Inter 1st Year

Andhra Pradesh BIEAP AP Inter 1st Year Economics Study Material 4th Lesson Theory of Production Textbook Questions and Answers.

AP Inter 1st Year Economics Study Material 4th Lesson Theory of Production

Essay Questions

Question 1.
Explain the law of variable proportions. [March 18, 17]
Answer:
The law of variable proportions has been developed by the 19^th century economists David Ricardo and Marshall. The law is associated with the names of these two economists. The law states that by increasing one variable factor and keeping other factors constant, how to change the level of output, total output first increases at increasing rate, then at a diminishing rate and later decreases. Hence this law is also known as the “Law of Diminishing returns”.

Marshall stated in the following words.

“An increase in capital and labour applied in the cultivation of land causes in general less than proportionate increase in the amount of produce raised, unless it happens to coincide with an improvement in the arts of agriculture”.
Assumptions :

1. The state of technology remain constant.
2. The analysis relates to short period.
3. The law assumes labour in homogeneous.
4. Input prices remain unchanged.

Explanation of the Law :
Suppose a farmer has ‘4’ acres of land he wants to increase output by increasing the number of labourers, keeping other factors constant. The changes in total production, average product and marginal product can be observed in the following table.

In the above table total product refers to the total output produced per unit of time by all the labourers employed.

Average product refers to the product per unit of labour marginal product refers to additional product obtained by employing an additional labour.
In the above table there are three stages of production.

1^st stage i.e., increasing returns at 2 units total output increases average product increases and marginal product reaches maximum.
2^nd stage i.e., diminishing returns from 3^rd unit onwards TP increases diminishing rate and reaches maximum, MP becomes zero, AP continuously decreases.
3^rd stage i.e., negative returns from 8^th unit TP decreases AP declines and MP becomes negative.
This can be explained in the following diagram.

In the diagram on ‘OX’, axis shown units labourer and OY’ axis show TP, MP, and AP. 1^st stage TP, AP increases MP is maximum. In the 2^nd stage TP maximum, AP MP is zero. At 3^rd stage TP declines, AP also declines, MP becomes negative.

Question 2.
Explain the law of returns to scale.
Answer:
The law of returns to scale relate to long run production function. In the long run it is possible to alter the quantities of all the factors of production. If all factors of production are increased in given proportion the total output has to increase in the same proportion. Ex : The amounts of all the factors are doubled, the total output has to be doubled increasing all factors in the same proportion is increasing the scale of operation. When all inputs are changed in a given proportion, then the output is changed in the same proportion. We have constant returns to scale and finally arises diminishing returns. Hence as a result of change in the scale of production, total product increases at increasing rate, then at a constant rate and finally at a diminishing rate.
Assumptions :

1. All inputs are variable.
2. It assumes that state of technology remain the same. The returns to scale can be shown in the following table.

The above table reveals the three patterns of returns to scale. In the 1^th place, when the scale is expanded upto 3 units, the returns are increasing. Later and upto 4^th units, it remains constant and finally from 5^th on words the returns go on diminishing.

In the diagram on ‘OX’ axis shown scale of production, on OY’ axis shown total product. RR₁ represents increasing returns R₁S – Constant returns; SS₁ represents diminishing returns.

Question 3.
Distinguish between internal and external economies.
Answer:
Economies of large scale production can be grouped into two types.

1. Internal economies
2. External economies.

1. Internal Economies:
Internal economies are those which arise from the expansion of the plant, size or from its own growth. These are enjoyed by that firm only.
“Internal economies are those which are open to a single factory or a single firm independently of the action of other firms”. – Cairncross
i) Technological Economies : The firm may be running many productive establishments. As the size of the productive establishments increase, some mechanical advantages may be obtained. Economies can be obtained from linking process to another process i.e. paper making and pulp making can be combined. It also used superior techniques and increased specialization.

ii) Managerial Economies : Managerial economies arises from specialisation of management and mechanisation of managerial functions. For a large size firm it becomes possible for the management to divide itself into specialised departments under specialised personnel. This increases efficiency of management at all levels. Large firms have the opportunity to use advanced techniques of communication, computers etc. All these things help in saving of time and improve the efficiency of the management.

iii) Marketing Economies : The large firm can buy raw materials cheaply, because it buys in bulk. It can secure special concession rates from transport agencies. The product can be advertise better. It will be able to sell better.

iv) Financial Economies : A large firm can arise funds more easily and cheaply a small one. It can borrow from bankers upon better security.

v) Risk bearing Economies : A large firm incurs unrisk and it can also reduce risks. It can spread risks in different ways. It can undertake diversifications of output. It can buy raw materials from several firms.

vi) Labour Economies : A big firm employs a large number of workers. Each worker is given the kind of job he is fit for.

2. External Economies :
An external economy is one which is available to all the firms in an industry. External economies are available as an industry grows in size.

1. Economies of Concentration : When a number pf firms producing an identical product are localised in one place, certain facilities become available to all. Ex : Cheap transport facility, availability of skilled labour etc.
2. Economies of Information : When the number of firms in an industry increases collective action and co-operative effort becomes possible. Research work can be carried on jointly. Scientific journal can be published. There is possibility for exchange of ideas.
3. Economies of Disintegration : When the number of firms increases, the firm may agree to specialise. They may divide among themselves the type of products of stages of production. Ex : Cotton industry.

Question 4.
Explain short-run cost structure of a Firm.
Answer:
Costs are divided into two categories i.e.,

1. Short run cost curves
2. Long run cost curves.

In short run by increasing only one factor i.e., (labour) and keeping other factor constant. The short run cost are again divided into two types.

1. General costs
2. Economic costs.
   1. General costs :

i) Money costs : Production is the outcome of the efforts of factors of production like land, labour, capital and organisation. So, rent to land, wage to labour, interest to capital and profits to entrepreneur has to paid in the form of money is called money cost.

ii) Real cost: Adam Smith regarded pains and sacrifices of labour as real cost. So it cannot be measured interms of money.

iii) Opportunity cost: Factors of production are scarce and have alternative uses. The opportunity cost of a factor is the benefit that is foregone from the next best alternative use.

2. Economic costs :
i) Fixed costs : The cost of production which remains constant even the production may be increase or decrease is known as fixed cost. The amount spent by the cost of plant and equipment, permanent staff are treated as fixed costs.

ii) Variable cost: The cost of production which is changing according to changes in the production is said to be variable cost. In the long period all costs are variable costs. It include price of raw materials, payment of fuel, excise taxes etc. Marshall called “Prime cost”.

iii) Average cost: Average cost means cost per units of output. If we divided total cost by the number of units produced, we will get average cost.
AC = \(\frac{\text { Total cost }}{\text { Output }}\)
iv) Marginal cost: Marginal cost is the additional cost of production producing one more unit.
MC = \(\frac{\Delta \mathrm{TC}}{\Delta \mathrm{Q}}\)
v) Total cost: Total cost is the sum of total fixed cost and total variable cost.
TC = FC + VC
The short term cost in relation to output are explained with the help of a table.

In the above table shows that as output is increased in the 1^st column, fixed cost remains constant. Variable costs have changed as and when there are changes in output. To produce more output in the short period, more variable factors have to be employed. By adding FC & VC we get total cost different levels of output. AC falls output increases, reaches its minimum and then rises MC also change in the total cost associated with a change in output. This can be shown in the diagram.

In the above diagram on ‘OX’ axis taken by output and ‘OY taken by costs. The shapes of different cost curves explain the relationship between output and different costs. TFC is horizontal to ‘X’ axis. It indicates that increase in output has no effect on fixed cost. TVC on the other side increases along with level of output. TC curve rises as output increases.

Additional Questions

Question 1.
Explain Production function.
Answer:
Production function is technical concept. It explains the physical relationship between input and output at any period of time. It represents functional relationship between inputs and the amount of output produced.
According to Stigler “Production is the name given to the relationship between rates of inputs of productive services and the rate of output of product”.
The production function can be expressed mathematically as follows.
G_x = f (L, K, R, N, T)
G_x = Output
f = Functional relationship
L = Amount of Labour
K = Amount of Capital
R = Raw material
N = Natural resources or land
T = State of Technology.
Where G_x is dependent variable and is determined by the inputs used, whereas L, K, R, N, T are independent variables.

Question 2.
Explain Law of supply.
Answer:
The law of supply explains the functional relationship between price of a commodity and its quantity supplied. The law of supply can be stated as follows “Other things remaining the same, as the price of a commodity rises its supply is extended and as the price falls its supply is contracted”.

The law of supply can be explained with the help of supply schedule and supply curve.

Supply Schedule : Supply schedule explains various amounts of good that the seller offers for sale at different prices. It represents the functional relationship between price and quantities supplied. There is direct relationship between price and supply. This can be shown in the following schedule.

The above schedule high price i.e, ₹ 5.00 per unit 1000 units are supplied and at ₹ 1 per 200 units are supplied. It means high price indicate high supply and low price indicates low supply. So, it shows the direct relationship between price and supply.

Supply curve :
A supply curve can be drawn with the help of above supply schedule to explain the direct relationship between price and supply.

In the above diagram supply is shown on ‘OX’ axis and price is shown ‘OY axis. SS is supply curve. It slopes upwards from left to right. The slope of supply curve is always positive. Because there is direct relationship between the price and supply.’

Question 3.
Define Factors of production.
Answer:
Factors that help in the production process are called factors of production.
Ex : land, labour, capital and organization
1) Land : Land stands in economics for natural resources. There are nature given resources like soil, earth, water, rainfall, forests, mines, clijnate etc. All these come under land. Land is the productive equipment given by nature. The remuneration to land is called rent.

2) Labour : Labour is man’s effort. It may be physical or mental labour. Any service offered for a price is considered as labour in economics. All services rendered to produce scarce goods come under labour. The remuneration to labour is called wage.

3) Capital : Capital is man made production equipment. Factories, buildings, vehicles, rail-roads, roads, irrigation dam etc., come under capital. The remuneration to capital is interest.

4) Organisation or Entrepreneurship : The entrepreneur or businessmen brings together all the factors of production required for production. He bears the risk is doing so. He coordinates the functions of different factors. Profit is the remuneration for organisation or entrepreneurship.

Question 4.
Define land and explain the characteristics of land.
Answer:
Land stands in economics for natural resources. There are nature given resources like soil, earth, water, rainfall, forests, mines, climate etc., all these come under land. Land is the productive, equipment given by nature. The remuneration to land is called rent. Characteristics of land:

1. Land is a free gift of nature.
2. The supply of land is perfectly inelastic from the point of view of economy.
3. Land cannot be shifted from one place to another place.
4. It does not yield any result unless human effort’s are employed.

Question 5.
Define labour. Explain the characteristics of labour.
Answer:
Labour is man’s effort. It may be physical (or) mental labour. Any’service offered for a price is considered as labour in economics. The remuneration to labour is called wage. Characteristics of labour:

1. Labour is inseparable from the labourer himself.
2. Labour is highly ‘perishable’. It means labour lost cannot store his labour.
3. Labour has a very weak bargaining power.
4. Labour power differs from labourer to labourer of their skills.
5. The supply curve of a labourer is backward bending.

Question 6.
What is supply ? What are the determinants of supply.
Answer:
The supply of a commodity means the total quantity of the commodity that sellers offer to sell at different prices from the stock of that commodity existing at any given time. The supply of commodity depends upon the following factors.
Determinants of supply :

1. Price of the commodity: The supply of the commodity depends upon the price of that commodity. When price falls, supply falls and when price rises, supply also rises. Thus price and supply are directly related.
2. Factor prices : The cost of production of a commodity depends upon the prices of various factors of production.
3. Prices of related goods : The supply of the commodity depends upon the prices of related goods. If the price of a substitute good goes up, the producer will be induced to
4. State of technology: Technological improvements determine supply of a commodity. Progress in technology leads to reduction in the cost of production which will increase supply.
5. Government policy : Imposition of heavy taxes as a commodity discourages its production. Hence production decreases.

Question 7.
Explain the nature of revenue curves untier perfect competition.
Answer:
If competition is perfect in market, the market is known as perfectly competitive market. Due to the existing features the total demand and total supply pf a commodity interact to determine the price in the industry and individual firm. Hence, an individual seller or firm is a price taker but not a price maker in perfect competition, ‘therefore he can sell any quantity at the ruling price. Thus, the demand curve is parallel to X-axis. The nature of Average Revenue (AR) and Marginal Revenue (MR) and their relationship under perfect competition can be better understood from the following schedule.
Revenue Schedule :

The above table indicates that in perfect competition price remains the same irrespective of the number of units sold. Therefore, the total revenue increases at a constant rate. AR and MR are equal. There is no difference among price AR and MR. This can be reveals the following diagram.

In the diagram X-axis represents number of units (output) and Y-axis represents revenue. DP is demand curve.

In diagram, the Average Revenue curve (AR) is horizontal straight line parallel to the X-axis and Marginal Revenue Curve MR coincides with it. Because the seller can sell number of units given the price, the AR curve facing the seller is a horizontal line.

Very Short Answer Questions

Question 1.
Production function [March 18]
Answer:
The production function is the none for the relation between the physical inputs and the physical outputs of a film. Production of a firm, production function explains the functional relationship between inputs and outputs this can be as follows G_x = f (L, K, R, N, T).

Question 2.
Law of supply [March 16]
Answer:
The law of supply explains the functional relationship between price of a commodity and its quantity supplied. The law of supply can be stated as follows. “Other things remaining the same, as the price of a commodity rises its supply is extended and as the price falls its supply is contracted”.

Question 3.
Factors of production
Answer:
Factors that help in the production process are called factors of production.
Ex : land, labour, capital and organization.

Question 4.
Average cost
Answer:
If we divided total cost, by the number of units produced. We will get average cost. Average cost means cost per unit of output.
AC = \(\frac{\mathrm{TC}}{\text { Output }}\)

Question 5.
Marginal cost
Answer:
Marginal cost is the additional cost of production producing one more unit in other words it is the addition made to total cost by producing one more unit of a commodity.
MC = \(\frac{\Delta \mathrm{TC}}{\Delta \mathrm{Q}}\)

Additional Questions

Question 6.
Production
Answer:
Production is the process that converts inputs into output in economies production includes services along with physical goods.

Question 7.
Short period
Answer:
Short period is a period in which a producer is unable to change factors of production to increase output. This relates to law of variable proportions.

Question 8.
Long period
Answer:
Long period is a period in which a producer is unable to change factors of production to increase output. This relates to returns to scale.

Question 9.
Average product
Answer:
It refers to the product per unit of labour it is Obtained by dividing total product by the number of labourers employed.
AP = \(\frac{\mathrm{TP}}{\mathrm{L}}\)

Question 10.
Marginal product
Answer:
It is the additional product by employing an additional labour.
MP = \(\frac{\Delta \mathrm{TP}}{\Delta \mathrm{L}}\)

Question 11.
Fixed factor
Answer:
Fixed factors are those costs which can’t be changed by the producer in the short period.
Ex : Buildings, Machinery etc.

Question 12.
Variable factors
Answer:
The factors of production which are possible to change in relation to a change in output is known as variable factors in the long run all factors of production are variable.
Ex : Labour, Raw materials.

Question 13.
Change in scale of production
Answer:
It refers that output will be increase by increasing inputs in the long period.

Question 14.
Internal economies
Answer:
It refers that when a firm expands output by increasing all inputs.

Question 15.
External economics
Answer:
It refers to one which is available to all the firms in an industry. External economies are available as an industry grows in size.

Question 16.
Supply
Answer:
The quantity of a commodity that a seller is prepared to sell at a particular price and at a particular time is known as supply. The supply curve slopes upwards from left to right.

Question 17.
Supply function
Answer:
It explains the functional relation between supply and the factors of production of a good.

Question 18.
Opportunity cost
Answer:
The opportunity cost of a factor is the benefit i.e, forgone from the next best alternative use.

Question 19.
Fixed cost
Answer:
The cost of production which remains constant even the production may be increase (or) decrease is known as fixed cost.
Ex : Machinery, permanent staff salaries.

Question 20.
Variable cost
Answer:
The cost of production which is changed according to changes in the production is said to be variable cost. In the long period all costs are variable costs it include price of raw materials, wages of labour, power transport charges etc.

Question 21.
Total Revenue
Answer:
Total revenue of a producer depends on the price and the quantity of output sold in the market.
Total Revenue = Price × Quantity of output
TR = P × Q

Question 22.
Average Revenue
Answer:
Average revenue is the revenue per unit of output. Average revenue is obtained by dividing to revenue by the number of unit sold.
AR = \(\frac{\text { TR }}{\mathrm{Q}}\)

Question 23.
Marginal Revenue
Answer:
Marginal revenue is the additional revenue earned by selling one more unit of the product. In other words change in total revenue arising from the sale of an additional unit of output is called marginal revenue.
MR = \(\frac{\Delta \mathrm{TR}}{\Delta \mathrm{Q}}\)

Question 24.
Total Cost
Answer:
Total cost can be obtained by adding total fixed costs and variable costs.

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 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 and the marginal utility of the commodity.

According to Marshall “The additional benefit which a person derives from a given increase of his stock thing diminishes with every increase in the 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 go 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 Physics Study Material 14th Lesson Kinetic Theory Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 14th Lesson Kinetic Theory

Very Short Answer Questions

Question 1.
Define mean free path. [A.P. Mar. 18, 17, 15; T.S. Mar. 17, 15]
Answer:
The average distance covered by a molecule between two successive collisions is called the mean free path.

Question 2.
How does kinetic theory justify Avogadro’s hypothesis and show that Avogadro’s Number in different gases is the same?
Answer:
For two different gases, we have \(\frac{P_1 V_1}{N_1 T_1}=\frac{P_2 V_2}{N_2 T_2}\) = K_B (constant)
If P, V, T are the same, then N is also the same for the two gases.
N is called Avogadro’s number. According to Avogadro’s hypothesis, the number of molecules per unit volume is the same for all gases at a fixed temperature and pressure.
In this way, kinetic theory justify Avogadro’s hypothesis.

Question 3.
When does a real gas behave like an ideal gas ? [T.S. Mar. 16; Mar. 14]
Answer:
At low pressures and high temperatures real gases behave like an ideal gas.

Question 4.
State Boyle’s Law and Charle’s Law. [A.P. Mar. 18; T.S. Mar. 15]
Answer:
Boyle’s law : At constant temperature, the pressure of a given mass of gas varies inversely with volume.
Charle’s law: At constant pressure, the volume of the gas is proportional to its absolute temperature.

Question 5.
State Dalton’s law of partial pressures. [T.S. Mar. 18; A.P. Mar. 16; Mar. 14]
Answer:
The total pressure exerted by a mixture of non-reacting gases occupying a vessel is equal to the sum of the individual pressures which each gas would exert if it is alone occupied the whole vessel.
i.e., P = P₁ + P₂ + ………….

Question 6.
Pressure of an ideal gas in container is independent of shape of the container – explain. [T.S. Mar. 17]
Answer:
The kinetic theory expression for the pressure of a given mass of an ideal gas in a container is 1 -2.
P = \(\frac{1}{3}\) nm\(\bar{v}^2\), where \(\overline{\mathrm{V}}\) indicates mean-square-speed, n is number of molecules, m is mass of molecule. From the above equation, shape of the container is immeterial. Hence pressure of an ideal gas is independent of shape of the container.

Question 7.
Explain the concept of degrees of freedom for molecules of a gas.
Answer:
The degrees of freedom of a particle indicate the number of independent motions which the particle can undergo, or the number of independent methods of exchanging energy.

A monoatomic molecule (He) has three degrees of freedom, a diatomic (H₂, O₂) one has five, while a triatomic (H₂O) one has six.

Question 8.
What is the expression between pressure and kinetic energy of a gas molecules ? [AP – Mar. ’17, ’16, ’15]
Answer:
By kinetic theory pressure, P = \(\frac{1}{3} m n \overline{V}^2\) and kinetic energy = \(\frac{1}{2} m n \overline{V}^2\) where m is the mass of the molecules, n is the number of moles per unit volume, v is the mean-square-speed.
∴ P = \(\frac{2}{3}\left(\frac{1}{2} m n \overline{V}^2\right)\) = \(\frac{2}{3}\) (kineticenergy)

Question 9.
The absolute temperature of a gas is increased 3 times. What will be the increase in rms velocity of the gas molecule ?
Answer:
Case (i) : The r.m.s velocity of gas molecule, \(\overline{V}_1=\sqrt{\frac{3 R T_1}{M}}\)
Case (ii) : The r.m.s velocity of gas molecule, \(\overline{V}_2=\sqrt{\frac{3 R T_2}{M}}\)
\(\frac{\overline{\mathrm{V}}_2}{\overline{\mathrm{V}}_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}}=\sqrt{\frac{3 \mathrm{~T}_1}{\mathrm{~T}_1}}\)  [∵ T₂ = 3T₁] ⇒ \(\overline{\mathrm{V}}_2=\sqrt{3} \overline{\mathrm{V}}_1=1.732 \overline{\mathrm{V}}_1\)
∴ Increase in r.m.s velocity of gas molecule = \(\overline{\mathrm{V}}_2-\overline{\mathrm{V}}_1=1.732 \overline{\mathrm{V}}_1-\overline{\mathrm{V}}_1=0.732 \overline{\mathrm{V}}_1\)

Short Answer Questions

Question 1.
Explain the kinetic interpretation of Temperature.
Answer:
Kinetic interpretation of Temperature : Since the pressure of the gas is given by
P = \(\frac{1}{3} \mathrm{mn} \overline{\mathrm{V}}^2\), where m = mass of the gas, n = \(\frac{\mathrm{N}}{\mathrm{V}}\) = Number of molecules per unit volume
\(\overline{V}\) = r.m.s velocity of gas, P = \(\frac{1}{3} m \frac{N}{V} \overline{V}^2\) ⇒ PV = \(\frac{1}{3} \mathrm{mN} \overline{\mathrm{V}}^2\)
Also for a gram molecule of the gas, PV = RT
Hence RT = \(\frac{1}{3} m N \overline{V}^2 \Rightarrow 3 \frac{R T}{N}=m \overline{V}^2 \text { or } \frac{1}{2} m \overline{V}^2=\frac{3}{2} K_B T\) [∵ \(\frac{R}{N}\) = K_B]
Here K_B is Boltzman constant. So mean kinetic energy of a molecule is \(\frac{3}{2}\) K_B T, which depends upon the temperature. As temperature increases mean kinetic energy of the molecules also increases.

Question 2.
How specific heat capacity of mono atomic, diatomic and poly atomic gases can be explained on the basis of Law of equipartition of Energy ? [Mar. 13]
Answer:
Specific heat capacity:
1) Mono atomic gases : According to law of equipartition of energy, a molecule of mono atomic gas has only 3 (translational) degrees of freedom, i.e., f = 3.
The molar specific heat of the gas at constant volume is given by C_v = \(\frac{f}{2}\) R where f is degree of freedom.
∴ C_v = \(\frac{3}{2}\)R = 3 cal/mole k [∵ R = 2 cal/mole – k]
The molar specific heat at constant pressure is given by
C_p = (\(\frac{f}{2}\) + 1)R = (\(\frac{3}{2}\) + 1) R = \(\frac{5}{2}\)R = 5 cal/ m0le – k

2) Diatomic gases : A molecule of diatomic gas has 5 degrees of freedom, 3 translational and 2 rotational i.e., f = 5
Therefore C_v = \(\frac{f}{2}\) R = \(\frac{5}{2}\)R = 5 cal/mole – k, C_p = (\(\frac{f}{2}\) + 1) R = \(\frac{7}{2}\) R = 7 cal/mole – k

3) Polyatomic gases : Polyatomic molecule has 3 translational, 3 rotational degrees of freedom, i. e., f = 6
Therefore C_v = \(\frac{f}{2}\)R = 3R = 6 cal / mole – k, C_p = (\(\frac{f}{2}\) + 1) = 4R = 8 cal / mole – k.

Question 3.
Explain the concept of absolute zero of temperature on the basis of kinetic theory.
Answer:
Concept of absolute zero on the basis of kinetic theory :
Since the pressure of the gas is given by
P = \(\frac{1}{3} \mathrm{mn} \overline{\mathrm{V}}^2\), where m = mass of the gas, n = \(\frac{\mathrm{N}}{\mathrm{V}}\) = number of molecules per unit volume.
V = r.m.s velocity of gas, P = \(\frac{1}{3} \frac{\mathrm{mN}}{\overline{\mathrm{V}}} \overline{\mathrm{V}}^2\) ⇒ PV = \(\frac{1}{3} m N \overline{V}^2\)
Also for a gram molecule of the gas, PV = RT
Hence RT = \(\frac{1}{3} M \overline{V}^2\) [∵ M = mN = molecular weight]
\(\overline{\mathrm{V}}^2=\frac{3 R T}{M}\)
∴ \(\overline{v}=\sqrt{\frac{3 R T}{M}}\)
Putting T = 0 in the above equation \(\overline{\mathrm{V}}\) = 0. Hence pressure becomes zero. Then the gas converts into liquids. Thus, this temperature is called absolute zero.

Question 4.
Prove that the average kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.
Answer:
Since the pressure of the gas is given by
P = \(\frac{1}{3} \mathrm{mn} \overline{\mathrm{V}}^2\), where m = mass of the gas, n = \(\frac{\mathrm{N}}{\mathrm{V}}\) = number of molecules per unit volume
\(\) = r.m.s velocity of gas, P = \(\frac{1}{3} m \frac{N}{V} \overline{V}^2\) ⇒ PV = \(\frac{1}{3} \mathrm{mNV}^2\)
Also for a gram molecule of the gas, PV = RT
Hence RT = \(\frac{1}{3}\) mNV²
3\(\frac{\mathrm{R}}{\mathrm{V}}\)T = \(\mathrm{m} \overline{\mathrm{V}}^2\) or \(\frac{1}{2}\) mV² = \(\frac{3}{2}\)K_BT [∵ \(\frac{\mathrm{R}}{\mathrm{V}}\) = K_B]
Where KB is the Boltzmann constant and T is the absolute temperature. ∴ E = \(\frac{3}{2}\)K_BT
Hence the average kinetic energy of a molecule is proportional to the absolute temperature of the gas.

Question 5.
Two thermally insulated vessels 1 and 2 of volumes V₁ and V₂ are joined with a valve and filled with air at temperatures (T₁, T₂) and pressures (P₁, P₂) respectively. If the valve joining the two vessels is opened, what will be the temperature inside the vessels at equilibrium.
Answer:
During adiabatic process, there is no loss of energy, i.e. K.E_T of molecules before mixing = K.E_T of molecules after mixing.

Question 6.
What is the ratio of r.m.s speed of Oxygen and Hydrogen molecules at the same temperature ?
Answer:
The r.m.s speed of gas is \(\overline{\mathrm{V}}=\sqrt{\frac{3 R T}{M}} \Rightarrow \frac{\overline{\mathrm{V}}_0}{\mathrm{~V}_{\mathrm{H}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{H}}}{\mathrm{M}_{\mathrm{O}}}}\), Given M_H = 2 and M₀ = 32
\(\frac{\overline{\mathrm{V}}_0}{\overline{\mathrm{V}}_{\mathrm{H}}}=\sqrt{\frac{2}{32}}=\frac{1}{4}\)
∴ V₀ : V_H = 1 : 4

Question 7.
Four molecules of a gas have speeds 1, 2, 3 and 4 km/s. Find the rms speed of the gas molecule.
Answer:
Given V₁ = 1 km/s; V₂= 2 km/s; V₃ = 3 km/s; V₄ = 4 km/s; = ?
V_rms = \(\sqrt{\frac{V_1^2+V_2^2+V_3^2+V_4^2}{n}}=\sqrt{\frac{1^2+2^2+3^2+4^2}{4}}\)
= \(\sqrt{\frac{1+4+9+16}{4}}=\sqrt{\frac{30}{4}}=\sqrt{7.5}\)
= 2735 kms^-1

Question 8.
If a gas has ‘f’ degrees of freedom, find the ratio of C_P and C_V.
Answer:
If a gas has ‘f’ degrees of freedom, then
C_V = \(\frac{f}{2}\)R and C_P = C_V + R = \(\frac{f}{2}\) R + R = (\(\frac{f}{2}\) + 1)R
Hence the ratio of the two specific heats of a gas is, r = \(\frac{C_p}{C_V}=\left[\frac{\frac{f}{2}+1}{\frac{f}{2} R}\right]=1+\frac{2}{f}\)

Question 9.
Calculate the molecular K.E of 1 gram of Helium (Molecular weight 4) at 127°C. Given R = 8.31 J mol^-1 K^-1.
Answer:
Given, t = 127°C, T = 273 + 127 = 400; R = 8.31 J mol^-1 K^-1
K.E. = \(\frac{3}{2}\)K_BT = \(\frac{3}{2}\) × 1.38 × 10^-23 × 400 = 8.28 × 10^-21 J.

Question 10.
When pressure increases by 2%, what is the percentage decrease in the volume of a gas, assuming Boyle’s law is obeyed ?
Answer:
The gas obeys the Boyles law PV = constant, on differentiating, PdV + VdP = 0
PdV = – V dP ⇒ \(\frac{\mathrm{dV}}{\mathrm{V}}=\frac{-\mathrm{dP}}{\mathrm{P}}\) ⇒ \(\frac{\mathrm{dV}}{\mathrm{V}}\) × 100% = –\(\frac{-\mathrm{dP}}{\mathrm{P}}\) × 100%
% change in volume = – percentage change in pressure
∴ % change in volume = – 2%
Here negative sign indicates decrease in volume.

Long Answer Questions

Question 1.
Derive an expression for the Pressure of an ideal gas in a container from Kinetic Theory and hence give Kinetic Interpretation of Temperature.
Answer:
Let us consider cubical vessel of side T with perfectly elastic wall containing gas molecules. Let the three sides of the cube be taken as co-ordinate axis. Consider a molecule moving with velocity V, in any direction at any instant. The components of V₁ along the three sides are V_x, V_y and V_z respectively. Then
\(\overline{V}^2=V_x^2+V_y^2+V_z^2\) ……………………. (1)
If ‘m is the mass of this molecule, it transfers a momentum mV when it strikes the face ABCD of the cube. Since the wall is perfectly elastic, this molecule is reflected back with a velocity – V_x and momentum – mV_x. So the change in momentum = mV_x – (- mV_x)
= 2mV_x
This molecule then travels towards the opposite face, collide with it, rebounds and travels again towards the face ABCD. The distance travelled between two successive collisions is 2l. Time taken between two successive collisions is \(\frac{2 l}{\mathrm{~V}_{\mathrm{x}}}\)

∴ Number of collisions per second = \(\frac{V_x}{2 l}\)
Change in momentum per second = (2mV_x) × \(\left(\frac{\mathrm{V}_{\mathrm{x}}}{2 l}\right)=\frac{\mathrm{mV}_{\mathrm{x}}^2}{l}\)
Thus force exerted by this molecule = \(\frac{\mathrm{mV}_{\mathrm{x}}^2}{l}\) [∵ F = \(\frac{\mathrm{dP}}{\mathrm{dt}}\)]
Hence, force exerted by N such molecule in the X-direction.
f₁ = \(\mathrm{N} \frac{\mathrm{mV}_{\mathrm{x}}^2}{l}=\frac{\mathrm{mNV}_{\mathrm{x}}^2}{l}\)
∴ Pressure exerted by the molecules in the x-direction is P_x = \(\frac{f_1}{l^2}=\frac{m N \overline{V}_x^2}{l^3}\)
Similarly, pressure exerted by the molecules in the y and z-directions are
P_y = \(\frac{\mathrm{m}}{l^3} \mathrm{~N} \overline{\mathrm{V}}_{\mathrm{y}}^2\) and P_z = \(\frac{\mathrm{m}}{l^3} \mathrm{~N} \overline{\mathrm{V}}_{\mathrm{z}}^2\)
Since the pressure exerted by a gas in all the directions is same, the average pressure
P = \(\frac{P_x+P_y+P_z}{3}=\frac{m N}{3 l^3}\left[V_x^2+v_y^2+V_z^2\right]=\frac{m N}{3 V} \overline{V}^2\) [ from (1) and V = l³]
Here V² is the mean square velocity of the molecule, V is the volume of the vessel.
If M is the mass of the gas then M = mN
∴ P = \(\frac{1}{3} \frac{M^2}{V}=\frac{1}{3} m n \overline{V}^2\) [∵ n = \(\frac{N}{V}\)]
This pressure is actually the pressure exerted by an ideal gas.

b) Kinetic interpretation of temperature :
Since the pressure of the gas is given by
P = \(\frac{1}{3} m n \overline{V}^2\), where m = mass of the gas, n = \(\frac{N}{V}\) = Number of molecules per unit volume
\(\overline{V}\) = r.m.s velocity of gas

Here K_B is Boltzmann constant. So mean kinetic energy of a molecule is \(\frac{3}{2}\) K_BT, which depends upon the temperature. As temperature increases means kinetic energy of the molecules also increases.

Additional Problems

Question 1.
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.
Solution:
Here, diameter,
d = 3A°, r = \(\frac{\mathrm{d}}{2}=\frac{3}{2}\) A° = \(\frac{3}{2}\) × 10^-8 cm.
Molecular volume,
v = \(\frac{4}{3}\) πr³N, where N is Avagadros No.
= \(\frac{4}{3}\) × \(\frac{22}{7}\) (1.5 × 10^-8)³ × (6.023 × 10²³)
= 8.52 cc.
Actual volume occupied by 1 mole of oxygen at STP v = 22400 cc
= \(\frac{v}{v^{\prime}}=\frac{8.52}{22400}\) = 3.8 × 10^-4 = 4 × 10^-4.

Question 2.
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0 °C.) Show that it is 22.4 litres.
Solution:
For one mole of an ideal gas, ρv = RT
∴ v = \(\frac{\mathrm{RT}}{\rho}\)
Put R = 8.31 J mole^-1 k^-1, T = 273k, ρ = 1 atm = 1.013 × 10⁵ Nm^-2
v = \(\frac{8.31 \times 273}{1.013 \times 10^5}\) = o.0224m³ = 0.0224 × 10⁶
cc = 22400 cc = 22.4 lit.

Question 3.
Figure 14.8 shows plot of PV/T versus P for 1.00 × 10^-3 kg of oxygen gas at two different temperatures.

a) What does the dotted plot signify ?
b) Which is true: T₁ > T₂ or T₁ < T₂ ?
c) What is the value of PV/T where the curves meet on they y-axis ?
d) If we obtained similar plots for 1.00 × 10^-3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis ? If not, what mass of hydrogen yields the same value of PV/T (for low pressurehigh temperature region of the plot) ? (Molecular mass of H₂ = 2.02 u, of O₂ = 32.0 u, R = 8.31 J mole^-1 K^-1.)
Solution:
a) The dotted plot shows that \(\frac{\mathrm{PV}}{\mathrm{T}}\) (= μR) is a constant quantity, independent of pressure p. This sign signifies the ideal gas behaviour.

b) The curve at temperature T₁ is closer to dotted plot than the curve at temperature T₂. As the behaviour of a real gas approaches the behaviour of a perfect gas when temperature is increased, therefore T₁ > T₂.

c) Where the two curves meet, the value of
\(\frac{\mathrm{PV}}{\mathrm{T}}\) on y-axis is equal to μR.
As mass of oxygen gas = 1.00 × 10^-3 kg = 1g
\(\frac{\mathrm{PV}}{\mathrm{T}}\) = nR (\(\frac{1}{32}\)) × 8.31 Jk^-1.

d) If we obtained similar plots for 1.00 × 10^-3 kg of hydrogen, we will not get the same value of \(\frac{\mathrm{PV}}{\mathrm{T}}\) at the point, where the curve meets on the y-axis. This is because molecular mass of hydrogen is different from that of oxygen.
For same value of \(\frac{\mathrm{PV}}{\mathrm{T}}\) mass of hydrogen required is obtained from.
\(\frac{\mathrm{PV}}{\mathrm{T}}\) = nR = \(\frac{\mathrm{M}}{2.02}\) × 8.31 = 0.26
M = \(\frac{2.02 \times 0.26}{8.31}\) gm = 6.32 × 10^-2 grams.

Question 4.
All oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol^-1 K^-1, molecular mass of O2 = 32 u).
Solution:
Initially in the oxygen cylinder,
v₁ = 30 lit = 30 × 10^-3 m³
P₁ = 15 atm = 15 × 1.01 × 10⁵ Pa;
T₁ = 27 + 273 = 300 k.
If the cylinder contains n₁ mole of oxygen gas, then
P₁v₁ = nRT₁
(or) n₁ = \(\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{RT}_1}=\frac{\left(15 \times 1.01 \times 10^5\right) \times\left(30 \times 10^{-3}\right)}{8.3 \times 300}\)
= 18.253 .
For oxygen, molecular weight, M = 32 g
Initial mass of cylinder,
M₁ = n₁ M = 18.253 × 32 = 584.1 g
Finally in the oxygen cylinder, let n₂ moles of O₂ be left, were,
v₂ = 30 × 10^-3 m³, P₂ = 11 × 1.01 × 10⁵ Pa, T₂ = 17 + 273 = 29K
Now
n₂ = \(\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{RT}_2}=\frac{\left(11 \times 1.01 \times 10^5\right) \times\left(30 \times 10^{-3}\right)}{8.3 \times 290}\)
= 13.847
∴ Final mass of oxygen gas in the cylinder,
m₂ = 13.847 × 32 = 453.1 g
∴ Mass of oxygen taken out = m₁ – m₂ = 631.0 g.

Question 5.
An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grown when it reaches the surface, which is at a temperature of 35 °C ?
Solution:
v₁ = 1.0 cm³ = 1.0 × 10^-6 m³,
T₁ = 12 °C = 12 + 273 = 285 k
P₁ = 1 atm + h₁ ρg = 1.01 × 10⁵ + 40 × 10³ × 9.8
= 493000 Pa.
When the air bubble reaches at the surface of lake, then
v₂ = 2, T₂ = 35 °C = 35 + 273 = 308 K,
P₂ = 10 atm = 1.01 × 10⁵ pa.
Now \(\frac{P_1 v_1}{T_1}=\frac{P_2 v_2}{T_2}\) (or) v₂ = \(\frac{P_1 v_1 T_2}{T_1 P_2}\)
∴ v₂ = \(\frac{(493000) \times\left(1.0 \times 10^{-6}\right) \times 308}{285 \times 1.01 \times 10^5}\)
= 5.27 × 10^-6m³.

Question 6.
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m³ at a temperature of 27 °C and 1 atm pressure.
Solution:
Here, v = 25.0 m³, T = 27 + 273 = 300 k,
k = 1.38 × 10^-23 Jk^-1
Now Pv = nRT = n(NK) T = (nN) kT = N’kT
When nN = N’ = total no. of air molecules in the given gas.
N’ = \(\frac{\mathrm{Pv}}{\mathrm{kT}}=\frac{\left(1.01 \times 10^5\right) \times 25}{\left(1.38 \times 10^{-23}\right) \times 300}\) = 6.10 × 10²⁶

Question 7.
Estimate the average thermal energy of helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).
Solution:
i) Here, T = 27 °C = 27 + 273 = 300 k
Average thermal energy
= \(\frac{3}{2}\) kT = \(\frac{3}{2}\) × 1.38 × 10^-23 × 300
= 6.2 × 10^-21 J

ii) At T = 6000 k. Average thermal energy = \(\frac{3}{2}\)kT = \(\frac{3}{2}\) × 1.38 × 10^-23 × 6000
= 1.24 × 10^-19 J.

iii) At T = 10 million K = 10⁷ K.
Average thermal energy
= \(\frac{3}{2}\)kT = \(\frac{3}{2}\) × 1.38 × 10^-23 × 10⁷
= 2.1 × 10^-16 J.

Question 8.
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (poly-atomic). Do the vessels contain equal number of respective molecules ? Is the root mean square speed of molecules the same in the three cases ? If not, in which case is u_rms the largest ?
Solution:
All the three vessels (at the same temperature and pressure) have same volume. So in accordance with the avagadro’s law, the three vessels will contain equal number of respective molecules, being equal to
Avagadro’s number N = 6.023 × 10²³
As v_rms = \(\sqrt{\frac{3 k T}{m}}\)
i.e., v_rms ∝ \(\frac{1}{\sqrt{m}}\) at a given temp, therefore, rms speed of molecules will not be same in the three cases.
As neon has the smallest mass, therefore, rms speed will be largest in case of neon.

Question 9.
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C ? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Solution:
Let c and c’ be the rms velocity of argon and a helium gas atoms at temperature TK and TK’ respectively.
Here μ = 39.9, μ = 4.0,
T = 2, T’ = -20 + 273 = 253 K.

Question 10.
Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N₂ = 28.0 u).
Solution:
Here, λ = 2, f = 2
P = 2 atm = 2 × 1.013 × 10⁵ Nm^-2
T = 17 °C = (17 + 273) K = 290 k
σ = 2 × 1 = 2A° = 2 × 10^-10 m
K = 1.38 × 10^-23 J molecule^-1 k^-1,
μ = 28 × 10^-3 kg
λ = \(\frac{\mathrm{KT}}{\sqrt{2} \pi \sigma^2 \rho}\)
= \(\frac{1.38 \times 10^{-23} \times 290}{1.414 \times 3.14\left(2 \times 10^{-10}\right)^2 \times 2.026 \times 10^5}\)
λ = \(\frac{1.38 \times 29 \times 10^{-7}}{1.44 \times 3.14 \times 2.06}\) = 1.11 × 10^-7 m.
v_rms = \(\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 \times 8.31 \times 290}{28 \times 10^{-3}}}\) = 508.24 m/s.
Collision frequency = no. of collisions per second = \(\frac{v_{r m s}}{\lambda}=\frac{508.24}{1.11 \times 10^{-7}}\) = 4.58 × 10⁹

Question 11.
A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom ?
Solution:
When the tube is held horizontally, the mercury of length 76 cm traps length of air = 15 cm. A length of 9 cm of the tube will be left at the open end. Fig. (a). The pressure Of air enclosed in tube will be atmospheric pressure. Let area of cross section of the tube be 1 sq.cm.
∴ P₁ = 76 cm and v₁ = 15 cm³.
When the tube is held vertically, 15 cm air gets another 9 cm of air (filled in the right hand side position) and let h cm of the mercury flows out to balance the atmospheric pressure. Fig. (b). Then the heights of air column and mercury column are (24 + h) cm and (76 – h) cm respectively.
The pressure of air = 76 – (76 – h) = h cm of mercury

∴ v₂ = (24 + h) cm³ and P₂ = hcm.
If we assume that temperature remains constant, then
P₁v₁ = P₂v₁ or 76 × 15 = h × (24 + h) or h² + 24h – 1140 = 0
(or) h = \(\frac{-24 \pm \sqrt{(24)^2+4 \times 1140}}{2}\) = 23.8 cm
(or) -47.8 cm.
Since h cannot be negative, therefore h = 23.8 cm.
In vertical position it flows out (23.8) cm.

Question 12.
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³s^-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³s^-1. Identify the gas.
[Hint: Use Graham’s law of diffusion : R₁/R₂ = (M₂/M₁) where R₁, R₂ are diffusion rates of gases 1 and 2 and M₁ and M₂ their respective molecular masses. The law is a simple consequence of kinetic theory.]
Solution:
According to Graham’s law of diffusion
\(\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}\)
Where
r₁ = diffusion rate of hydrogen = 28.7 cm³s^-1
r₂ = diffusion rate of unknown gas = 7.2 cm³s^-1
M₁ = molecule wt. of hydrogen = 2u M2 = ?
M₂ = ?
∴ \(\frac{28.7}{7.2}=\sqrt{\frac{\mu_2}{\mu_1}} \text { or } M_2=\left(\frac{28.7}{7.2}\right)^2 \times 2\) = 31.78
≈ 32 µ

Question 13.
A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so- called law of atmospheres.
n₂ = n₁ exp [-mg(h₂ – h₁)/K_BT] Where n₂, n₁ refer to number density at heights h₂ and h₁ respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column :
n₂ = n₁ exp [-mg N_A (ρ – P) (h₂ – h₁)/(ρRT]
Where ρ is the density of the suspended particle and ρ that of surrounding medium. [N_A is Avogadro’s number and R the universal gas constant.]
[Hint: Use Archimedes principle to f ind the apparent weight of the suspended particle.]
Solution:
According to the law of atmospheres n₂ = n₁ exp. [-\(\frac{m g}{K_B T}\) (h₂ – h₁)] ………… (1)
Where n₂ n₁ refer to number density of particles at heights h₂ and h₁ respectively. If we consider the sedimentation equilibrium of suspended particles in a liquid, then in place of mg, we will have to take effective weight of suspended particles.
Let
v = average vol. of suspended particle
ρ = density of suspended particle
ρ’ = density of liquid
m = mass of one suspended particle
m’ = mass of equal vol. of liquid displaced
According to Archimedis principle, effective weight of one suspended particle.
= actual weight – weight of liquid displaced
= mg – m’ g
= mg – v P’g = mg – (m/P) P’g = mg \(\left(\frac{1-\rho^{\prime}}{\rho}\right)\)
Also boltzman constant, K_B = \(\frac{R}{N A}\) ………….. (i)
Where R is gas constant and NA is Avagadros number.
Putting mg \(\left(\frac{1-\rho^{\prime}}{\rho}\right)\) in place of mg and value of K_B in (i).
We get
n₂ = n₁ exp \(\left[-\frac{\text { mgNA }}{R T} \cdot\left(1-\frac{\rho^{\prime}}{\rho}\right)\left(h_2-h_1\right)\right]\)
Which is the required relation.

Question 14.
Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :

[Hint: Assume the atoms to be tightly packed1 in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of few Å].
Solution:
If r is the radius of the atom, then volume of each atom = \(\frac{4}{3}\) πr³
Volume of all atoms in one mole of substance
= \(\frac{4}{3}\) πr³ × N = μ/P
r = \(\left[\frac{3 \mu}{4 \pi \mathrm{PN}}\right]^{\frac{1}{3}}\)
For carbon p = 12.01 × 10^-3 kg,
P = 2.22 × 10³ kg m^-3
r = \(\frac{3 \times 12.01 \times 10^{-3}}{4 \times \frac{22}{7} \times\left(2.2210^3\right) \times\left(6.02310^{23}\right)}\)
r = 1.29 × 10^-10 m = 1.29A°
Similarly for gold
r = 1.59A°
For liquid nitrogen r = 1.77A°
For lithium r = 1.73A°
For liquid fluorine r = 1.88A°

Textual Examples

Question 1.
The density of water is 1000 kg m^-3. The density of water vapour at 100° C and 1 atm pressure is 0.6 kg m^-3. The volume of a molecule multiplied by the total number gives. What is called, molecular volume. Estimate the ratio (or fraction) of the molecular volume to the total volume occupied by the water vapour under the above conditions of temperature and pressure.
Answer:
For a given mass of water molecules, the density is less if volume is large. So the volume of the vapour is 1000/0.6 = /(6 × 10^-4) times, large. If densities of bulk water and water molecules are same, then the fraction of molecular volume to the total volume in liquid state is 1. As volume in vapour state has increased, the fractional volume is less by the same amount, i.e. 6 × 10^-4.

Question 2.
Estimate the volume of a water molecule using the data in Example 1.
Answer:
In the liquid (or solid) phase, the molecules of water are quite closely packed. The density of water molecule may therefore, be regarded as roughly equal to the density of bulk water = 1000 kgm^-3. We know that 1 mole of water has a mass approximately equal to (2 + 16)g = 18 g = 0.018 kg.
Since 1 mole contains about 6 × 10²³ molecules (Avogadro’s number).
The mass of a molecule of water is (0.018) / (6 × 10²³) kg = 3 × 10^-26 kg.
Volume of a water molecule = (3 × 10^-26 kg) (1000 kg m^-3)
= 3 × 10^-29 m³
= (4/3)π (Radius)³
∴ Radius » 2 × 10^-10 m = 2Å

Question 3.
What is the average distance between atoms (interatomic distance) in water ? Use the data given in Examples 1 and 2.
Answer:
A given mass of water in vapour state has 1.67 × 10³ times the volume of the same mass of water inliquid state (e.g. 1). When volume increases by 10³ times the radius increases by V^1/3 or 10 times,
i.e., = 10 × 12 = 2Å.
So the average distanceis 2 × 10 40 Å.

Question 4.
A vessel contains two non-reactive gases : neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is 3 : 2. Estimate the ratio of (i) number of molecules and (ii) mass density of neon and oxygen in the vessel. Atomic mass of Ne = 20.2 u, molecular mass of O₂ = 32.0 u.
Answer:
Each gas (assumed ideal) obeys the gas law. Since V and T are common to the two gases, we have P₁V = μ₁ RT and P₂V = μ₂ RT. i.e., (P₁ / P₂) = (μ₁ / p₂). Here 1 and 2 refer to neon and oxygen respectively.
Since (P₁ / P₂) = (3/2) (given), (μ₁ / μ₂) = 3/2.

i) By definition μ₁ = (N₁ /N_A) and μ₂ = (N₂/N_A) where N₁ and N₂ are the number of molecules of 1 and 2, and N_A is the Avogadro’s number. Therefore,
(N₁/N₂) = (μ₁/μ₂) = 3/2.

ii) We can also write μ₁ = (m₁/ M₁) and μ₂ = (m₂ / M₂) where m₁ and m₂ are their molecular masses. (Both m₁ and M₁; as well as m₂ and M₂ should be expressed in the same units). If ρ₁ and ρ₂ are the mass densities of 1 and 2 respectively, we helve
\(\frac{\rho_1}{\rho_2}=\frac{m_1 / V}{m_2 / V}=\frac{m_1}{m_2}=\frac{\mu_1}{\mu_2} \times\left[\frac{M_1}{M_2}\right]\)
= \(\frac{3}{2} \times \frac{20.2}{30.2}\) = 0.947

Question 5.
A flask contains argon and chlorine in the ratio of 2 : 1 by mass. The temperature of the mixtures is 27° C. Obtain the ratio of (i) average kinetic energy per molecule and (ii) root mean square speed v_rms of the molecules of the two gases. Atomtic mass of argon = 39.9 u. Molecular mass of chlorine = 70.9 u.
Answer:
The important point to remember is that the average kinetic energy (per molecule) of any (ideal) gas (be it monatomic like argon, diatomic like chlorine or polyatomic) is always equal to (3/2) k_BT. It depends only on temperature and is independent of the nature of the gas.
i) Since argon and chlorine both have the same temperature in the flask, the ratio of average kinetic energy (per molecule) of the two gases is 1 : 1.
ii) Now 1/2 mv_rms² = average kinetic energy . per molecule = (3/2) K_BT where m is the mass of a molecule of the gas. Therefore,
\(\frac{\left(\mathrm{V}_{\mathrm{rms}}^2\right)_{A \mathrm{C}}}{\left(\mathrm{V}_{\mathrm{ms}}^2\right)_{\mathrm{cl}}}=\frac{(\mathrm{m})_{\mathrm{C}}}{(\mathrm{m})_{\mathrm{Ar}}}=\frac{(\mathrm{M})_{\mathrm{Cl}}}{(\mathrm{M})_{\mathrm{Ar}}}=\frac{70.9}{39.9}\) = 1.77
where M denotes the molecular mass of the gas.
\(\frac{\left(\mathrm{V}_{\text {rms }}\right)_{\mathrm{Ar}}}{\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{cl}}}\) = 1.33

Question 6.
a) When a molecule (or an elastic ball) hits a (massive) wall, it rebounds with the same speed. When a ball hits a massive bat held firmly, the same thing happens. However, when the bat is moving towards the ball, the ball rebounds with a different speed. Does the ball move faster or slower ? (ch. 6 will refresh your memory on elastic collisions).
b) When gas in a cylinder is compressed by pushing in a piston, its temperature rises. Guess at an explanation of this in terms of kinetic theory using (a) above.
c) What happens when a compressed gas pushes a piston out and expands. What would you observe ?
d) Sachin Tendulkar uses a heavy cricket bat while playing. Does it help him in anyway ?
Answer:
a) Let the speed of the ball be u relative to the wicket behind the bat. If the bat is moving towards the ball with a speed V relative to the wicket, then the relative speed of the ball to bat is V + u towards the bat. When the ball rebounds (after hitting the massive bat) its speed, relative to bat, is V + u moving away from the bat. So relative to the wicket the speed of the rebounding ball is V + (V + u) = 2V + u, moving away from the wicket. So the ball speeds up after the collision with the bat. The rebound speed will be less than u if the bat is not massive. For a molecule this would imply an increase in temperature.
You should be able to answer (b) (c) and (d) based on the answer to (a).

Question 7.
A cylinder of fixed capacity 44.8 litres contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by 15.0° C ? (R = 8.31 J mol^-1 K^-1).
Answer:
Using the gas law PV = μRT, you can easily show that 1 mol of any (ideal) gas at standard temperature (273 K) and pressure (1 atm = 1.01 × 10⁵ Pa) occupies a volume of 22.4 litres. This universal volume is called molar volume. Thus the cylinder in this example contains 2 mol of helium. Further, since helium is monatomic, its predicted (and observed) molar specific heat at constant volume, C_v = (3/2) R, and molar specific heat at constant pressure, C_p = (3/2) R + R =(5/2) R. Since the volume of the cylinder is fixed, the heat required is determined by C_v.
Therefore,
Heat required = no. of moles × molar specific heat × rise in temperature
= 2 × 1.5 R × 15.0 = 45
R = 45 × 8.31 = 374 J.

Question 8.
Estimate the mean free path for water molecule in water vapour at 373 K. Use information from Exercises 1 and above. l = 2.9 × 10^-7 m = 1500d.
Ans. The d for water vapour is same as that of air. The number density is inversely proportional to absolute temperature.
So n = 2.7 × 10²⁵ × \(\frac{273}{373}\) = 2 × 10²⁵ m^-3
Hence, mean free path l = 4 × 10^-7 m.

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 13th Lesson Thermodynamics Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 13th Lesson Thermodynamics

Very Short Answer Questions

Question 1.
Define Thermal equilibrium. How does it lead to the Zeroth Law of Thermodynamics?
Answer:
If the temperatures of the two systems are equal, then they are said to be in thermal equilibrium.
Zeroth law of thermodynamics: “If two systems (A, B) are in thermal equilibrium with the third system (c) separately, then two systems (A, B) thermal equilibrium with each other.”

Question 2.
Define Calorie. What is the relation between calorie and mechanical equivalent of heat?
Answer:
Calorie : It is defined to be the amount of heat required to raise the temperature of 1g of water from 14.5°C to 15.5°C.
The relation between calorie and mechanical equivalent of heat is given by 1 cal = 4186 J Kg^-1K^-1 or 1 cal = 4.186 J g^-1K^-1.

Qeustion 3.
What thermodynamic variables can be defined by
a) Zeroth Law
b) First Law ?
Answer:
a) Temperature
b) Internal energy.

Question 4.
Define specific heat capacity of the substance. On which factors does it depend ?
Answer:
Specific heat capacity :
The amount of heat required to raise the 1 gm of substance through 1°c or 1 k is called specific heat capacity (or) the heat capacity per unit mass is called specific heat.
S = \(\left(\frac{1}{m}\right) \frac{\Delta Q}{\Delta T}\)
It depends on 1) natuture of the substance 2) temperature.

Question 5.
Define molar specific heat capacity.
Answer:
Molar specific heat capacity : The amount of heat required to raise the 1gm mole of substance through 1°C or 1°K is called molar specific heat capacity.

Question 6.
For a solid, what is the total energy of an oscillator ?
Answer:
For a mole of a solid, the total energy of an oscillator, U = 3K_B T × N_A = 3RT.

Question 7.
Indicate the graph showing the variation of specific heat of water with temperature. What does it signify ?
Ans.

Significance : the specific heat of water varies slightly with temperature in the range 0 to 100°C.

Question 8.
Define state variables and equation of state.
Answer:
State variables : The state of a system is described by pressure P, temparature T, density p (intensive variables) and internal energy U, Volume V, total mass M (Extensive variables). These are called state variables.

Equation of state : The connection between the state variables is called the equation of state.

Question 9.
Why a heat engine with 100% efficiency can never be realised in practise ?
Answer:
The efficiency of heat engine is η = 1 – \(\frac{Q_2}{Q_1}\)
For Q₂ = 0, η = 1, i.e., the engine will have 100% efficiency in converting heat into work.
The first law of thermodynamics does not rule out such an engine. But experience shows that such an ideal engine with η = 1 is never possible.

Question 10.
In summer, when the valve of a bicycle tube is opened, the escaping air appears cold.
Answer:
This happens due to adiabatic expansion of the air of the tube of the bicycle.

Question 11.
Why does the brake drum of an automobile get heated up while moving down at constant speed ?
Answer:
The work done by the brake drum on wheel is converted into heat due to friction.

Question 12.
Can a room be cooled by leaving the door of an electric refrigerator open ?
Answer:
No, a room can not be cooled by leaving the door of a refrigerator open, rather it will get slightly heated.

Question 13.
Which of the’ two will increase the pressure more, an adiabatic or an isothermal process, in reducing the volume to 50% ?
Answer:
Isothermal process obeys , P₁V₁ = P₂V₂
But V₂ = \(\frac{\mathrm{v_1}}{2}\) ⇒ P₁V₁, = P₂ \(\frac{\mathrm{v_1}}{2}\)
∴ P₂ = 2P₁
Adiabatic process obeys, P₁V₁^γ = P₂V₂^γ
P₂ = \(P_1\left[\frac{v_1}{v_2}\right]^\gamma=p_1\left[\frac{v_1}{\frac{v_1}{2}}\right]\)
∴ P₂ = 2^γ P₁ Where r = \(\frac{C_p}{C_v}\)
∴ pressure is more in adiabatic than isothermal.

Question 14.
A thermoflask containing a liquid is shaken vigorously. What happens to its temperature ? .
Answer:
Work is done by the liquid on the wall of flask, since it is vigorously shaken. Hence Internal energy and temperature of the liquid increases.

Question 15.
A sound wave is sent into a gas pipe. Does its internal energy change ?
Answer:
Yes, a sound wave is sent into a gas pipe internal energy increases.

Question 16.
How much will be the internal energy change in

1. isothermal process
2. adiabatic process

Answer:

1. Change in internal energy during isothermal process is dU = 0 [∵ U is constant)]
2. a) Change in internal energy during adiabatic compression increases,
   b) Change in internal energy during adiabatic expansion decreases.

Question 17.
The coolant in a chemical or a nuclear plant should have.high specific heat. Why ?
Answer:
In chemical and nuclear plants more heat is liberated. To absorb this heat, the coolant should have a property of high specific heat for small rise in temperature.
Since S = \(\frac{1}{m} \frac{d Q}{d t}\) ⇒ S ∝ \(\frac{1}{d t}\)

Question 18.
Explain the following processes

1. Isochoric process
2. Isobaric process

Answer:

1. Isochoric process : A process takes place at constant volume is called Isochoric process. During this process, no work is done on or by the gas, its internal energy and its temperature changes.
2.  Isoboric process : A process takes place at constant pressure is called isobaric process During this process temperature, internal energy changes. The heat absorbed goes partly to increase internal energy and partly to do work.

Short Answer Questions

Question 1.
State and explain first law of thermodynamics.
Answer:
Statement: The amount of heat supplied to system is equal to the algebraic sum of the change in internal energy of the system and the amount of external work-done.

Explanation : When ∆Q is the quantity of heat is supplied to system, a part of it is used to increase its internal energy ∆U and the rest is used in doing external work ∆W. The mathematical expressions of this law is ∆Q = ∆U – ∆W. It is a special case of law of conservation of energy.

Question 2.
Define two principal specific heats of a gas. Which is greater and why ?
Answer:
Two principal specific heats of a gas are

1. Molar specific heat capacity at constant pressure
2. molar specific heat capacity at constant volume.

1. Molar Specific heat capacity at constant pressure (C_p): The amount of heat required to raise the temperature of 1 gm – mole of a gas through 1°C at constant pressure is called molar specific heat at constant pressure.
i.e., C_p = \(\frac{1}{\mu} \frac{\Delta Q}{\Delta T}\) where μ is no of moles.

2. Molar Specific heat capacity at constant volume (C_v): The amount of heat required to raise the temperature of 1gm – mole of a gas through 1°C at constant volume is called molar specific heat at constant volume i.e., C_v = \(\frac{1}{\mu} \frac{\Delta Q}{\Delta T}\)

Explanation of C_p is greater that C_v: When a gas is heated at a constant pressure, it expands. The heat supplied to it is used partly in raising its temperature and partly in doing work against the external pressure. If, on the other hand, the gas is heated at constant volume, no work is done. Therefore, the heat supplied is to be used only in raising the temperature. Hence the amount of heat required to be supplied to a gas to raise its temperature by 1 °C (say) at constant pressure will be greater than the amount required at constant volume.

Question 3.
Derive a relation between the two specific heat capacities of gas on the basis of first law of thermodynamics
Answer:
The mathematical equation of first law of thermodynamics for 1 mole of gas is given by
∆Q = ∆U + P∆V
if ∆Q is absorbed heat at constant volume, ∆V = 0
C_v = \(\left[\frac{\Delta Q}{\Delta T}\right]_V=\left[\frac{\Delta U}{\Delta T}\right]_V=\left[\frac{\Delta U}{\Delta T}\right]\) ……………. (1)
Where the subscript V is dropped in the last step, since U of an ideal gas depends only on temperature. (The subscript denotes the quantity kept fixed.) If, on the other hand, ∆Q is absorbed heat at constant pressure,
C_p = \(\left[\frac{\Delta Q}{\Delta T}\right]_P=\left[\frac{\Delta U}{\Delta T}\right]_P+P\left[\frac{\Delta V}{\Delta T}\right]_P \Rightarrow\left[\frac{\Delta Q}{\Delta T}\right]_P=\left[\frac{\Delta U}{\Delta T}\right]+P\left[\frac{\Delta V}{\Delta T}\right]_P\) ………………… (2)
The subscript p.can be dropped from the first term since U of an ideal gas depends only on T. Now, for a mole of an ideal gas.
PV = RT
Which gives P\(\left[\frac{\Delta V}{\Delta T}\right]_P\) = R
Putting the equations (1) and (3) in (2), we get
C_p = C_v + R
∴ C_p – C_v = R

Question 4.
Obtain an expression for the work done by an ideal gas during isothermal change.
Answer:
Expression for the work done by an ideal gas during isothermal change : Let a certain mass of gas expands from volume V₁ to V₂ isothermally at constanst temperature T. Let the pressure changes form P₁ to P₂.
The work done during the small change in volume dv at a pressure ‘p’ is dw = pdv
The total work done during the change of volume from V₁ to V₂ is W = \(\int d w=\int_{V_1}^{V_2} p d v\) …………. (1)
The Isothermal change is represented by
PV = constant
PV = μRT ⇒ P =\(\frac{\mu \mathrm{RT}}{\mathrm{V}}\) where μ = no. of moles
substituting this value in equation (1)
W = \(\int_{V_1}^{v_2} p d v=\int_{V_1}^{V_2} \frac{\mu R T}{V} d V=\mu R T \int_{V_1}^{V_2} \frac{d v}{V}\)
W = μRT log_e \(\frac{V_2}{V_1}\)
∴ Work done during the isothermal change W = 2.3026 RT log₁₀ \(\frac{V_2}{V_1}\) .

Question 5.
Obtain an expression for the work done by an ideal gas during adiabatic change and explain.
Answer:
Expression for the work done by an ideal gas during adiabatic change :
During an adiabatic chnage the state of an ideal gas changes from (P₁, V₁, T₁) to (P₂, V₂, T₂)
The workdone during a small change in volume dV at a pressure p is dw = pdV
Total work done by gas from volume V₁ to V₂ is ,
W = \(\int\) dw = \(\int_{v_1}^{v_2}\) pdv ………………. (1)
The adiabatic relation between pressure and volume is given by PV^r = constant K …………. (2)
P = \(\frac{k}{v^r}\) …………… (3)
and P₁V₁^r = P₂V₂^r = K
Substituting eg. (3) in (1) becomes

This is equal to the work done during the adiabatic change.

Question 6.
Compare isothermal and an adiabatic process.
Answer:
Isothermal change

1. Changes in volume and pressure of a gas taking place at constant temperature are called isothermal changes,
2. Temperature of the gas remains constant.
3. The gas remains in good thermal contact with the surroundings and heat is exchanged.
4. Internal energy remains constant. Change in internal energy ∆U = 0
5. This process takes place slowly.
6. Boyle’s law PV = Constant holds good.
7. Work done W = RT log_e\(\frac{V_2}{V_1}\) .

Adiabatic change

1. Changes in volume and pressure of a gas taking place in a thermally isolated system are called adiabatic changes.
2. Temperature of the gas changes.
3. The gas is isolated from the surroundings and heat is not exchanged ∆Q = 0.
4. Internal energy changes.
5. This process takes place quickly.
6. PV^r = Constant
7. Work done W = \(\frac{R}{(r-1)}\)(T₁ – T₂)

Question 7.
Explain the following processes
i) Cyclic process with example
ii) Non cyclic process with example
Answer:
i) Cyclic process: “A process in which the system after passing through various stages (of pressure, volume and temperature changes) returns to its initial state1′ is called as a cyclic process.
For cyclic process P.V graph is a closed curve. The area under P-V graph gives work done by the substance. In a cyclic process there will be no change in the internal energy i.e., ∆U = 0
The first law of thermodynamics states that
∆Q = ∆U + ∆W
∴ For cyclic process ∆Q = ∆W
In cyclic process, the total heat absorbed by the system equals the work done by the system. Example : Fleat engine is a device by which a system is made to undergo a cyclic process that results in conversion of heat to work.

ii) Non cyclic process : A process which is not cyclic is called Non-cyclic process. In Non-cydic process, the system does not attains its initial state. Irreversible process is a Non cyclic process. Workdone in non-cyclic process is given by the area between the curve and volume axis.

Example :

1. Diffusion of liquids or gases
2. Free expansion of a perfect gas.

Question 8.
Write a short note on Quasi – static process.
Answer:
A quasi-static process is defined as “an infinitesimally slow process in which at each and every intermediate stage the system remains in thermal and mechanical (thermodynamic) equilibrium with the surroundings through out the entire process.”

In this process the pressure and temperature of the surroundings at each and every stage will differ from those of the system only in a very small amount.

Any process taking place sufficiently slowly, not involving acceleration motions and large temperature gradients can be considered as quasi -static process.
Example : isothermal process, adiabatic process.

Question 9.
Explain qualitatively the working of a heat engine.
Answer:
“A device used to convert heat energy into work (or mechanical energy)” is called heat engine. A heat engine converts heat into work.
A heat engine consists of the following essential parts :

1. A body at a higher temperature T₁. Heat Q₁ is extracted from this body and hence it is called a “hot reservior” (or source).
2. The engine contains working substance. In a steam engine the working substance is the steam. In a diesel engine the working substance is a mixture of fuel vapour and air.
3. A body at lower temperature T₂. Heat Q₂ is rejected by the working substance to this body and hence it is called a cold reservoir (or sink).

The difference in heat absorbed (Q₁) and heat rejected (Q₂) is equal to the workdone by the system (W). This is because the engine is operated in a cyclic process. W = Q₁ – Q₂.

The heat engine operates in a cyclic process and in each cycle, the work done (W) by the system is transferred to the environment via some arrangement (shaft) to drive vehicles.

The efficiency (η) of aheat engine is defined as the ratio of the workdone (W) by the engine to the amount of heat absorbed (Q₁) by the engine.
Efficiency (η) = \(\frac{\mathrm{W}}{\mathrm{Q}_1}=\frac{\mathrm{Q}_1-\mathrm{Q}_2}{\mathrm{Q}_1}=1-\frac{\mathrm{Q}_2}{\mathrm{Q}_1}\)

Long Answer Questions

Question 1.
Explain reversible and irreversible processes. Describe the working of Carnot engine. Obtain an expression for the effieciency. [AP – Mar. ’18, ’17, ’16, 15; TS – Mar. ’17, ’15, ’14]
Answer:
Reversible process : A process that can be retraced back in the opposite direction in such away that the system passes through the same states as in the direct process, and finally the system and the surroundings return in their original states, is called a reversible process.
A reversible process is only a purely idealised process.
Examples:

1. Slow isothermal and slow adiabatic changes
2. Peltier effect and seeback effect.
3. Fusion of ice and vaporisation of water.

Irreversible process : “A process that cannot be retraced back in the opposite direction” is called as irreversible process.

In this process the system does not pass through the same intermediate states as in the reversible process.
All natural processes such as conduction, radiation, radioactivity etc, are irreversible.
Example:

1. Work done against friction.
2. Heat produced in conductors by passing a current through it. (Joule heating)
3. Diffusion of gases.

Carnot Engine : A reversible heat engine operating between two temperatures is called a carnot engine. The cycle operating it is known as carnot’s cycle. In this cycle the working substance(say an ideal gas) is taken through a cycle by means of two isothermals and two adiabatics. The four operations are shown in P-V (Indicator) diagram.

Step 1 → 2 : Isothermal expansion of the gas taking its state from (P₁, V₁, T₁) to (P₂, V₂, T₂). It is shown in curve (a).
The heat absorbed by the gas (Q₁) from the reservoir at temperature
T1 equals the workdone by the gas. i.e.
W_{1 → 2} = Q₁ = μRT₁ log_e \(\frac{V_2}{V_1}\)

Step 2 →3 : Adiabatic expansion of the gas from (P₂, V₂, T₁) to (P₃, V₃, T₂). It is shown in curve (b). The work done by the gas is gÑen by
W_{2 → 3} = \(\frac{\mu R\left(T_1-T_2\right)}{(r-1)}\)

Step 3 → 4: Isothermal compression of the gas from (P₃, V₃, T₂) to (P₄, V₄, T₂). It is shown in curve (c).
Heat released (Q₂) by the gas to the reservoir at temperature T₂ equals the work done on the goes.
i.e., W_{3 → 4};= μRT₂ log_e \(\frac{V_4}{V_3}\) = -μRT₁ log_e \(\frac{V_3}{V_4}\)

Step 4 → 1: Adiabatic compression of the gas from (P₄, V₄, T₂) to (P₁, V₁, T₁). It is shown in curve (d) Work done on the gas is given by.
W_{4 → 1} = μR \(\frac{\left(T_2-T_1\right)}{(r-1)}=-\mu R \frac{\left(T_1-T_2\right)}{(r-1)}\)
Total work done by the gas in one complete cycle is
W = W_{1 → 2} + W_{2 → 3} + W_{3 → 4} + W_{4 → 1}
= \(\mu R T_1 \log _e \frac{V_2}{V_1}+\frac{\mu R\left(T_1-T_2\right)}{(r-1)}-\mu R T_2 \log _e \frac{V_3}{V_4}-\frac{\mu R\left(T_1-T_2\right)}{(r-1)}\)
∴ W = μRT₁ log_e \(\frac{V_2}{V_1}\) – μRT₂ log_e \(\frac{3}{V_4}\) = Q₁ – Q₂ …………… (2)
The efficiency of the cannot engine is
η = \(\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1}\) ………… (3)
Now since step 2 → 3 is an adiabatic process,
T₁V₂^{r – 1} ≈ T₂V₃^{r – 1}
Similarly, since step 4 → 1 is an adiabatic process,
T₂V₂^{r – 1} = T₁V₁^{r – 1} ⇒ T₁V₁^{r – 1} = T₂V₂^{r – 1}

Question 2.
State second law of thermodynamics. How is heat engine different form a refrigerator. [TS – Mar. ’18, ’16; AP – Mar. ’15, ’13]
Answer:
Second law of thermodynamics gives the direction of flow of heat. Second law consists two statements.

1. Kelvin – Planck statement: “No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work”.
   (or)
   “It is impossible to derive a continuous supply of engines in cooling a body below the coldest of its surroundings1′.
2. Clausius statement : “No process is possible whose sole result is the transfer of heat from a colder object to a hotter object”.
   (or)
   “Heat can not itself flow from cold body to hot body”.

Heat engine : A device which converts heat energy into work is called heat engine.
A heat engine consists of the essential parts.

1. Source or hot reservoir: It is maintained at a very high temperature T₁. Heat is extracted from this body.
2. Working substance : In a steam engine working substance is steam. In a diesel engine working substance is mixture of fuel vapour and air.
3. Sink or cold reservoir: It is maintained at a very low temperature T₂. It absorbs heat energy, rejected by working substance.

Work done: The difference of heat absorbed from source and heat rejected to sink is equal to work done by the engine, i.e., W = Q₁ – Q₂.
Efficiency : The ratio of workdone (W) by the engine to the amount of heat absorbed (Q₁) by the engine is called efficiency.
i.e., η = \(\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1}=1-\frac{Q_2}{Q_1}\)
Refrigerator : Refrigerator is a heat pump which is the reverse of a heat engine.

In refrigerator the working substance extracts (Q₂) an amount of heat from sink of lower temperature and a external work W is done on the working substance and finally amount of heat (Q₁) is transfered to source.
The coefficient of performance (α) of a refrigerator is given by
α = \(\frac{\mathrm{Q}_2}{\mathrm{~W}}=\frac{\mathrm{Q}_2}{\mathrm{Q}_1-\mathrm{Q}_2}\) (∴ heat released Q₁ = Q₂ + W)
For heat engine η can never exceed 1. For refregirator a can be greater than 1.
Therefore working of refrigerator is opposite to that of heat engine.

Problems

Question 1.
If a monoatomic ideal gas of volume 1 litre at N.T.P. is compressed (i) adia- batically to half of its volume, find the work done on the gas. Also find (ii) the work done if the cdmpression is iso- thermal, (γ = 5/3)
Solution:
i) During an adiabatic process
T₁V₁^{r – 1} = T₂V₂^{r – 1}
Here T₁ = 273 K
V₂ = \(\frac{V_1}{2}\)
r = \(\frac{5}{3}\)
T₂ = \(\left[\frac{V_1}{V_2}\right]^{\frac{5}{3}-1}=273\left[\frac{V_1}{V_1}\right]^{\frac{2}{3}}\)
T₂ = (2)^{\(\frac{2}{3}\)} = 273 = 431.6 K
Number of moles = \(\frac{1 \text { litre }}{22.4 \text { litre }}=\frac{1}{22.4}\)
Work done = \(\frac{\mu R}{(r-1)}\) [T₁ – T₂]
= \(\frac{8.314}{22.4\left[\frac{5}{3}-1\right]}\) × (273 – 431.6)
= \(\frac{8.314 \times 3}{22.4 \times 2}\) (-158.6)
= -89 J

ii) Work done during isothermal compression is
w = 2.3026 μRT log₁₀\(\frac{V_2}{V_1}\)
μ = Number of moles = \(\frac{1}{22.4}\)
T = 273K
R = 8.314 Jmol^-1K^-2
\(\frac{V_2}{V_1}=\frac{1}{2}\) = 0.5
∴ W = \(\frac{2.3026 \times 8.314 \times 273 \log _{10}(0.5)}{22.4}\)
Or w = -70J

Question 2.
Five moles of hydrogren when heated through 20 K expand by an amount of 8.3 × 10^-3m³ under a constant pressure of 10⁵ N/m². If C_v = 20.1/mole K, find C_p.
Solution:
Mayer’s relation C_p – C_v = R
Multiplying throughout by μ∆T
μC_p∆T – μC_v∆T = μ R∆T
μ ∆T(C_p – C_v) = P∆T [∴ μ R∆T = P∆V]
5 × 20 (C_p – 20) = 105 × 8.3 × (10 – 3)
[∴ μ = 5, ∆T = 20 K, P = 1 × 10⁵ N/m² C_v = 20 J/mole K and ∆V = 8.3 × 10³ M³]
C_p – 20 = 8.3
∴ C_p = 28.3 J/mole-K

Additional Problems

Question 1.
A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operate on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 10⁴ J/g ?
Solution:
Here volume of water heated = 3.0 lit/min.
Mass of water heated m = 3000 g/min
Rise in temperature ∆T = 77 – 27 = 50°C
Sp. heat of water C = 4.2 Jg^-1C^-1
Amount of heat used
∆Q = mc∆T = 3000 × 4.2 × 50
= 63 × 10⁴ J/min
Heat of combustion = 4 × 10⁴ J/g
Rate of combustion of fuel = \(\frac{63 \times 10^4}{4 \times 10^4}\)
= 15.75 g/min

Question 2.
What amount of heat must be supplied t 2.0 × 10^-2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure ? (Molecular mass of N₂ = 28; R = 8.3 J mol^-1 K^-1.)
Solution:
Here mass of gas m = 2 × 10^-2kg = 20g
Rise in temperature ∆T = 45°C
Heat required ∆Q = ?
Molecular mass M = 28
Number of moles n = \(\frac{m}{M}=\frac{20}{28}\) = 0.714
As nitrogen is a diatomic gas, molecular specific heat of constant pressure is
C_p = \(\frac{7}{2}\) R = \(\frac{7}{2}\) × 8.3 J mole^-1 K^-1
As ∆Q = nC_p∆T
∆Q = 0.714 × \(\frac{7}{2}\) × 8.3 × 45 J = 933.4 J

Question 3.
Explain why
a) Two bodies at different temperatures T₁ and T₂ if brought in thermal contact do not necessarily settle to the mean temperature (T₁ + T₂)/2.
b) The coolant in a chemical or nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
c) Air pressure in a car tyre increases during driving.
d) The climate of a harbour town is more temperate then that of a town in a desert at the same latitude.
Solution:
a) In thermal contact heat flows from the body at higher temp to the body at lower temperature till temperatures becomes equal. The final temperature can be the mean temp (T₁ + T₂)/2 only when thermal capacities of the two bodies are equal.

b) This is because heat absorbed by a substance is directly proportional to the specific heat of the substance.

c) During, driving the temperature of air inside the tyre increases due to motion. According to Charles law, P α T. Therefore, air pressure inside the tyre increases.

d) This is because in a harbour town, the relative humidity is more than in a desert town. Hence the climate of a harbour town is without extremes of hot and cold.

Question 4.
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume ?
Solution:
As no heat is allowed to be exchanged, the process is adiabatic.
∴ P₂V₂^r = P₁V₁^r (or) \(\frac{P_2}{P_1}=\left(\frac{V_1}{V_2}\right)^r\)
As V₂ = \(\frac{V_1}{2}\)
∴ P₂/P₁ = \(\left(\frac{V_1}{\frac{1}{2} V_1}\right)^{1 .4}\)
= 2^1.4 = 2.64

Question 5.
In changing the state of a gas adiabatically form an equilibrium state A to another equilibrium state B, and amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal. how much is the net work done by the system in the latter case ? (Take 1 cal = 4.19 J)
Solution:
Here, when the change is adiabatic, ∆Q = 0, ∆w = -22.3 J
If ∆u is change in internal energy of the system, then as
∆Q = ∆u + ∆w
O = ∆u – 22.3 (or) ∆u = 22.3 J
In the second case ∆Q = 9.35 cal
= 9.35 × 4.2 J = 39.3 J
∆w = ?
As ∆u + ∆w = ∆Q
∆w = ∆Q – ∆u
= 39.3 – 22.3 = 17.0 J

Question 6.
Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following :
a) What is the final pressure of the gas in A and B ?
b) What is the change in internal energy of the gas ?
c) What is the change in the temperature of the gas ?
d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface ?
Solution:
a) When the stopcock is suddenly opened, the volume available to the gas at 1 atm pressure will become two times. Therefore pressure will decrease to one-half i.e., 0.5 atm.

b) There will be no change in the internal energy of the gas as no work is done on/ by the gas.

c) Also there will be no change in temp of the gas as gas does no work in expansion.

d) No, because the process called free expansion is rapid and cannot be controlled. The intermediate states are non equilibrium states and do not satisfy the gas equation. In due course, the gas does return to an equilibrium state.

Question 7.
A steam engine delivers 5.4 × 10⁸J . of work per minute and services 3.6 × 10⁹J of heat per minute form its boiler, what is the efficiency of the engine ? How much heat is wasted per minute ? .
Solution:
Output i.e., useful workdone per min
= 5.4 × 10⁸ J
Input i.e., heat absorbed per min
= 3.6 × 10⁹ J
Efficiency = \(\frac{\text { Output }}{\text { Input }}=\frac{5.4 \times 10^8}{3.6 \times 10^9}\) = 0.15
= 0.15 × 100% = 15%
heat energy wasted / minute = Heat absorbed / minute – useful work done / minute
= 3.6 × 10⁹ – 5.4 × 10⁸ = 109 (3.6 – 0.54)
= 3.06 × 10⁹

Question 8.
An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing ?
Solution:
Heat supplied, ∆Q = 100 w = 100 J/s
Useful workdone ∆w = 75 J/s
Increase in internal energy/ sec, ∆u = ?
As ∆Q = ∆u + ∆w
∴ ∆u = ∆Q – ∆w
= 100 – 75 = 25 J/S

Question 9.
A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig.

Its volume is then reduced to the original value form E to F by an isobaric process. Calculate the total work done by the gas from D to E to F
Solution:
A is clear from fig.
Change in pressure dp = EF = 5.0 – 2.0
= 3.0 atm = 3.0 × 10¹⁵ Nm^-2
Change in volume dv = DF = 600 – 300
= 300 cc = 300 × 10^-6 m³
Workdone by the gas from D to E to F = area of ∆DEF
w = \(\frac{1}{2}\) × DF × EF
= \(\frac{1}{2}\) × (300 × 10^-6 ) × (3.0 × 10⁵ ) = 45 J

Question 10.
A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36°C. calculate the coefficient of performance.
Solution:
Here T₁ = 36°C = 36 + 273 = 309 K
T₂ = 10°C = 10 + 273 = 283 K
COP = \(\frac{T_2}{T_1-T_2}=\frac{283}{309-283}\)
= \(\frac{283}{26}\) = 10.9

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter

Very Short Answer Questions

Question 1.
Distinguish between heat and temperature. [T.S. Mar. 15]
Answer:
Heat

1. It is a form of energy.
2. It is the cause on the body.
3. It is determined with calorimeter.
4. It’s unit is Joule and Calories.

Temperature

1. It is a degree of hotness or coldness of a body.
2. It is an effect on a body.
3. It is measured with thermometer.
4. It’s unit is degree’Celsius or Kelvin or degree Fahrenheit.

Question 2.
What are the lower and upper fixing points in Celsius and Fahrenheit scales ? [T.S. Mar. 16]
Answer:
In Celsius scale, lower fixed point is ice point or 0°C and upper fixed point is steam point or 100°C. In Fahrenheit scale, lower fixed point is 32°F and upper fixed point is 212°F.

Question 3.
Do the values of coefficients of expansion differ, when the temperatures are measured on Centigrade scale or on Fahrenheit scale ?
Answer:
Yes. The coefficients of expansion depends on scale of temperature because
\(\frac{\alpha}{{ }^{\circ} \mathrm{C}}=\frac{9}{5} \alpha /{ }^{\circ} \mathrm{F}\)

Question 4.
Can a substance contract on heating ? Give an example.
Answer:
Yes. Rubber, type metal, cast iron contract on heating.

Question 5.
Why gaps are left between rails on a railway track ? [A.P. Mar. 17, 16]
Answer:
The length of the rails increases in summer due to high temperature. Therefore a gap is left to allow this expansion.

Question 6.
Why do liquids have no linear and areal expansions ?
Answer:
Liquids occupy the same shape of vessel. They do not have individual length and area. Hence, liquids have no linear and areal expansions.

Question 7.
What is latent heat of fusion ?
Answer:
The amount of heat per unit mass required to change a substance from solid into liquid at the same temperature and pressure is called the latent heat of fusion (L_f).

Question 8.
What is latent heat of vapourisation ? [Mar. 13]
Answer:
The amount of heat per unit mass required to change a substance from the liquid to the vapour state at the same temperature and pressure is called the latent heat of vaporisation (L_v).

Question 9.
What is specific gas constant ? Is it same for all gases ?
Answer:
Universal gas constant per molecular mass is called specific gas constant.
i.e., r = \(\frac{\mathrm{R}}{\mathrm{m}}\). No, it is different for different gases.

Question 10.
What are the units and dimensions of specific gas constant ?
Answer:
The units of specific gas constant is J/Kg/K. Dimensional formula is (m⁰.L²t^-2K¹).

Question 11.
Why utensils are coated black ? Why the bottom of the utensils are made of copper ?
Answer:

1. Utensils are coated black, because it is a good absorber of heat.
2. Copper is a good conductor of heat. So, copper is used at the bottom of cooking utensils.

Question 12.
State Weins displacement law. [A.P. Mar. 17]
Answer:
The wavelength (λ_m) corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature
i.e., λ_m ∝ \(\frac{1}{T}\)

Question 13.
Ventilators are provided in rooms just below the roof. Why ? [Mar. 14]
Answer:
Ventilators are provided in rooms just below the roof, because the hot air escapes out and fresh air enters into the rooms due to convection.

Question 14.
Does a body radiate heat at 0 K ? Does it radiate heat at 0°C ?
Answer:

1. No. A body does not radiate heat at 0k.
2. Yes. A body radiate heat at 0°C.

Question 15.
State the different modes of transmission of heat. Which of these modes require medium ?
Answer:
The different modes of transmission are : 1) conduction 2) convection and 3) radiation.
Among three modes conduction and convection require medium.

Question 16.
Define coefficient of thermal conductivity and temperature gradient.
Answer:
Coefficient of thermal conductivity : It is defined as the quantity of heat energy flowing per second between the opposite faces of cube of unit side, which are maintained at unit temperature difference.
Temperature gradient: The change of temperature per unit length of conductor is called temperature gradient.

Question 17.
What is thermal resistance of a conductor ? On what factors does it depend ?
Answer:
The resistance offered by the conductor for the flow of heat is called thermal resistance (R) of the conductor. R = \(\frac{\mathrm{d}}{\mathrm{kA}}\). It depends on

1. The nature of the material, i.e., thermal conductivity (K).
2. d; length of conductor along which heat flows and
3. A; area of cross section of the conductor.

Question 18.
State the units and dimensions of coefficient of convection.
Answer:
The unit of coefficient of convection is, wm^-2k^-1.
Dimensional formula (m.L⁰T³θ^-1).

Question 19.
Define emissive power and emissivity ?
Answer:
Emissive power : The energy flux emitted by unit surface area of a radiating body is known as emissive power (e_λ).

Missivity (e): It is defined as the ratio of the emissive power of the body to that of black body at the same temperature.

Question 20.
What is greenhouse effect ? Explain global warming. [A.P. Mar. 15, 13]
Answer:
Greenhouse effect: When the earth receives sun light, it gets heated up and emits infrared radiation into air. CO₂, CH₄, N₂O, O₃, Chlorofluoro carbon (green house gases) present in the air absorbs the heat content of infrared radiation and keeps the earth warm. This is called green house effect.

Global warming : As CO₂ content increases, more heat is retained in the atmosphere and the temperatures all over the world increases. This is called global warming.

Effects : a) Polar ice caps melt and fed more water in rivers and seas results in severe floods, b) In some areas, water resources dry up leading to drought conditions.

Question 21.
Define absorptive power of a body. What is the absorptive power of a perfect black body ?
Answer:
Absorptive power : At a given temperature and wavelength, the ratio of the amount of radiant energy absorbed to the amount of radiant energy incident in a wavelength range is called the absorptive power at that temperature and wavelength. Absorptive power of a perfect black body is 1.

Question 22.
State Newton’s law of cooling. [A.P. Mar. 18, 16; T.S. Mar. 18]
Answer:
The rate of loss of heat is directly proportional to the difference in temperature between the body and its surroundings provided the temperature difference is small. i.e., – \(\frac{\mathrm{dQ}}{\mathrm{dt}} \alpha\left(\mathrm{T}_B-\mathrm{T}_{\mathrm{S}}\right)\)

Question 23.
State the conditions under which Newton’s law of cooling is applicable.
Answer:

1. Loss of heat is negligible by conduction and only when it is due to convection.
2. Loss of heat occurs in a streamlined flow of air i.e., forced convection.
3. Temperature of the body is uniformly distributed over it.

Question 24.
The roof of buildings are often painted white during summer. Why ? [T.S. Mar. 17, 15]
Answer:
The roof of buildings are painted white during summer, because it reflects more heat radiations and absorbs less heat radiations. Hence we feel comfort inside the house.

Short Answer Questions

Question 1.
Explain Celsius and Fahrenheit scales of temperature. Obtain the relation between Celsius and Fahrenheit scales of temperature.
Answer:
Centigrade (Celsius) scale of temperature: In the Celsius scale of temperature, the lower fixed point is called the ice point and is assigned the value 0°C. The upper fixed point is called the steam point and is assigned the value 100°C. The interval between these two points (i.e., 100°C – 0 = 100°C) is subdivided into 100 equal parts each one corresponding to 1°C.

Fahrenheit scale of temperature : In the Fahrenheit scale of temperature, the lower fixed point is the ice point and is assigned a value 32°F and the upper fixed point is the staempoint and is assigned a value 212°F. The interval between these two points (i.e., 212°F – 32°F = 180°F) is subdivided into 180 equal parts, each one corresponding to 1°F.

Fahrenheit scale of temperature : In the Fahrenheit scale of temperature, the lower fixed point is the ice point and is assigned a value 32°F and the upper fixed point is the staempoint and is assigned a value 212°F. The interval between these two points (i.e., 212°F – 32°F = 180°F) is subdivided into 180 equal parts, each one corresponding to 1°F.

Relation between Celsius and Fahreinheit scales of temperature:
Difference of 100 Celsius degrees = Difference of 180 Fahrenheit degress
When the temperature of a body is measured on both the Celsius and Fahrenheit scales, let the readings be t_C and t_F respectively. Then
\(\frac{t_c-0}{t^{100}}=\frac{t_F-32}{180} \Rightarrow \frac{t_C}{5}=\frac{t_F-32}{9}\)
C = \(\frac{5}{9}\) (F – 32)

Question 2.
Two identical rectangular strips, one of copper and the other of steel, are riveted together to form a compound bar. What will happen on heating ?
Answer:
Bimetallic strip:

Two similar strips of copper and steel are revitted side by side as a compound strip called bimetallic strip. It is straight at ordinary temperatures. When the bimetallic strip is heated, copper expands more than steel. So, it bends with copper on convex side. When the bimetallic strip is cooled below room temperature, copper contracts more than steel and it bends with copper on concave side. Bimetallic strip is used in refrigerators, automatic iron, fire alarm etc.

Question 3.
Pendulum clocks generally go fast in winter and slow in summer. Why ?
Answer:
The time period of a pendulum at t₁°C is T₁ = 2π \(\sqrt{\frac{L_1}{g}}\) ………….. (1)
Where L₁ is length of pendulum at t₁°C.
If L₂ is length of pendulum at t₂°C,
then T₂ = 2π \(\sqrt{\frac{L_2}{g}}\) ………………. (2)
\(\frac{(2)}{(1)}\) gives \(\frac{T_2}{T_1}=\sqrt{\frac{L_2}{L_1}}=\sqrt{\frac{L_1(1+\alpha t)}{L_1}}\)
Where a is the linear coefficient of expansion of pendulum clock. Where t = t₂ – t₁
\(\frac{T_2}{T_1}=\sqrt{(1+\alpha t)}=(1+\alpha t)^{\frac{1}{2}}\)
\(\frac{T_2}{T_1}=1+\frac{\alpha t}{2}\)
\(\frac{T_2}{T_1}-1=\frac{\alpha t}{2}\)
\(\frac{T_2-T_1}{T_1}=\frac{\alpha t}{2}\)
\(\frac{T_2-T_1}{T_1}\) = time lost by pendulum per second.
Due to expansion in length, during summer, time period increases or the clock looses time in summer. In winter due to fall in temperature, the length contracts, time period decreases, hence clock gains time.

Question 4.
In what way is the anomalous behaviour of water advantageous to acquatic animals ? [A.P. Mar. 18, Mar. 14]
Answer:
In cold countries, as atmospheric temperature decreases, the upper layers of the lakes, rivers etc., cool, contract and sink to the bottom (fig). This goes on until the whole of the water reaches the temperature of 4°C. When the top layers cool further temperature falls below 4°C, it expands and becomes lighter. It does not sink downwards and remains at the top. With further cooling the top layer gradually form ice at the top (fig). Ice and water are bad conductors of heat. So the lower layers are protected against freezing by the layers of ice and cold water at 1°C, 2°C and 3°C. This results in water remaining at the bottom at 4°C. So that aquatic animals survive in those layers of water.

Question 5.
Explain conduction, convection and radiation with examples. [TS – Mar. ’18, ’16, ’15; AP – Mar. ’15]
Answer:
The heat is transmitted in three types. They are 1) Conduction 2) Convection 3) Radiation.

1. Conduction : The process of transmission of heat from one place to other without actual movement of the particles of the medium is called conduction.
   Ex : When long iron rod is heated at one end, heat transmits to the other end.
2. Convection : The process of transmission of heat from one place to another by the actual
   movement of the particles is called convection.
   Ex. : If water in a beaker is heated, the particles of water at the bottom receive the heat first. These particles expand, become lighter and rise up. At the same time colder and denser particles reach the bottom. They get in their turn heated and move up. This process is known as convection.
3. Radiation : The process of transmission of heat from one place to another without any intervening medium is called radiation.
   Ex. : Earth receives heat radiations from the sun.

Long Answer Questions

Question 1.
State Boyle’s law and charle’s law. Hence, derive ideal gas equation which of two laws is better for the purpose of Thermometry and why ?
Answer:
P ∝ \(\frac{1}{\mathrm{V}}\)
⇒ PV = constants ⇒ P₁V₁ – P₂V₂
Charles law at constant volume : At constant volume the pressure of given mass of a gas is directly proportional to absolute temperature of gas.
P ∝ T
⇒ \(\frac{P}{T}\) = constant ⇒ \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\)
Charles law at constant pressure : At constant pressure the volume of given mass of gas is directly proportional to absolute temperature.
V ∝ T
⇒ \(\frac{V}{T}\) = constant ⇒ \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\)
Ideal gas equation : A gas which obeys all the gas laws at all temperature and all pressures is called an ideal gas.

Consider are mole of a gas initially at a pressure P₁ volume V₁ and temperature T₁ and Let P₂, V₂ and T₂ be the final pressure, volume and absolute temperature T₂.
From Boyle’s law, P₁V₁ = P₂V₂
⇒ V = \(\frac{P_1 V_1}{P_2}\) ………….. (1)
Now the gas is heated at constant pressure P₂, then its volume changes from V to V₂ and temperature changes from T₁ and T₂.
From charles law, \(\frac{V}{T_1}=\frac{V_2}{T_2} \Rightarrow V=\frac{V_2 T_1}{T_2}\) ………….. (2)
From (1) & (2) \(\frac{P_1 V_1}{P_2}=\frac{V_2 T_1}{T_2} \Rightarrow \frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)
⇒ \(\frac{P V}{T}\) = constant ⇒ \(\frac{P V}{T}\) = R
Where ‘R’ is coniversal gas constant.
⇒ PV = RT
From ‘n’ gram moles PV = nRT. This law is called ideal gas equation.
Out of Boyle’s law, charles law is better for the purpose of thermometry because with increasing temperature, pressure and volume of gas also increase.

Question 2.
Explain thermal conductivity and coefficient of thermal conductivity. A copper bar of thermal conductivity 401 W/(mK) has one end at 104°C and the other end at 24°C. The length of the bar is 0.10 m and the cross – sectional area is 1.0 × 10^-6 m^-2. What is the rate of heat conduction, along the bar ?
Answer:
Thermal conductivity : Thermal conductivity of a solid body is ability to conduct heat in the body. Coefficient of thermal conductivity: The coefficient of thermal conductivity of a material may be defined as the quantity of heat energy flowing per second between the opposite faces of the cube of unit side, which are maintained at unit temperature difference.

Explanation of conduction : In steady state, conduction between the opposite faces which are at temperatures θ₁°C and θ₂°C separated by distance d is

1. Directly proportional to area of cross section of the rod.
   Directly proportional to temperature difference (θ₂ – θ₁) between the opposite faces.
2. Time of flow of heat, t
3. Inversely proportional to the separation of faces ‘d’.
   
   ∴ Q ∝ \(\frac{\mathrm{A}\left(\theta_2-\theta_1\right) t}{\mathrm{~d}}\)
   or Q = \(\frac{K A\left(\theta_2-\theta_1\right) t}{d}\)
   K is called coefficient of thermal conductivity of the material.
   When A = 1; (θ₂ – θ₁) = 1; t = 1; d = 1
   ∴ K = Q

Problem:
K_c = 401 W/mK, θ₂ = 104°C, θ₂ = 24°C, d = 0.10 m, A = 1.6 × 10^-6 m^-2
Rate of heat conduction = \(\frac{Q}{t}=K_c A \frac{\left(\theta_2-\theta_1\right)}{d}\)
= 401 × 1 × 10^-6 × (104-24) = 0.32 W

Question 3.
State and explain Newton’s law of cooling. State the conditions under which Newton’s law of cooling is applicable. A body cools down from 60°C to 50°C in 5 minutes and to 40°C in another 8 minutes. Find the temperature of the surroundings.
Answer:
Expression for Newton’s law of cooling : Consider a hot body at temperature T. Let T₀ be the temperature of its surroundings. According to Newton’s law of cooling,
Rate of loss of heat ∝ Temperature difference between the body and its surroundings.
\(\frac{-\mathrm{dQ}}{\mathrm{dt}}\) ∝ (T – T₀)
\(\frac{-\mathrm{dQ}}{\mathrm{dt}}\) = k (T – T₀) ⇒ (1) where k is proportionality constant. Let m be the mass and C the specific heat of the body at temperature T. If the temperature of the body falls by small amount dT in time dt, then the amount of heat lost is
dQ = mc dT
∴ Rate of loss of heat is given by
\(\frac{\mathrm{dQ}}{\mathrm{dt}}\) = mc \(\frac{\mathrm{dT}}{\mathrm{dt}}\) ………….. (2)
putting (2) in (1) ⇒
-mc \(\frac{\mathrm{dT}}{\mathrm{dt}}\) = k(T – T₀)
\(\frac{\mathrm{dT}}{\mathrm{dt}}=\frac{\mathrm{k}}{\mathrm{mc}}\) (T – T₀) = -K(T – T₀)
where K = \(\frac{\mathrm{k}}{\mathrm{mc}}\) is another constant.
⇒ \(\frac{d T}{\left(T-T_0\right)}\) = – K dt
on integrating both sides, we get
\(\int \frac{d T}{\left(T-T_0\right)}=-K \int d t\)
log_e (T – T₀) = – Kt + c …………… (3)
T – T₀ = e^{-Kt + c}
T = T₀ + e^c e^-Kt
T = T₀ + Ce^-Kt ……………… (4)
Where c is integration constant and C = e^c
equations (1), (2), (3) and (4) are the different expressions for Newton’s law of cooling.

Explanation of Graphs:
1) If we plot a graph by taking equation (4) different values of temperature difference ∆T = T – T₀ along y – axis and the corresponding values of t along X – axis, we get a curve of the form shown in figure. It clearly shows that the rate of cooling is higher initially and then decreases as the temperature of the body falls.
Curve showing cooling of hot water with time.

2) More over, the equation (3) is of the form y = mx + c. So if we plot a graph, by taking log_e(T – T₀) along Y – axis and time t along x – axis, we must get a straight line, as shown in figure. It has a negative slope equal to – K and intercept on Y – axis equal to C.
In both of the above situations, Newton’s law of cooling stands verified.

Newton’s law of cooling is applicable :

1. Loss of heat is negligible by conduction and only when it is due to convection.
2. Loss of heat occurs in a stream lined flow of air i.e., forced convection.
3. Temperature of the body is uniformly distributed over it.
4. Temperature differences are moderate i.e., upto 30 K, however if heat body is due to forced convection the law is valid for large differences of temperature also.
   
   ∴ Surrounding temperature, θ₀ = \(\frac{85^{\circ} \mathrm{C}}{3}\) = 28.33°C

Problems

Question 1.
What is the temperature for which the readings on Kelvin and Fahrenheit scales are same?
Answer:
The relation between Kelvin and Fahrenheit scales is given by
\(\frac{K-273.15}{100}=\frac{F-32}{180}\)
but K = F
\(\frac{F-273.15}{100}=\frac{F-32}{180}\)
F – 273.15 = \(\frac{5}{9}\) F – \(\frac{160}{9}\)
F – \(\frac{5}{9}\) F = 273.15 – \(\frac{160}{9}\)
= 273.15 – 17.77
\(\frac{4F}{9}\) = 255.38
∴ F = \(\frac{9}{4}\) (255.38)
= 574.6°F

Question 2.
Find the increase in temperature of aluminium rod if its length ¡s to be increased by 1%. (a for aluminium = 25 × 10^-6/0 C). [A.P (Mar. ‘15)]
Answer:
Percentage increase in length = \(\frac{l_2-l_1}{l_1}\) × 100
= α(t₂ – t₁) × 100 [∵ \(\frac{l_2-l_1}{l_1}\) = α(t₂ – t₁)]
Here, percentage increase in length = 1,
α = 25 × 10^-6/ °C
1 = 25 × 10^-6(t₂ – t₁) × 100
t₂ – t₁ = \(\frac{1}{25 \times 10^{-4}}=\frac{10^4}{25}\) = 400°C

Question 3.
How much steam at 100°C is to be passed into water of mass 100 g at 20°C to raise its temperature by 5°C ? (Latent heat of steam is 540 cal/g and specific heat of water is 1 cal/g°C).
Answer:
In the method of mixtures,
Heat lost by steam = Heat gained by water
m_SL_S + m_SS(100 – 1) = m_WS (t – 20)
Where m_S is the mass of steam, L_S is the latent heat of steam, S is the specific heat of steam and m_W is the mass of water.
Here, L_S = 540 cal/g
S = 1 cal/g°C;
m_W = 100 g
t = 20 + 5 = 25°C
m_S × 540 + m_S × 1 × (100 – 25) = 100 × 1 × (25 – 20)
615 m_S = 500
m_S = \(\frac{500}{615}\)
= 0.813 g.

Question 4.
2 kg of air is heated at constant volume. The temperature of air is increased from 293 K to 313 K. If the specific heat of air at constant volume is 0.718 kJ/kgK, find the amount of heat absorbed in kJ and kcal.
(J = 4.2 kJ/kcal).
Answer:
Here, M = 2 kg
dT = 313- 293 = 20 K
C_V = 0.718 × 10³ J/Kg – K
J = 4.2 KJ/Kcal
C_V = \(\frac{1}{\mathrm{~m}} \frac{\mathrm{d} \theta}{\mathrm{dt}}\)
dθ = mC_VdT
= 0.718 × 10³ × 2 × 20
= 28720 J
dθ = 28.72 KJ
dθ = \(\frac{28720}{4.2 \times 10^3}\)
= 6.838 Kcal.

Question 5.
A dock, with a brass pendulum, keeps correct time at 20°C, but loses 8.212 s per day, when the temperature rises to 30°C. Calculate the coefficient of linear expansion of brass.
Answer:
Here t₁ = 20°C, t₂ = 30°C
Loss of time per day = 8.212 sec
Loss of time per day = \(\frac{1}{2}\) α (t₂ – t₁) × 86,400
8.212 = \(\frac{1}{2}\) α (30 – 20) × 86,400
∴ α = \(\frac{16.424}{864 \times 10^3}\)
= 19 × 10^-6/C°

Question 6.
A body cools from 60°C to 40°C in 7 minutes. What will be its temperature after next 7 minutes if the temperature of its surroundings is 10°C ?
Answer:
Rate of cooling \(\frac{\mathrm{d} \theta}{\mathrm{dt}}\) = K(θ_av – θ₀)
Case (i) :
Given, .
θ₁ = 60°C, θ₂ = 40°C, θ₀ = 10°C, t = 7 min
⇒ \(\frac{60^{\circ} \mathrm{C}-40^{\circ} \mathrm{C}}{7}=K\left[\frac{60^{\circ} \mathrm{C}+40^{\circ} \mathrm{C}}{2}-10^{\circ} \mathrm{C}\right]\)
\(\frac{20}{7}\) = K[50 – 10]
⇒ \(\frac{20}{7}\) = K × 40
∴ K = \(\frac{1}{14}\) ……………. (1)

Case (ii) :
Given,
θ₁ = 40°C, θ₀ = 10°C, t = 7 min, θ₂ = ?
⇒ \(\frac{40-\theta_2}{7}=K\left[\frac{40+\theta_2}{2}-10\right]\)
\(\frac{40-\theta_2}{7}=\frac{1}{14}\left[\frac{40+\theta_2}{2}-10\right]\)
80 – 2θ₂ = \(\frac{40+\theta_2-20}{2}\)
160 – 4θ₂ = 20 + θ₂
5θ₂ = 140
∴ θ₂ = \(\frac{140}{5}\) = 28°C

Question 7.
If the maximum intensity of radiation for a black body is found at 2.65 pm what is the temperature of the radiating body ? (Wien’s constant = 2.9 × 10^-3 mK)
Answer:
λ_max = 2.65 pm = 2.65 × 10^-6m
Wien’s constant (b) = 2.90 × 10^-3 mk
λ_max T = b(constant)
T = \(\frac{b}{\lambda_{\max }}=\frac{2.9 \times 10^{-3}}{2.65 \times 10^{-6}}\)
= 1094 K.

Additional Problems

Question 1.
The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Answer:
Relation between kelvin scale and Celsius scale is T_C = T_K – 273.15
Where T_C.T_K = temperature on Celsius and kelvin scales respectively
For Neon T_C = 24.57 – 273.15 = – 248.58°C
For CO₂ T_C = 216.55 – 273.15 = -56.60°C
Relation between Kelvin and Fahrenheit scales is
\(\frac{T_F-32}{180}=\frac{T_K-273.15}{100}\)
T_F = \(\frac{180}{100}\) (T_K – 273.15) + 32
FQ or Neon T_F
= \(\frac{180}{100}\)(24.57 – 273.15) + 32 = – 415.44° F
FQ or CO₂ T_F = \(\frac{180}{100}\) (216.55 – 273.15) + 32
= – 69.88° F

Question 2.
Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between T_A and T_B ?
Answer:
Given triple point of water on scale A = 200
A triple point of water on scale B = 350 B
As per question 200 A = 350, B = 273.16 K
(Or)
IA = \(\frac{273.16}{200}\) K and IB = \(\frac{273.16}{350}\) K
If T_A and T_B represent the triple point of water on two scales A and B then
\(\frac{273.16}{200}\) T_A = \(\frac{273.16}{350}\)T_B (Or) \(\frac{T_A}{T_B}=\frac{200}{350}=\frac{4}{7}\)
(Or) T_A = \(\frac{4}{7}\) T_B

Question 3.
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:
R = R₀[1 + α(T – T₀)]
The resistance is 101.6 Ω at the triple-point of water 273.16 K and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω ?
Answer:
Here, R₀ = 101. 6 Ω, T₀ = 273.16 K
Case (i) R₁ = 165.5 Ω, T₁ = 600.5 K
Case (ii) R₂ = 123.4 Ω; T₂ = ?
Using the relation R = R₀(1 + α(T – T₀))
Case (i):
165.5 = 101.6(1 + α(600.5-273.16))
α = \(\frac{165.5-101.6}{101.6 \times(600.5-273.16)}\)
= \(\frac{63.9}{101.6 \times 327.34}\)

Case (ii): 123.4 = 101.6(1 + α(T₂ – 273.16))
(Or)
123.4 = 101.6
(1 + \(\frac{63.9}{101.6 \times 327.34}\) (T₂ – 273.16))
= 101.6 + \(\frac{63.9}{327.34}\) (T₂ – 273.16)
T₂ = \(\frac{(123.4-101.6) \times 327.34}{63.9}\) + 273.16
= 111.67 + 273.16 = 384.83 K

Question 4.
Answer the following :
a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale) ?
b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0°C and 100°C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale ?
c) The absolute temperature (Kelvin scale) T is related to the temperature t_C on the Celsius scale by
t_c = T – 273.15
Why do we have 273.15 in this relation and not 273.16 ?
Answer:
a) This is on account of the fact that the triple point of water has a unique value i.e., 273.16 K at a unique point, where exists unique values of pressure and volume. On the other hand, the melting point of ice and boiling point of water do not have unique set of value as they change in pressure and volume.

b) The other fixed point on the absolute scale is the absolute zero itself.

c) On Celsius 0°C corresponds to melting points of ice of normal pressure. The corresponding value of absolute temperature is 273.15 K. The temperature 273.16 K corresponds to the triple point of water.
From the given relation the corresponding value of triple point of water on Celsius scale
= 273.16 – 273. 15 = 0.01°C.

Question 5.
Two. ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made :

a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?
b) What do you think is the reason behind the slight difference in answers of thermoineters A and B ? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings ?
Answer:
a). Let T be the melting point of sulphur, for water T_tr = 273.16 K
For thermometer A, T = P/P_tr × 273.16
= \(\frac{1.797 \times 10^5}{1.250 \times 10^5}\) × 273.16 = 392.69 K
For thermometer B, T = P/P_tr × 273.16
= \(\frac{0.287 \times 10^5}{0.200 \times 10^5}\) × 273.16 = 391.98 K

b) The cause of slight different answers is that the oxygen and hydrogen gases are not perfectly ideal. To reduce this discrepancy, the readings should be at lower and lower pressure as in that case, the gases approach to the ideal gas behaviour.

Question 6.
A steel tape 1 m long is correctly calibrated for a temperature of 27.0°C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0°C. What is the actual length of the steel rod on that day ? What is the length of the same steel rod on a day when the temperature is 27.0°C ? Coefficient of linear expansion of steel = 1.20 × 10^-5 K^-1.
Answer:
Length of steell tape at 27°C is 100 cm i.e.
L = 100 cm and T = 27°C
The length of steel tape at 45°C is L¹ = L + ∆L
= L + αL∆T
= 100 + (1.20 × 10^-5) × 100 × (45° – 27)
= 100.0216 cm
Length of 1 cm mark at 27°C on this scale, at 45°C
= 100.0216/100 cm
Length of 63 cm measured’ by this tape at 45°C will be
= \(\frac{100.0216}{100}\) × 63 = 63.0136 cm
Length of the same steel rod on a day when the temperature is 27°C = 63 × 1 = 63 cm.

Question 7.
A large steel wheel is to be fitted on to a shaft of the same material. At 27°C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft ? Assume coefficient of linear expansion of the steel to be constant over the required temperature range :
α_steel = 1.20 × 10^-5K^-1.
Answer:
Here T₁ = 27°C = 27 + 273 = 300 K
Length at temperature, T₁K = L_T1 = 8.70 cm
Length at temperature, T₂K = L_T2 – 8.69 cm
Change in length = L_T2 – L_T1 = L_T2 α(T₂ – T₂)
Or 8.69 – 8.70 = 8.70 × (1.20 × 10^-5) × (T₂ – 300)
Or T₂ – 300 = \(\frac{0.01}{8.70 \times 1.2 \times 10^{-5}}\) = -95.8
Or T₂ = 300 – 95.8 = 204.2 K = -68.8°C

Question 8.
A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227°C ? Coefficient of linear expansion of copper = 1.70 × 10^-5K^-1.
Answer:
In this problem superfical expansion of copper sheet will be involved on heating. Here, area of hole at 227°C, then area of the hole at 227°C,
S₂ = \(\frac{\pi \mathrm{D}_2^2}{4}\)cm².
Coefficient of superficial expansion of copper is β = 2α = 2 × 1.70 × 10^-5 = 3.4 × 10^-5 c¹
Increase in area = S₂ – S₁ = βS₁ ∆T (or)
S₂ = S₁ + βS₁ ∆T
= S₁(1 + β∆T)
(Or)
\(\frac{\pi \mathrm{D}_2^2}{4}=\frac{\pi}{4}\) (4.24)² [1 + 3.4 × 10^-5 (228 – 27)]
(Or)
D₂² = (4.24)² × 1.0068
D₂ = 4.2544 cm
Change in diameter = D₂ – D₁
= 4.2544 – 4.24
= 0.0144 cm.

Question 9.
A brass wire 1.8 m long at 27°C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of -39°C, what is the tension developed in the wire, if its diameter is 2.0 mm ? Co-efficient of linear expansion of brass = 2.0 × 10^-5K^-1; Young’s modulus of brass = 0.91 × 10¹¹ Pa.
Answer:
Here L = 1.8m, T₁ = 27°C, T₂ = -39°C, r = 1
mm = 10^-3m, F = ?
α = 2 × 10^-5C^-1, y = 0.91 × 10¹¹ N/m²
From y = \(\frac{\mathrm{FL}}{a \Delta L}\), ⇒ ∆L = \(\frac{\mathrm{FL}}{\mathrm{ay}}\)
Also ∆L = αL∆T = ∴ \(\frac{\mathrm{FL}}{\mathrm{ay}}\) = αL∆T
(Or) F = α∆Tay = α(T₂ – T₁) πr²y
= 2 × 10^-5 × (-39 – 27) × (10^-3)² × 0.91 × 10¹¹
= -3.77 × 10^-2 N
Negative sign indicates that the force is inward due to contraction of the wire.

Question 10.
A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C ? Is there a ‘thermal stress’ developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2,0 × 10^-5K^-1, steel = 1.2 × 10^-5K^-1.
Answer:
∆L₁ = L₁α₁∆T = 50 × (2.10 × 10^-5) (250 – 40) = 0.2205 cm
∆L₂ = L₂α₂∆T
= 50 × (1.2 × 10^-5) (250 – 40) = 0.216 cm
∴ Change in length of combined rod
= ∆L₁ + ∆L₂
= 0.220 + 0.126 = 0.346 cm

Question 11.
The coefficient of volume expansion of glycerin is 49 × 10^-5K^-1. What is the fractional change in its density for a 30°C rise in temperature ?
Answer:
Here r = 49 × 10^-5C^-1, ∆T = 30°C .
As V¹ = V + ∆V = V(1 + r∆T)
∴ V¹ = V(1 + 49 + 10^-5 × 30) = 1.0147 V
As P = \(\frac{m}{V}\), p¹ = \(\frac{m}{V^1}\) = \(\frac{\mathrm{m}}{1.0147 \mathrm{~V}}\)= 09855 P
Fractional change in density = \(\frac{\rho-\rho^{\prime}}{\rho}\)
= \(\frac{\rho-0.9855 \rho}{\rho}\)
= 0.0145

Question 12.
A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 Jg^-1K^-1.
Answer:
Here p = 10 kw = 10⁵w,
Mass m = 8.0 kg = 8 × 10³ g
Rise in temp; ∆T = ?
time t = 2.5 min = 2.5 × 60 sec
Sp. heat C = 0.91 Jg^-1°C^-1
Total energy = p × t = 10⁴ × 150
= 15 × 10⁵ J
As 50% of energy is lost
∴ Energy available ∆Q = \(\frac{1}{2}\) × 15 × 10⁵
= 7.5 × 10⁵ J
As ∆Q = mc ∆T
∴ ∆T = \(\frac{\Delta Q}{\mathrm{mc}}=\frac{7.5 \times 10^5}{8 \times 10^3 \times 0.91}\) = 103°C

Question 13.
A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt ? (Specific heat of copper = 0.39 Jg^-1K^-1 heat of fusion of water = 335 Jg^-1.
Answer:
Here, mass of copper block m = 2.5 kg = 2500 kg
Fall in temperature ∆T = 500 – 0 = 500°C
Specific heat of copper C = 0.39 Jg^-1°C^-1
Latent heat of fusion L = 335 Jg^-1
Let the mass of ice melted by m¹
As heat gained by ice = Heat lost by copper
m^-1L = Mc∆T
m^-1 = \(\frac{\mathrm{mC} \Delta \mathrm{T}}{\mathrm{L}}\)
m¹ = \(\frac{2500 \times 0.39 \times 500}{335}\)
= 1500 g = 1.5 kg

Question 14.
In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27°C. The final temperature is 40°C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal ?
Answer:
Here, mass of metal, m = 0.20 kg = 200 g
Fall in temperature of metal ∆T = 150 – 40
= 110°C
If L is specific heat of metal, then heat lost by the metal
∆Q = mc∆T = 200 × L × 110
Volume of water = 150 C.C
Mass of water m¹ = 150 g
Water equivalent of calorimeter, w = 0.025 kg = 25 kg
Rise in temp of water and calorimeter
∆T¹ = 40 – 27 = 13°C
Heat gained by water and calorimeter,
∆Q¹ = (m¹ + w)T¹
= (150 + 25) × 13 = 175 × 13
As ∆Q = ∆Q¹
∴ From (i) and (ii)
200 × C × 110 = 175 × 13
C = \(\frac{175 \times 13}{200 \times 110}\) ≈ 0.1
(Or)
If some heat is lost to the sorroundings, value of L is so obtained will be less than the actual value of L.

Question 15.
Given below are observations on molar specific heats at room temperature of some common gases.

The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine ?
Answer:
The gases which are listed in the above table are diatomic gases and not mono atomic gases. For diatomic gases, molar specific heat
= \(\frac{5}{2}\) R = \(\frac{5}{2}\) × 1.98 = 4.95, which agrees fairly well with all observations listed in the . table except for chlorine. A monoatomic gas molecules has only the transnational motion. A diatomic gas molecules, a part from translation motion. The vibrational as well as rotational motion is also possible. Therefore to raise the temperature of 1 mole of diatomic gas through 1°C heat is also to be supplied increase not only translation energy but also rotational and vibrational energies. Hence molar specific heat of diatomic gas is greater than that for monoatomic gas. The higher value of molar specific heat of chlorine as compared to hydrogen, nitrogen, oxygen etc. Shows that for chlorine molecule, at room temperature vibrational motion. Also occurs along with translational and rotational motions. Where as other diatomic molecules at room temperature usually have rotational motion apart from their translational motion. This the reason that chlorine has some what larger value of molar specific heat.

Question 16.
Answer the following questions based on the P-T phase diagram of carbon dioxide :
a) At what temperature and pressure can the solid, liquid and vapour phases of CO₂ co-exist in equilibrium ?
b) What is the effect of decrease of pressure on the fusion and boiling point of CO₂ ?
c) What are the critical temperature and pressure for CO₂ ? What is their significance ?
d) Is CO₂ solid, liquid or gas at (a) – 70°C under 1 atm, (b) -60°C under 10 atm, (c) 15°C under 56 atm ?
Answer:
a) The solid, liquid and vapour phase of carbon dioxide exist in equilibrium at the point i.e., temperature = -56.6°C and pressure = 5.11 atm.

b) With the decrease in pressure, both the fusion and boiling point of carbondioxide will decrease.

c) For carbondioxide, the critical tempera-ture is 31,1°C and critical pressure is 73.0 atm. If the temp of CO₂ is more than 31.1°C. It cannot be liquified, however large pressure we may apply.

d) Carbondioxide will be (a) a vapour at – 70°C under 1 atm (b) a solid at 6°C under 10 atm (c) a liquid at 15°C under 56 atm.

Question 17.
Answer the following questions based on the P – T phase diagram of CO₂ :
a) CO₂ at 1 atm pressure and temperature – 60°C is compressed isothermally. Does it go through a liquid phase ?
b) What happens when CO₂ at 4 atm pressure is cooled from room tem-perature at constant pressure ?
c) Describe qualitatively the changes in a given mass of solid CO₂ at 10 atm pressure and temperature – 65°C as it is heated up to room temperature at constant pressure.
d) CO₂ is heated to a temperature 70°C and compressed isothermally. What changes in its properties, do you expect to observe ?
Answer:
a) Since the temp – 60°C lies to the left of 56.6°C on the curve i.e. lies in the region of vapour and solid phase. So CO₂ will condense directly into solid with out becoming liquid.

b) Since the pressure 4 atm is less than 5.11 atm. The carbondioxde will condenses directly into solid without becoming liquid.

c) When a solid CO₂ at 10 atm. Pressure and -65°C temp is heated, It is first converted into a liquid. A further increase in temperature brings it to the vapour phase. At P = 10 atm. If a horizontal line is drawn parallel to the T-axis. Then the points of intersection of this line with the fusion and vapourisation curve will give the fu-sion and boiling point of CO₂ at 10 atm.

d) Since 70°C is higher than the critical temperature of CO₂. So the CO₂ gas can not be converted into liquid state on being compressed isothermally at 70°C. It will remain in the vapour state. Nowever the gas will depart more and more now its perfect gas behavious with the increase in pressure.

Question 18.
A child running a temperature of 101 °F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 min, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water and latent heat of evaporation of water at that temperature is about 580 cal g^-1.
Answer:
Here fall in temp = ∆T = 101 – 98 = 3°F
= 3 × \(\frac{5}{3}\)°C = \(\frac{5}{3}\)°C
Mass of child M = 30 kg
Sp. heat of human body = Sp heat of water
C = 1000 cal.kg^-1C^-1
∴ Heat last by the body of child, ∆Q = mC∆T
= 30 × 1000 × \(\frac{5}{3}\) = 5000 calories
If m be the mass of water evapourated in 20 min then m¹L = ∆Q
m¹ = \(\frac{\Delta Q}{L}=\frac{5000}{580}\) = 86.2 g
∴ Average rate of extra evapouration = \(\frac{86.2}{20}\)
= 4.31 gmin^-1

Question 19.
A ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45°C and co-efficient of thermal conductivity of thermacole is 0.01 Js^-1m^-1 K^-1. [Heat of fusion of wafer = 335 × 10³ J kg^-1]
Answer:
Here length of each side l = 30 cm = 0.3 m
Thickness of each side ∆x = 5 cm = 0.05 m
Total surface area through which heat enters into the box
A = 6 l² = 6 × 0.3 × 0.3 = 0 Jum²
Temp.diff ∆T = 45 – 0 = 45°C,
K = 0.01 JS^-1m^-1°C^-1
Time ∆T = 6 hrs = 6 × 60 × 60 S
Latent heat of fusion L = 335 × 10³ J/kg
Let m be the mass of ice melted in this time
∆Q = mL = KA \(\left(\frac{\Delta \mathrm{T}}{\Delta \mathrm{x}}\right)\) ∆t
m = kA \(\left(\frac{\Delta T}{\Delta \mathrm{T}}\right) \frac{\Delta \mathrm{t}}{\mathrm{L}}\) = 0.01 × 0.54 × \(\frac{45}{0.05}\) × \(\frac{6 \times 60 \times 60}{335 \times 10^3}\) = 0.313 kg
Mass of ice left = 4 – 0.313 = 3.687 kg

Question 20.
A brass boiler has a base area of 0.15 m² and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 Js^-1m^-1 K^-1: Heat of vaporisation of water = 2256 × 10³ Jkg^-1.
Answer:
Were A = 0.15 m² ∆x = 1.0 m = 10^-2 m
\(\frac{\Delta Q}{\Delta t}=\frac{6 \times 10^3 \times 2256}{60}\) JS^-1 = 2256 × 10² JS^-1
K = 609 JS^-1m^-1°C^-1, ∆T = (t – 100)
From \(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\mathrm{KA}\left(\frac{\Delta \mathrm{T}}{\Delta \mathrm{x}}\right)\)
2256 × 10² = 609 × 0.15 \(\frac{(t-100)}{10^{-2}}\)
t – 100 = \(\frac{2256}{609 \times 0.15}\) = 2470 t
t = 24.70

Question 21.
Explain why :
a) a body with large reflectivity is a poor emitter.
b) a brass tumbler feels much colder than a wooden tray on a chilly day.
c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace.
d) the earth without its atmosphere would be inhospitably cold.
e) healting systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.
Answer:
a) This is because a body with large refelectivity is a poor absorber of heat and poor absorbers of heat are poor imilters.

b) When we touch a brass tumbler on a chill day; heat flows from our body to the tumbler quickly cas thermal conductivity or brass is very high and as a result, it appears colder. On the other hand as the wood is bad conductor, heat does not flow to the wooden tray from our body, on touching it.

c) When the red hot iron pieces is in the oven, its temperature TK is the given by the relation.
E = αT⁴ But if the red hot iron piece is in the open having the surrounding temperature T₀, its energy is radiated according to relation E¹ = α(T⁴ – T⁴₀). As the working principle of optical pysometer is based on the fact that the brightness of – a glowing surface of a body depends , upon its temperature. Therefore, pysometes gives too low a value for the temperature of red iron in the open.

d) The lower layer’s of Earth’s atmosphere reflect infrared radiations from earth back to the surface of the earth. Thus the heat radiation recieved by the earth from the sun during the day are kept trapped by the atm. If atmosphere of earth were not there, its surface would become too cold to live.

e) Steam at 100°C possess more heat than the same mass of water at 100°C possess 540 calories of heat more than possessed by 1 gm of ware at 100°C. That is why heating systems based on circulation of stream are more efficient than those based on circulation of hot water.

Question 22.
A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C.
Answer:
\(\frac{\mathrm{d} \theta}{\mathrm{dt}}\) = K [θ_av – θ₀]
\(\frac{80-50}{5}\) = K(65 – 20)
⇒ \(\frac{30}{5}\) = K × 45 = K = \(\frac{2}{15}\)
\(\frac{60-30}{t}=\frac{2}{15}[45-20]\)
\(\frac{30}{t}=\frac{2}{15} \times 15\)
t = 5 min
Time of cooling is given by t = \(\frac{2.3026}{k}\) log₁₀
\(\frac{T_1-T_0}{T_2-T_0}\)
As per question condition (i) T₁ = 80°C, T₂ = 50°C, T₀ = 20°C, t = 5 min
5 × 60 = \(\frac{2.3026}{K} \log _{10} \frac{80-20}{50-20}\)
= \(\frac{2.3026}{k} \log _{10^2}\) ……………. (2)
As per question condition (i) T₁ = 60°C, T₂ = 30°C, T₀ = 20°C, t = ?
t = \(\frac{2.3026}{K} \log _{10} \frac{60-20}{30-20}\)
= \(\frac{2.3026}{k} \log _{10^4}\) ……………. (3)
Div (3) by (2) we get
\(\frac{t}{5 \times 60}=\frac{\log _{10^4}}{\log _{10^2}}=\frac{0.6021}{0.3010}\) = 2
(Or) t = 5 × 60 × 2 = 10 × 60 s
= 10 min

Textual Examples

Question 1.
Show that the coefficient of areal expansions. (∆A/A)/∆T, of rectangular sheet of the solid is twice its linear expansivity, α₁.
Answer:
∆A₃ = (∆a) (∆b)

When the temperature increases by ∆T, a increases by ∆a = α₁a ∆T and b increases by ∆b = α₁b ∆T. From fig. the increase in area
∆A = ∆A₁ + ∆A₂ + ∆A₃
∆A = a ∆b + b ∆a + (∆a) (∆b)
= a α₁b ∆T + b α₁a ∆T + (α₁)² ab(∆T)²
= α₁ab ∆T(2 + α₁∆T)
= α₁A ∆T(2 + α₁∆T)
Since α₁ = 10^-5 K^-1, from Table 12.1, the product α₁∆T for fractional temperature is small in comparision with 2 and may be neglected.
Hence,
\(\left(\frac{\Delta \mathrm{A}}{\mathrm{A}}\right) \frac{1}{\Delta \mathrm{T}}\) ≈ 2α₁

Question 2.
A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diametbr of the rim and the iron ring are 5.243 m and 5.231 m respectively at 27°C. To what temperature should the ring be heated so as to fit the rim of the wheel ?
Answer:
Given,
T₁ = 27°C
L_T1 = 5.231 m
L_T2 = 5.243 m
So,
L_T2 = L_T1 [1 + α₁(T₂ – T₁)]
5.243 m = 5.231 m[1 + 1.20 × 10^-5 K^-1 (T₂ – 27°C)]
or T₂ = 218°C

Question 3.
A sphere of aluminium of 0.047 kg is placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100°C. It is then immediately transfered to 0.14 kg copper calorimeter containing 0.25 kg of water at 20°C. The temperature of water rises and attains a steady state at 23°C. Calculate the specific heat capacity of aluminium.
Answer:
Mass of aluminium sphere (m.,) = 0.047 kg
Initial temp, of aluminium sphere = 100°C
Final temp. = 23°C
Change in temp. (∆T) = (100°C – 23°C)
= 77°C
Let specific heat capacity of aluminium be s_Al. The amount of heat lost by the aluminium sphere = m₁s_Al ∆T = 0.047 kg × s_Al × 77°C s ……………. (i)
Mass of water (m₂) = 0.25 kg
Mass of calorimeter (m₃) = 0.14 kg
Initial temp, of water and calorimeter = 20°C
Final temp, of the mixture = 23°C
Change in temp. (∆T₂) = 23°C – 20°C = 3°C
Specific heat capacity of water (s_w) from table
12.3 = 4.18 × 10³ J kg^-1 K^-1
Specific heat capacity of copper calorimeter = 0.386 × 10³ J kg^-1 K^-1
The amount of heat gained by water and calorimeter = m₂s_w ∆T₂ + m₃s_cu ∆T₂ = (m₂s_w + m₃s_cu) (∆T₂)
= (0.25 kg × 4.18 × 10³ J kg^-1 K^-1 + 0.14 kg × 0.386 × 10³ J kg^-1 K^-1) (23°C – 20°C) ………………… (ii)
In the steady state heat lost by the aluminium sphere = heat gained by water + heat gained by calorimeter.
So, from (i) and (ii)
0.047 kg × s_Al × 77°C = (0.25 kg × 4.18 × 10³ J kg^-1 K^-1 + 0.14 kg × 0.386 × 10³ J kg^-1 K^-1) (3°C)
s_Al = 0.911 kJ kg^-1 K^-1

Question 4.
When 0.15 kg of ice of 0°C mixed with 0.30 kg of water at 50°C in a container, the resulting temperature is 6.7°C. Calculate the heat of fusion of ice.
(s_water = 4186 J kg^-1 K^-1)
Answer:
Heat lost by water = ms_w (θ_f – θ_i)_w
= (0.30 kg) (4186 kg^-1 K^-1) (50.0°C – 6.7°C)
= 54376.14 J
Heat required to melt ice = m₂L_f = (0.15 kg) L_f
Heat required to raise temperature of ice water to final temperature
= m_Is_w (θ_f – θ_i)_I
= (0.15 kg) (4186 J kg^-1 K^-1) (6.7°C – 0°C)
= 4206.93 J
Heat lost = heat gained
54376.14 J = (0.15 Kg)L_f + 4206.93 J
L_f = 3.34 × 10⁵ J kg^-1

Question 5.
Calculate the heat required to convert 3 kg of ice at -12°C kept in a calorimeter-to steam at 100°C at atmospheric pressure. Given specific heat capacity of ice = 2100 J kg^-1 K^-1, specific heat capacity of water = 4186 J kg^-1 K^-1, latent heat of fusion of ice = 3.35 × 10⁵ J kg^-1 and latent heat of steam = 2.256 × 10⁶ J kg^-1.
Answer:
We have
Mass of the ice, m = 3 kg
Specific heat capacity of ice, s_ice
= 2100 J kg^-1 K^-1
Specific heat capacity of water, s_water
= 4186 J kg^-1 K^-1
Latent heat of fusion ice, L_{f ice}
= 3.35 × 10⁵ J kg^-1
Latent heat of steam, L_steam
= 2.256 × 10⁶ J kg^-1
Now,
Q = Heat required to convert 3 kg of ice at -12°C to steam at 100°C.
Q₁ = Heat required to convert ice at – 12°C to ice at 0°C. .
ms_ice ∆T₁ = (3 kg) (2100 J kg^-1 K^-1) [0 – (-12)]°C = 75600 J
Q₂ = Heat required to melt ice at – 0°C to water at 0°C.
mL_{f ice} = (3 kg) (3.35 × 10⁵ J kg^-1) = 1005000 J
Q₃ = Heat required to convert water at 0°C to water at 100°C.
ms_W ∆T₂ = (3 kg) (4186 J kg^-1 K^-1) (100°C)
= 1255800 J
Q₄ = Heat required to convert water at 100°C to steam at 100°C.
mL_steam = (3 kg) (2-256 × 10⁶ J kg^-1)
= 6768000 J
So,
Q = Q₁ + Q₂ + Q₃ + Q₄
= 75600 J + 1005000 J + 1255800 J + 6768000 J
= 9.1 × 10⁶ J

Question 6.
What is the temperature of the steel- copper junction in the steady state of the system shown in fig. Length of the steel rod = 15.0 cm, length of the copper rod = 10.0 cm, temperature of the furnace = 300°C, temperature of the other end = 0°C. The area of cross section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel = 50.2 J s^-1m^-1K^-1 and of copper = 385 J s^-1m^-1K^-1].

Answer:
Let T be the temperature of the steel-copper junction in the steady state. Then,
\(\frac{K_1 A_1(300-T)}{L_2}=\frac{K_2 A_2(T-0)}{L_2}\)
Where 1 and 2 refer to the steel and copper rod respectively. For A₁ = 2, A₂, L₁ = 15.0 cm, L₂ = 10.0 cm, K₁ = 50.2 J s^-1m^-1K^-1,
K₂ = 385 J s^-1m^-1K^-1, we have
\(\frac{50.2 \times 2(300-\mathrm{T})}{15}=\frac{385 \mathrm{~T}}{10}\)
which gives T = 44.4°C

Question 7.
An iron bar (L₁ = 0.1 m, A₁ = 0.02 m², K₁ = 79 W m^-1 K^-1) and a brass bar (L₂ = 0.1 m, A₂ = 0.02 m², K₂ = 109 W m^-1K^-1) are soldered end to end as shown in fig. The free ends of the iron bar and brass bar are maintained at 373 K and 273 K respectively. Obtain expressions for and hence compute (i) the temperature of the junction of the two bars, (ii) the equivalent thermal conductivity of the compound bar and (iii) the heat current through the compound bar.

Answer:
Given,
L₁ = L₂ = L = 0.1 m, A₁ = A₂ = A = 0.02 m², K₁ = 79 W m^-1 K^-1, K₂ = 109 W m^-1K^-1, T₁ = 373 K and T₂ = 273 K.
Under steady state condition, the heat current (H₁) through iron bar is equal to the heat current (H₂) through brass bar.
So, H = H₁ = H₂
= \(\frac{K_1 A_1\left(T_1-T_0\right)}{L_1}=\frac{K_2 A_2\left(T_0-T_2\right)}{L_2}\)
For A₁ = A₂ = A and L₁ = L₂ = L this equation leads to
K₁(T₁ – T₀) = K₂(T₀ – T₂)
Thus the junction temperature T₀ of the two bars is T₀ = \(\frac{\left(K_1 T_1+K_2 T_2\right)}{\left(K_1+K_2\right)}\) ………….. (a)
Using this equation, the heat current H through either bar is

Using these equations, the heat current H’ through the compound bar of length L₁ + L₂ = 2L and the equivalent thermal conductivity K, of the compound bar are given by

Question 8.
A pan filled with hot food cools from 94°C to 86°C in 2 minutes when the room temperature is at 20°C. How long will it take to cool from 71°C to 69°C ?
Answer:
The average temperature of 94°C and 86°C is 90°C, which is 70°C above the room temperature. Under these conditions the pan cools 8°C in 2 minutes.
Using \(\frac{\mathrm{dT}_2}{\mathrm{~T}_2-\mathrm{T}_1}=-\frac{\mathrm{k}}{\mathrm{ms}}\) dt = – K dt we have
The average of 69°C and 71°C is 70°C, which is 50°C above room temperature. K is the same for this situation as for the original.

The average of 69°C and 71°C is 70°C, which is 50°C above room temperature. K is the same for this situation as for the original.
\(\frac{2^{\circ} \mathrm{C}}{\text { Time }}\) = K(50°C)
When we divide above two equations, we have
\(\frac{8^{\circ} \mathrm{C} / 2 \min }{2^{\circ} \mathrm{C} / \text { time }}=\frac{K\left(70^{\circ} \mathrm{C}\right)}{\mathrm{K}\left(50^{\circ} \mathrm{C}\right)}\)
Time = 0.7 min = 42 s

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 11th Lesson Mechanical Properties of Fluids Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 11th Lesson Mechanical Properties of Fluids

Very Short Answer Questions

Question 1.
Define average pressure. Mention its unit and dimensional formula. Is it a scalar or a vector? [A.P. Mar. 17]
Answer:
Average pressure (P_av) : Average power is defined as the normal force acting per unit area.
P_av = \(\frac{F}{A}\)
units → N/m² (or) pascal
Dimensional formula → [ML^-1 T^-2]
Pressure is a scalar quantity.

Question 2.
Define Viscosity. What is it’s units and dimensions?
Answer:
Viscosity: The property of a liquid that opposes the relative motion between its layers is called viscosity. .
G.G.S unit poise
S.I unit → Nm^-2s
Dimensional formula = [M¹L^-1T^-1]

Question 3.
What is the principle behind the carburetor of an automobile ? [AP – Mar. 15; TS – Mar. ’18, ’17]
Answer:
The carburetor of automobile has a venturi channel (nozzle) through which air flows with a large speed. The pressure is then lowered at the narrow neck and the petrol is sucked up in the chamber to provide the correct mixture of air to fuel necessary for combustion. ”

Question 4.
What is magnus effect ? [T.S. Mar. 16; A.P. Mar. 15]
Answer:
The difference in the velocities of air results in the pressure difference between the lower and upper faces and there is a net upward force on the ball. This dynamic lift due to spinning is called magnus effect.

Question 5.
Why are drops and bubbles spherical ? [A.P. Mar. 18, 17, 16, 14]
Answer:
The surface tension of a liquid tends to have minimum surface area. For a given volume, the surface area is minimum for a sphere. Hence rain drops are spherical shape.

Question 6.
Give the expression for the excess pressure in a liquid drop. [T.S. Mar. 17]
Answer:
Excess pressure in a liquid drop, p_i – p₀ = \(\frac{2 s}{r}\)
where s = Surface tension; r = Radius of the liquid drop.

Question 7.
Give the expression for the excess pressure in an air bubble inside the liquid.
Answer:
Excess pressure in an air bubble inside the liquid, p_i – p₀ = \(\frac{2 s}{r}\)
where s = Surface tension
r = Radius of the air bubble.
Air bubble forms inside the liquid, hence it has one liquid surface.

Question 8.
Give the expression for the excess pressure soap bubble in air. [T.S. Mar. 16]
Answer:
Soap bubble have two interfaces, hence excess pressure inside a soap bubble is p_i – p₀ = \(\frac{4 s}{r}\)
where s = Surface tension
r = Radius of the soap bubble.

Question 9.
What are water proofing agents and water wetting agents ? What do they do ?
Answer:
Water proofing agents are added, to create a large angle of contact between water and fibres.
Soaps, detergents and dying substances are wetting agents. When they are added, the angle of contact becomes small. So that they may penetrate well and become effective.

Question 10.
What is angle of contact ? [A.P. Mar. 16]
Answer:
The angle between tangent to the liquid surface at the point of contact and solid surface inside the liquid is termed as angle of contact (0).

Question 11.
Mention any two examples that obey Bernoullis theorem and justify them. [A.P. Mar. 18]
Answer:

1. In heavy winds house roof’s are blown off. When the velocity of the wind is greater on the roof top than inside the house, then the pressure on the roof top becomes less than that inside the house. This pressure difference causes the dynamic lift.
2. When a fan is rotating, papers are blown off from .the table top. The velocity of wind on the paper increases due to fan and hence pressure decreases. Due to this pressure difference papers are blown off.

Question 12.
When water flows through a pipe, which of the layers moves fastest and slowest ?
Answer:
Water flows through a pipe, the layers near the axis of the tube are fastest and at the walls of the tube are slowest

Question 13.
“Terminal velocity is more if surface area of the body is more”. Give reasons in support of your answer.
Answer:
Surface area (A) = 4πr² and terminal velocity (υ_t) αr²
As surface area increases, r2 is also increases. Then terminal velocity is also increases.
∴ Terminal velocity is more, if surface area of the body is more.

Short Answer Questions

Question 1.
What is atmospheric pressure and how is it determined using Barometer ?
Answer:
Atmospheric pressure : Atmospheric pressure at any point is equal to the weight of a column of air of unit cross sectional area extending from that point to the top of the earth’s atmosphere.
1 atm = 1.013 × 10⁵ pa
Determination of atmospheric pressure using barometer: A long glass tube closed at one end and filled with mercury is inverted into a trough of mercury. This device is known as mercury barometer.

The space above the mercury column in the tube contains only mercury vapour whose pressure Pis so small, that it may be neglected. The pressure inside the column at point A must equal the pressure at B.
∴ Pressure at B = Atmospheric pressure = P_a
P_a = ρgh = Pressure at A …………………… (1)
Where ρ is density of mercury and h is the height of the mercury column in the tube. In the experiment it is found that the mercury column in the barometer has a height of about 76cm at the sea level equivalent to one atmosphere.

Question 2.
What is guage pressure and how is a manometer used for measuring pressure differences ?
Answer:
Guage pressure : The guage pressure is the difference of the actual pressure and the atmospheric pressure. P_g = P – P_a
measurement of pressure differences :

1. The manometer consists of a U-shaped tube, which is filled with a low density liquid (oil) for measuring small pressure difference and high density liquid
   (mercury) for measuring large pressure difference.
2. One end of the tube is connected to the vessel D whose pressure of air measure and the other end of the tube is open.
3. If pressure of air in vessel D is more than the earth’s atmosphere, the level of liquid in arm I will go down up to point A and level of liquid in arm II rises up to C.
4. Then the pressure of air in vessel is equal to pressure at point A.
   
5. Note the difference of liquid levels in the two arms of U-tube.(Say h). p be density of the liquid. Pa is the atmospheric pressure.
6. Pressure at point A (P_A) = Pressure at point B = Pressure at point C + Pressure due to column of liquid.
   P_A = P_c + hρg (or) P_A – P_c = hρg
   Here P_c = P_a, P_A = P, ∴ P – P_a = hρg
   P – P_a = P_g = guage pressure = hρg

Question 3.
State Pascal’s law and verify it with the help of an experiment.
Answer:
Pascal’s law : It states that if gravity effect is neglected, the pressure at every point of liquid in equilibrium of rest is same.
Proof :

1. Imagine a circular cylinder of uniform cross-sectional area A, such that points C and D lie on flat faces of the cylinder.
2. The liquid inside the cylinder is in equilibrium under the action of forces exerted by the liquid outside the cylinder.
3. These forces are acting every where perpendicular to the surface of the cylinder.
4. Thus the forces on the flat faces of the cylinder at C and D will be perpendicular to the forces on the curved surface of the cylinder.
5. Since liquid is in equilibrium, the sum of the forces acting on the curved surface of the cylinder must be zero.
   
6. If P₁ and P₂ are the pressures at points C and D respectively. F₁ and F₂ are the forces acting on the flat surfaces of the cylinder due to liquid, then
   F₁ = P₁ A and F₂ = P₂A Since liquid is in equilibrium, therefore
   F₁ = F₂
   P₁ A = P₂A
   (or) P₁ = P₂
   It means the pressures at C and D are the same. This proves the pascal’s law.

Question 4.
Explain hydraulic lift and hydraulic brakes.
Answer:
Hydraulic lift and hydraulic brakes are based on the Pascal’s law.

Hydraulic lift : Here C and D are two cylinders of different areas of cross section. They are connected to each other with a pipe E. Each cylinder is provided with airtight frictionless piston. Let a, A be the area’s of cross-sections of the piston at C and D (a < < A)

The cylinders are filled with an incompressible liquid.
Let f be the applied force at C. Pressure exerted on the liquid
P = \(\frac{f}{a}\) ……………. (1)
According to pascal’s law, this pressure is transmitted to piston of cylinder D. Upward force at D will be
F = PA = \(\frac{f}{a}\) A = f \(\frac{A}{a}\) …………… (2)
As A > > a
∴ F > > f .
∴ Heavy load placed on the larger piston is easily lifted.

Hydraulic Brakes :
When the brake pedal is pressed, the piston (P) of the master cylinder is pushed inwards. There will be increased pressure on liquid at P, which is transmitted equally to P₁ and P₂ of wheel cylinder in accordance with pascal’s law. Due to which P₁ and P₂ move outwards. Breakshoes to move away from each other which in turn press against the inner rim of the wheel. The brake becomes operative.

Question 5.
What is hydrostatic paradox ?
Answer:
Consider three vessels A, B and C of different shapes. They are connected at the bottom by a horizontal pipe. On filling with the level in the three vessels is the same, though they hold different amounts of water. This is so because water at the bottom has the same pressure below each section of the vessel. It means the liquid pressure at a point is independent of the quantity of liquid but depends upon the depth of point below the liquid surface. This is known as hydrostatic paradox.

Question 6.
Explain how pressure varies with depth.
Answer:
Consider a fluid at rest in a container. In figure point 1 is at a height h above a point 2. The pressure at points 1 and 2 are P₁ and P₂. As the fluid is at rest, the horizontal forces should be zero. The resultant vertical forces should balance the weight of the element. Pressure at top (P₁A) acting downward, pressure at bottom (P₂A) acting upward.

(P₂ – P₁) A = mg ………………… (1)
Mass of the fluid (m) = ρv = ρhA
(P₂ – P₁) = ρgh ………………. (2)
Pressure difference depends on the vertical distance h.
If the point 1 under discussion is shifted to the top of the fluid (water), which is open to the atmosphere, P₁ may be replaced by atmospheric pressure (P_a) and P₂ by P. Then eq (2) gives
P – P_a = ρgh
P = P_a + ρgh ……………… (3)
Thus the pressure P, at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount ρgh.

Question 7.
What is Torricelli’s law ? Explain how the speed of efflux is determined with an experiment.
Answer:
Torricelli’s theorem : The velocity of efflux i.e., the velocity with which the liquid flows out of an orifice (i.e., a narrow hole) is equal to that which is freely falling body would acquire in falling through a vertical distance equal to the depth of orifice below the free surface of liquid.

Consider ideal liquid of density p contained in a tank provided with a narrow hole.
Let h = Height of free surface of liquid above O.
P = Atmospheric pressure
v = Velocity of efflux
Applying Bernoulli’s theorem at A and O
(P + ρgh + O)_atA = [P + 0 + \(\frac{1}{2}\)ρv²]_atO
P + ρgh = P + \(\frac{1}{2}\) ρv² = ρgh = \(\frac{1}{2}\)ρv²
v = \(\sqrt{2 g h}\)

Question 8.
What is Venturi-meter? Explain how it is used.
Answer:
Venturi meter: The Venturi-meter is a device to measure the flow speed of incompressible fluid.

1. It consists of a tube with a broad diameter and a small constriction at the middle.
2. A manometer in the form of a U-tube is also attached to it, with one of arm at the broad neck point of the tube and the other at constriction as shown in figure.
   
3. The manometer contains a liquid of density ρ_m.
4. The pressure difference causes the fluid in the U-tube connected at the narrow neck to rise in comparison to the other arm.
5. Filter pumps, sprayers used for perfumes, carburetor of automobile has used on this principle.

Question 9.
What is Reynolds number ? What is it’s significance ?
Answer:
Reynolds number : “Reynold number is a pure number which determines the nature of flow of liquid through a pipe”.
Reynold number (R_e) = \(\frac{\rho \mathrm{vd}}{\eta}\)
Where ρ is density of the fluid
v is speed of the fluid
d is diameter of the pipe

1. If the flow is stream line (or) laminor R_e < 1000
2. If the flow is turbulent, R_e > 2000
3. If the flow becomes unsteady, 1000 < R_e < 2000

Physical significance of Reynolds number : Reynold’s number describes the ratio of the inertial force per unit area to the viscous force per unit area for a flowing fluid.

Question 10.
Explain dynamic lift with examples.
Answer:
Dynamic lift: Dynamic lift is the force that act on a body, by virtue of its motion through a fluid.

e.g.1 : Fig (a) shows, ball moving without spin. Stream lines are equally distributed above and below the ball. The velocity above and below the ball is same resulting zero pressure difference. There no upward (or) downward force on the ball.

Fig (b) shows, ball moving with spin stream lines are more crowded above the ball than below. The velocity of air above the ball is large (v + v_r) and below it is smaller (v – v_r). As a result, there is a pressure difference between lower and upper faces. Pressure is less at top of the ball and pressure is morebelow the ball. There is a net upward force on the ball.
e.g. 2 : Dynamic lift also acts on the an aeroplane wing.

Question 11.
Explain Surface Tension and Surface energy. [Mar. 13]
Answer:
Surface tension (S): The force acting per unit length of an imaginary line drawn on the surface of a liquid, normal to it and parallel to the surface is called surface tension.
T = \(\frac{\mathrm{F}}{l}\)
S.l unit → N/m
D.F → [MT^-2]
Surface energy (E): The additional potential energy due to molecular forces per unit surface area is called surface tension.
Surface energy = \(\)
S.l Unit → J/m²
D.F → (MT^-2 )

Consider a horizontal liquid film ending in bar free to slide over parallel guides. We move the bar by a small distance d. The area of the surface increases, the system now has more energy, this means – that some work has to be done against an internal force F.
Work done (W) = F.d
If the surface energy of the film is S per unit area, the extra area is 2dl. A film has two sides and the liquid in between them.
So there are two surfaces and the extra energy is
S (2dl) = Fd
S = \(\frac{\mathrm{F}}{2 l}\)
Surface tension is equal to the surface energy and is also equal to the force per unit length exerted by the fluid on the movable bar.

Question 12.
Explain how surface tension can be measured experimentally.
Answer:
A flat vertical glass plate, below which a vessel of some liquid is kept, forms one arm of the balance. The plate is balanced by weights on the otherside, with its horizontal edge just over water. The vessel is raised slightly till the liquid just touches the glass plate and pulls it down a little because of surface tension. Weights are added till the plate just clears water. Suppose the additional weight requires is W.

Surface tension of liquid air interface is
S_la = \(\frac{\mathrm{W}}{2 l}=\frac{\mathrm{mg}}{2 l}\)
Where l is length of the plate edge, m is extra mass.

Long Answer Questions

Question 1.
State Bernoulli’s principle. From conservation of energy in a fluid flow through a tube, arrive at Bernoulli’s equation. Give an application of Bernoulli’s theorem.
Answer:
Bernoulli’s principle : Bernoulli principle states that in a stream line flow, the sum of the pressure, the K.E per unit volume and the P.E per unit volume remains a constant.
P + \(\frac{1}{2}\) ρv² + ρgh = constant

1. Consider a non-viscous, incompressible fluid is flowing through the pipe in a steady flow.
2. Consider the flow at two regions BC and DE. Initially the fluid lying between B and D.
   
3. During short time interval At, this fluid would have moved. Suppose V₁ is the speed at B and V₂ is the speed at D.
4. In time ∆t, distance moved from B to C is V₁ ∆t in the same interval (∆t) distance moved from D to E is V₂∆t.
5. Let P₁ and P₂ be pressure act at area’s of cross¬sections A₁ and A₂ of the two regions.
6. The work done on the fluid at left end (BC)
   = Force × displacement
   = Pressure × Area × displacement
   = P₁A₁ × V₁∆t (∆V = A₁V₁∆t) = P₁∆V ………………. (1)
7. Similarly work done by the fluid at right end (DE)
   = P₂A₂ × V₂∆t = P₂∆V …………………. (2)
8. Work done on the fluid is taken as positive and workdone by the fluid is taken as negative.
   ∴ Total work done (W) = (P₁ – P₁) ∆V ………………. (3)
   Part of this work goes into changing the K.E of the fluid and part goes into changing gravitational P.E.
9. Mass of the fluid(∆m) passing through the pipe in time (∆t) is ∆m = ρA₁V₁∆t
   where ρ is the density of the fluid.
   ∆m = ρ∆V ………………….. (4)
10. Gravitational P.E = ρg∆V (h₂ – h₁) ………………… (5)
    Change in K.E (∆K) = \(\frac{1}{2}\) ρ∆V (V₂² – V₁²) …………….. (6)
11. According to law of conservation of energy
    (P₁ – P₂) ∆V = \(\frac{1}{2}\) ρ ∆V (V₂² – V₁²) + ρg∆V (h₂ – h₁)
    P₁ – P₂ = \(\frac{1}{2}\) ρ (V₂² – V₁²) + ρg (h₂ – h₁)
    P₁ + \(\frac{1}{2}\) ρ V₁² + ρgh₁ = P₁ + \(\frac{1}{2}\) ρ V₂² + ρgh₂
    P + \(\frac{1}{2}\) ρ v² + ρgh = constant ………………. (7)

∴ Sum of the pressure, K.E per unit volume and P.E per unit volume remains constant. Application of Bernoulli’s theorem :

1. It explains the dynamic lift on the wings of aeroplanes.
2. It explains the dynamic lift experienced by a spinning cricket ball.

Question 2.
Define coefficient of viscosity. Explain Stoke’s law and explain the conditions under which a rain drop attains terminal velocity, υ_t. Give the expression for υ_t.
Answer:
Coefficient of viscosity (η): The coefficient of viscosity is defined as the tangential force per unit area of the layer, required to maintain unit velocity gradient.
η = \(\frac{F}{A\left(\frac{\Delta V}{\Delta x}\right)}\)
S.l. unit → Nm^-2 s (or) P_aS
C.G.S unit → Poise
Dimensional formula = [ML^-1T^-1]
Stoke’s law : According to this law the viscous force acting on a moving body which is spherical in shape is directly proportional to

1. Coefficient of viscosity of fluid (η)
2. Radius of the spherical body (r)
3. Velocity of the body (v)

∴ F ∝ ηrv
F = Kηrv
Where K is a constant of proportionality. Experimentally it was found to be 6π.
∴ F = 6πηrv
When rain drops falling through air from the clouds reach the surface with almost constant speed through they are moving under gravitational force. This velocity is called terminal velocity. After attaining the terminal velocity, net force acting on the rain drop is zero.
According to stokes law, F ∝ ηrv
F = 6πηrv (∵ 6π = K = Proportionality constant)
Let ρ, r be the density and radius of the sphere.
The fluid density be σ.
The forces acting on the sphere are

1. Weight of the sphere W = mg
   W = Vρg = \(\frac{4}{3}\)πr³ρg …………….. (1)
2. The force of buoyancy (B) = V σ g = \(\frac{4}{3}\)πr³σg ……………. (2)
3. Viscous force (f) = 6πηv ………………… (3)
   When the sphere attains terminal velocity (v_t), the net force on the body becomes zero.
   
   ∴ At terminal velocity
   Net downward force = Net upward force
   W = B + f, W – B = f
   \(\frac{4}{3}\)πr³ρg – \(\frac{4}{3}\)πr³σg = 6πηrv_t;
   ∴ \(\frac{4}{3}\)πr³g (ρ – σ) = 6πηrv_t
   V_t = \(\frac{2}{9} \frac{r^2 g(\rho-\sigma)}{\eta}\) ………………. (4)

Problems

Question 1.
Calculate the work done in blowing a soap bubble of diameter 0.6 cm against the surface tension force. (Surface tension of soap solution = 2.5 × 10² Nm^-1).
Solution:
D = 0.6 cm = 0.6 × 10² m
r = \(\frac{D}{2}=\frac{0.6 \times 10^{-2}}{2}\) = 0.3 × 10² m
S = 2.5 × 10² N/m
W = 8πr²s
= 8 × 3.14 × (0.3 × 10^-2)₂ × 2.5 × 10^-2
W = 5.652 × 10^-6 J

Question 2.
How high does methyl alcohol rise in a glass tube of diameter 0.06 cm ? (Surface tension of methyl alcohol = 0.023 Nm^-1 and density = 0.8 gmcm^-3. Assume that the angle of contact is zero)
Solution:
D = 0.06cm, θ = 0°
r = \(\frac{D}{2}=\frac{0.06}{2}\) = 0.03 cm = 0.03 × 10^-2 m
= 3 × 10^-4 m
S = 0.023 N/m, Density = 0.8 gm/c.c.
= 0.8 × \(\frac{10^{-3}}{10^{-6}}\)
ρ = 0.8 × 10³ kg/m³
S = \(\frac{\mathrm{hr \rho g}}{2 \cos \theta}\)
h = \(\frac{2 \cos \theta}{r \rho g}\)
= \(\frac{2 \times 0.023}{3 \times 10^{-4} \times 0.8 \times 10^3 \times 9.8}\) (∵ cos 0° = 1)
= 0.0019 × 10^-1 m
≈ 0.002 m
h = 2 cm .

Question 3.
What should be the radius of a capillay tube if water has to rise to a height pf 6 cm in it ? (Surface tension of water = 7.2 × 10^-2 Nm^-1).
Solution:
h = 6 × 10^-2 m
S = 7.2 × 10^-2 N/m
Density of water (ρ) = 10³ kg/m³
S = \(\frac{\mathrm{hr \rho g}}{2}\)
r = \(\frac{2 \mathrm{~S}}{\mathrm{h \rho g}}=\frac{2 \times 7.2 \times 10^{-3}}{6 \times 10^{-2} \times 10^3 \times 9.8}\)
r = \(\frac{14.4}{58.8} \) × 10^-3
r = 0.24 × 10^-3 m
r = 0.24 mm

Question 4.
Find the depression of the meniscus in the capillary tube of diameter 0.4 mm dipped in a beaker containing , mercury. (Density of mercury = 13.6 × 10³ Kg m^-3 and surface tension of mercury = 0.49 Nm^-1 and angle of contact = 135°).
Solution:
D = 4 m.m
r = \(\frac{D}{2}=\frac{4}{2}\) = 2m.m = 2 × 10^-3 m
ρ = 13.6 × 10³ kg/m³
θ = 135°, S = 0.49 Nm
cosθ = cos 135°
= – sin 45° = – \(\frac{1}{\sqrt{2}}\)
S = \(\frac{\mathrm{hr \rho g}}{2 \cos \theta}\)
h = \(\frac{2 \mathrm{~s} \cos \theta}{\mathrm{r \rho g}}\)
= 2 × 0.49 \(\left(\frac{-1}{\sqrt{2}}\right) \frac{1}{2 \times 10^{-3} \times 13.6 \times 10^3 \times 9.8}\)
h = \(\frac{-0.49}{13.6 \times 9.8 \times \sqrt{2}}\)
h = -0.024m.

Question 5.
If the diameter of a soap bubble is 10 mm and its surface tension is 0.04 Nm^-1, find the excess pressure inside the bubble. [Mar. 14]
Solution:
D = 10 m.m
r = \(\frac{D}{2}=\frac{10}{2}\) = 5 m.m
= 5 × 10^-3
S = 0.04 N/m
P_i – P₀ = \(\frac{4 S}{r}=\frac{4 \times 0.04}{5 \times 10^{-3}}\)
= 0.032 × 10³
P_i – P₀ = 32 N/m² (or) Pascal.

Question 6.
If work done by an agent to form a bubble of radius R is W, then how much energy is required to increase its radius to 2R ?
Solution:
R₁ = R, R₂ = 2R
Initial work done (W) = 8πR²S
Final work done (W) = 8π[R₂² – R₁²]S
= 8π [4R² – R²]S
= 3 × 8π R²S
W = 3W.

Question 7.
If two soap bubbles of radii R₁ and R₂ (in vacuum) coalasce under isothermal conditions, what is the radius of the new bubble. Take T as the surface tension of soap solution.
Solution:
R₁, R₂ and R be the radii of first, second and resultant bubble. The soap bubbles coalesce in vacuum, so surface energy do not change
E = E₁ + E₂
8π R²T = 8π R₁²T + 8π R₂² T
R² = R₁² + R₂²
R = \(\sqrt{R_1^2+R_2^2}\)

Additional Problems

Question 1.
Explain why
a) The blood pressure in humans is greater at the feet than at the brain.
b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km.
c) Hydrostatic pressure is a scalar quantities even though pressure is force divided by area.
Answer:
a) The height of the blood column in the human body is more at feet than at the brain. That is why, the blood exerts more pressure at the feet than at the brain (∴ pressure = hρg)

b) We know that the density of air is maximum near the surface of earth and decreases rapidly with height and at a height of about 6 km, it decreases to near by half its value at the sea level. Beyond 6 km height the density of air decreases very slowly with height. Due to this reason, the atmospheric pressure at a height of about 6 km decreases to nearby half of its value at sea level.

c) Since due to applied force on liquid, the pressure is transmitted equally in all directions inside the liquid. That is why there is no fixed direction for the pressure due to liquid. Hence hydrostatic pressure is a scalar quantity.

Question 2.
Explain why
a) The angle of contact of mercurywith glass is obtuse, while that of water with glass is acute.
b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.) [T.S. Mar. 15]
c) Surface tension of a liquid is independent of the area of the surface.
d) Water with detergent disolved in it should have small angles of contact.
e) A drop of liquid under no external forces is always spherical in shape.
Answer:
a) When a small quantity of liquid is poured on solid, three interfaces, namely liquid- air, solid-air and solid-liquid are formed. The surface tensions corresponding to these three interfaces are S_LA, S_SA and S_SL respectively. Let 0 be the angle of contact between the liquid and solid. The molecules in the region, where the three interfaces meet are in equilibrium. It means net force acting on them is zero. For the molecule at 0 to be in equilibrium, we have.

In case of mercury glass, S_SA < S_SL, therefore cosθ is negative or θ > 90° i.e. obtuse. In case of water-glass, S_SA > S_SL, therefore cosθ is positive or θ < 90° i,e. acute.

b) For mercury glass, angle of contact is obtuse. In order to achieve this obtuse value of angle of contact, the mercury tends to form a drop. In case of water glass, the angle of contact is acute. To achieve this acute value of angle of contact, the water tends to spread.

c) Surface tension of liquid is the force acting per unit length on a line drawn tangentially to the liquid surface at rest. Since this force is independent of the area of liquid surface, therefore surface tension is also independent of the area of the liquid surface.

d) We know that the cloth has narrow spaces in the form of capillaries. The rise of liquid in a capillary tude is directly proportional to cos0. if 0 is small cos0 will be large. Due to which capillary rise will be more and so the detergent will penetrate more in cloth.

e) In the absence of external forces, the surface of the liquid drop tends to acquire the minimum surface area due to surface tension. Since for a given volume, the surface area of sphere is least, hence the liquid drop takes the spherical shape.

Question 3.
Fill in the blanks using the word (S) from the list appended with each statement ;
a) Surface tension of liquids generally … with temperatures (increases / decreases)
b) Viscosity of gases… with temperature, whereas viscosity of liquids … with temperature (increases / decreases)
c) For solids with elastic modulus of rigidity, the shearing force is proportional to…. while for fluids it is proportional to … (shear strain / rate of shear strain)
d) For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / Bernoulli’s principle)
e) For the model of a plane in a wind tunnel, turbulence occurs at a… speed for turbulence for an actual plane (greater / smaller)
Answer:
a) Decreases
b) increases; decreases
c) Shear strain; rate of shear strain.
d) Conservation of mass; Bernoullis principle.
e) Greater.

Question 4.
Explain why
a) To keep a piece of paper horizontal, you should blow over, not under, it.
b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers.
c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection.
d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel.
e) A spinning cricket ball in air does not follow a parabolic trajectory.
Answer:
a) When we blow over the paper, the velocity of air blow increases and hence pressure of air on it decreases (according to Bernoullis theorem), whereas pressure of air below the paper is atmosphere. Hence the paper stays horizontal.

b) By doing so the area of the outlet of water jet is reduced, so velocity of water increases according to equation of continuity av = a constant.

c) When a fluid is flowing out of a small hole in a vessel it acquires a large velocity and hence possesses large momentum. Since no external force is acting on the system, a backward velocity must be attained by the vessel (according to law of conservation of momentum). As a result of it, impulse (backward thrust) is experienced by the vessel.

d) There, size of the needle controls velocity of flow and thumb pressure controls pressure. According to Bernoulli’s theorem, P + ρgh + \(\frac{1}{2}\) ρV² = a constant shows that P occurs with power one and V occurs with power two, hence the velocity has more influence. That is why the needle has a better control over flow.

e) If the ball is spinning well as moving linearly, the streamlines at the top of ball due to two types of motion are opposed to each other and those below are in the same direction. As a result of it, the velocity of air flow is greater below than above the ball. Now, according to Bernoullis principle, the pressure on the upper side of the ball becomes more than the pressure on the lower side of ball. Due to it, a resultant force F acts upon the ball at right angle to linear motion in the downward, direction, resulting the ball to move along a curved path.

Question 5.
A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor ?
Answer:
Here, m = 50kg; r =D/2 = 1/2 cm = \(\frac{1}{200}\) m
Pressure = \(\frac{\text { Force }}{\text { area }}=\frac{m g}{\pi r^2}=\frac{50 \times 9.8}{(22 / 7)(1 / 200)^2}\)
= 6.24 × 10⁶ Nm^-2.

Question 6.
Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m^-3. Determine the height of the wine column for normal atmosphere pressure.
Answer:
P = 0.76 × (13.6 × 10³) × 9.8
= h × 984 × 9.8 or
h = \(\frac{0.76 \times 13.6 \times 10^3 \times 9.8}{984 \times 9.8}\)
= 10.5 m.

Question 7.
A vertical off-shore structure is built to withstand a maximum stress of 10⁹ Pa. Is the structure suitable for putting up on top of an oil well in the ocean ? Take the depth of the ocean to be roughly 3 km and ignore ocean currents.
Answer:
Here, maximum stress = 10⁹Pa,
h = 3km = 3 × 10³m;
ρ(water) = 10³kg/m³ and g = 9.8 m/s²
The structure will be suitable for putting upon top of an oil well provided the pressure exerted by sea water is less than the maximum stress it can bear.
Pressure due to sea water, P = hρg
= 3 × 10³ × 10³ × 9.8
= 2.94 × 10⁷ Pa
Since the pressure of sea water is less than the maximum stress of 10⁹ Pa, the structure will be suitable for putting upon top of the oil well.

Question 8.
A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm². What maximum pressure would the smaller piston have to bear ?
Answer:
The maximum force, which the bigger piston can bear,
F = 3000 kgf = 3000 × 9.8 N
∴ Area of piston, A = 425 cm²
= 425 × 10^-4 m²
∴ maximum pressure on the bigger piston.
P = \(\frac{F}{A}=\frac{3000 \times 9.8}{425 \times 10^{-4}}\) = 6.92 × 10⁵ Pa
Since the liquid transmits pressure equally, therefore the maximum pressure the smaller piston can bear is 6.92 × 10⁵ Pa.

Question 9.
A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit ?
Answer:
For water column in one arm of U-tube,
h₁ = 10.0 cm; ρ₁ (density) = 1g cm^-3
For spirit column in other arm of U-tube, h₂ = 12.5 cm; ρ₁ = ?
As the mercury column in the two arms of U-tube are in level, therefore pressure exerted by each is equal. Hence h₁ρ₁g = h₂ρ₂g or
P₂ = \(\frac{h_1 \rho_1}{h_2}=\frac{10 \times 1}{12.5}\) = 0.8 g cm^-3
Therefore, relative density of spirit = ρ₂/ρ₁
= \(\frac{0.8}{1}\) = 0.8

Question 10.
In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms ? (Specific, gravity of mercury = 13.6)
Answer:
On pouring 15.0 cm of water and spirit each into the respective arms of U-tube, the mercury level will rise in the arm containing spirit. Let h be the difference in the levels of mercury in two arms of U-tube and p be the density of mercury.
∴ The pressure exerted by h cm of mercury column = difference in pressure exerted by water and spirit.
∴ h₁ρ₁g = h₂ρ₂g ……………. (1)
Here h = ? ρ =13.6 g cm^-3
ρ₁ =1 cm^-3
ρ₂ = 0.8 g cm^-3
h₁ = 15 + 10 = 25 cm
h₂ = 15 + 12.5 = 27.5 cm
Putting values in (i) we get h × 13.6 × g
= 25 × 1 × g – 27.5 × 0.8 × g = 3g
or h = 3/13.6 = 0.22cm

Question 11.
Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river ? Explain.
Answer:
No, Bernoulli’s theorem is used only for stream – line flow.

Question 12.
Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation ? Explain.
Answer:
No, it does not matter if one uses gauge instead of absolute pressures in applying Bernoulli’s’ equation, provided the atmospheric pressure at the two points where Bernoulli’s equation is applied are significantly different.

Question 13.
Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10^-3 kg s^-1, what is the pressure difference between the two ends of the tube ? (Density of glycerine = 1.3 × 10³ kg m^-3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].
Answer:
Here, l = 1.5 m, r = 1.0 cm = 10^-2 m, ρ = 1; 1.3 × 10^-3 kg/m3; η = 0.83 Nsm^-2.
Mass of glycerine flouring per sec, M = 4 × 10^-3 kg/s
Volume of glycerine flouring per sec, V = \(\frac{M}{\rho}\)
V = \(\frac{4 \times 10^{-3}}{1.3 \times 10^3} \mathrm{~m}^3 \mathrm{~s}^{-1}\) m³s^-1
If ρ is the difference of pressure between two ends of the tube,then using poisevilles formula we have
V = \(\frac{\pi \rho r^4}{8 \eta l}\) or P = \(\frac{V \times 8 \eta l}{\pi r^4}\)
P = \(\left(\frac{4 \times 10^{-3}}{1.3 \times 10^3}\right) \times \frac{8 \times 0.83 \times 1.5}{3.142 \times\left(10^{-2}\right)^4}\)
= 975.37 Pa

Question 14.
In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s^-1 and 63 m s^-1 respectively. What is the lift on the wing if its area is 2.5 m² ? Take the density of air to be 1.3 kg m^-3.
Answer:
Let V₁, V₂ be the speeds on the upper and lower surfaces of the wing of aeroplane and P₁ and P₂ be the pressures on upper and lower surfaces of the wing respectively.
Then V₁ = 70ms^-1; V₂ = 63ms^-1,
P = 1.3kg m^-3.
This difference of pressure provides the lift to the aeroplane.
\(\frac{P_1}{\rho}\) + gh + \(\frac{1}{2} V_1^2\) = \(\frac{P_2}{\rho}\) + gh + \(\frac{1}{2} V_2^2\)
∴ \(\frac{P_1}{\rho}-\frac{P_2}{\rho}=\frac{1}{2}\left(V_2^2-V_1^2\right)\)
or P₁ – P₂ = \(\frac{1}{2} \rho\left(V_2^2-V_1^2\right)\)
= \(\frac{1}{2}\) × 1.3 [(70)² – (63)²]
= 605.15 Pa
This difference of pressure provides the lift to the aeroplane.
So, lift on the aeroplane = Pressure difference × area of wings
= 605.15 × 2.5
= 1512.875 N
= 1.51 × 10³N.

Question 15.
Figures (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?

Answer:
Figure a is incorrect. According to equation of continuity i.e. av = a constant, where area of cross – section of tube is less, the velocity of liquid flow is more than the other portion of tube. According to Bernoulli’s theorem,
P + \(\frac{1}{2}\) ρv² = a constant i.e. where V is more P is less and avice versa.

Question 16.
The cylindrical tube of a spray pump has a cross-section of 8.0 cm² one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min^-1, what is the speed of ejection of the liquid through the holes ?
Answer:
Area of cross – section of tube, a₁ = 8.0 cm² = 8 × 10^-4 m²
No. of holes = 40, Diameter of each hole, D = 1 mm = 10^-3 m
∴ Radius of hole, r = \(\frac{D}{2}=\frac{1}{2}\) × 10^-3 m
= 5 × 10^-4 m
Area of cross – section of each hole = πr²
= π(5 × 10^-4)²m²
Total area of cross – section of 40 holes,
a₂ = 40 × π (5 × 10^-4)²m²
Speed of liquid inside the tube,
V₁ = 1.5m/min
= \(\frac{1.5}{60}\) ms^-1
If V₂ is the velocity of ejection of the liquid through the holes, then
a₁V₁ = a₂V₂ or V₂ = \(\frac{a_1 V_1}{a_2}\)
V₂ = \(\frac{\left(8 \times 10^{-4}\right) \times 1.5}{60 \times 40 \times \pi \times\left(5 \times 10^4\right)^2}\)
= 0.637 ms^-1

Question 17.
A U-shaped wire is dipped in a soap solution and removed. The thin shaped film formed between the surfaces and the light slider supports a weight of 1.5 × 10^-2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film ?
Answer:
We know that soap film has two free surfaces, so total length of the film to be supported,
= 2l = 2 × 30 = 60 cm
= 0.6 m
Total force on the slider due to surface tension will be,
F = S × 2l
= S × 0.6 N
In equilibrium position, the force F on slider due to surface tension must balance the weight mg
(1.5 × 10^-2)N i.e. F = 1.5 × 10^-2
= S × 0.6 Or
S = \(\frac{1.5 \times 10^{-2}}{0.6}\)
2.5 × 10^-2Nm^-1

Question 18.
Figure (a) shows a thin liquid film supporting a small weight = 4.5 × 10^-2 N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c) ? Explain your answer physically.

Answer:
a) Here, length of film supporting the weight
= 40cm = 0.4 m
Total liquid film supported (or force)
= 4.5 × 10^-2 N
film has two free surfaces, ∴ surface tension,
S = \(\frac{4.5 \times 10^{-2}}{2 \times 0.4}\)
S = 5.625 × 10^-2 Nm^-1
Since the liquid is same for all the cases (a), (b), (c) and temperature is also same, therefore surface tension for cases (b) and (c) will also be the same = 5.625 × 10^-2. In figure (b) and (c) the length of the film supporting the weight is also the same as that of(a), hence the total weight supported in each case is 4.5 × 10^-2 N.

Question 19.
What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature ? Surface tension of mercury at that temperature (20°C) is 4.65 × 10^-1 Nm^-1. The atmospheric pressure is 1.01 × 10⁵ Pa. Also give the excess pressure inside the drop.
Answer:
Here, r = 3.0 mm = 3 × 10^-3 m;
S = 4.65 × 10^-1 Nm^-1;
P = 1.01 × 10⁵ Pa
Excess of pressure inside the drop of mercury is given by
P = \(\frac{2 S}{r}=\frac{2 \times 4.65 \times 10^{-1}}{3 \times 10^{-3}}\)
= 310 Pa
Total pressure inside the drop = P + ρ
= 1.01 × 10⁵ + 310
= 1.01 31 × 10⁵ Pa

Question 20.
What is the excess pressure inside a bubble of soap solution of radius 5.0 mm, given that the surface tension of soap solution at the temperature (20°C) is 2.50 × 10^-2 Nm^-1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble ? (1 atmospheric pressure is 1.01 × 10⁵ Pa).
Answer:
Here, S = 2.5 × 10^-2 Nm^-1, r = 5.00 mm = 5 × 10^-3 m.
Density of soap solution, ρ = 1.2 × 10³ kg m^-3
Excess pressure inside the soap bubble,
P = \(\frac{4 s}{r}=\frac{4 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}\) = 20 Pa
Excess pressure inside the air bubble, P’ = \(\frac{2S}{r}\)
= \(\frac{2 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}\) = 10 Pa
Total pressure inside the air bubble at depth h in soap solution – ρ’ + atmospheric pressure + hρg
= 10 + 1.01 × 10⁵ + 0.4 × 1.2 × 10³ × 9.8
= 1.06 × 10³ Pa

Question 21.
A tank with a square base of area 1.0 m² is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm². The tank is filled with water in one compartment and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. compute the force necessary to keep the door close.
Answer:
For compartment containing water,
h₁ = 4m, ρ₁ = 10³ kg m^-3
The pressure exerted by water at the door provided at bottom

P₁ = h₁ρ₁g
= 4 × 10³ × 9.8
= 3.92 × 10⁴ Pa
For compartment containing acid,
h₂ = 4m,
ρ₂ = 1.7 × 1.0³ kg/m³
The pressure exerted by acid at the door provided at bottom.
P₂ = h₂ρ₂g
= 4 × 1.7 × 10³ × 9.8
= 6.664 × 10⁴ Pa
∴ Difference of pressure = P₂P₁
= 6.664 × 10⁴ – 3.92 × 10⁴
= 2.774 × 10⁴Pa
Given, area of door, A = 20cm²
= 20 × 10^-4m²
Force on the door = difference in pressure × area
= (P₂ – P₁) × A
= (2.774 × 10⁴) × (20 × 10⁴)
= 54.88N
≈ 55N
To keep the door closed, the force equal to 55 N should be applied horizontally on the door from compartment containing water to that containing acid.

Question 22.
A manometer reads the pressure of a gas in an enclosure as shown in Fig. (a) when a pump removes some of the gas, the manometer reads as in Fig. (b) The liquid used in the manometers is mercury and the atmo-spheric pressure is 76 cm of mercury.
a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and b) in units of cm of mercury.
b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury, are poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).

Answer:
a) Here, atomospheric presure, p = 76 cm of mercury
In Fig (a) Pressure head
h = + 20 cm
∴ Absolute pressure = p + h
= 76 + 20
= 96 cm of mercury
Gauge pressure = h = 20 cm of mercury.
In fig (b) pressure head, h = -18 cm
Absolute pressure = p + h
= 76+ (-18)
= 58 cm of mercury
Gauge pressure = h = -18cm of mercury.

b) Here 13.6 cm of water added in right limb is equiralent to
\(\frac{13.6}{13.6}\) = 1 cm of mercury column.
i.e., h¹ = 1 cm of mercury column.
Now pressure at A, PA = P + h¹ = 76 + 1
= 77 cm
Let the difference in mercury levels in the two lumbs be h₁, then pressure at B,
P_B = 58 + h₁ or
As P_A = P_B = 77 = 58 + h₁
h₁ = 77 – 58 = 19 cm of mercury column.

Question 23.
Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill upto a particular comon height. Is the force exerted by the water on the base of the vessel the same in the two cases ? If so, why do the vessels filled with water to that same height give different readings on a weighing scale ?
Answer:

Since the pressure depends upon the height of water column and the height of the water column in the two vessels of different shapes is the same.hence there will be same pressure due to water on the base of each vessel. As the base area of each vessel is same, hence there will be equal force acting on the two base areas due to water pressure. The water exerts force on the walls of the vessel also. In case, the walls of the vessel are not perpendicular to base, the force exerted by water on the walls has a net non-zero vertical component which is more in first vessel than that of second vessel. That is why, the two vessel:; filled with water to same vertical height show different readings on a weighing machine.

Question 24.
During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein ? Use the density of whole blood from Table 1.
Answer:
h = \(\frac{p}{p g}=\frac{200}{1.06 \times 10^3}\) × 9.8 = 0.1925 m
The blood may just enter the vein if the height at which the blood container be kept must be slightly greater than 0.1925 m i.e. 0.2 m.

Question 25.
In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy.
a) What is the largest average velocity of blood flow in an artery of diameter 2 × 10^-3 m if the flow must remain laminar ?
b) Do the dissipative forces become more important as the fluid velocity increases ? Discuss qualitatively.
Answer:
a) If dissipative forces are present, then some forces in liquid flow due to pressure different is spent against dissipative forces. Due to which the pressure drop becomes large.

b) The dissipative forces become more important with increasing flow velocity, because of tubulence.

Question 26.
a) What is the largest average velocity of blood flow in an artery of radius 2 × 10^-3m if the flow must remain lanimar ?
Answer:
Here, r = 2 × 10^-3m ;
D = 2r = 2 × 2 × 10^-3 = 4 × 10^-3m
η = 2.084 × 10^-3 Pa s;
p = 1.06 × 10³ kgm³
For flow tobe laminar, N_r = 2000
a) Now, V_c = \(\frac{N_r \eta}{\rho D}\)
= \(\frac{2000 \times\left(2.084 \times 10^{-3}\right)}{\left(1.06 \times 10^3\right) \times\left(4 \times 10^{-3}\right)}\) = 0.98m/s.

b) What is the corresponding flow rate ? (Take viscosity of blood to be 2.084 × 10^-3 Pa s).
Answer:
Volume flowing per second = πr²Vc
= \(\frac{22}{7}\) × (2 × 10^-3)² × 0.98
= 1.23 × 10^-5 m³s^-1

Question 27.
A plane is in level flight at constant speed and each of its two wings has an area of 25m². If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg m^-3)
Answer:
Here, V₁ = 180 km/h = 50m/s, V₂ = 234 km/ h = 65 m/s;
A = 2 × 25 = 50m²; P = 1kg/m³
P₁ – P₂ = \(\frac{1}{2}\) p (V₂² – V₁²)
= \(\frac{1}{2}\) × 1 × (65² – 50²)
Upward force = (P₁ – P₂) A = \(\frac{1}{2}\) × (65² – 50²) × 50N
As the plane is in level flight, so
mg = (P₁ – P₂)A
or m = \(\frac{\left(P_1-P_2\right) A}{g}=\frac{1 \times\left(65^2-50^2\right) \times 50}{2 \times 9.8}\)
= 4.4 × 10³N

Question 28.
In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 × 10^-5 m and density 1.2 × 10³ kg m^-3. Take the viscosity of air at the temperature of the experiment to be 1.8 × 10⁵ Pa s. How much is the viscous force on the drop at that speed ? Neglect buoyancy of the drop due to air.
Answer:
Here, r = 2.0 × 10^-5m; ρ = 1.2 × 10³ kg m^-3;
η = 1.8 × 10⁵ Ns m^-2
P₀ = 0, V = ?, F = ?
Terminal velocity V = \(\frac{2 r^2\left(\rho-\rho_0\right) g}{9 \eta}\)
= \(\frac{2 \times\left(2.0 \times 10^{-5}\right)^2 \times\left(1.2 \times 10^3-0\right) \times 9.8}{9 \times 1.8 \times 10^{-5}}\)
= 5.8 × 10^-2ms^-1 = 5.8 cms^-1
Viscous force on the drop, F = 6πηrv
= 6 × \(\frac{22}{7}\) × (1.8 × 10^-5) × (2.0 × 10^-5) × (5.8 × 10^-2)
= 3.93 × 10^-10N.

Question 29.
Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside ? Surface tension of mercury at the temperature of the experiment is 0.456 N m^-1. Density of mercury = 1.36 × 10³ kg m^-3.
Answer:
Here, θ = 140°, r = 1 × 10^-3 m;
S = 0.465 Nm^-1, ρ = 13.6 × 10³ kg, h = ?]
Cos = 140° = – cos40° = -0.7660
Now h = \(\frac{2 S \cos \theta}{r \rho g}\)
= \(\frac{2 \times 0.465 \times \cos 140^{\circ}}{10^{-3} \times 13.6 \times 10^3 \times 9.8}\)
= \(\frac{2 \times 0.465 \times(-0.7660)}{10^{-3} \times 13.6 \times 10^3 \times 9.8}\)
= -5.34 × 10^-3m
= -5.34mm
Negative value of h shows that the mercury levels is depressed in the tube.

Question 30.
Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to forma a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube ? Surface tension of water at the temperature of the experiment is 7.3 × 10^-2 N m^-1. Take the angle of contact to be zero and density of water to be 1.0 × 10³ kg m^-3 (g = 9.8 m^-2).
Answer:
Here, S = 7.3 ^-2 Nm^-1, ρ = 1.0 × 10³ kg m^-3; θ = 0°
For narrow tube, 2r₁ = 3.00 mm = 3 × 10^-3 m or r₁ =1.5 × 10^-3 m .
For wider tube, 2r₂ = 6.00 mm = 6 × 10^-3 m or r₂ = 3 × 10^-3 m
let h₁, h₂ be the heights to which water rises in narrow tube and wider tube respectively.
Then, h₁ = \(\frac{2 s \cos \theta}{r_1 \rho g}\) and h₂ = \(\frac{2 s \cos \theta}{r_2 \rho g}\)
∴ Difference in levels of water in two limbs of U-tube is

Question 31.
a) It is known that density of air decreases with height y as
ρ = ρ₀e^-y/y₀
Where p₀ = 1.25 kg m^-3 is the density at sea level, and y₀ is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of g remains constant.
Answer:
We know that the rate of decrease of density p of air is directly proportional to height y i.e.
\(\frac{-d \rho}{d y}\) ∝ p or \(\frac{d \rho}{d y}\) = – Kρ
Where K is a constant of proportionality. Here – ve sign shows that ρ decreases as y increases.
\(\frac{d y}{\rho}\) = – Kρ
Integrating it with in the conditions, as y changes fromotoy density changes from ρ₀ to ρ, we have
\(\left.\int_{P_0}^p \frac{d P}{P}=-\int_0^y k d y=\left[\log e^p\right)\right]_{\rho_0}^p=k y\)
\(\frac{\rho}{\rho_0}=e^{-k y}=\rho=\rho_0 e^{-\mathrm{y} / y_0}\)

b) A large He balloon of volume 1425 m³ is used to lift a payload of 400 kg. Assum that the balloon maintains constant radius as it rises. How high does it rise ?
(Take y₀ = 8000 m and P_He = 0.18 kgm^-3)
Answer:
The balloon will rise to aheight, where its density becomes equal to the air at that height.

y = 8000 × 1 = 8000 m
= 8 km.

Textual Examples

Question 1.
The two thigh bones (femurs), each of cross-sectional area 10 cm² mass 40 kg. Estimate the average pressure sustained by the femurs.
Answer:
Total cross-sectional area of the femurs is A = 2 × 10 cm² 20 × 10^-4 m². The force acting on them is F = 40 kg wt = 400 N (taking g 10 ms^-2). This force is acting vertically down and hence, normally on the femurs. Thus, the average pressure is
P_av = \(\frac{F}{A}\) = 2 × 10⁵ N m^-2

Question 2.
What is the pressure on a swimmer 10m below the surface of a lake ?
Answer:
Here
h = 10 m and p = 1000 kg m^-3.
Take g 10 m S^-2
From Equation = P = P_a + ρgh
P = P_a + ρgh
= 1.01 × 10⁵ Pa + 1000 kgm^-3 × 10m S^-2 × 10 m
= 2.10 × 10⁵ Pa
≈ 2 atm
This is a 100% increase in pressure from surface level. At a depth of 1 km the increase in pressure is 100 atm! Submarines are designed to withstand such enormous pressures.

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 atmosphee extend ?
Answer:
We use P – P_a = ρgh
ρgh = 1.29 kg m^-3 × 9.8 ms² × hm
= 1.01 × 10⁵ pa
∴ h = 7989 m ≈ 8 km
In reality the density of air decreases with height. So does the value of g. The atmospheric cover extends with decreasing press-ure over 100 km. We should also note that the sea level atmospheric pressure is not always 760 mm of Hg. A drop in the Hg level by 10 mm or more is a sign of an approaching storm.

Question 4.
At a depth of 1000 m in an ocean
(a) What is the absolute pressure ?
b) What is the guage Pressure ?
c) Find the force acting on the window of area 20 cm × 20 cm of a submarine at this depth, the interior of which is maintained at sealevel atmospheric pressure. (The density of sea water is 1.03 × 10³ kg m^-3, g = 10ms^-2.
Answer:
Here h = 1000 m and ρ = 1.03 × 10³ kg m^-3
a) From P₂ – P₁ = ρgh absolute pressure
P = P_a + ρgh
= 1.01 × 10⁵ Pa + 1.03 × 10³ kg m^-3 × 10 m s^-2 × 1000 m
= 104.01 × 10⁵ Pa .
= 104 atm

b) Gauge pressure is P – P_a = ρgh = P_g
P_g = 1.03 × 10³ kg m^-3 × 10 m s² × 1000 m
= 103 × 10⁵ Pa
≈ 103 atm

c) The pressure outside the submarine is P = P_a + ρgh and the pressure inside it is P_a. Hence, the net pressure acting on the window is gauge pressure, P_g = ρgh. Since the area of the window is A = 0.04 m², the force acting on it is
F = P_gA = 103 × 10⁵ Pa × 0.04m²
= 4.12 × 10⁵ N
≈ 103 atm

Question 5.
Two syringes of different cross sections A₁, A₂ and lengths L₁, L₂ (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1.0 cm and 3.0 cm respectively.
a) Find the force exerted on the larger piston when a force of ION is applied to the smaller piston, b) if the smaller piston is pushed in through 6.0 cm,’how much does the larger piston move out ?
Answer:
a) Since pressure is transmitted undiminished throughout the fluid.
F₂ = \(\frac{A_2}{A_1} F_1=\frac{\pi\left(3 / 2 \times 10^{-2} m\right)^2}{\pi\left(1 / 2 \times 10^{-2} \mathrm{~m}\right)^2} \times 10 \mathrm{~N}\)
= 90 N

b) Water is considered to be perfectly incompressible. Volume covered by the move-ment of smaller piston inwards is equal to volume moved outwards due to the larger piston.
L₁A₁ = L₂A₂
L₂ = \(\frac{A_1}{A_2} L_1=\frac{\pi\left(1 / 2 \times 10^{-2} \mathrm{~m}\right)^2}{\pi\left(3 / 2 \times 10^{-2} \mathrm{~m}\right)^2}\)
= × 6 × 10^-2m
≃ 0.67 × 10^-2m = 0.67 cm.
Note, atmospheric pressure is common to both pistons and has been ignored.

Question 6.
In a car lift compressed air exerts a force F₁ on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston or radius 15 cm (Fig). If the mass of the car to be lifted is 1350 kg, calculate F₁. What is the pressure necessary to accomplish this task ? (g = 9.8 ms^-2).

Answer:
Since pressure is transmitted undiminished throughout the fluid.
F₁ = \(\frac{A_1}{A_2} F_2=\frac{\pi\left(5 \times 10^{-2} \mathrm{~m}\right)^2}{\pi\left(15 \times 10^{-2} \mathrm{~m}\right)^2}\)
= (1350 N × 9.8m s^-2) = 1470 N
= 1.5 × 10³N
The air pressure that will produce this force is
P = \(\frac{F_1}{A_1}=\frac{1.5 \times 10^{-3} \mathrm{~N}}{p\left(5 \times 10^{-2}\right)^2 \mathrm{~m}}\) = 1.9 × 10⁵ Pa
This is almost double the atmospheric pressure.

Question 7.
Blooc velocity : The flow of blood in a large artery of an anesthetised dog is diverted through a Venturi meter. The wider part of the meter has a cross-sectional area equal to that of the artery. A = 8 mm². The harrower part has an area = 4mn². The pressure drop in the artery is 24 Pa. What is the speed of the blood in the artery ?
Answer:
We take the density of blood from whole blood to be 1.06 × 10³ kg m³. The ratio of the areas is \(\left(\frac{\mathrm{A}}{\mathrm{a}}\right)\) = 2
Using v₁ = \(\left(\sqrt{\frac{2 \rho_m g h}{\rho}}\right)\left(\left(\frac{A}{2}\right)^2-1\right)^{-1 / 2}\)
we obtain
v₁ = \(\sqrt{\frac{2 \times 24 \mathrm{pA}}{1060 \mathrm{~kg} \mathrm{~m}^{-3} \times\left(2^2-1\right)}=0.125 \mathrm{~m} \mathrm{~s}^{-1}}\)

Question 8.
A fully loaded Boeing aircraft has a mass of 3.3 × 10⁵ kg. Its total wing area is 500 m2. It is in level flight with a speed of 960 km/h. a) Estimate the pressure difference between the lower and upper surfaces of the wings (b) Estimate the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface. [The density of air is ρ = 1.2 kgm^-3].
Answer:
a) The weight of’the Boeing aircraft is balanced by the upward force due to the pressure difference
∆P × A – 3.3 × 10⁵ kg × 9.8 = mg.
∆P = (3.3 × 10⁵ kg × 9.8 m s^-2) / 500 m²
= 6.6 × 10³ N m^-2

b) We ignore the small height difference between the top and bottom sides in
P₁ + (\(\frac{1}{2}\)) ρv₁² + ρgh₁ = P₂ + (\(\frac{1}{2}\))ρv₂² + ρgh₂
The pressure difference between them is then
∆P = \(\frac{\rho}{2}\left(v_2^2-v_1^2\right)\)
Where v₂ is the speed of air over the upper surface and v₁ is the speed under the bottom surface.
(v₂ – v₁) = \(\frac{2 \Delta P}{\rho\left(v_2+v_1\right)}\)
Taking the average speed v_av (v₂ + v₁)/2 – 960 km/h = 267 m s^-1, we have
(v₂ – v₁)/v_av = \(\frac{\Delta \mathrm{P}}{\rho v_{\mathrm{av}}^2}\) ≈ 0.08
The speed of air above the wing needs to be only 8% higher than that below.

Question 9.
A metal block of area 0.10 m² is connected to a 0.010 kg mass via a string that passes over an ideal pulley (considered massless and frictionless), as in Fig. A liquid with a film thickness of 0.30 mm is placed between the block and the table. When released, the block moves to the right with a constant speed of 0.085 ms^-1. Find the co-efficient of viscosity of the liquid.

Answer:
The metal block moves to the right because of the tension in the string. The tension T is equal in magnitude to the weight of the suspended mass m. Thus the shear force F is
F = T = mg = 0.010 kg × 9.8 ms^-2
= 9.8 × 10^-2 N
Shear stress on the fluid = F/A = \(\frac{9.8 \times 10^{-2}}{0.10}\)
Strain rate = \(\frac{v}{l}=\frac{0.085 \mathrm{~ms}^{-1}}{0.3 \times 10^{-3} \mathrm{~m}}\)
η = \(\frac{\text { stress }}{\text { strain rate }}\)
= \(\frac{\left(9.8 \times 10^{-2} \mathrm{~N}\right)\left(0.30 \times 10^{-3} \mathrm{~m}\right)}{\left(0.085 \mathrm{~ms}^{-1}\right)\left(0.10 \mathrm{~m}^2\right)}\)
= 3.45 × 10^-3 Pa s

Question 10.
The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at 20°C is 6.5 cm s^-1. Compute the viscosity of the oil at 20°C. Density of oil is 1.5 × 10³ kg m^-3, density of copper is 8.9 × 10³ kg m^-3.
Answer:
We have v_t = 6.5 × 10^-2 ms^-1,
a = 2 × 10^-3m,
g = 9.8 ms^-2, ρ = 8.9 × 10³ kg m^-3,
σ = 1.5 × 10³ kg m^-3. From
v_t = \(\frac{2 a^2(\rho-\sigma) g}{(9 \eta)}\)
η = \(\frac{2}{9} \times \frac{\left(2 \times 10^{-3}\right) \mathrm{m}^2 \times 9.8 \mathrm{~ms}^{-2}}{6.5 \times 10^{-2} \mathrm{~ms}^{-1}}\) × 7.4 × 10³ kg m^-3
= 9.9 × 10^-1 kg m^-1s^-1

Question 11.
a) The flow rate of water from a tap of diameter 1.25 cm is L7min. The co-efficient of viscosity of water is 10^-3 Pa s.
b) After sometime the flow rate is increased to 3L / min. Characterise the flow for both the flow rates.
Answer:
a) Let the speed of the flow be v and the diameter of the tap be d = 1.25 cm. The volume of the water flowing out per second is
Q = v × π d²/4
v = 4Q / d² π
We then estimate the Reynolds number to be
R_e = 4vQ / πdη
= 4 × 10³ kg m^-3 × Q / (3.14 × 1.25 × 10^-2 m × 10^-3 Pa S)
= 1.019 × 10⁸ m^-3 SQ
Since initially (a)
Q = 0.48 L/min = 8cm³/s
= 8 × 10^-6 m³ s^-1, we obtain,
R_e = 815
Since this is below 1000, the flow is steady.
After some time

b) When Q = 3L/ min = 50 cm³, s = 5 × 10^-5 m³ s^-1 we obtain,
R_e = 5095
The flow will be turbulent. You may carry out an experiment in your washbasin to determine the ransition from laminar to turbulent flow.

Question 12.
The lower end of a capillary tube of diameter 2.00 mm is dipped 8.00 cm below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water ? The surface tension of water at temperature of the experiments is 7.30 × 10^-2 Nm¹. 1 atmospheric pressure = 1.01 × 10⁵ Pa, density of water = 1000 kg/m³, g = 9.8 × ms^-2. Also calculate the excess pressure.
Answer:
The excess pressure in a bubble of gas in a liquid is given by 2S / r, where S is the surface tension of the liquid gas interface. You should note there is only one liquid surface in this case. (For a bubble of liquid in a gas, there are two liquid surfaces, so the formula for excess pressure in that case is 4S / r.) The radius of the bubble is r. Now the pressure outside the bubble, within water, P₀ equals atmospheric pressure plus the pressure due to 8.00 cm of water column. That is
P₀ = (1.01 × 10⁵ Pa + 0.08 m × 1000 kg m^-3 × 9.80 m s^-2)
= 1.01784 × 10⁵ Pa. .
Therefore, the pressure inside the bubble is
P₁ = P₀ + 2S / r (as r = 1 mm)
= 1.01784 × 10⁵ Pa + (2 × 7.3 × 10^-2 Pa m/10^-3 m)
= (1.01784 + 0.00146) × 10⁵ Pa
= 1.02 × 10⁵ Pa
where the radius of the bubble is taken to be equal to the radius of the capillary tube, since the bubble is hemispherical ! (The answer has been rounded off to three significant figures). The excess pressure in the bubble is 146 Pa from (1.0178 + 0.00146) × 10⁵ Pa.

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 8th Lesson Oscillations Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 8th Lesson Oscillations

Very Short Answer Questions

Question 1.
Give two examples of periodic motion which are not oscillatory.
Answer:

1. The motion of planets around the sun.
2. The motion of an electron round the nucleus.

Question 2.
The displacement in S.H.M. is given by y = a sin (20t + 4). What is the displacement when it is increased by 2π/ω?
Answer:
The displacement in S.H.M. is y = a sin (20t + 4)
The time period T = \(\frac{2 \pi}{\omega}\) is increased, the displacement of the particle remains the same.

Question 3.
A girl is swinging seated in a swing. What is the effect on the frequency of oscillation if she stands?
Answer:
Frequency (n) = \(\frac{1}{2 \pi} \sqrt{\frac{g}{l}}\), n ∝ \(\frac{1}{\sqrt{l}}\)
A girl swinging in standing position location of centre of mass shifts upwards l decreases, frequency of oscillation increases.

Question 4.
The bob of a simple pendulum is a hollow sphere filled with water. How will the period of oscillation change, if the water begins to drain out of the hollow sphere ?
Answer:
The period of the pendulum is same, when the bob is hollow (or) completely filled with water. As water flows out from the bob, the centre of gravity of the bob lowered. The pendulum length increases. Hence time period also increases. When the bob becomes empty, again centre of gravity shifts upwards. The pendulum length decreases. The time period also decreases.

Question 5.
The bob of a simple pendulum is made of wood. What will be the effect on the time period if the wooden bob is replaced by an identical bob of aluminum ?
Answer:
Time period (T) = \(2 \pi \sqrt{\frac{l}{g}}\), Time period is independent of mass of the bob.
Hence wooden bob is replaced by an identical aluminium bob, Time period remains constant.

Question 6.
Will a pendulum clock gain or lose time when taken to the top of a mountain.?
Answer:
T ∝ \(\frac{1}{\sqrt{g}}\), At the mountain top, the value of g decreases, hence time period increases i.e. the pendulum will take longer time to complete one vibration. Hence pendulum clock will loose time on the mountain top.

Question 7.
A pendulum dock gives correct time at the equator. Will it gain or lose time if it is taken to the poles ? If so, why ?
Answer:
Time period (T) = \(2 \pi \sqrt{\frac{l}{g}}\)
g value at poles is greater than at equator.
If it is taken to the poles g value increases, time period decreases.
So pendulum clock gains time.

Question 8.
What fraction of the total energy is K.E when the displacement is one half of a amplitude of a particle executing S.H.M.
Answer:
Total energy (E) = \(\frac{1}{2}\) mω² A²
Given y = \(\frac{\mathrm{A}}{2}\), Kinetic energy = \(\frac{1}{2}\)mω²(A² – y²) = \(\frac{1}{2}\) mω²\(\left(A^2-\frac{A^2}{4}\right)\) = \(\frac{3}{4}\) × \(\frac{1}{2}\)mω²A²
K.E = \(\frac{3}{4}\) × E
∴ \(\frac{K \cdot E}{E}\) = \(\frac{3}{4}\)

Question 9.
What happens to the energy of a simple harmonic oscillator if its amplitude is doubled ?
Answer:
Total energy (E) = \(\frac{1}{2}\)mω²A²
Given Amplitude A is doubled
E’ = \(\frac{1}{2}\)mω²(2A)²
E’ = 4 × \(\frac{1}{2}\)mω²A²
E’ = 4E
∴ Energy becomes four times.

Question 10.
Can a simple pendulum be used in an artificial satellite? (T.S. Mar. 16)
Answer:
No, in an artificial satellite acceleration due to gravity is zero. Hence we cannot use simple pendulum in an artificial satellite.

Short Answer Questions

Question 1.
Define simple harmonic motion. Give two examples.
Answer:
Simple harmonic motion : “A body is said to be in simple harmonic motion, if it moves to and fro along a straight line, about its mean position such that, at any point its acceleration is proportional to its displacement but opposite in direction and is directed always towards the mean position”.
w ∝ -x

If a is the acceleration of the body at any given displacement x from the mean position, the time for the body to be in S.H.M.
Displacement of a particle in S.H.M is given by
x(t) = A cos (ωt + ϕ)

Example:

1. Motion of a simple pendulum.
2. Motion of mass attached to a spring.
3. Motion of atoms in solids.
4. Cork floating on water.

Question 2.
Present graphically the variations of displacement, velocity and acceleration with time for a particle in S.H.M.
Answer:
Let us put ϕ = 0 and write the expressions for x(t),
υ(t) and a(t).
x(t) = A cos ωt. υ(t) = -Aωsinωt
a(t) = -ω²A cos ωt. The corresponding plots are shown in figure. All quantities varies sinusoidally with time.

x(t) varies between – A to A; υ(t) varies from -ωA to ωA and a(t) varies from – ω²A to ω²A with respect to displacement plot, velocity plot has a phase difference of \(\frac{\pi}{2}\) and acceleration plot has a phase difference of π.

Question 3.
What is phase ? Discuss the phase relations between displacement, velocity and acceleration in simple harmonic motion.
Answer:
Phase : The phase of a particle executing S.H.M. at any instant is defined as its state (or) condition as regards to its position and direction of motion at that instant.

1. Displacement : x = A cos (ωt – ϕ), (ωt – ϕ) is called phase and ϕ is epoch.
2. Velocity : V = -Aω sin (ωt – ϕ), Here also (ωt – ϕ) is phase angle
3. Acceleration : a = -ω²A cos (ωt – ϕ), Here also (ωt – ϕ) is phase angle.

Phase difference between displacement and velocity = \(\frac{\pi}{2}\)
Phase difference between velocity and acceleration = \(\frac{\pi}{2}\)
Phase difference between displacement and acceleration = π.

Question 4.
Obtain an equation for the frequency of oscillation of spring of force constant k to which a mass m is attached. Answer:
Let us consider a spring suspended vertically from a rigid support and loaded with a mass m. If it is now pulled down and released, it executes vertical oscillations about mean position.
Restoring force is directly proportional to the displacement, but oppositely directed

F ∝ -y
F = -ky —–> (1)
Where k is force constant
Ma = -ky (∴ F = Ma)
a = \(-\left(\frac{K}{M}\right) y\) —-> (2)
Since K and M are constant. We can write a ∝ -y
ie Acceleration is directly proportional to the displacement but oppositely directed. Hence oscillations of a loaded spring is S.H.M.
Comparing eq. (2) with a = -ω²y
ω² = \(\frac{K}{M}\) ; ω = \(\sqrt{\frac{\mathrm{K}}{\mathrm{M}}}\)
T = \(\frac{2 \pi}{\omega}\) ; T = \(2 \pi \sqrt{\frac{\mathrm{M}}{\mathrm{K}}}\)
Frequency of oscillation (n) = \(\frac{1}{2 \pi} \sqrt{\frac{K}{M}}\) —> (3)

Question 5.
Derive expressions for the kinetic energy and potential energy of a simple harmonic oscillator.
Answer:
Kinetic energy of simple harmonic oscillator : The velocity of a particle in S.H.M. is given by
v = \(\omega \sqrt{A^2-y^2}\) ∴ Kinetic energy = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\) mω²(A² – y²)
When y = 0, (K.E)_Max = \(\frac{1}{2}\) mω²A² (Mean position)
When y = A, (K.E)_Min = 0 (Extreme position)
K.E is maximum at mean position and minimum at extreme position.

Potential energy of simple harmonic oscillator: When the displacement of a particle executing simple Harmonic oscillations increases, the restoring force also increases. The restoring force is in the opposite direction to the displacement. Therefore work is done in moving through the displacement, against restoring force. If F is the restoring force at the displacement y.
The average force against which work is done = \(\frac{\mathrm{O}+\mathrm{F}}{2}\) = \(\frac{F}{2}\)

∴ Workdone on the particle for the displacement (y) = Average force × displacement F
i.e. w = \(\frac{\mathrm{F}}{2}\) × y
w = \(\frac{\max y}{2}\) —- (1) (∵ F = ma)
But acceleration of a particle in S.H.M. is given by
a = ω²y —– (2)
Using eq’s (1) and (2). we get
The work done (w) = \(\frac{1}{2} m \omega^2 y^2\) mw2y2
This work done is stored in the form of P.E.
∴ P.E. = \(\frac{1}{2} m \omega^2 y^2\) —– (3)
If y = 0, (P.E)_Min = 0 (At mean position)
If y = A, (P.E)_Max = \(\frac{1}{2} m \omega^2 A^2\) (At extreme position).
∴ P.E. is maximum at extreme position and minimum at mean position.

Question 6.
How does the energy of a simple pendulum vary as it moves from one extreme position to the other during its oscillations?
Answer:
The total energy associated with a particle executing S.H.M. at any point is the sum of potential energy and kinetic energy at that point.

Total energy (E) = K.E + P.E
K.E = \(\frac{1}{2} m \omega^2\left(A^2-y^2\right)\)
P.E. = \(\frac{1}{2} m \omega^2 y^2\)
T.E. = \(\frac{1}{2}\)mω²(A² – y²) + \(\frac{1}{2}\)mω²y² = \(\frac{1}{2}\)mω²A²
At mean position y = 0, P.E = 0
and (K.E)_Max = \(\frac{1}{2} m \omega^2 A^2\)
∴ T.E. = 0 + \(\frac{1}{2} m \omega^2 A^2\) = \(\frac{1}{2} m \omega^2 A^2\)
At extreme position y = A, K.E = 0 and P.E = \(\frac{1}{2} m \omega^2 A^2\)
∴ (P.E)_Max = \(\frac{1}{2} m \omega^2 A^2\)
From mean position to extreme position K.E. is to be converted into P.E.

Question 7.
Derive the expressions for displacement, velocity and acceleration of a particle executes S.H.M.
Answer:
Consider a particle P moves on the circumference of a circle of radius A with uniform angular velocity ω. Let PN be the perpendicular drawn to the diameter yy’ from P.
As P moves on the circumference of the circle, N moves on the diameter yy’ to and fro about the centre O.

Velocity : The velocity of a partcile executing SHM is given by
v = \(\frac{d y}{d t}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\) (A sin ωt)
= Aω cos ωt = Aω\(\sqrt{1-\sin ^2 \omega t}\)
v = \(A \omega \sqrt{1-\left(\frac{y}{A}\right)^2}\) (∵ sin ωt = \(\frac{y}{A}\))
v = ω\(\omega \sqrt{A^2-y^2}\) —- (2)

Acceleration : As the rate of change of velocity gives acceleration of the particle executing S.H.M is given by
a = \(\frac{d v}{d t}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\)(Aω cos ωt) = -Aω² sin ωt
∴ a = -ω²y —- (3)

Long Answer Questions

Question 1.
Define simple harmonic motion. Show that the motion of (point) projection of a particle performing uniform circular motion, on any diameter, is simple harmonic. (AP – Mar. ‘18; TS – Mar. ‘16)
Answer:
Simple harmonic motion : A body is said to be in simple harmonic motion, if it moves to and fro along a straight line, about its mean position such that, at any point its acceleration is proportional to its displacement but opposite in direction and directed always towards the mean position.

Show that the projection of uniform circular motion on any diameter is simple harmonic : Consider a particle P moving on the circumference of a circle of radius A with uniform angular velocity ω. Let O be the centre of the circle. XX’ and YY’ are two mutually perpendicular diameters of the circle as shown in the figure. Let PN be drawn perpendicular to the diameter YY’ from P. As P moves on the circumference of the circle, N moves on the diameter YY’ to and fro about the centre O. Let us consider the position of N at any time t, after leaving the point ‘O’, during its motion. The corresponding angular displacement of the particle P is

Hence acceleration is directly proportional to the displacement and opposite direction. Hence motion of N is simple harmonic.

Question 2.
Show that the motion of a simple pendulum is simple harmonic and hence derive an equation for its time period. What is seconds pendulum ? (TS – Mar. ’18, ’17, ’15, ’14, ’13; AP – Mar. ’17, ’16, ’15, ’14, ’13)
Answer:

1. Consider simple pendulum, a small bob of mass m tied to an inextensible mass less string of length L and other end of the string is fixed from a rigid support.
2. Once the bob is slightly displaced and released, it begins to oscillate about mean position.
3. Let θ be the angular displacement and T be the tension in the string.
4. The forces acting on the bob are (a) tension T along the string (b) weight mg acts vertically downwards.
5. The force mg can be resolved into two components (1) mg cos θ along the PA and (2) mg sin θ acts along PB.
   
6. From figure at point P,
   T = mg cos θ —– (1)
7. The force mg sin θ will provide the restoring torque, which tends to bring the bob back to its mean position O.
8. The restoring torque is given by
   \(\tau\) = Restoring force × ⊥^lar distance
   \(\tau\) = -mg sin θ × L —– (2)
   Here negative sign shows that the torque acts to reduce θ.
   Then sin θ is replaced by θ i.e., sin θ ≈ θ
   x = -mg L θ —– (3) (∵ sin θ = θ – \(\frac{\theta^3}{3 !}\) + \(\frac{\theta^3}{5 !}\) …..)
9. From equation (3), we note that \(\tau\) ∝ θ. and This \(\tau\) will bring the bob back towards its equilibrium position.
   So, if the bob is left free, it will execute angular simple harmonic motion.
   Comparing equation (3), with the equation \(\tau\) = -kθ, we have Spring factor, k = mgL.
10. Here inertia factor = Moment of inertia of the bob about the point of suspension = mL2
11. In S.H.M, Time period (T) = \(2 \pi \sqrt{\frac{\text { Inertia factor }}{\text { Spring factor }}}\)
    T = \(2 \pi \sqrt{\frac{m L^2}{m g L}}\)
    T = \(2 \pi \sqrt{\frac{L}{g}}\) — (4)
    Seconds pendulum : A pendulum whose time period is 2 seconds is called seconds pendulum.
    T = 2 seconds.

Question 3.
Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion is constant at any point on its path.
Answer:
Kinetic energy : The velocity of a particle in S.H.M is given by v = ω\(\sqrt{A^2-y^2}\)
∴ Kinetic energy = \(\frac{1}{2} \mathrm{mv}^2\) = \(\frac{1}{2} m \omega^2\left(A^2-y^2\right)\) —— (1)
We know that y = A sin ωt
K.E = \(\frac{1}{2}\)mω²A²[1 – sin²ωt) —— (2)
When y = 0, (K.E)_Max = –\(\frac{1}{2}\)mω²A² (Mean position)
When y = A, (K.E)_Min = 0 (At extreme position)
∴ K.E is maximum at mean position and minimum at extreme position.

Potential energy : When the displacement of a particle executing simple harmonic oscillations increase, the restoring force also increases. The restoring force is in the opposite direction to the displacement. Therefore work is done in moving through the displacement, against restoring force. If F is restoring force at the displacement y.
The average force against which work is done = \(\frac{\mathrm{O}+\mathrm{F}}{2}\) = \(\frac{F}{2}\)

∴ Workdone on the particle for the displacement (y) = Average force × displacement.
i.e., w = \(\frac{1}{2}\) × y
w = \(\frac{\max y}{2}\) —– (3) (∵ F = ma)
But acceleration of a particle in S.H.M is given by
a = -ω²y —– (4)
Using equations (3) and (4), we get
The work done (W) = \(\frac{1}{2}\)mω²y²
This work done is stored in the form of P.E
∴ P.E = \(\frac{1}{2}\)mω²y² —– (5)
∴ P.E = \(\frac{1}{2}\)mω²A² sin²ωt —– (6) (∵ y = A sin ωt)
If y = 0, (P.E)_Min = 0 (At mean position)
If y = A, (P.E)_Max = – \(\frac{1}{2}\)mω²A² (At extreme position)
∴ P.E. is maximum at extreme position and minimum at mean position.

Total energy (E) : The total energy associated with a particle executing S.H.M at any point is the sum of RE and K.E at that point.

Total energy (E) = K.E + P.E
K.E. = \(\frac{1}{2}\)mω²y²
∴ T.E. = \(\frac{1}{2}\)mω² (A² – y²) + \(\frac{1}{2}\)mω²y² = \(\frac{1}{2}\)mω²A²
At mean position y = 0, RE = 0, (KE)_Max = – \(\frac{1}{2}\)mω²A²
∴ T.E = K.E + P.E
T.E = \(\frac{1}{2}\)mω²A² + 0 = \(\frac{1}{2}\)mω²A²
At extreme position, y = A, K.E = 0 and,
(P.E)_Max = \(\frac{1}{2}\)mω²A².
∴ T.E = K.E + P.E

T.E. = O + \(\frac{1}{2}\)mω²A² = \(\frac{1}{2}\)mω²A²
From mean position to extreme position K.E is to be converted into RE.

Problems

Question 1.
The bob of a pendulum is made of a hollow brass sphere. What happens to the time period of the pendulum, if the bob is filled with water completely ? Why?
Solution:
Time period (T) = \(2 \pi \sqrt{\frac{1}{g}}\)
The period of the pendulum is same. When the bob is hollow (or) completely filled with water. As water flows out from the bob, the centre of gravity of the bob lowers.

The pendulum length increases. Hence time period also increases. When the bob becomes empty, again centre of gravity shifts upwards. The pendulum length decreases. The time period also decreases.

Question 2.
Two identical springs of force constant “k” are joined one at the end of the other (in series). Find the effective force constant of the combination.
Solution:
k₁ = k₂ = k
If two springs are connected in series

Question 3.
What are the physical quantities having maximum value at the mean position in SHM ?
Solution:

1. Velocity, V_max = Aω
2. Kinetic energy, (K.E)_Max = \(\frac{1}{2}\)mω²A².

Question 4.
A particle executes SHM such that, the maximum velocity during the oscillation is numerically equal to half the maximum acceleration. What is the time period ?
Solution:
Given V_Max = \(\frac{1}{2}\)a_Max
Aω = \(\frac{1}{2}\)ω²
ω = 2
T = \(\frac{2 \pi}{\omega}\) = \(\frac{2 \pi}{2}\) = π sec.

Question 5.
A mass of 2 kg attached to a spring of force constant 260 Nm^-1 makes 100 oscillations. What is the time taken ?
Solution:
m = 2 kg, k = 260 N/m
T = \(2 \pi \sqrt{\frac{m}{k}}\) = 2 × 3.14\(\sqrt{\frac{2}{260}}\) = 0.5508sec.
∴ Time for 100 oscillations = 100 × 0.5508
= 55.08 sec.

Question 6.
A simple pendulum in a stationery lift has time period T. What would be the effect on the time period when the lift

(i) moves up with uniform velocity
(ii) moves down with uniform velocity
(iii) moves up with uniform acceleration a
(iv) moves down with uniform acceleration ‘a’
(v) begins to fall freely under gravity ?

Solution:

i) When the lift moves up with uniform velocity
T = \(2 \pi \sqrt{\frac{1}{g}}\)
No change in time period.

ii) When the lift moves down with uniform velocity. No change in the time period.

iii) When the lift moves up with acceleration.
T = \(2 \pi \sqrt{\frac{1}{g+a}}\)
Time period decreases.

iv) When the lift moves down with acceleration.
T = 2π\(\sqrt{\frac{1}{g-a}}\)
Time period increases.

v) Lift falls freely, a = g
T = 2π\(\sqrt{\frac{l}{g-g}}\) = 2π\(\sqrt{\frac{1}{0}}\) = ∝
Time period becomes infinity.

Question 7.
A particle executing SHM has amplitude of 4 cm and its acceleration at a distance of 1 cm from the mean position is 3 cm s^-2. What will is velocity be when it is at a distance of 2 cm from its mean position ?
Solution:
A = 4 cm, x₁ = 1 cm, a = 3 cm/s²
a = ω²x₁
3 = ω² × ₁
ω = \(\sqrt{3}\)
Velocity v = ω\(\sqrt{A^2-x_2^2}\)
(∵ x2 = 2 cm)
v = \(\sqrt{3} \sqrt{4^2-2^2}\)
v = \(\sqrt{3} \times \sqrt{12}\)
v = \(\sqrt{36}\) = 6 cm/s.

Question 8.
A simple harmonic oscillator has a time period of 2s. What will be the change in the phase 0.25 s after leaving the mean position ?
Solution:
T = 2 sec
t = 0.25 sec
sin ωt = sin\(\left(\frac{2 \pi}{T}\right) t\)
ϕ = ωt = \(\frac{2 \pi}{T}\) × t
= \(\frac{2 \pi}{2}\) × 0.25
ϕ = \(\frac{\pi}{4}\)

Question 9.
A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. find the acceleration and velocity of the body when the displacement is
(a) 5 cm.
(b) 3 cm.
(c) 0 cm.
Solution:
A = 5 cm = 5 × 10^-2 m
T = 0.2 sec.

i) y = 5 cm = 5 × 10^-2 m
ω = \(\frac{2 \pi}{T}\) = \(\frac{2 \pi}{0.2}\) = 10π
Acceleration (a) = -ω²2y = -(10π)² × 5 × 10^-2
a = -5π² m/s²
Velocity (v) = ω\(\sqrt{A^2-y^2}\)
= 10π\(\sqrt{\left(5 \times 10^{-2}\right)^2-\left(5 \times 10^{-2}\right)^2}\)

ii) y = 3 cm = 3 × 10^-2 m
Acceleration (a) = -ω²y = -(10π)² × 3 × 10^-2
= -3π² m/s²
Velocity (v) = ω\(\sqrt{A^2-y^2}\)
= 10π\(\sqrt{\left(5 \times 10^{-2}\right)^2-\left(3 \times 10^{-2}\right)^2}\)
= 10π\(\sqrt{25-9}\) × 10^-2
v = 0.4π m/s.

iii) y = 0 cm
a = -ω²y = -(10π)² × 0
Velocity (v) = ω\(\sqrt{A^2-y^2}\)
= 10π\(\sqrt{\left(5 \times 10^{-2}\right)^2-0}\)
= 0.5π m/s

Question 10.
The mass and radius of a planet are double that of the earth. If the time period of a simple pendulum on the earth is T, find the time period on the planet.
Solution:
g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\)
g ∝ \(\frac{M}{R^2}\)

Question 11.
Calculate the change in the length of a simple pendulum of length 1 m, when its period of oscillation changes from 2 s to 1.5 s. (T.S. Mar. ’18)
Solution:

Question 12.
A freely falling body takes 2 seconds to reach the ground on a plane, when it is dropped from a height of 8 m. If the period of a simple pendulum is seconds on the planet, calculate the length of the pendulum.
Solution:
u = 0, t = 2 sec, s = h = 8 m
s = ut + \(\frac{1}{2}\)at
s = 0 × t + \(\frac{1}{2}\) × g × 2²
g = 4m/s²
T = \(2 \pi \sqrt{\frac{1}{g}}\)
π = \(2 \pi \sqrt{\frac{1}{4}}\)
l = 1 m = 100 cm.

Question 13.
The period of a simple pendulum is found to increase by 50% when the length of the pendulum is increased by 0.6 m. Calculate the initial length and the initial period of oscillation at a place where g = 9.8 m s^-2.
Solution:

Question 14.
A clock regulated by a second’s pendulum keeps correct time. During summer the length of the pendulum increases to 1.02 m. How much will the clock gain or lose in one day ?
Solution:
T = 2π\(\sqrt{\frac{1}{g}}\)
T ∝ \(\sqrt{l}\)
\(\frac{d T}{T}\) = \(\frac{1}{2} \frac{\mathrm{d} l}{l}\)
T = 2 sec, l = \(\frac{\mathrm{g}}{\pi^2}\) = 0.9927
dl = 1.02 – 0.9927 = 0.0273
\(\frac{d T}{2}\) = \(\frac{1}{2}\left(\frac{0.0273}{0.9927}\right)\)
dT = \(\frac{0.0273}{0.9927}\)
No. of oscillations performed per day by seconds pendulum = \(\frac{\text { One day }}{2 \text { Sec. }}\)
= \(\frac{86,400}{2}\) = 43,200.

The gain (or) loss of time per day = No. of oscillations in one day to the change in time for one oscillation = 43,200 × \(\frac{0.0273}{0.9927}\)
= 1180 sec.

Question 15.
The time period of a body suspended from a spring is T. What will be the new time period if the spring is cut into two equal parts and the mass is suspended
(i) from one part
(ii) simultaneously from both the parts ?
Answer:
T = \(2 \pi \sqrt{\frac{m}{k}}\)
i) Spring is cut into two parts, k’ = 2k
T’ = \(2 \pi \sqrt{\frac{m}{k^{\prime}}}\) = \(\frac{T}{\sqrt{2}}\)
ii) When the mass is suspended from both the parts
T = \(2 \pi \sqrt{\frac{m}{2 k+2 k}}\) = \(2 \pi \sqrt{\frac{m}{4 k}}\) = \(\frac{T}{2}\)

Additional Problems

Question 1.
Which of the following examples represent periodic motion ?
a) A swimmer completing one (return) trip from one bank of a river to the other and back. .
b) A freely suspended bar magnet displaced from its N – S direction and released.
c) A hydrogen molecule rotating about its centre of mass.
d) An arrow released from a bow.
Solution:
a) It is not a periodic motion. Though the motion of a swimmer is to and fro but will not have a definite period.
b) It is a periodic motion because a freely suspended magnet if once displaced from N-S direction and let it go, it oscillation about this position. Hence it is simple harmonic motion also.
c) It is also a periodic motion.
d) It is not a periodic motion.

Question 2.
Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion ?
a) The rotation of earth about its axis.
b) Motion of an oscillating mercury column in a U-tube.
c) Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.
d) General vibrations of a polyatomic molecule about its equilibrium position.
Solution:
a) It is periodic but not SHM because it is not to and fro motion about a fixed point.
b) It is SHM
c) It is SHM
d) It is a periodic but not SHM. A polyatomic gas molecule has a number of natural frequencies and its general motion is the resultant of S.H.M’s of a number of different frequencies. The resultant motion is periodic but not SHM.

Question 3.
Fig. depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion ? What is the period of motion (in case of periodic motion)?

Solution:
a) Does not represent periodic motion, as the motion neither repeats nor comes to mean position.
b) Represents periodic motion with period equal to 2s.
c) Does not represent periodic motion, because it is not identically repeated.
d) Represents periodic motion with periodic equal to 2s.

Question 4.
Which of the following functions of time represent
(a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion ? Give period for each case of periodic motion (ω is any positive constant)
a) sin ωt – cos ωt
b) sin³ ωt
c) 3 cos (π/4 – 2ωt)
d) cos ωt + cos 3ωt + cos 5 ωt
e) exp (-ω²t²)
f) 1 + ωt + ω²t².
Solution:
The function will represent a periodic motion, if it is identically repeated after a fixed interval of time and will represent SHM. If it can be written uniquely in the form of a
cos \(\left(\frac{2 \pi \mathrm{t}}{\mathrm{T}}+\phi\right)\) or a
sin \(\left(\frac{2 \pi \mathrm{t}}{\mathrm{T}}+\phi\right)\), where T is the time period.

a) sin ωt – cos ωt
= \(\sqrt{2}\)(\(\frac{1}{\sqrt{2}}\)sin ωt – \(\frac{1}{\sqrt{2}}\)cos ωt)
= \(\sqrt{2}\)(sin ωt cos \(\frac{\pi}{4}\) – cos ωt sin \(\frac{\pi}{4}\))
= \(\sqrt{2}\)(ωt\(\frac{\pi}{4}\))
It is a S.H.M and its period is 2π/ω.

b) sin³ ωt = \(\frac{1}{4}\)(3sin ωt – sin 3ωt)
Here each term sin ωt and sin 3ωt individually represents S.H.M. But (ii) which is the out come of the super position of two S.H.Ms will only be periodic but not S.H.M. its time period is 2π/ω.
(∵ cos (-θ) = cos θ).

c) 3 cos \(\left(\frac{\pi}{4}-2 \omega t\right)\) = 3 cos \(\left(2 \omega t-\frac{\pi}{4}\right)\)
Clearly it represents S.H.M. and its time period is 2π/2ω.

d) cos ωt + cos 3ωt + cos 5ωt it represents the periodic but not S.H.M its time period is 2π/ω.

e) \(\mathrm{e}^{-\omega^2 t^2}\) it is an exponential function which never repeats itself. Therefore it represents non-periodic motion.

f) 1 + ωt + ω²t² also represents non periodic motion.

Question 5.
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is
a) at the end A
b) at end B
c) at the mid-point of AB going towards A
d) at 2 cm away from B going towards A
e) at 3 cm away from A going towards B and
f) at 4 cm away from B going towards A.
Solution:

Refer figure here A and B represent the two extreme positions of a S.H.M. For velocity, the direction from A to B is taken as positive for acceleration and the force, the direction is taken positive if directed along AP and negative if directed along BP.
a) At the end A, the artice executing S.H.M is momentarily at rest being its extreme position of motion, hence its velocity is zero, acceleration is +ve because directed along AP. Force is also +ve since the force is directed towards AP i.e + ve direction.
b) At the end B, velocity is zero. Here acceleration and force are negative as they are directed along BP i.e. along negative direction.
c) At the mid point AB going towards A, the particle is at its mean position P, with a tendency to move along PA i.e. -ve direction. Hence velocity is -ve both acceleration and force are zero.
d) At 2 cm away from B going towards A, the particle is at Q, with a tendency to move along QP, which is a negative direction there velocity, acceleration and force are all -ve.
e) At 3 cm away from A going towards B, the particle is at R with a tendency to move along RP, which is positive direction, there, velocity, acceleration and force are all +ve.
f) At 4 cm away from A going towards A, the particle is at S, with a tendency to move along SA which is negative direction for velocity. Therefore velocity is negative but acceleration is directed towards mean position i.e., along SP, hence +ve. Similarly force is also +ve.

Question 6.
Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion ?
a) a = 0.7x
b) a = -200x²
c) a = -10x
d) a = 100x³
Solution:
In S.H.M acceleration a is related to displacement by the related of the form a = -kx which is for relation (c).

Question 7.
The motion of a particle executing simple harmonic motion is described by the displacement function.
x(t) = A cos (ωt + ϕ).
If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle ? The angular frequency of the particle is πs^-1. If instead of the cosine function, we choose the sine function to describe the SHM : x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.
Solution:
Here, at t = 0, x = 1 cm and v = ω cm s^-1, ϕ = ?; ω = πs^-1
Given x = A cos (ωt + ϕ)
1 = A cos (π × 0 + ϕ) or
= A cos ϕ —– (i)
Velocity, v = \(\frac{d x}{d t}\) = – Aω(sin ωt + ϕ)
∴ ω = -Aω sin (π × 0 + ϕ) or 1 = – A sin ϕ
or A sin ϕ = -1 —— (ii)
Squaring and adding (i) and (ii)
A²(cos² ϕ + sin² ϕ) = 1 + 1 = 2 or A² = 2 or A = \(\sqrt{2}\) cm
Dividing (ii) by (i), we get
tan ϕ = -1 or ϕ = \(\frac{3 \pi}{4}\) or \(\frac{7 \pi}{4}\)
For, x = B sin (ωt + α) —— (iii)
At t = 0, x = 1, so,
1 = B sin (ω × 0 + α) = B sin α —– (iv)
Differentiating (iii), w.r.t, t we have dx
velocity, v = \(\frac{d x}{d t}\) = Bω cos (ωt + α)
Applying initial conditions i.e. at t = 0, v = ω
ω = Bωcos (π × 0 + α)
or 1 = B cos α —– (v)
Squaring and adding (iv) and (v) we get
B²sin²α + B² cos² α = 1² + 1² = 2 or B² = 2 or B = \(\sqrt{2}\) cm
Dividing (iv) by (v), we have
\(\frac{B \sin \alpha}{B \cos \alpha}\) = \(\frac{1}{1}\) or tan α = 1 or α = \(\frac{\pi}{4}\) or \(\frac{5 \pi}{4}\).

Question 8.
A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body ?
Solution:
Here. m = 50 kg, max. extension.
y = 20 – 0 = 20 cm = 0.2 m; T = 0.65
Max. Force, F = mg = 5 × 9.8 N
K = \(\frac{F}{y}\) = \(\frac{50 \times 9.8}{0.2}\) = 2450 Nm^-1
As T = \(2 \pi \sqrt{\frac{m}{k}}\)
m = \(\frac{T^2 k}{4 \pi^2}\)
= \(\frac{(0.6)^2 \times 2450}{4 \times(3.14)^2}\)
= 22.36 kg
∴ Weight of body = mg = 22.36 × 9.8
= 219.1 N
= 22.36 kgf

Question 9.
A spring having with a spring constant 1200 Nm^-1 is mounted on a horizontal table as shown in fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.

Determine
(i) the frequency of oscillations,
(ii) maximum acceleration of the mass, and
(iii) the maximum speed of the mass.
Solution:
Here, k = 1200 Nm^-1, m = 3.0 kg, a = 2.0 cm = 0.02 m
a)
Frequency,
υ = \(\frac{1}{T}\) = \(\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\)
= \(\frac{1}{2 \times 3.14} \sqrt{\frac{1200}{3}}\)
= 3.2s^-1

b) Acceleration A = ω²y = \(\frac{k}{m}\)y
Acceleration will be maximum when y is maximum i.e. y = a
Max. acceleration, A_max = \(\frac{k_a}{m}\) = \(\frac{1200 \times 0.02}{3}\)
= 8 ms^-2

c) Max. speed of the mass will be when it is passing through the mean position, which is given by
V_max = aω = a\(\sqrt{\frac{k}{m}}\)
= 0.02 × \(\sqrt{\frac{1200}{3}}\) = 0.4 ms^-1

Question 10.
In exercise, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is
a) at the mean position
b) at the maximum stretched position and
c) at the maximum compressed position
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase ?
Solution:
Here, a = 2.0 cm; ω = \(\sqrt{\frac{k}{m}}\) = \(\sqrt{\frac{1200}{3}}\) = 20s^-1

a) As time is noted from the mean position, hence using
x = a sin ωt, we have x = 2 sin 20t

b) At maximum stretched position, the body is the extreme right position, with an initial phase of \(\frac{\pi}{2}\) rad. Then
x = a sin \(\left(\omega t+\frac{\pi}{2}\right)\)
= a cos ωt = 2 cos 20t

c) At maximum compressed position, the body is at the extreme left position, with an initial phase of \(\frac{3 \pi}{2}\) rad.
Then x = a sin (ωt + \(\frac{3 \pi}{2}\))
= -a cos ωt
= -2 cos 20t

Question 11.
Figures correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure.

Obtain the corresponding simple ‘ harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
Solution:
In Fig. (a) T = 2s ; a = 3 cm
At t = 0, OP makes an angle \(\frac{\pi}{2}\) with X-axis
i.e., ϕ = \(\frac{\pi}{2}\) radian.
While moving clockwise, here ϕ = +\(\frac{\pi}{2}\). Thus the X-projection of OP at time t will give us the equation of S.H.M. given by
x = a cos \(\left(\frac{2 \pi t}{T}+\phi\right)\)
= 3cos \(\left(\frac{2 \pi t}{2}+\frac{\pi}{2}\right)\) or x = -3
sin πt (x is in cm)
In Fig. (b) T = 4s ; a = 2m
At t = 0, OP makes an angle π with the positive direction of X-axis i.e., ϕ = π. While moving anticlockwise, here ϕ = +π.
Thus the X-projection of OP at time t will give us the equation of S.H.M. as
x = a cos \(\left(\frac{2 \pi t}{T}+\phi\right)\)
= 2 cos \(\left(\frac{2 \pi t}{4}+\pi\right)\)
= -2 cos \(\left(\frac{\pi}{2} t\right)\) (x is in m)

Question 12.
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case : (x is in cm and t is in s).
a) x = -2 sin (3t + π/3)
b) x = cos (π/6 – t)
c) x = 3 sin (2πt + π/4)
d) x = 2 cos πt.
Solution:
If we express each function of the form
x = a cos (cot + ϕ) —— (i)
Where ϕ is the initial phase i.e., ϕ represents the angle which the initial radius vector of the particle makes with the -l-ve direction of X-axis.

a) x = -2 sin (3t + \(\frac{\pi}{3}\)) = 2 cos (\(\frac{\pi}{2}\) + 3t + \(\frac{\pi}{3}\))
or x = 2 cos (3t + \(\frac{5 \pi}{6}\))
Comparing it with equation (i), we note
that a = 2, ω = 3 and ϕ = \(\frac{5 \pi}{6}\)
Hence, the reference circle will be shown in Fig. (a).

b) x = cos\(\left(\frac{\pi}{6}-t\right)\) = cos \(\left(t-\frac{\pi}{6}\right)\)
(∵ cos (-θ) = cos θ))
Comparing it with equation (i), we note
that a = 1, ω and ϕ = \(\frac{\pi}{6}\)
The reference circle will be as shown in Fig. (b).

c) x = 3 sin \(\left(2 \pi t+\frac{\pi}{4}\right)\)
= 3 cos \(\left(2 \pi t+\frac{3 \pi}{2}+\frac{\pi}{4}\right)\)
Comparing it with equation (i), we note that a = 3, ω = 2π and ϕ = \(\frac{3 \pi}{2}\) + \(\frac{\pi}{4}\) = \(\frac{7 \pi}{4}\)
The reference circle will be as in Fig. (c).

d) x = 2 cos πt
Comparing it with equation (i), we note
that a = 2, = π and ϕ = 0.
The reference circle will be as shown Fig. (d)

Question 13.
Figures (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass at either end. Each end of the spring in Fig. (b) is stretched by the same force F.

a) What is the maximum extension of the spring in two cases ?
b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case ?
Solution:

a) The maximum extension of the spring in both the cases will be = \(\frac{F}{K}\), where K is the spring constant of the spring used.
b) In Fig. (a), if x is the extension in the spring, when mass m is returning to its mean posi-tion after being released free, then restoring force on the mass is F = -Kx i.e., F ∝ x.
As this F is directed towards mean position of the mass, hence the mass attached to the spring will execute SHM. Here, spring factor = spring constant
= K
Inertia factor = mass of the given mass
= m

In Fig.(b). we have a two body system of spring constant K and reduced mass,
μ = \(\frac{m \times m}{m+m}\) = \(\frac{m}{2}\)
Here, inertia factor = \(\frac{m}{2}\)
and spring factor = K
∴ Time period, T = 2π\(\sqrt{\frac{(m / 2)}{k}}\)
= 2π\(\sqrt{\frac{m}{2 K}}\)

Question 14.
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min. what is its maximum speed ?
Solution:
Given a = \(\frac{1}{2}\)m; ω = 200 rev/min ;
V_max = aω
= \(\frac{1}{2}\) × 200
= 100 m/min.

Question 15.
The acceleration due to gravity on the surface of moon is 1.7 ms^-2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 ms^-2).
Solution:

Question 16.
Answer the following questions :
a) Time period of a particle in SHM depends on the force constant k and mass m of the particle :
T = \(2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}\). A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum ?
b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than \(2 \pi \sqrt{\frac{l}{g}}\). Think of a qualitative argument to appreciate this result.
c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall ?
d) What is the frequency of oscillation of a simple pendulum mounted in cabin that is freely falling under gravity ?
Solution:
a) For a simple pendulum, force constant or spring factor K is proportional to mass m, therefore, m cancels out in denominator as well as in numerator. That is why the time period of simple pendulum is independent of the mass of the bob.

b) The effective restoring force acting on the bob of simple pendulum in displaced position is
F = -mg sin θ. When θ is small, sin θ = θ. Then the expression for time period of simple pendulum is given by T = \(2 \pi \sqrt{\mu \mathrm{g}}\) When θ is large sin θ < θ, if the restoring force mg sin θ is replaced by mgθ, this amounts to effective reduction in the value of ‘g’ for large angles and hence an increase in the value of time period T.

c) Yes, because the working of the wrist watch depends on spring action and it has nothing to do with gravity.

d) We know that gravity disappears for a man under free fall, so frequency is zero.

Question 17.
A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period ?
Solution:
Centripetal acceleration, a_c = \(\frac{v^2}{R}\), it is acting horizontally.
Acceleration due to gravity = g acting vertically downwards.
Effective acceleration due to gravity
g’ = \(\sqrt{g^2+\frac{v^4}{R^2}}\)
∴ Time period, T = 2π\(\sqrt{\frac{1}{g}}\)
= 2π\(\sqrt{\frac{1}{g^2+v^4 / R^2}}\)

Question 18.
A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρ₁. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period.
T = \(2 \pi \sqrt{\frac{h \rho}{\rho_1 g}}\)
where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Solution:
Mass of the cylinder (m) = volume × density
= Ahρ ——- (1)
F₁ = weight of the liquid displaced by the length l of the cylinder
= (Al)ρ₁g —— (2)
Weight of the cylinder = mg —– (3)
In equilibrium position, mg = Alρ₁g
m = Alρ₁ —– (4)
F₂ = A(l + y)ρ₁g —– (5)

Restoring force (F) = -(F₂ – mg)
= -[A(l + y)ρ₁g – Alρ₁g]
F = Ayρ₁g = -(Aρ₁g)y —— (6)
In S.H.M. F = -Ky —– (7)
From eqs. (6) & (7),
spring factor (K) = Aρ₁g ——(8)
Inertia factor, m = Ahρ ——- (9)

Question 19.
One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
Solution:
Consider a liquid of density ρ contained in a vertical U-tube of cross-sectional area A. Total length of the liquid column from P to P₁ is L.

Mass (m) = LAρ
PQ = y, P₁Q₁ = y, QQ₁ = 2y
Restoring force (F) = -(A2y)ρg
= -(2Aρg)y —– (1)
F ∝ -y
Hence oscillations in U-tube is S.H.M.

Question 20.
An air chamber of volume V has a neck area of cross section a into which a ball of mass m just fits and can move up and down without any friction (Fig.). Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal (see Fig.).

Solution:
Consider an air chamber of volume V with a long neck of uniform area of cross-section A, and a frictionless ball of mass m filled smoothly in the neck at position C. The pressure of air below the ball inside the chamber is equal to the atmospheric pressure. Increase the pressure on the ball by a little amount P, so that the ball is depressed to position D, where CD = y.
There will be decrease in volume and hence increases in pressure of air inside the chamber. The decrease in volume of the air inside the chamber.

Here, negative sign shows that the increase in pressure will decrease the volume of air in the chamber.
Now, ρ = \(\frac{-E A y}{V}\)
Due to this excess pressure, the restoring force acting on the ball is
F = P × A = \(\frac{-E A y}{V} \cdot A\)
= \(\frac{-E A^2}{V} y\) —- (1)
Since F ∝ y and negative sign shows that the force is directed towards equilibrium position, if the applied increased pressure is removed from the ball, the ball with start executing linear SHM in the neck of chamber with C as mean position.
In a S.H.M., the restoring force,
F = -Ky —- (2)
Comparing (1) and (2), we have spring factor.
K = \(\frac{E A^2}{V}\)
Here, inertia factor = mass of ball = m
Inertia factor Spring factor

Note : If the ball oscillates in the neck of chamber under isothermal conditions, then E = P = Pressure of air inside the chamber, when ball is at equilibrium position. If the ball oscillates in the neck of chamber under adiabatic conditions, then E = υP, where v = \(\frac{c_p}{c_v}\).

Question 21.
You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.
Solution:
Here, M = 3000 kg ; x = 0.15 cm ; if K is the spring constant of each spring, then spring constant of 4 springs in parallel to support the whole mass is, K = 4 K.
4 kx = Mg
k = \(\frac{M g}{4 x}\)
= \(\frac{3000 \times 10}{4 \times 0.15}\)
= 5 × 10⁴ N/m.

b) If m is the mass supported by each spring, then m = \(\frac{3000}{4}\) = 750 kg.

Question 22.
Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
Solution:
Consider a particle of mass m be executing S.H.M. with period T. The displacement of the particle at an instant t, when time period is noted from the mean position is given by
Y = a sin ωt
∴ Velocity 1 V = \(\frac{\mathrm{dy}}{\mathrm{d} \mathrm{t}}\) = aω cos ωt
K.E., E_K = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\)ma²ω² cos² ωt

Average P.E. over one cycle is

Question 23.
A circular disc of mass 10 kg is suspended by a wire attached to is centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant a is defined by the relation J = -α θ, where J is the restoring couple and θ the angle of twist).
Solution:
Here, m = 10kg; R = 15 cm = 0.15m; T = 1.55, ∝ = ?
Moment of inertia of disc,
I = \(\frac{1}{2} m R^2\)
= \(\frac{1}{2}\) × 10 × (0.15)² kgm²
Now T = \(2 \pi \sqrt{\frac{1}{\alpha}}\)
so, α = \(\frac{4 \pi^2 1}{T^2}\)
= 4 × \(\left(\frac{22}{7}\right)^2\) × \(\frac{1}{2}\) × \(\frac{10 \times(0.15)^2}{(1.5)^2}\)
= 1.97Nm/rad.

Question 24.
A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is
(a) 5 cm
(b) 3 cm
(c) 0 cm.
Solution:
Here, r = 5 cm = 0.05 m ; T = 0.25 ;
ω = \(\frac{2 \pi}{\mathrm{T}}\) = \(\frac{2 \pi}{0.2}\)
= 10p rad/s
When displacement is y, then acceleration
A = -ω²y
Velocity, v = ω\(\sqrt{r^2-y^2}\)

Case (a) : When y = x cm = 0.05 m
A = -(10π)² × 0.05
= -5π² m/s²
V = 10π\(\sqrt{(0.05)^2-(0.05)^2}\) = 0

Case (b) : When y = 3 cm = 0.03 m
A = -(10π)² × 0.03
= -3π² m/s²
V= 10π × \(\sqrt{(0.05)^2-(0.03)^2}\)
= 10π × 0.04
= 0.4π m/s

Case (c): When y = 0,
A = -(10π)² × 0 = 0
V= 10π\(\sqrt{(0.05)^2-(0)^2}\)
= 10π × 0.05
= 0.5π m/s.

Question 25.
A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x₀ and pushed towards the centre with a velocity v₀ at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x₀ and V₀.
[Hint : Start with the equation x = a cos (ωt + θ) and note that the initial velocity is negative.]
Solution:
x = A cos (ωt + θ) dx
Velocity, \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = -Aω sin (ωt + θ) dt
When t = 0, x = x₀, and \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = -V₀
∴ x₀ = A cos θ
-V₀ = -Aω sin θ or A sin θ = \(\frac{V_0}{\omega}\)
Squaring and adding- (i) and (ii), we get
A²(sin² θ + cos² θ) = \(\left(\frac{v_0^2}{\omega^2}\right)+x_0^2\)
A = \(\left[\frac{v_0^2}{\omega^2}+x_0^2\right]^{1 / 2}\)

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 the other hands (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 because of the 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.