the trolly F’ = ma = 15 × 0.5 = 7.5 N
Force of limiting friction on the block
= F = μR = μmg
= 0.18 × 15 × 9.8 = 26.46 N
This opposes the motion of the block. The block shall not move. The force of static friction F will adjust itself equal and opposite to F’, the applied force.

Flence to a stationary observer on the ground, the block will appear to be at rest relative to the trolly. When trolly moves with uniform velocity, the block will continue to be stationary. Because in that case, forward force is zero. Force of friction alone is acting on the block.

b) An observer moving with the trolly has accelerated motion. The observer is therefore non-inertial.. The law of inertia is no longer valid.

Question 36.
The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 ms^-2. At what distance from the starting point does the box fall of the truck?

Answer:
Here mass of the box m = 40 kg
Acceleration of truck a = 2 ms^-2
Distance of box from open end S = 5m
Coeff. of friction μ = 0.15
Force on the box due to accelarated motion of the truck, F = ma = 40 × 2 = 80 N
This force F is in the forward direction Reaction F’ on the box is equal to F = 80 N in the backward direction. This is opposed by force of limiting friction
f = μ R = μ mg
= 0.15 × 40 × 9.8
= 58.8 N in the forward direction
∴ Net force on the box in the backward direction is p = F’ – F = 80 – 58.8 = 21.2 N
Backward acceleration produced in the box
a = \(\frac{p}{m}=\frac{21.2}{40}\) = 0.53 ms^-2
It t is time taken by the box to travel S = 5 metre and fall off the truck, then from
S = ut + \(\frac{1}{2}\) at²
5 = 0 × t + \(\frac{1}{2}\) × 0.53 t²
t = \(\frac{\sqrt{5 \times 2}}{0.53}\) = 4.345
If the truck travels a distance x during this time, then again from
S = ut + \(\frac{1}{2}\) at²
x = 0 × 4.34 + \(\frac{1}{2}\) × 2(4.34)² = 18.84 m

Question 37.
A disc revolves with a speed of 33\(\frac{1}{3}\) rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record ?
Answer:
The coin revolves with the record in the case when the force of friction is enough to provide the necessary centripetal force. If this force is not sufficient to provide centripetal force, the coin slips on the record.
Now the frictional force is µR where R is the
normal reaction and R = mg
Hence force of friction = µ mg and centripetal force required is \(\frac{m v^2}{r}\) or mrw²
µ₁w are same for both the coins and we have different values of r for the two coins.
So to prevent slipping i.e., causing coins to rotate µ mg > mrω² or µg > rω² …………….. (1)
For 1^st coin
r = 4 cm = \(\frac{4}{100}\) m
n = 33 \(\frac{1}{2}\) rev/min = \(\frac{100}{3 \times 60}\) rev/sec
w = 2πn = 2π × \(\frac{100}{180}\) = 3.49 S^-1
∴ rw² = \(\frac{4}{100}\) × (3.49)² = 0.49 ms^-2 and
µg = 0.15 × 9.8 = 1.47 ms^-2
As µg > rw², there fore this coin Will revolve with the record.
Note: We have nothing to do with the radius of the record = 15 cm

Question 38.
You may have seen in a circus a motor-cyclist driving in verticle loops inside a ‘death-well’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a verticle loop if the radius of the chamber is 25 m ?
Answer:
At the uppermost point of the death well, with no support from below, the motorcyclist does not drop down. This is because his weight is being balance by the centrifugal force. Infact, the weight of the motorcyclist is spent up in providing the necessary centripetal force to the motorcyclist and hence he does not drop drown.

At the uppermost point, R + mg = \(\frac{m v^2}{r}\),
where R is the normal reaction (downwards) on the motor cyclist by the ceiling of the chamber.
Speed will be minimum, when N = 0
∴ mg = \(\frac{m v^2}{r}\) or
v = \(\sqrt{\mathrm{rg}}=\sqrt{25 \times 10}\)
= 15.8 m/s

Question 39.
A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed ?
Answer:
Here m = 70 kg, r = 3
n = 200 rpm = \(\frac{200}{60}\) rps
µ = 0.15
w = ?
The horizontal force N by the wall on the man provides the necessary centripetal force = mrω². The frictional force (f) in this case is vertically upwards opposing the weight (mg) of the man.
After the floor is removed, the man will remain stuck to the wall, when
mg = f < µ N i.e., mg < µ mr ω² or g < µ r ω²
∴ Minimum angular speed of rotation of the cylinder is ω = \(\sqrt{\frac{g}{\mu \mathrm{r}}}=\sqrt{\frac{10}{0.15 \times 3}}\)
= 4.7 rad/s

Question 40.
A thin circular loop of radius R rotates about its vertical diameter with an angular frequency no. Show that a small bead on the wire loop remains at its
lowermost point for ω ≤ \(\sqrt{g / R}\) . What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω = 2g/R ?
Answer:
In fig. we have show that radius vector joining the bead to the centre of the wire makes as angle θ with the vertical downward direction. It N is normal reaction, then as it is clear from the figure,

mg = N cos θ ………… (1)
m rω² = N sin θ …………….. (2)
Or m(R sin θ) ω² = N sin θ
Or mRω² = N
From (i) mg = mRω² cos θ
Or cos θ = \(\frac{\mathrm{g}}{\mathrm{R} \omega^2}\) ……………….. (3)
As |cos θ| ≤ 1, therefore, bead will remain at its lower most point for
\(\frac{\mathrm{g}}{\mathrm{R} \omega^2}\) ≤ 1 or w ≤ \(\sqrt{\frac{g}{R}}\)
When ω = \(\sqrt{\frac{2 g}{R}}\), from (iii),
cos θ = \(\frac{g}{R}\left(\frac{R}{2 g}\right)=\frac{1}{2}\)
∴ θ = 60°

Textual Examples

Question 1.
An astronaut accidentally gets separated out of his small spaceship accelerating in inter stellar space at a constant rate of 100 ms^-2. What is the acceleration of the astronaut the instant after he is outside the spaceship ? (Assume that there are no nearby stars to exert gravitational force on him).
Answer:
Since there are no nearby starts to exert gravitational force on him and the small spaceship exerts negligible gravitational attraction on him, the net force acting on the astronaut, once he is out of the spaceship, is zero. By the first law of motion the acceleration of the astronaut is zero.

Question 2.
A bullet of mass 0.04 kg moving with a speed of 90 m s^-1 enters a heavy wooden block and is stopped after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet ?
Answer:
The retardation ‘a’ of the bullet (assumed constant) is given by
a = \(\frac{-u^2}{2 s}=\frac{-90 \times 90}{2 \times 0.6}\) m s^-2 = -6750 m s^-2
The retarding force, by the second law of motion, is = 0.04 kg × 6750 ms^-2 = 270 N
The actual resistive force, and therefore, retardation of the bullet may not be uniform. The answer therefore, only indicates the average resistive force.

Question 3.
The motion of a particle of mass m is described by y = ut + \(\frac{1}{2}\)at². Find the force acting on the particle.
Answer:
We know, y = ut + \(\frac{1}{2}\) gt²
Now, υ = \(\frac{\mathrm{dy}}{\mathrm{dt}}\)u + gt
acceleration, a = \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = g
Then the force is given by F = \(\frac{\mathrm{dp}}{\mathrm{dt}}\) = ma
F = ma = mg

Question 4.
A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m s^-1. If the mass of the ball is 0.15 kg determine the impulse imparted to the ball. (Assume linerar motion of the ball) [A.P. Mar. 17]
Answer:
Change in momentum
= 0.15 × 12-(-0.15 × 12) = 3.6 N s
Impulse = 3.6 N s, in the direction from the batsman to the bowler.
This is an example where the force on the ball by the batsman and the time of contact of the ball and the bat are difficult to know, but the impulse is readily calculated.

Question 5.
Two identical billiard balls strike a rigid wall with the same speed but at different angles and get reflected without any change in speed, as shown in Fig. What is (i) the direction of the force on the wall due to each ball ? (ii) the ratio of the magnitudes of impulses imparted to the balls by the wall ?

Answer:
An instinctive answer to (i) might be that the force on the wall in case (a) is normal to the wall, while that in case (b) is inclined at 30° to the normal. This answer is wrong. The force on the wall is normal to the wall in both cases. How to find the force on the wall ? The trick . is to consider the force on the wall ? The trick is to consider the force (or impulse) on the ball due to the wall using the second law, and then use the third law to answer (i). Let u be the speed of each ball before and after collision with the wall and m the mass of each bah. Choose the x and y axes as shown in the figure, and consider the change in momentum of the ball in each case :
Case (a) :
(P_x)_initial = mu
(P_x)_initial = 0
(P_x)_final = – mu
(P_y)_final = 0
Impulse is the change in momentum vector.
Therefore,
x-component of impulse = -2mu
y-component of impulse – 0
Impulse and force are in the same direction. Clearly, from above, the force on the ball due to the wall is normal to the wall, along the negative x-direction. Using Newton’s third law of motion, the force on the wall due to the ball is normal to the wall along the positive x-direction. The magnitude of force cannot be ascertained since the small time taken for the collision has not been specified in the problem. .
Case (b) :
(P_x)_initial = mu cos 30°
(P_y)_initial = – m u sin 30°
(P_x)_final = – mu cos 30°
(P_y)_final = -m u sin 30°
Note, while p_x changes sign after collision, p_y does not. Therefore,
x-component of impulse = -2 m u cos 30°
y-component of impulse 0
The direction of impulse (and force) is the same as in (a) and is normal to the wall along the negative x direction. As before, using Newton’s third law, the force on the wall due to the ball is normal to the wall along the positive x direction.
The ratio of the magnitudes of the impulses imparted to the balls in(a) and (b) is
2 mu / (2 m u cos 30°) = \(\frac{2}{\sqrt{3}}\) ≈ 1.2

Question 6.
See Fig. A mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the horizontal direction is applied at the mid point P of the rope, as show. What is the angle the rope makes with the vertical in equilibrium ? (Take g = 10 ms^-2). Neglect the mass of the rope.

Answer:
Figures (b) and (c) are known as free-body diagrams. Figure (b) is the free-body diagram of W and Fig. (c) is the free-body diagram of point P.
Consider the equilibrium of the weight W.
Clearly, T₂ = 6 × 10 = 60 N.
Consider the equilibrium of the point P under the action of three forces – the tensions T₁ and T₂, and the horizontal force 50 N. The horizontal and vertical components of the resultant force must vanish separately :
T₁ cos θ = T₂ = 60N
T₂ sin θ = T₂ = 50N
which gives that tan θ = \(\frac{5}{6}\) or
θ = tan^-1 (\(\frac{5}{6}\)) = 40°
Note the answer does not depend on the length of the rope (assumed massless) nor on the point at which the horizontal force is applied.

Question 7.
Determine the maximum acceleration of the train in which a box lying on the floor will remain stationary, given that the coefficient of static friction between the box and the train’s floor is 0.15.
Answer:
Since the acceleration of the box is due to the static friction,
ma = f_s ≤ μ_s N = μ_s m g
i.e. a ≤ μ_s g
∴ a_maximum = μ_s g = 0.15 × 10m s^-2
= 1.5 ms^-2

Question 8.
See Fig. A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined unti at an angle θ = 15° with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface ?

Answer:
The forces acting on a block of mass …. at rest on an inclined plane are (i) the weight mg acting vertically downwards (ii) the normal force N of the plane on the block, and (iii) the static frictional force fs opposing the impending motion. In equilibrium, the resultant of these forces must be zero. Resolving the weight mg along the two directions shown, we have
m g sin θ = f_s, m g cos θ = N
As θ increases, the self-adjusting frictional force f_s increases until at θ = θ_max, f_s achieves its maximum value, (f_s)max = μ_s N.
Therefore,
tan θ_max = μ_s, or θ_max = tan^-1 μ_s
When θ becomes just a little more than there is a small net force on the block and θ_nm begins to slide. Note that θ_max depends only on μ_s and is independent of the mass of the block
θ_max = 15°, μ_s = tan 15° = t = 0.27

Question 9.
What is the acceleration of the book and trolley system shown in the Fig., if the coefficient of kinetic friction between the trolley and the suriace is 0.04 what is the tension in the string. Take g 10 ms^-2). Neglect the mass of the string.

Answer:
As the string is inextensible, and the pully is smooth, the 3 kg block and the 20 kg trolley both have same magnitude of acceleration. Applying second law to motion of the block (Fig. (b).
30 – T = 3a
Apply the second law to motion of the trolley (Fig. (c))
T – f_k = 20a
Now f_k = μ_k N,
Here μ_k = 0.04
N = 20 × 10 = 200N
Thus the equation for the motion of the trolley is
T – 0.04 × 200 = 20 a or T – 8 = 20a
These equations give a = \(\frac{22}{23}\) m s^-2 = 0.96 m s^-2 and T = 27.1 N.

Question 10.
A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The co-efficient of static friction between the tyres and the road is 0.1. Will the cyclist slip while taking the turn ?
Answer:
On an unbanked road, frictional force alone can provide the centripetal force needed to keep the cyclist moving on a circular turn without slipping. If the speed is too large, or if the turn is too sharp (i.e. of too small a radius) or both, the frictional force is not sufficient to provide the necessary centripetal force, and the cyclist slips. The condition for the cyclist not to slip is given by
v_max = \(\sqrt{\mu_s R_g}\) v₂ ≤ μ_s Rg
Now, R = 3m, g = 9.8 m s^-2, μ_s = 0.1.
That is, μ_s R g = 2.94 m² s^-2. v = 18 km/h = 5 m s^-1; i.e., = 25 v². The condition is not obeyed. The cyclist will slip while taking the circular turn.

Question 11.
A circular racetrack of radius 300 m is banked at an angle of 15°. If the coefficient of friction between the wheels of a race-car and the road is 0.2, what is the(a) optimum speed of the race-car to avoid wear and tear on its tyres, and (b) maximum permissible speed to avoid slipping?
Answer:
On a banked road, the horizontal component of the normal force and the frictional force contribute to provide centripetal force to keep the car moving on a circular turn without slipping. At the optimum speed, the normal reaction’s component is enough to provide the needed centripetal force, and the frictional force is not needed. The optimum speed v₀ is given by
v₀ = (R g tan θ)^1/2–
Here R = 300 m, θ = 15°, g = 9.8 m s^-2; we have
v₀ = 28.1 ms^-1
The maximum permissible speed v_max is given by
v_max = R g\(\left(\frac{\mu_s+\tan \theta}{1-\mu_s \tan \theta}\right)^{1 / 2}\) = 38.1 m s^-1

Question 12.
See (Fig.) A wooden block of mass 2 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg is placed on top of the block, the floor yields steadily and the block and the cylinder together go down with an acceleration of 0.1 ms^-2. What is the action of the block on the floor (a) before and (b) after the floor yields ? Take g = 10 m s^-2. Identify the action- reaction pairs in the problem.

Answer:
(a) The block is at rest on the floor. Its free- body diagram shows two forces on the block, the force of gravitational attraction by the earth equal to 2 × 10 = 20 N; and the normal force R of the floor on the block. By the First Law, the net force on the block must be zero i.e., R = 20 N. Using third law the action of the block (i.e. the force exerted on the floor by the block) is equal to 20 N and directed vertically downwards.

(b)The system (block cylinder) accelerates downwards with 0.1 ms^-2. The free – body diagram of the system shows two forces on the system : the force of gravity due to the earth (270 N); and the normal force R’ by the floor. Note, the free-body diagram of the system does not show the internal forces between the block and the cylinder. Applying the second law to the system,
270 – R’ = 27 × 0.1 N
i.e, R’ = 267.3 N
By the third law, the action of the system on the floor is equal to 267.3 N vertically downward.
Action-reaction pairs
For (a) :
(i) the force of gravity (20N) on the block by the earth (say, action); the force of gravity on the earth by the block (reaction) equal to 20 N directed upwards (not shown in the figure).
(ii) the force on the floor by the block (action); the force on the block by the floor (reaction).

For (b):
(i) the force of gravity (270 N) on the system by the earth (say, action), the force of gravity on the earth by the system (reaction), equal to 270 N, directed upwards (not shown in the figure).

(ii) the force on the floor by the system (action); the force on the system by the floor (reaction). In addition, for (b), the force on the block by the cylinder and the force on the cylinder by the block also constitute an action-reaction pair.

The important thing to remember is that an action-reaction pair consists of mutual fortes which are always equal and opposite between two bodies. Two forces on the same body which happen to be. equal and opposite can never constitute an action-reaction pair. The force of gravity on the mass in (a) or (b) and the normal force on the mass by the floor are not action-reaction pairs. These forces happen to be equal and opposite for (a) since the mass is at rest. They are not so for case (b), as seen already. The weight of the system is 270 N, while the normal force R’ is 267.3 N.

Andhra Pradesh BIEAP AP Inter 1st Year Accountancy Study Material 4th Lesson Journal Textbook Questions and Answers.

AP Inter 1st Year Accountancy Study Material 4th Lesson Journal

Very Short Answer Questions

Question 1.
What is Journal ?
Answer:
The word ‘Journal, is derived from the Latin word Journ’ which means a day. Therefore journal means a day book where in day-to-day business transactions are recorded in chronological order. This is also known as ‘Book of original entry’ or ‘Book of primary entry’.

Question 2.
What is Journalizing ?
Answer:
The process of recording the transaction in the Journal is called “Journalizing”.

Question 3.
What is Journal Entry ?
Answer:
The entry made in the Journal is called “Journal entry”.

Question 4.
What is Narration ?
Answer:
Narration means a brief explanation of the transaction for which the entry is recorded is written within the brackets known as ’narration’.

Problems

Question 1.
Mr. Anil started business with cash Rs. 75,000 on 1^st January 2014. The details of business transactions for the month of January are as follows.

Answer:

Question 2.
Pass journal entries in the books of Mr. Varan.

Answer:

Question 3.
Journalize the following transactions.

Answer:

Question 4.
Journalize the following transactions in the books of Bhagat.

Answer:

Question 5.
Journalize the following transactions.

Answer:

Question 6.
Journalize the following transactions in the books of Atma Ram.

Answer:
Journal Entries in the books of Atma Ram

Question 7.
Journalize the following transactions.
March 01 Commenced business with cash Rs. 13,000 and stock 7,000
March 02 Bought plant Rs. 5,000
March 03 Paid for postage Rs. 500
March 04 Withdrew cash for office use Rs. 1000
March 05 Paid for Sundry Expenses Rs. 500
March 06 Paid into bank Rs. 10,000
March 07 Paid salaries Rs 5,000
Answer:

Question 8.
Write journal entries in the books of Sudha & Co.

Answer:
Journal Entries in the book of Sudha & Co

Student Activity

Visit any organisation and find journal entries from its journal.

Andhra Pradesh BIEAP AP Inter 1st Year Accountancy Study Material 3rd Lesson Double Entry Book Keeping System Textbook Questions and Answers.

AP Inter 1st Year Accountancy Study Material 3rd Lesson Double Entry Book Keeping System

Short Answer Questions

Question 1.
Define double entry system Of accounting and explain its features.
Answer:
According to J.R. Batliboi “Every business transaction has two-fold effect and that it affects two accounts in opposite directions and if a complete record were to be made of each such transaction, it would be necessary to debit one account and credit another account. This recording of the two-fold effect of every transaction has given rise to the term Double Entry System”.

Features:

1. Every business transaction affects two accounts.
2. Each transaction has two aspects i.e. debit and credit.
3. It is based upon accounting assumptions, concepts and principles.
4. It helps in preparing balance which is a test arithemetical accuracy in accounting.
5. Finally it helps in preparation of final accounts with the help of trial balance.

Question 2.
Explain different types of accounts along with their debit, credit rules. (Mar. 2019; May ’17 – A.P. & T.S.) (Mar. 2018, ’17 – A.P.)
Answer:
All the transactions are divided into two types of accounts. Personal accounts and Impersonal accounts. The impersonal accounts are further subdivided into Real accounts and Nominal accounts. There are three types of accounts in all. They are Personal accounts, Real accounts and Nominal accounts.

1) Personal accounts : The accounts which relate to individuals or persons are known as “Personal accounts’. Personal accounts are of two types. They are Natural persons and Artificial persons.
Ex: Ramesh a/c, Sita a/c, Andhra bank a/c, capital a/c, drawing a/c, outstanding salaries a/c.
Rule :
Debit the receiver
Credit the giver

2) Real accounts : Accounts relating to properties and assets which are owned by the business concern are, real accounts which include tangible and intangible assets.
Ex: Cash a/c, bills receivable a/c, goods a/c, furniture a/c, goodwill a/c.
Rule : Debit what comes in
Credit what goes out

3) Nominal accounts : These accounts do not have any existence, form or shape. They relate to incomes and expenses or gains and loss of a business concern.
Ex : Salaries a/c, commission a/c, rent a/c, discount a/c, bad debts a/c.
Rule : Debit all expenses and losses
Credit all incomes and gains

Question 3.
Explain the advantages of double entry system.
Answer:
Advantages : The following are the main advantages of double entry system.

1. Complete record of transactions : Double entry system maintains a complete record of all business transactions, because it records both the aspects of financial transaction.
2. Scientific system : This is the only scientific system of recording business transactions. It helps to attain the objectives of accounting.
3. Ascertainment of profit or loss : It helps in ascertainment of profit or loss for a particular period by preparing the profit and account.
4. Ascertainment of the financial position : The financial position of the concern can be as¬certained at the end of each period by preparing balance sheet.
5. Full details for control : This system permits accounts to be kept in a very detailed form, and thereby provides sufficient information for the purpose of control.
6. Comparative study: The results of one year may be compared with those of previous years and the reasons for change may be ascertained.
7. Helps in decision making: This system provides sufficient information to the management for making decision.

Very Short Answer Questions

Question 1.
Double entry book keeping system (May. ’17 – A.P. ; Mar. ’15 – T.S.)
Answer:
According to J R. Batliboi Every business transaction has a two-fold effect that it affects two accounts in opposite directions and if a complete record were to be made of each such transaction, it would be necessary to debit one account and credit another account. This recording of the two-fold effect of every -transaction has given rise to the term Double Entry System”.

Question 2.
What is an account ? (Mar. 2019 – T.S.)
Answer:
Every transaction has two aspects and each aspect has an account. It is stated that an account is ‘a summary or relevant transactions at one place relating to a particular head’.

The common form of an account has three parts.

1. A Title’ that describes the name of the asset, liability or equity account.
2. A ‘left side’ or the “debit” side
3. A right side’ or the “credit” side

Format of Account

Question 3.
Explain accounting equation.
Answer:
Accounting equation is based on dual aspect concept (Debit and Credit). The accounting equa¬tion shows the relationship between the economic resources of a business and the claims against those resources.
Economic Resources = claims
Another term for economic resources is assets.
The claims consist of liabilities and owners claims or equity.
Assets = Equities or Capital + Liabilities

Question 4.
Impersonal accounts
Answer:
Impersonal accounts are two types i.e. Real accounts and Nominal accounts.

1. Real accounts: Accounts relating to properties and assets which are owned by the business concern are, Real accounts, which include tangible and intangible accounts.
   Ex: Cash a/c, bill receivable a/c, goodwill a/c, goods a/c, furniture a/c.
2. Nominal accounts: These accounts do not have any existence, form or shape. They relate to incomes and expenses or gains, and losses of a business concern.
   Ex: Salaries a/c, bad debts, rent a/c, commission a/c, discount a/c etc.

Student Activity

Visit a nearest business organization and make a list of various personal, real and nominal accounts from the books of accounts.

Andhra Pradesh BIEAP AP Inter 1st Year Accountancy Study Material 2nd Lesson Accounting Principles Textbook Questions and Answers.

AP Inter 1st Year Accountancy Study Material 2nd Lesson Accounting Principles

Essay Type Questions

Question 1.
What are accounting concepts ? Explain any four accounting concepts in detail.
Answer:
The term concept means an idea or thought. Basic accounting concepts are the fundamental ideas underlying in the theory and practice of financial accounting. The Ifnpdftant accounting concepts are –

1. Business Entity Concept: Business is treated as separate form the proprietor. All the transactions are recorded in the books of business and not in the books of the proprietor. The accounting system gives information only about the business and not its owner. The proprietor is also treated as a creditor for the business.
2. Dual Aspect Concept: Dual aspect concept principle is the basis for Double Entry System of book-keeping. All business transactions recorded in accounts have two aspects : receiving benefit and giving benefit.
   For example, when a business acquires as asset: receiving of benefit. It must pay cash : giving of benefit.
3. Going Concern Concept: According to this concept it is assumed that the business will continue for long time. All transactions are recorded from this point of view. The investors lend money and the creditors supply goods with the expectation that the enterprise would continue for long period of time. Hence financial statements are prepared on a going concern basis and not on liquidation basis.
4. Money Measurement Concept: This concept suggests that accountancy should record only the transactions which can be measured in terms of money. The transactions which cannot be measured in terms of money fall beyond the scope of accountancy.
   For example the events of machinery breakdown is not recorded as it does not have monetary value. However, the expenditure incurred for the repair of the machinery can be measured in monetary value.

Question 2.
What are accounting conventions ? Explain them briefly.
Answer:
Conventions mean customs or traditions. These conventions provide useful guidance in preparing accounts or financial statements. The important accounting conventions are –

1. Convention of Disclosure : Accounting statements should disclose fully and completely and significant information, based on which, decisions can be taken by various interested parties. It involves proper classification and explanations of accounting information which are published in the financial statements.
2. Convention of Materiality : According to this convention only those events should be recorded which have a significant bearing and insignificant things should be ignored. The avoidance of insignificant things will not materially off the records of the business.
3. Convention of Consistency: The accounting practices should be remained same from one year to another year. This is useful to the entrepreneur to compare the financial statements of one year with that of the other years. For example, an organization should not change its method of depreciation every year.
4. Convention of Conservation : According to this convention the accountant has to record the actual financial position. We should be careful in calculating profits. Profits should never be anticipated or exaggerated. But losses should be anticipated and provided for. Example : While taking the value of the closing stock the market cost or actual cost which ever is less is to be taken into the books of accounts.

Short Answer Questions

Question 1.
Explain business entity concept of accounting. (Mar. 2018 – A.P. & T.S.; Mar. ’17 – A.P.; May 17 – T.S.)
Answer:
Business is treated as separate from the proprietor. All the transactions are recorded in the books of business and not in the books of the proprietor. The accounting system gives information only about the business and not its owner. The proprietor is also treated as a creditor for the business.

Question 2.
Explain money measurement concept. (Mar. 2019 – T.S.)
Answer:
This concept suggests that accountancy should record only the transactions which can be measured in terms of money. The transactions which cannot be measured in terms of money fall beyond the scope of accountancy.
For example the event of machinery breakdown is not recorded as it does not have monetary value. However, the expenditure incurred for the repair of the machinery can be measured in monetary value.

Question 3.
Explain convention of conservatism. (Mar. ’17 – T.S.)
Answer:
According to this convention the accountant has to record the actual financial position. We should be careful in calculating profits. Profits should never be anticipated or exaggerated. But losses should be anticipated and provided for.
Example : While taking the value of the closing stock the market cost or actual cost whichever is less is to be taken into the books of accounts.

1. Convention of Disclosure : Accounting statements should disclose fully and completely and significant information, based on which, decisions can be taken by various interested parties. It involves proper classification and explanations of accounting information which are published in the financial statements.
2. Convention of Materiality : According to this convention only those events should be recorded which have a significant bearing and insignificant things should be ignored. The avoidance of insignificant things will not materially off the records of the business.
3. Convention of Consistency: The accounting practices should be remained same from one year to another year. This is useful to the entrepreneur to compare the financial statements of one year with that of the other years. For example, an organization should not change its method of depreciation every year.
4. Convention of Conservation : According to this convention the accountant has to record the actual financial position. We should be careful in calculating profits. Profits should never be anticipated or exaggerated. But losses should be anticipated and provided for.
   Example : While taking the value of the closing stock the market cost or actual cost which ever is less is to be taken into the books of accounts.

Short Answer Questions

Question 1.
Explain business entity concept of accounting. (Mar. 2018 – A.P. & T.S.; Mar. ’17 – A.P.; May 17 – T.S.)
Answer:
Business is treated as separate from the proprietor. All the transactions are recorded in the books of business and not in the books of the proprietor. The accounting system gives information only about the business and not its owner. The proprietor is also treated as a creditor for the business.

Question 2.
Explain money measurement concept. (Mar. 2019 – T.S.)
Answer:
This concept suggests that accountancy should record only the transactions which can be measured in terms of money. The transactions which cannot be measured in terms of money fall beyond the scope of accountancy.
For example the event of machinery breakdown is not recorded as it does not have monetary value. However, the expenditure incurred for the repair of the machinery can be measured in monetary value.

Question 3.
Explain convention of conservatism. (Mar. ’17 – T.S.)
Answer:
According to this convention the accountant has to record the actual financial position. We should be careful in calculating profits. Profits should never be anticipated or exaggerated. But losses should be anticipated and provided for.
Example : While taking the value of the closing stock the market cost or actual cost whichever is less is to be taken into the books of accounts.

Question 4.
Explain convention of consistency.
Answer:
The accounting practices should be remained same from one year to another year. This is useful to the entrepreneur to compare the financial statements of one year with that of other years. For example, an organization should not change its method of depreciation every year.

Question 5.
Write a brief note on accounting standards.
Answer:
Accounting standard is a principle that guides and standardizes accounting practices. The Generally Accepted Accounting Principles (GAAP) are a group of accounting standards that are widely accepted as appropriate to the field of accounting. Accounting standards are necessary so that financial statements are meaningful across a wide variety of businesses; otherwise, the accounting rules of different companies would make comparative analysis almost impossible.

Student Activity

Visit a nearest business organization and observe the book keeping and accounting systems adopted by it.

Andhra Pradesh BIEAP AP Inter 1st Year Accountancy Study Material 1st Lesson Book Keeping and Accounting Textbook Questions and Answers.

AP Inter 1st Year Accountancy Study Material 1st Lesson Book Keeping and Accounting

Short Answer Questions

Question 1.
Explain the advantages and limitations of Accounting.
Answer:
Advantages of Accounting: The following are the main advantages of accounting.

1. Permanent and Reliable Record : Accounting provides permanent record for all business transactions and provides reliable information to different interested parties.
2. Net Result of Business Operations: Accounting provides the final result (Profit or Loss) of business for a given period of time.
3. Ascertainment of Financial Position: It is not enough to know only the profit or loss, but the proprietor requires a full picture of his financial position to plan for the next year’s business.
4. Facility of Comparative Study: Accounting provides the facility of comparative study of the various aspects of the business such as profits, sales, expenses, etc.
5. Control over Assets: In the course of business, the proprietor acquires various assets like building, machinery, furniture, etc. which are well protected by generating records.

Limitations of Accounting: The following are the limitations of Accounting.

1. Records only monetary transactions : Accounting considers monetary transactions only, non-monetary transactions like quality, organization culture, units of production sales, etc. are ignored in accounting.
2. Historical in nature : Accounting considers only historical transactions, i.e transactions which have occurred in the past only recorded in accounting books.
3. Price level changes are not considered: Accounting does not consider price level changes which may occur from time to time, thus, it does not reflect the current position.
4. Does not provide realistic information: While preparing the books of account, subjectivity of the accountant may influence the final results of the business enterprise. This may not provide realistic information which in turn affects the overall results of the business concern.

Question 2.
Distinguish between Book-keeping and Accounting.
Answer:

Question 3.
Explain the steps involved in Accounting process.
Answer:
Accounting process:

1. Identifying: Identifying the business transactions from the source documents.
2. Recording: The next function of accounting is to keep a systematic record of all business transactions, which are identified in an orderly manner, soon after their occurrence in the journal or subsidiary books.
3. Classifying : This is concerned with the classification of the recorded business transactions so as to group the transactions of similar type at one place.
4. Summarizing : It is the process of finding the totals of balances of all accounts so as to prepare trial balance.
5. Reporting : The classified information available from the trial balance is used to prepare – profit and loss account and balance sheet in a manner useful to the users of accounting
   information.
6. Analysing: It establishes the relationship between the items of the profit and loss account and the balance sheet.
7. Interpreting : It is concerned with explaining the meaning and significance of the relationship so established by the analysis. Interpretation should be useful to the users.

Very Short Answer Questions

Question 1.
What is Book-keeping ? (Mar. 2018 – A.P. ; Mar. ’17 – T.S. ; May ’17 – A.P. & T.S.)
Answer:
Book-keeping is the art of recording business transactions in a systematic manner. It covers four activities i.e. identifying the transactions, measuring the identified transactions, recording in a chronological order and classifying the recorded transactions.

Question 2.
Define Accounting. (Mar. 2019, ’17 – A.P.)
Answer:
The American Institute of Certified Public Accountants has defined the Accounting as “the art of recording, classifying and summarizing in a significant manner in terms of money transactions and events which in part, at least of a financial character and interpreting the results thereof”.

Question 3.
What is an Accounting cycle ? (Mar. 2018 – T.S. ; Mar. ’15 – A.P. & T.S.)
Answer:
An accounting cycle is a complete sequence beginning with the recording of transaction and ending with preparation of financial statements. It involves journalizing, ledger posting, balancing, trial balance, preparation of trading, profit and loss account and balance sheet.

Student Activity

1. Goto nearest business organizations and observe the book keeping and accounting systems adopted by them.
2. Observe the accounting procedures adopted by large size or medium size organizations located nearby them.