Hc Verma II for Class 12 Science Physics Chapter 23 - Heat And Temperature

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Page No 11:

Answer:

If two bodies are in thermal equilibrium in one frame, they will be in thermal equilibrium in all the frames. In case there is any change in temperature of one body due to change in frame, the same change will be acquired by the other body.

Page No 11:

Answer:

No, the temperature of a body is not dependent on the frame from which it is observed. This is because atoms /molecules of matter move or vibrate in all possible directions. Increase in velocity at a particular direction of the container/ matter does not increase or decrease the overall velocity of the molecules/atoms because of the random collisions the entities suffer. So, there is no net rise in temperature of the system.

Page No 11:

Answer:

It is not illogical to say that mercury expands uniformly before temperature scale was defined. It's uniform expansion can be studied by comparing the expansion of mercury with expansion of other substances (like alcohol water etc).

Page No 11:

Answer:

The ideal gas thermometer is based on the ideal gas equation, PV=nRT, where P is pressure of the gas at constant volume V with n number of moles at temperature T. Therefore, P = constant $\times$ T. According to this relation, if the volume of the gas used is constant, the pressure will be directly proportional to the temperature of the gas. We need not use kinetic theory of gases or any experimental results.

Page No 11:

Answer:

The bulb of a thermometer plays an important role in measuring the temperature of the surrounding body. It is put in contact with the body whose temperature is to be measured. The bulb attains the temperature of the body, which allows calibration of temperature. If the bulb is made of an adiabatic wall, then no heat will be transferred through the wall and the bulb cannot attain thermal equilibrium with the surrounding body. Therefore, the bulb cannot be made of an adiabatic wall.

Page No 11:

Answer:

Water possesses an anomalous behavour. The volume of a given amount of water decreases as it is cooled from room temperature, until its temperature reaches 4 °C. Below 4 °C, the volume increases, and therefore the density decreases.

When the temperature of the surface of lake falls in winter, the water at the surface becomes denser and sinks. As, the temperature reaches below 4^oC , the density of the water at surface becomes less. Thus, it remains at surface and freezes. As, the ice is a bad conductor of heat, it traps the heat present in the lake's water beneath itself. Hence, no further cooling of water takes place once the top layer of the lake is completely covered by ice. Thus the life of the marine animals inside the lake is possible.

Page No 11:

Answer:

On a hot summer day, metals tend to expand due to the heat. Different metals have different expansion coefficients. The coefficient of linear expansion of aluminium is more than that of brass. Therefore, it'll expand more than brass, leading to an apparent decrease in length of the brass rod, as measured by the aluminium scale. So, we cannot conclude that brass shrinks on heating. Instead, aluminium expands more than brass on heating.

Page No 11:

Answer:

Yes, we can make a mercury thermometer in a glass tube. Mercury and glass have equal coefficients of volume expansion. So, when temperature changes, the increase in the volume of the glass tube as which is equal to the real increase in volume minus the increase in the volume of the container, would be zero. Hence, it will give correct reading at every temperature.

Page No 11:

Answer:

At sea level, the pressure is around 1 atmosphere and at high altitude, the density of air reduces.
Pressure of liquid,
$P = h ρ g, where ρ = density of fluid$
The above equation shows that pressure depends on density. Therefore at 4^oC, the density of water will be less at high altitude, compared to the density at sea level.

Page No 11:

Answer:

When a bottle with a tightly-closed metal lid is put in hot water for sometime, its lid can be opened easily because metals have greater coefficient of expansion than glass. Therefore, when the metal lid comes in contact with hot water, it'll expand more than the glass container. As a result, it will be easier to open the bottle.

Page No 11:

Answer:

In a hot engine the hot parts are expanded because of heat, if cold water is poured suddenly then there will be uneven thermal contraction in the parts. This will result in a stress to develop between the various parts of the engine and may let the engine to crack down.

Page No 11:

Answer:

Two bodies are said to be in thermal equilibrium if they are at the same temperature. Consider two bodies A and B that are not in contact with each other but in contact with a heat reservoir. Since both the bodies will attain the temperature of the reservoir, they will be at the same temperature and, hence, in thermal equilibrium. Therefore, it is possible to have two bodies in thermal equilibrium even though they are not in contact.

Page No 11:

Answer:

When a spherical shell is heated, its volume changes according to the equation, $V_{θ} = V_{0} (1 + γ ∆ θ)$ . The volume referred to here is the volume of the material used to make up the shell, as its volume expands with the rise of temperature with coefficient of expansion of volume, $γ$ .

Page No 11:

Answer:

(c) may be in thermal equilibrium

The given data in the question is insufficient to specify the relation between the physical conditions of systems Y and Z. As system X is not in thermal equilibrium with Y and Z, systems Y and Z may be at the same temperature or they may or may not be in thermal equilibrium with each other. So, the only possible option is (c).

Page No 11:

Answer:

(a)

Celsius and Fahrenheit temperatures are related in the following way:
$C = \frac{5}{9} F - \frac{160}{9}$
Here, F = temperature in Fahrenheit
C = temperature in Celsius
If this equation is plotted on the graph, then the curve will be represented by curve 'a' lying in the fourth quadrant with slope 5/9.
So, the correct option is (a).

Page No 11:

Answer:

(a) Fahrenheit and Kelvin

Let θ be the temperature in Fahrenheit and Kelvin scales.

We know that the relation between the temperature in Fahrenheit and Kelvin scales is given by
$\frac{T_{F} - 32}{180} = \frac{T_{K} - 273.15}{100}$
T_F = T_K = θ

Therefore,
$\frac{θ - 32}{180} = \frac{θ - 273.15}{100} 5 θ - 160 = 9 θ - 2458.5 4 θ = 2298.35 θ = 574.59^{o}$

If we consider the same for Celsius and Kelvin scales
$\frac{T_{C} - 0}{100} = \frac{T_{K} - 273.15}{100}$

Let the temperature be t
$\frac{t - 0}{100} = \frac{t - 273.15}{100} t = t - 273.15$
Thus, t does not exist.

The Kelvin scale uses mercury as thermometric substance, whereas the platinum scale uses platinum as thermometric substance. The scale depends on the properties of the thermometric substance used to define the scale. The platinum and Kelvin scales do not agree with each other. Therefore, there is no such temperature that has same numerical value in the platinum and Kelvin scale.

Page No 11:

Answer:

(c) high temperature and low pressure.

A constant-volume gas thermometer should be filled with an ideal gas in which particles don't interact with each other and are free to move anywhere, so that the thermometer functions properly. An ideal gas is only a theoretical possibility. Therefore, the gas that is filled in the thermometer should be at high temperature and low pressure, as under these conditions, a gas behaves as an ideal gas.

Page No 11:

Answer:

(a) A and B are correct.

The coefficient of linear expansion,
$α = \frac{1}{L} \frac{△ L}{△ T}$
$= \frac{[L]}{[L T]} = K^{- 1}$
Here, L = initial length
$△$ L = change in length
$△$ T = change in temperature
On the other hand, the coefficient of volume expansion,
$γ = \frac{1}{V} \frac{△ V}{△ T} = \frac{[L^{3}]}{[L^{3} T]} = K^{- 1}$
Here, V = initial volume
$△$ V = change in volume
$△$ T = change in temperature
K = kelvin, the S.I. unit of temperature

Page No 12:

Answer:

(a) gets larger

When a metal sheet is heated, it starts expanding and its surface area will start increasing, which will lead to an increase in the radius of the hole. Hence, the circular hole will become larger.

Page No 12:

Answer:

(b) bend with copper on convex side

We are provided with two metal strips of copper and steel. On heating, both of them will expand. Expansion coefficient of copper is more than that of steel. So, the copper metal strip will expand more, causing the bimetallic strip to bend with copper at the convex side, as it'll have more surface area compared to the steel sheet, which will be on the concave side.

Page No 12:

Answer:

(c) 2αI∆t

The change in moment of inertia of uniform rod with change in temperature is given by,
$I' = I (1 + 2 α Δ t)$
Here, I = initial moment of inertia
I' = new moment of inertia due to change in temperature
$α$ = expansion coefficient
$∆$ t = change in temperature
So, $I' - I = 2 α I Δ t$

Page No 12:

Answer:

(c) 2αI∆t

The moment of inertia of a solid body of any shape changes with temperature as
$I' = I (1 + 2 α Δ t)$
Here, I = initial moment of inertia
           I' = new moment of inertia due to change in temperature
           $α$ = expansion coefficient
          $Δ$ t = change in temperature
So, $I' - I = 2 α I Δ t$ I−I0=2αΔt

Page No 12:

Answer:

(c) 4^oC

The density of water is maximum at 4^oC, and the water at the bottom of the lake is most dense, compared to the layers of water above. Therefore, the temperature expected at the bottom is 4^oC.

Page No 12:

Answer:

(b) will decrease

When an aluminium sphere is dipped in water and the temperature of water is increased, the aluminium will start expanding leading to increase in its volume. This will lead to increase in the surface area of the shell and it'll exert less pressure on the water such that the volume of the sphere submerged in water will decrease and it'll start float easily on water. Now, the volume of water displaced will be less compared to what was displaced initially. Therefore, the force of buoyancy will decrease, as it is directly proportional to the volume of water displaced.

Page No 12:

Answer:

(a) Kinetic
(c) Mechanical

The kinetic energy of a body depends on its speed. Since when a spinning wheel is slowed down, its speed decreases leading to reduction in its kinetic energy. The mechanical energy of a body is defined as the sum of its potential and kinetic energies. Since the kinetic energy of the wheel has been decreased, it'll lead to decrease in its mechanical energy. When the wheel slows down due to friction, its mechanical energy gets converted into heat energy, leading to increase in internal energy, which increases with increase in temperature.

Page No 12:

Answer:

(a) Kinetic
(b) Total
(c) Mechanical
(d) Internal

When the wheel B starts spinning because of the friction at contact, it will gain kinetic energy and, hence, mechanical energy (kinetic + potential energies). Also, internal energy will increase, which increases with rise in temperature. Along with it, the generation of heat energy due to friction will lead to increase in the net sum of all the energies, i.e. total energy.

Page No 12:

Answer:

(a) Kinetic
(b) Total
(c) Mechanical

As body A is at rest on the ground, it possesses only potential energy, whereas body B, being placed inside a moving train, possesses kinetic energy due to its motion along with the train. Therefore, body B will have greater kinetic, mechanical (energy possessed by the body by virtue of its position and motion = kinetic energy+potential energy) energy and, hence, total (sum of all the energies) energy. No information is given about the temperature of the body so we can not say wheather body B^' s internal energy will be or will not be greater than that of body A.

Page No 12:

Answer:

(c) Celsius scale and ideal gas scale
(d) Ideal gas scale and absolute scale

Celsius scale and ideal gas scale measure temperature in kelvin (K) and the ideal gas scale is sometimes also called the absolute scale. A mercury scale gives reading in degrees and its size of degree, which depends on length of mercury column, doesn't match any of the above-mentioned scales.

Page No 12:

Answer:

(a) must have increased.

The whole system (water + solid object) is enclosed in an adiabatic container from which no heat can escape. After some time, the temperature of water falls, which implies that the heat from the water has been transferred to the object, leading to increase in its temperature.

Page No 12:

Answer:

(b) increases
In general, the time period of a pendulum,t, is given by
$t = \frac{1}{2 π} \sqrt{\frac{l}{g}}$ .
When the temperature (T) is increased, the length of the pendulum (l) is given by,
$l = l_{0} (1 + α T)$ ,
where l₀ = length at 0 ^oC
$α$ = linear coefficient of expansion.
Therefore, the time period of a pendulum will be
$t = \frac{1}{2 π} \sqrt{\frac{l_{0} (1 + α T)}{g}}$
Hence, time period of a pendulum will increase with increase in temperature.

Page No 12:

Answer:

Given:
Ice point of a mercury thermometer, T₀ = 20° C
Steam point of a mercury thermometer, T₁₀₀ = 80° C
Temperature on thermometer that is to be calculated in centigrade scale, T₁ = 32° C
Temperature on a centigrade mercury scale, T, is given as:
$T = \frac{T_{1} - T_{0}}{T_{100} - T_{0}} \times 100 \Rightarrow T = \frac{32 - 20}{80 - 20} \times 100 \Rightarrow T = \frac{12}{60} \times 100 \Rightarrow T = \frac{120}{6} \Rightarrow T = 20 ° C$
Therefore, the temperature on a centigrade mercury scale will be 20^oC.

Page No 12:

Answer:

Given:
Pressure registered by a constant-volume thermometer at the triple point, P_tr = 1.500 × 10⁴ Pa
Pressure registered by the thermometer at the normal boiling point, P = 2.050 × 10⁴ Pa
We know that for a constant-volume gas thermometer, temperature (T) at the normal boiling point is given as:
$T = \frac{P}{P_{t r}} \times 273.16 K \Rightarrow T = \frac{2.050 \times 10^{4}}{1.500 \times 10^{4}} \times 273.16 K \Rightarrow T = 373.31 K$
Therefore, the temperature at the normal point (T) is 373.31 K.

Page No 12:

Answer:

Given:
In a gas thermometer, the pressure measured at the melting point of lead, P = 2.20 × Pressure at triple point(P_tr)
So the melting point of lead,(T) is given as:
$T = \frac{P}{P_{t r}} \times 273.16 K \Rightarrow T = \frac{2.20 \times P_{t r}}{P_{t r}} \times 273.16 K \Rightarrow T = 2.20 \times 273.16 K \Rightarrow T = 600.952 K \Rightarrow T ≃ 601 K$
Therefore, the melting point of lead is 601 K.

Page No 12:

Answer:

Given:
Pressure measured by a constant volume gas thermometer at the triple point of water, P_tr = 40 kPa = 40 × 10³ Pa
Boiling point of water, T = 100°C = 373.16 K
Let the pressure measured at the boiling point of water be P.
For a constant volume gas thermometer, temperature-pressure relation is given below:
$T = \frac{P}{P_{t r}} \times 273.16 K \Rightarrow P = \frac{T \times P_{t r}}{273.16} \Rightarrow P = \frac{373.16 \times 40 \times 10^{3}}{273.16} \Rightarrow P = 54643 Pa \Rightarrow P = 54.6 \times 10^{3} Pa \Rightarrow P ≃ 55 kPa$
Therefore, the pressure measured at the boiling point of water is 55 kPa.

Page No 12:

Answer:

Given:
Temperature of ice point, T₁ = 273.15 K
Temperature of steam point, T₂ = 373.15 K
Pressure of the gas in a constant volume thermometer at the ice point, P₁ = 70 kPa,
Let P_trbe the pressure at the triple point and P₂be the pressure at the steam point.
The temperature-pressure relations for ice point and steam point are given below:
For ice point,
$T_{1} = \frac{P_{1}}{P_{t r}} \times 273.16 K$
$\Rightarrow 273.15 = \frac{70}{P_{t r}} \times 10^{3} \times 273.16 \Rightarrow P_{t r} = \frac{70 \times 273.16 \times 10^{3}}{273.15} Pa$

For steam point,
$T_{2} = \frac{P_{2} \times 273.16}{P_{t r}} K$
On substituting the value of P_tr ,we get:
$373.15 = \frac{P_{2} \times 273.15 \times 273.16}{70 \times 273.16 \times 10^{3}} \Rightarrow P_{2} = \frac{373.15 \times 70 \times 10^{3}}{273.15} \Rightarrow P_{2} = 95.626 \times 10^{3} Pa \Rightarrow P_{2} ≃ 96 kPa$
Therefore, the pressure at steam point is 96 kPa.

Page No 12:

Answer:

Given:
In a constant volume gas thermometer,
Pressure of the gas at the ice point, P₀ = 80 cm of Hg
Pressure of the gas at the steam point, P₁₀₀ = 90 cm of Hg
Pressure of the gas in a heated wax bath, P = 100 cm of Hg
The temperature of the wax bath $(T)$ is given by:
$T = \frac{P - P_{0}}{P_{100} - P_{0}} \times 100 ° C \Rightarrow T = \frac{100 - 80}{90 - 80} \times 100 \Rightarrow T = \frac{20}{10} \times 100 \Rightarrow T = 200 ° C$
Therefore, the temperature of the wax bath is 200^o C.

Page No 12:

Answer:

Given:
Volume of the bulb in a Callender's compensated constant pressure air thermometer, (V) =
1800 cc
Volume of mercury that has to be poured out, V' = 200 cc
Temperature of ice bath, T_o = 273.15 K
So the temperature of the vessel(T') is given by:
$T' = \frac{V}{V - V'} \times T_{0} \Rightarrow T' = \frac{1800}{1600} \times 273.15 K \Rightarrow T' = 307.293 \Rightarrow T' ≃ 307 K$
Therefore, the temperature of the vessel is 307 K.

Page No 12:

Answer:

Given:
Resistance at 0^oC, R₀ = 80 $Ω$
Resistance at 100^oC, R₁₀₀ = 90 $Ω$
Let t be the temperature at which the resistance (R_t) is 86 $Ω$ .
$t = \frac{R_{t} - R_{0}}{R_{100} - R_{0}} \times 100 \Rightarrow t = \frac{86 - 80}{90 - 80} \times 100 \Rightarrow t = \frac{6}{10} \times 100 \Rightarrow t = 60 °$
Therefore, the resistance is 86 $Ω$ at 60^oC.

Page No 13:

Answer:

Given:
Reading on resistance thermometer at ice point, R₀ = 20 Ω
Reading on resistance thermometer at steam point, R₁₀₀ = 27.5 Ω
Reading on resistance thermometer at zinc point, R₄₂₀ = 50 Ω
The variation of resistance with temperature in Celsius scale,θ, is given as:
$R_{100} = R_{0} (1 + α θ + β θ^{2}) \Rightarrow R_{100} = R_{0} + R_{0} α θ + R_{0} β θ^{2} \Rightarrow R_{100} = R_{0} + R_{0} α θ + R_{0} β θ^{2} \Rightarrow \frac{(R_{100} - R_{0})}{R_{0}} = α θ + β θ^{2} \Rightarrow \frac{27.5 - 20}{20} = α θ + β θ^{2} \Rightarrow \frac{7.5}{20} = α \times 100 + β \times 10000 . . . (i) Also, R_{420} = R_{0} (1 + α θ + β θ^{2}) \Rightarrow R_{420} = R_{0} + R_{0} α θ + R_{0} β θ^{2} \Rightarrow \frac{R_{420} - R_{0}}{R_{0}} = α θ + β θ^{2} \Rightarrow \frac{50 - 20}{20} = 420 α + 176400 β \Rightarrow \frac{3}{2} = 420 α + 176400 β . . . (i i)$
Solving (i) and (ii), we get:
α = 3.8 ×10^–3°C^-1
β = –5.6 ×10^–7°C^-1
Therefore, resistance R₀is 20 Ω and the value of α is 3.8 ×10^–3°C^-1and that of β is –5.6 ×10^–7°C^-1.

Page No 13:

Answer:

Given:
Length of the slab when the temperature is 0°C, L₀ = 10 m
Temperature on the summer day, t = 35 °C
Let L₁ be the length of the slab on a summer day when the temperature is 35°C.
The coefficient of linear expansion of concrete, α = 1 ×10^–5 °C^-1
$L_{1} = L_{0} (1 + α t)$
$\Rightarrow L_{1}$ = 10 (1 + 10^–5 × 35)
$\Rightarrow L_{1}$ = 10 + 35 × 10^–4
$\Rightarrow L_{1}$ = 10.0035 m
So, the length of the slab on summer day when the temperature is 35^oC is 10.0035 m.

Page No 13:

Answer:

Given:
Temperature at which the steel metre scale is calibrated, t₁ = 20^oC
Temperature at which the scale is used, t₂ = 10^oC
So, the change in temperature, $Δ$ t = (20^o $-$ 10^o) C
The distance to be measured by the metre scale, L_o = (51 $-$ 50) = 1 cm = 0.01 m
Coefficient of linear expansion of steel, $α_{s t e e l}$ = 1.1 × 10^–5 °C^–1
Let the new length measured by the scale due to expansion of steel be L₂, Change in length is given by,
$∆ L = L_{1} α_{s t e e l} Δ t \Rightarrow ∆ L = 1 \times 1.1 \times 10^{- 5} \times 10 \Rightarrow ∆ L = 0.00011 cm$
As the temperature is decreasing, therefore length will decrease by $∆ L$ .
Therefore, the new length measured by the scale due to expansion of steel (L₂) will be,
L₂ = 1 cm $-$ 0.00011 cm = 0.99989 cm

Page No 13:

Answer:

Given:
Length of the iron sections when there's no effect of temperature on them, L_o = 12.0 m
Temperature at which the iron track is laid in winter, t_w = 18^oC
Maximum temperature during summers, t_s = 48^oC
Coefficient of linear expansion of iron, $α$ = 11 × 10^–6 °C^–1
Let the new lengths attained by each section due to expansion of iron in winter and summer be L_wand L_s_,respectively, which can be calculated as follows:
$L_{w} = L_{0} (1 + α t_{w}) \Rightarrow L_{w} = 12 (1 + 11 \times 10^{- 6} \times 18) \Rightarrow L_{w} = 12.00237 m L_{s} = L_{0} (1 + α t_{s}) \Rightarrow L_{s} = 12 (1 + 11 \times 10^{- 6} \times 48) \Rightarrow L_{s} = 12.006336 m ∴ ∆ L = L_{s} - L_{w} \Rightarrow Δ L = 12.006336 - 12.002376 \Rightarrow Δ L = 0.00396 m \Rightarrow Δ L \approx 0.4 cm$
Therefore, the gap ( $Δ$ L) that should be left between two iron sections, so that there is no compression during summer, is 0.4 cm.

Page No 13:

Answer:

Given:
Diameter of a circular hole in an aluminium plate at 0°C, d₁ = 2 cm = 2 × 10^–2 m
Initial temperature, t₁ = 0 °C
Final temperature, t₂ = 100 °C
So, the change in temperature, ( $Δ$ t) = 100°C - 0°C = 100°C
The linear expansion coefficient of aluminium, α_al = 2.3 × 10^–5 °C^–1
Let the diameter of the circular hole in the plate at 100^oC be d₂, which can be written as:
$d_{2} = d_{1} (1 + α Δ t)$
$\Rightarrow d_{2}$ = 2 × 10^–2(1 + 2.3 × 10^–5× 10²)
$\Rightarrow d_{2}$ = 2 × 10^–2 (1 + 2.3 × 10^–3)
$\Rightarrow d_{2}$ = 2 × 10^–2 + 2.3 × 2 × 10^–5
$\Rightarrow d_{2}$ = 0.02 + 0.000046
$\Rightarrow d_{2}$ = 0.020046 m
$\Rightarrow d_{2}$ ≈ 2.0046 cm
Therefore, the diameter of the circular hole in the aluminium plate at 100^oC is 2.0046 cm.

Page No 13:

Answer:

Given:
At 20°C, length of the metre scale made up of steel, L_st= length of the metre scale made up of aluminium, L_al
Coefficient of linear expansion for aluminium, α_al = 2.3 × 10^–5°C^-1
Coefficient of linear expansion for steel, α_st = 1.1 × 10^–5°C^-1
Let the length of the aluminium scale at 0°C, 40°C and 100°C be L_0al,L₄_0aland L₁₀_0al.
And let the length of the steel scale at 0°C, 40°C and 100°C be L_0st,L₄_0stand L₁₀_0st.
(a) So, L_0st(1 – α_st × 20) = L_0al(1 – α_al × 20)
$\frac{L_{0 s t}}{L_{0 a l}} = \frac{(1 - α_{a l} \times 20)}{(1 - α_{s t} \times 20)} \Rightarrow \frac{L_{0 s t}}{L_{0 a l}} = \frac{1 - 2.3 \times {10^{-}}^{5} \times 20}{1 - 1.1 \times 10^{- 5} \times 20} \Rightarrow \frac{L_{0 s t}}{L_{0 a l}} = \frac{0.99954}{0.99978} \Rightarrow \frac{L_{0 s t}}{L_{0 a l}} = 0.999759$
(b)
$\frac{L_{40 a l}}{L_{40 s t}} = \frac{L_{0 a l} (1 + α_{a l} \times 40)}{L_{0 s t} (1 + α_{s t} \times 40)} \Rightarrow \frac{L_{40 a l}}{L_{40 s t}} = \frac{L_{0 a l}}{L_{0 s t}} \times \frac{(1 + 2.3 \times 10^{- 5} \times 40)}{(1 + 1.1 \times 10^{- 5} \times 40)} \Rightarrow \frac{L_{40 a l}}{L_{40 s t}} = \frac{0.99977 \times 1.00092}{1.00044} \Rightarrow \frac{L_{40 a l}}{L_{40 s t}} = 1.0002496$
(c)
$\frac{L_{100 a l}}{L_{100 s t}} = \frac{L_{0 a l} (1 + α_{a l} \times 100)}{L_{0 s t} (1 + α_{s t} \times 100)} \Rightarrow \frac{L_{100 a l}}{L_{100 s t}} = \frac{0.99977 \times 1.0023}{1.0023} \Rightarrow \frac{L_{100 a l}}{L_{100 s t}} = 1.00096$

Page No 13:

Answer:

(a) Let the correct length measured by a metre scale made up of steel 16 °C be L.
Initial temperature, t₁ = 16 °C
Temperature on a hot summer day, t₂ = 46 °C
So, change in temperature, Δθ = t₂ $-$ t₁= 30 °C
Coefficient of linear expansion of steel, $α$ = 1.1 × 10^–5°C^-1
Therefore, change in length,
ΔL = L αΔθ = L × 1.1 × 10^–5 × 30
$% of error = (\frac{∆ L}{L} \times 100) % = (\frac{L α Δ θ}{L} \times 100) % = [1.1 \times 10^{- 5} \times 30 \times 100] % = 3.3 \times 10^{- 2} %$

(b) Temperature on a winter day, t₂ = 6 °C
So, change in temperature, Δθ = t₁ $-$ t₂= 10 °C
ΔL = L₂ $-$ L₁= L αΔθ = L × 1.1 × 10^–5 × 10
$% of error = (\frac{∆ L}{L} \times 100) % = (\frac{L α Δ θ}{L} \times 100) % = 1.1 \times 10^{- 5} \times 10 \times 100 % = 1.1 \times 10^{- 2}$

Page No 13:

Answer:

Given:
Temperature at which a metre scale gives an accurate reading, T₁ = 20 °C
The value of variation admissible, ΔL = 0.055 mm = 0.055 × 10^–3 m, in the length, L₀ = 1 m
Coefficient of linear expansion of steel, α = 11 × 10^–6 °C^–1
Let the range of temperature in which the experiment can be performed be T₂.
We know: ΔL = L₀ αΔT
$\Rightarrow 0.055 \times 10^{- 3} = 1 \times 11 \times 10^{- 6} \times (T_{1} \pm T_{2}) \Rightarrow 5 \times 10^{- 3} = (20 \pm T_{2}) \times 10^{- 3} \Rightarrow 20 \pm T_{2} = 5 \Rightarrow E ither T_{2} = 20 + 5 = 25 ° C or T_{2} = 20 - 5 = 15 ° C$
Hence, the experiment can be performed in the temperature range of 15 °C to 25 °C .

Page No 13:

Answer:

Given:
Density of water at 0°C, ( f₀)= 0.998 g cm^-3
Density of water at 4°C, (f₄) = 1.000 g cm-3
Change in temperature, (Δt) = 4^oC
Let the average coefficient of volume expansion of water in the temperature range of 0 to 4°C be γ.
$We know : f_{4} = f_{0} (1 + γ ∆ t) \Rightarrow f_{0} = \frac{f_{4}}{1 + γ ∆ t} \Rightarrow 0.998 = \frac{1}{1 + γ . 4} \Rightarrow 1 + 4 γ = \frac{1}{0.998} \Rightarrow 4 γ = (\frac{1}{0.998}) - 1 \Rightarrow γ = 0.0005 = 5 \times 10^{- 4}^{o} C^{- 1}$
As the density decreases,
$γ = - 5 \times 10^{- 4}^{o} C^{-1}$
Therefore,the average coefficient of volume expansion of water in the temperature range of 0 to 4°C will be $γ = - 5 \times 10^{- 4}$ ^oC^-1.

Page No 13:

Answer:

Let the original length of iron rod be L_Fe and L^'_Febe its length when temperature is increased by ΔT.
Let the original length of aluminium rod be L_Al and L^'_Albe its length when temperature is increased by ΔT.

Coefficient of linear expansion of iron, $α_{F e}$ = 12 × 10^–6 °C^{$-$ 1}
Coefficient of linear expansion of aluminium, α_Al = 23 × 10^–6 °C^{$-$ 1}
Since the difference in length is independent of temperature, the difference is always constant.
$L'_{F e} = L_{F e} (1 + α_{F e} \times ∆ T) and L'_{A l} = L_{A l} (1 + α_{A l} \times ∆ T) \Rightarrow \begin{matrix} L'_{F e} - L'_{A l} = L_{F e} - L_{A l} + L_{F e} \times α_{F e} ∆ T - L_{A l} \times α_{A l} \times ∆ T & - (1) \end{matrix} Given : L'_{F e} - L'_{A l} = L_{F e} - L_{A l} Hence, L_{F e} α_{F e} = L_{A l} α_{A l} [using (1)] \Rightarrow \frac{L_{F e}}{L_{A l}} = \frac{23}{12}$
The ratio of the lengths of the iron to the aluminium rod is 23:12.

Page No 13:

Answer:

Given:
The temperature at which the pendulum shows the correct time, T₁ = 20 °C
Coefficient of linear expansion of steel, $α$ = 12 × 10^–6 °C^–1
Let T₂ be the temperature at which the value of g is 9.788 ms^–2and $Δ$ T be the change in temperature.
So, the time periods of pendulum at different values of g will be t₁ and t₂ , such that
$t_{1} = 2 π \sqrt{\frac{l_{1}}{g_{1}}} t_{2} = 2 π \sqrt{\frac{l_{2}}{g_{2}}} = \frac{2 π \sqrt{l_{1} (1 + α Δ T)}}{\sqrt{g_{2}}} (∵ l_{2} = l_{1} (1 + α ∆ T)) Given, t_{1} = t_{2} \Rightarrow \frac{2 π \sqrt{l_{1}}}{\sqrt{g_{1}}} = \frac{2 π \sqrt{l_{1} (1 + α ∆ T)}}{\sqrt{g_{2}}} \Rightarrow \sqrt{(\frac{l_{1}}{g_{1}})} = \frac{\sqrt{l_{1} (1 + α ∆ T)}}{\sqrt{g_{2}}} \Rightarrow \frac{1}{9.8} = \frac{1 + 12 \times 10^{- 6} \times ∆ T}{9.788} \Rightarrow \frac{9.788}{9.8} = 1 + 12 \times 10^{- 6} \times ∆ T \Rightarrow \frac{9.788}{9.8} - 1 = 12 \times 10^{- 6} \times ∆ T \Rightarrow ∆ T = \frac{- 0.00122}{12 \times 10^{- 6}} \Rightarrow T_{2} - 20 = - 102.4 \Rightarrow T_{2} = - 102.4 + 20 = - 82.4 \Rightarrow T_{2} \approx - 82 ° C$
Therefore, for a pendulum clock to give correct time, the temperature at which the value of g is 9.788 ms^–2 should be $-$ 82^oC.

Page No 13:

Answer:

Given:
Diameter of the steel sphere at temperature (T₁= 10 °C) , d_st = 2.005 cm
Diameter of the aluminium sphere, d_Al = 2.000 cm
Coefficient of linear expansion of steel, α_st = 11 × 10^{$- 6$} °C^{$- 1$}
Coefficient of linear expansion of aluminium, α_Al = 23 × 10^{$-$ 6} °C^{$-$ 1}
Let the temperature at which the ball will fall be T₂_,so that change in temperature be ΔT.
d'_st= 2.005(1 + α_st ΔT)
$\Rightarrow d'_{st} = 2.005 + 2.005 \times 11 \times 10^{- 6} \times ∆ T d'_{Al} = 2 (1 + α_{Al} \times ∆ T) \Rightarrow d'_{Al} = 2 + 2 \times 23 \times 10^{- 6} \times ∆ T$
The steel ball will fall when both the diameters become equal.
So, d'_st= d'_Al
$\Rightarrow 2.005 + 2.005 \times 11 \times 10^{- 6} ∆ T = 2 + 2 \times 23 \times 10^{- 6} ∆ T \Rightarrow (46 - 22.055) \times 10^{- 6} ∆ T = 0.005 \Rightarrow ∆ T = \frac{0.005 \times 10^{6}}{23.945} = 208.81 Now, ∆ T = T_{2} - T_{1} = T_{2} - 10 ° C \Rightarrow T_{2} = ∆ T + T_{1} = 208.81 + 10 \Rightarrow T_{2} = 218.8 ≅ 219 ° C$
Therefore, the temperature at which the ball will fall is 219 °C.

Page No 13:

Answer:

Given:
At 40^oC, the length and breadth of the glass window are 20 cm and 30 cm, respectively.
Coefficient of linear expansion of glass, $α_{g}$ = 9.0 × 10^–6 °C^–1
Coefficient of linear expansion for aluminium, $α_{A l}$ = 24 ×100^–6 °C^–1
The final length of aluminium should be equal to the final length of glass so that there is no stress on the glass in winter, even if the temperature drops to 0 °C.
Change in temperature, $Δ θ$ = 40 °C
Let the initial length of aluminium be l.
$l (1 - α_{A l} ∆ θ) = 20 (1 - α_{g} ∆ θ) \Rightarrow l (1 - 24 \times 10^{- 6} \times 40) = 20 (1 - 9 \times 10^{- 6} \times 40) \Rightarrow l (1 - 0.00096) = 20 (1 - 0.00036) \Rightarrow l = \frac{20 \times 0.99964}{1 - 0.00096} = \frac{20 \times 0.99964}{0.99904} \Rightarrow l = 20.012 cm$
Let the initial breadth of aluminium be b.
$b (1 - α_{A l} ∆ θ) = 30 (1 - α_{g} ∆ θ) \Rightarrow b = \frac{30 \times (1 - 9 \times 10^{- 6} \times 40)}{(1 - 24 \times 10^{- 6} \times 40)} = \frac{30 \times 0.99964}{0.99904} \Rightarrow b = 30.018 cm$
Therefore, the size of the aluminium frame should be 20.012 cm × 30.018 cm.

Page No 13:

Answer:

At T = 20°C, the volume of the glass vessel, V_g= 1000 cc.
Let the volume of mercury be V_Hg .
Coefficient of cubical expansion of mercury, γ_Hg = 1.8 × 10^–4 /°C
Coefficient of cubical expansion of glass, γ_g = 9 × 10^–6 /°C
Change in temperature, ΔT, is same for glass and mercury.
Let the volume of glass and mercury after rise in temperature be V'_g and V'_Hg respectively.
Volume of remaining space after change in temperature,(V'_g – V'_Hg) = Volume of the remaining space (initial),(V_g – V_Hg)
We know: V'_g = V_g (1 + γ_g ΔT) ...(1)
V'_Hg = V_Hg (1 + γ _Hg ΔT) ...(2)

Subtracting (2) from (1), we get:
$V'_{g} - V'_{Hg} = V_{g} - V_{Hg} + V_{g} γ_{g} ∆ T - V_{Hg} γ_{Hg} ∆ T \Rightarrow V_{g} γ_{g} ∆ T - V_{Hg} γ_{Hg} ∆ T = 0 \Rightarrow \frac{V_{g}}{V_{Hg}} = \frac{γ_{Hg}}{γ_{g}} \Rightarrow \frac{1000}{V_{Hg}} = \frac{1.8 \times 10^{- 4}}{9 \times 10^{- 6}} \Rightarrow V_{Hg} = \frac{9 \times 10^{- 3}}{1.8 \times 10^{- 4}} \Rightarrow V_{Hg} = 50 cc$
Therefore, the volume of mercury that should be poured into the glass vessel is 50 cc.

Page No 13:

Answer:

Given:
Volume of water contained in the aluminium can, V₀ = 500 cm³
Area of inner cross-section of the can, A = 125 cm²
Coefficient of volume expansion of water, γ = 3.2 × 10^–4 °C^–1
Coefficient of linear expansion of aluminium, $α_{A L}$ = 23 × 10^–6 °C^–1
If $∆ θ$ is the change in temperature, then final volume of water $(V)$ due to expansion,
V = V₀(1 + γΔθ)
   = 500 [1 + 3.2 × 10^–4 × (80 – 10)]
   = 500 [1 + 3.2 × 10^–4× 70]
   = 511.2 cm³
The aluminium vessel expands in its length only.
So, area of expansion of the base can be neglected.
Increase in volume of water = 11.2 cm³
Consider a cylinder of volume 11.2 cm³
∴ Increase in height of the water $= \frac{11.2}{125}$ = 0.0896
                                           = 0.089 cm

Page No 13:

Answer:

Given: At 0^oC, volume of glass vessel, V_g = 10 × 10 × 10 = 1000 cc = volume of mercury, V_Hg
Let the volume of mercury at 10°C be V'_Hgand that of glass be V'_g.
At 10^oC, the additional volume of mercury than glass, due to heating, V'_Hg – V'_g= 1.6 cm³
So change in temperature, ΔT = 10°C
Coefficient of linear expansion of glass, α_g = 6.5 × 10^–6°C^–1
Therefore, the coefficient of volume expansion of glass, γ_g = 3 × 6.5 × 10^–6°C^–1
Let the coefficient of volume expansion of mercury be γ_Hg.
We know that
V'_Hg = V_Hg (1 + γ _Hg ΔT)     ...(1)
V'_g= V_g (1 + γ_g ΔT)              ...(2)
Subtracting (2) from (1) we get,
V'_Hg– V'_g= V_H_g – V_g + V_Hg γ_Hg ΔT – V_g γ_g ΔT (as V_Hg = V_g)
$\Rightarrow 1.6 = 1000 \times γ_{Hg} \times 10 - 1000 \times 6.5 \times 3 \times 10^{- 6} \times 10 \Rightarrow γ_{Hg} = \frac{1.6 + 19.5 \times 10^{- 2}}{10000} \Rightarrow γ_{Hg} = \frac{1.6 + 0.195}{10000} \Rightarrow γ_{H g} = \frac{1.795}{10000} \Rightarrow γ_{H g} = 1.795 \times 10^{- 4} \Rightarrow γ_{Hg} ≅ 1.8 \times 10^{- 4} ° C^{- 1}$
Therefore, the coefficient of volume expansion of mercury is 1.8× 10^–4°C^–1.

Page No 13:

Answer:

Given:
Density of wood at 0 °C, f_w= 880 kgm^-3
Density of benzene at 0 °C, f_b = 900 kgm^-³
Coefficient of volume expansion for wood, γ_w = Coefficient of volume expansion for benzene, γ_b = 1.5 × 10^–3°C^–1
So, initial temperature, T₁ = 0 °C
Let T₂be the temperature at which the piece of wood will just sink in benzene and $Δ$ T = T₂ $-$ T₁.
The piece of wood begins to sink when its weight is equal to the weight of the benzene displaced.
Mass = volume $\times$ density
$Therefore, V f'_{w} g = V f'_{b} g \Rightarrow \frac{f_{w}}{1 + γ_{w} Δ T} = \frac{f_{b}}{1 + γ_{b} Δ T} \Rightarrow \frac{880}{1 + 1.2 \times 10^{- 3} Δ T} = \frac{900}{1 + 1.5 \times 10^{- 3} Δ T} \Rightarrow 880 + 880 \times 1.5 \times 10^{- 3} Δ T = 900 + 900 \times 1.2 \times 10^{- 3} Δ T \Rightarrow (1320 - 1080) \times 10^{- 3} Δ T = 20 \Rightarrow Δ T = 83.3 ° C \Rightarrow T_{2} - T_{1} ≅ 83^{o} \Rightarrow T_{2} - 0 ° ≅ 83^{o} \Rightarrow T_{2} ≅ 83 ° C$
Therefore, the piece of wood will just sink in benzene at 83 ^oC.

Page No 13:

Answer:

The steel rod is resting on a horizontal base at 0 °C. When the temperature is increased to 100 °C, it will lead to an increase in the length of the steel due to expansion on heating. Since, there is no opposition in expansion of length, no longitudinal strain will be developed.

Page No 13:

Answer:

Given:
Temperature at which rod is resting on a fixed horizontal base without any strain, T₁=20 °C. Then the rod is heated to temperature, T₂ = 50 °C
So change in temperature, $Δ T = T_{2} - T_{1} = 30^{o} C$
Coefficient of linear expansion of steel, α = 1.2 × 10^–5 °C^{$-$ 1}
Let L be the length of the rod without heating and L' be the length of the rod on heating.
Let longitudinal strain developed in the rod be S.
We know that
$L' = L (1 + α Δ T) \Rightarrow Δ L = L α Δ T Strain, S = \frac{Δ L}{L} = \frac{L α Δ T}{L} = α Δ T \Rightarrow S = 1.2 \times 10^{- 5} \times (50 - 20) = 1.2 \times 10^{- 5} \times 30 = 36 \times 10^{- 5} S = 3.6 \times 10^{- 4}$
The strain of 3.6 × 10^{$-$ 4}will be opposite to the direction of expansion.

Page No 13:

Answer:

Given:
Cross-sectional area of the steel wire, A = 0.5 mm² = 0.5 × 10^–6 m²
The wire is taut at a temperature, T₁= 20 °C,
After this, the temperature is reduced to T₂ = 0 °C
So, the change in temperature, Δθ = T₁ $-$ T₂= 20 °C
Coefficient of linear expansion of steel, α = 1.2 ×10^–5 °C^-1
Young's modulus, γ = 2 ×10¹¹ Nm^{$- 2$}
Let L be the initial length of the steel wire and L' be the length of the steel wire when temperature is reduced to 0°C.
Decrease in length due to compression, ΔL = L' $-$ L= LαΔθ ...(1)
Let the tension applied be F.

^{$γ = \frac{stress}{strain} = \frac{(\frac{F}{A})}{(\frac{Δ L}{L})} \Rightarrow γ = \frac{F}{A} \times \frac{L}{Δ L} \Rightarrow Δ L = \frac{F L}{A Y} . . . (2)$}
Change in length due to tension produced is given by (1) and (2).
So, on equating (1) and (2), we get:
$L α Δ θ = \frac{F L}{A Y} \Rightarrow F = α Δ θ A Y = 1.2 \times 10^{- 5} \times (20 - 0) \times 0.5 \times 10^{- 6} \times 2 \times 10^{11} = 1.2 \times 20 \Rightarrow F = 24 N$
Therefore, the tension produced when the temperature falls to 0°C is 24 N.

Page No 13:

Answer:

Given:
Temperature of the rod at zero tension, T₁ = 20 °C
Final temperature, T₂ = 100 °C
Change in temperature, $Δ θ$ = 80 °C
Cross-sectional area of the rod, A = 2 mm² = 2 × 10^{$-$ 6} m²
Coefficient of linear expansion of steel, α= 12 ×10^–6°C^{$-$ 1}
Young's modulus of steel, Y= 2 × 10¹¹Nm^{$- 2$}
Let L be the length of the steel rod at 20 °C and L' be the length of steel rod at 100 °C.
Change of length of the rod, $∆ L$ = L' $-$ L
If F be the force exerted by the rod on one of the clamps due to rise in temperature, then

$Y = \frac{stress}{strain} = \frac{F / A}{\frac{Δ L}{L}} \Rightarrow F = \frac{Y \times Δ L}{L} \times A Δ L = L α Δ θ \Rightarrow F = \frac{Y L α Δ θ A}{L} \Rightarrow F = Y A α Δ θ = 2 \times 10^{11} \times 2 \times 10^{- 6} \times 12 \times 10^{- 6} \times 80 = 48 \times 80 \times 10^{- 1} So, F = 384 N$
Therefore, the rod will exert a force of 384 N on one of the clamps.

Page No 14:

Answer:

Given:
Initial length of the system (two identical steel rods + an aluminium rod) at 0 °C = l₀
Coefficient of linear expansion of steel and aluminium are α_sand α_Al, respectively.
Temperature is raised by θ.
So, the change in temperature, $∆ θ$ = θ $-$ 0 °C = θ
Young's modulus of steel and aluminium are γ_s and γ_Al_,respectively.
If l be the final length of the system at temperature θ,
strain on the system $= \frac{(l - l_{0})}{l_{0}}$ ...(1)
Young's modulus = Stress/ Strain
Therefore, the total strain on the system $= \frac{T otal stress on the system}{T otal Young' s modulus of the system}$

Now, total stress = stress due to the two steel rods + stress due to the aluminium rod
$S tress = \frac{F}{A} = α Y ∆ θ$
Total stress = γ_sα_sθ + γ_sα_sθ + γ_Alα_Alθ
= 2γ_sα_sθ + γ_Alα_Alθ ...(2)
Young's modulus of the system,
Y = γ_s + γ_s + γ_Al = 2γ_s + γ_Al...(3)
Using (1), (2) and (3), we get:
$Strain on the system = \frac{2 γ_{s} α_{s} θ + γ_{A l} α_{A l} θ}{2 γ_{s} + γ_{A l}} \Rightarrow \frac{l - l_{0}}{l_{0}} = \frac{2 γ_{s} α_{s} θ + γ_{A l} α_{A l} θ}{2 γ_{s} + γ_{A l}} \Rightarrow l = l_{0} [1 + \frac{2 γ_{s} α_{s} θ + γ_{A l} α_{A l} θ}{2 γ_{s} + γ_{A l}}]$
Therefore, the final length of the system will be $l_{0} [1 + \frac{2 γ_{s} α_{s} θ + γ_{A l} α_{A l} θ}{2 γ_{s} + γ_{A l}}]$ , where l₀ is its initial length.

Page No 14:

Answer:

Given:
Initial pressure on the steel ball = 1.0 × 10⁵Pa
The ball is heated from 20 °C to 120 °C.
So, change in temperature, $Δ θ$ = 100 °C.
Coefficient of linear expansion of steel, $α$ = 12 × 10^–6 °C^–1
Bulk modulus of steel, B = 1.6 × 10¹¹ Nm^–2
Pressure is given as,
$\Rightarrow P = B \times γ ∆ θ \Rightarrow P = B \times 3 α ∆ θ (∵ γ = 3 α) \Rightarrow P = 1.6 \times 10^{11} \times 3 \times 12 \times 10^{- 6} \times (120 - 20) = 1.6 \times 3 \times 12 \times 10^{11} \times 10^{- 6} \times 10^{2} = 57.6 \times 10^{7} \Rightarrow P = 5.8 \times 10^{8} Pa$
Therefore, the pressure inside the ball is 5.8 × 10⁸Pa.

Page No 14:

Answer:

Given:
Coefficient of linear expansion of solid = α
Moment of inertia at 0 °C = I₀
If temperature changes to θ from 0 °C, then change in temperature, $(∆ T)$ = $θ$
Let I be the new moment of inertia attained due to rise in temperature.
Let R₀ be the radius of gyration at 0 °C.
We know that on heating, radius of gyration will change as
R= R₀(1 + αθ)
Here, R is the radius of gyration after heating.
I₀ = MR₀² , where M = mass of the body
Now, I = MR² = MR₀²(1 + αθ)²
Expanding binomially and neglecting the higher terms of order (αθ) that will be very small, we get
I = MR₀²(1 + 2 αθ)
So, I = I₀(1 + 2 αθ)
Hence, proved.

Page No 14:

Answer:

Given:
Coefficient of linear expansion of the wire, α = 2.4 × 10^–5°C^–1
Let I₀be the moment of inertia of the torsional pendulum at 0 °C.
If K is the torsional constant of the wire, then time period of torsional pendulum $(T)$ :
$T = 2 π \sqrt{\frac{I}{K}} . . . (1)$
Here, I = moment of inertia after change in temperature
When the temperature is changed by $∆ θ$ , moment of inertia $(I)$ ,
I = I₀(1+2 $α$ $∆ θ$ )
On substituting the value of I in equation(1), we get:
$T = 2 π \frac{\sqrt{I_{0} (1 + 2 α ∆ θ)}}{\sqrt{K}}$
In winter, $∆ θ$ = 5 °C
$∴$ Time period $(T_{1})$
$= 2 π \frac{\sqrt{I_{0} (1 + 2 α \times 5)}}{\sqrt{K}}$
In summer, $∆ θ$ = 45 °C
Time period $(T_{2})$
$= 2 π \frac{\sqrt{I_{0} (1 + 2 α \times 45)}}{\sqrt{K}}$
$So, \frac{T_{2}}{T_{1}} = \frac{\sqrt{1 + 90 α}}{\sqrt{1 + 10 α}} = \frac{\sqrt{1 + 90 \times 2.4 \times 10^{- 5}}}{\sqrt{1 + 10 \times 2.4 \times 10^{- 5}}} \Rightarrow \frac{T_{2}}{T_{1}} = \frac{\sqrt{1.00216}}{\sqrt{1.00024}} % change = (\frac{T_{2}}{T_{1}} - 1) \times 100 = 0.0959 % \Rightarrow % change in time period \approx 9.6 \times 10^{- 2} %$

Therefore, the percentage change in time period of a torsional pendulum between peak winters and peak summers is 9.6 × 10^–2% .

Page No 14:

Answer:

Let initial radius of the circular disc at 20^ₒC = $r_{20}$
Let final radius of the circular disc at 50^ₒC = $r_{50}$
Coefficient of linear expansion of iron, $α$ = 1.2 × 10^–5 °C^–1.
Change in temperature, $∆ T$ = 30 ^₀C
Let R' and R be the radius of the paricle at 50^ₒC and 20^ₒC respectively.
If v and v' be the linear speed of the particle at 50^ₒC and 20^ₒC respectively, as the angular velocity remains( $ω$ ) constant.
Therefore,
$ω = \frac{v}{R} = \frac{v'}{R'} . . . . (1)$
Now,
R' = R(1+ $α ∆ T$ )
$\Rightarrow$ R' = R + R $\times$ 1.2 × 10^–5 °C^–1 $\times ∆ T$ .
$\Rightarrow$ R' = 1.00036R
Using equation(1) we have,
$\frac{v}{R} = \frac{v'}{R'} \Rightarrow \frac{v}{R} = \frac{v'}{1.00036 R} \Rightarrow v' = 1.00036 v$
Percentage change in linear speed will be,
$= \frac{v' - v}{v} \times 100 = \frac{1.00036 v - v}{v} \times 100 = 3.6 \times 10^{- 2}$

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