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#### Question 1:

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.

#### Question 2:

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.

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.

#### Question 3:

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.

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).

#### Question 4:

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).

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$×$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.

#### Question 5:

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$×$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.

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.

#### Question 6:

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.

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.

#### Question 7:

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.

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.

#### Question 8:

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.

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.

#### Question 9:

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.

At sea level, the pressure is around 1 atmosphere and at high altitude, the density of air reduces.
Pressure of liquid,

The above equation shows that pressure depends on density. Therefore at 4oâ€‹C, the density of water will be less at high altitude, compared to the density at sea level.

#### Question 10:

At sea level, the pressure is around 1 atmosphere and at high altitude, the density of air reduces.
Pressure of liquid,

The above equation shows that pressure depends on density. Therefore at 4oâ€‹C, the density of water will be less at high altitude, compared to the density at sea level.

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.

#### Question 11:

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.

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.

#### Question 12:

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.

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.

#### Question 13:

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.

When a spherical shell is heated, its volume changes according to the equation, ${V}_{\theta }={V}_{0}\left(1+\gamma ∆\theta \right)$. 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, $\gamma$.

#### Question 1:

When a spherical shell is heated, its volume changes according to the equation, ${V}_{\theta }={V}_{0}\left(1+\gamma ∆\theta \right)$. 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, $\gamma$.

(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).

#### Question 2:

(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).

(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).

#### Question 3:

(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).

(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
$\frac{{T}_{\mathrm{F}}-32}{180}=\frac{{T}_{\mathrm{K}}-273.15}{100}$
TF = TK = θ

Therefore,

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

Let the temperature be t
$\frac{t-0}{100}=\frac{t-273.15}{100}\phantom{\rule{0ex}{0ex}}t=t-273.15$
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.

#### Question 4:

(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
$\frac{{T}_{\mathrm{F}}-32}{180}=\frac{{T}_{\mathrm{K}}-273.15}{100}$
TF = TK = θ

Therefore,

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

Let the temperature be t
$\frac{t-0}{100}=\frac{t-273.15}{100}\phantom{\rule{0ex}{0ex}}t=t-273.15$
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.

(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.

#### Question 5:

(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.

(a) A and B are correct.

The coefficient of linear expansion,
$\alpha =\frac{1}{L}\frac{△L}{△T}\phantom{\rule{0ex}{0ex}}$
$=\frac{\left[L\right]}{\left[LT\right]}={\mathrm{K}}^{-1}$
Here, L = initial length
$△$L = change in length
$△$T = change in temperature
On the other hand, the coefficient of volume expansion,
$\gamma =\frac{1}{V}\frac{△V}{△T}=\frac{\left[{L}^{3}\right]}{\left[{L}^{3}T\right]}={\mathrm{K}}^{-1}$
Here, V = initial volume
$△$V = change in volume
$△$T = change in temperature
K = kelvin, the S.I. unit of temperature

#### Question 6:

(a) A and B are correct.

The coefficient of linear expansion,
$\alpha =\frac{1}{L}\frac{△L}{△T}\phantom{\rule{0ex}{0ex}}$
$=\frac{\left[L\right]}{\left[LT\right]}={\mathrm{K}}^{-1}$
Here, L = initial length
$△$L = change in length
$△$T = change in temperature
On the other hand, the coefficient of volume expansion,
$\gamma =\frac{1}{V}\frac{△V}{△T}=\frac{\left[{L}^{3}\right]}{\left[{L}^{3}T\right]}={\mathrm{K}}^{-1}$
Here, V = initial volume
$△$V = change in volume
$△$T = change in temperature
K = kelvin, the S.I. unit of temperature

(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.

#### Question 7:

(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.

(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.

#### Question 8:

(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.

(c) 2αIâˆ†t

The change in moment of inertia of uniform rod with change in temperature is given by,
$I\text{'}=I\left(1+2\alpha \Delta t\right)$
Here, I = initial moment of inertia
I' = new moment of inertia due to change in temperature
= expansion coefficient
$∆$t = change in temperature
So, $I\text{'}-I=2\alpha I\Delta t$

#### Question 9:

(c) 2αIâˆ†t

The change in moment of inertia of uniform rod with change in temperature is given by,
$I\text{'}=I\left(1+2\alpha \Delta t\right)$
Here, I = initial moment of inertia
I' = new moment of inertia due to change in temperature
= expansion coefficient
$∆$t = change in temperature
So, $I\text{'}-I=2\alpha I\Delta t$

(c) Iâˆ†t

The moment of inertia of a solid body of any shape changes with temperature as
$I\text{'}=I\left(1+2\alpha \Delta t\right)$
Here, I = initial moment of inertia
I' = new moment of inertia due to change in temperature
$\alpha$ = expansion coefficient
$\Delta$t = change in temperature
So, $I\text{'}-I=2\alpha I\Delta t$II0=2αΔt

#### Question 10:

(c) Iâˆ†t

The moment of inertia of a solid body of any shape changes with temperature as
$I\text{'}=I\left(1+2\alpha \Delta t\right)$
Here, I = initial moment of inertia
I' = new moment of inertia due to change in temperature
$\alpha$ = expansion coefficient
$\Delta$t = change in temperature
So, $I\text{'}-I=2\alpha I\Delta t$II0=2αΔt

(c) 4oC

The density of water is maximum at 4oC, 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 4oC.

#### Question 11:

(c) 4oC

The density of water is maximum at 4oC, 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 4oC.

(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.

#### Question 1:

(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.

(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.

#### Question 2:

(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.

(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.

#### Question 3:

(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.

(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.

#### Question 4:

(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.

(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.

#### Question 5:

(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.

(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.

#### Question 6:

(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.

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

#### Question 1:

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

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° Câ€‹
Temperature on a centigrade mercury scale, T, is given as:

Therefore, the temperature on a centigrade mercury scale will be 20oC.

#### Question 2:

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° Câ€‹
Temperature on a centigrade mercury scale, T, is given as:

Therefore, the temperature on a centigrade mercury scale will be 20oC.

Given:
Pressure registered by a constant-volume thermometer at the triple point, Ptr = 1.500 × 10 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:

Therefore, the temperature at the normal point (T) is 373.31 K.

#### Question 3:

Given:
Pressure registered by a constant-volume thermometer at the triple point, Ptr = 1.500 × 10 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:

Therefore, the temperature at the normal point (T) is 373.31 K.

Given:
In a gas thermometer, the pressure measured at the melting point of lead, P = 2.20 × Pressure at triple point(Pâ€‹tr)
So the melting point of lead,(T) is given as:

Therefore, the melting point of lead is 601 K.

#### Question 4:

Given:
In a gas thermometer, the pressure measured at the melting point of lead, P = 2.20 × Pressure at triple point(Pâ€‹tr)
So the melting point of lead,(T) is given as:

Therefore, the melting point of lead is 601 K.

Given:
Pressure measured by a constant volume gas thermometer at the triple point of water, Ptr = 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.
â€‹For a constant volume gas thermometer, temperature-pressure relation is given below:

Therefore, the pressure measured at the boiling point of water is 55 kPa.

#### Question 5:

Given:
Pressure measured by a constant volume gas thermometer at the triple point of water, Ptr = 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.
â€‹For a constant volume gas thermometer, temperature-pressure relation is given below:

Therefore, the pressure measured at the boiling point of water is 55 kPa.

Given:
â€‹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, P1â€‹ = 70 kPa,
Let Ptrbe 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,

For steam point,

On substituting the value of Ptr ,we get:

Therefore, the pressure at steam point is 96 kPa.

#### Question 6:

Given:
â€‹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, P1â€‹ = 70 kPa,
Let Ptrbe 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,

For steam point,

On substituting the value of Ptr ,we get:

Therefore, the pressure at steam point is 96 kPa.

Given:
In a constant volume gas thermometer,
Pressure of the gas at the ice point, Pâ€‹â€‹â€‹0 = 80 cm of Hgâ€‹
Pressure of the gas at the steam point, P100 = 90 cm of Hg
â€‹Pressure of the gas in a heated wax bath, P = 100 cm of Hg
The temperature of the wax bath $\left(T\right)$ is given by:

Therefore, the temperature of the wax bath is 200o C.

#### Question 7:

Given:
In a constant volume gas thermometer,
Pressure of the gas at the ice point, Pâ€‹â€‹â€‹0 = 80 cm of Hgâ€‹
Pressure of the gas at the steam point, P100 = 90 cm of Hg
â€‹Pressure of the gas in a heated wax bath, P = 100 cm of Hg
The temperature of the wax bath $\left(T\right)$ is given by:

Therefore, the temperature of the wax bath is 200o C.

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, To = 273.15 K
â€‹So the temperature of the vessel(T') is given by:

Therefore, the temperature of the vessel is 307 K.

#### Question 8:

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, To = 273.15 K
â€‹So the temperature of the vessel(T') is given by:

Therefore, the temperature of the vessel is 307 K.

Given:
Resistance at 0oC, R0 = 80 $\mathrm{\Omega }$
Resistance at 100oC, R100 = 90 $\mathrm{\Omega }$
Let t be the temperature at which the resistance (Rt) is 86 $\mathrm{\Omega }$.
$t=\frac{{R}_{t}-{R}_{0}}{{R}_{100}-{R}_{0}}×100\phantom{\rule{0ex}{0ex}}⇒t=\frac{86-80}{90-80}×100\phantom{\rule{0ex}{0ex}}⇒t=\frac{6}{10}×100\phantom{\rule{0ex}{0ex}}⇒t=60°$
Therefore, the resistance is 86 $\mathrm{\Omega }$ at 60oC.

#### Question 9:

Given:
Resistance at 0oC, R0 = 80 $\mathrm{\Omega }$
Resistance at 100oC, R100 = 90 $\mathrm{\Omega }$
Let t be the temperature at which the resistance (Rt) is 86 $\mathrm{\Omega }$.
$t=\frac{{R}_{t}-{R}_{0}}{{R}_{100}-{R}_{0}}×100\phantom{\rule{0ex}{0ex}}⇒t=\frac{86-80}{90-80}×100\phantom{\rule{0ex}{0ex}}⇒t=\frac{6}{10}×100\phantom{\rule{0ex}{0ex}}⇒t=60°$
Therefore, the resistance is 86 $\mathrm{\Omega }$ at 60oC.

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:

Solving (i) and (ii), we get:
α = 3.8 ×10–3°Câ€‹-1
β = –5.6 ×10–7°C-1
Therefore, resistance R0 is 20 Ω and the value of α is 3.8 ×10–3°Câ€‹-1  and that of β is –5.6 ×10–7°C-1.

#### Question 10:

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:

Solving (i) and (ii), we get:
α = 3.8 ×10–3°Câ€‹-1
β = –5.6 ×10–7°C-1
Therefore, resistance R0 is 20 Ω and the value of α is 3.8 ×10–3°Câ€‹-1  and that of β is –5.6 ×10–7°C-1.

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 °Câ€‹-1
${L}_{1}={L}_{0}\left(1+\alpha t\right)$
$⇒{L}_{1}$= 10 (1 + 10–5 × 35)
$⇒{L}_{1}$ = 10 + 35 × 10–4
$⇒{L}_{1}$ = 10.0035 m
â€‹So, the length of the slab on summer day when the temperature is 35oC is 10.0035 m.

#### Question 11:

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 °Câ€‹-1
${L}_{1}={L}_{0}\left(1+\alpha t\right)$
$⇒{L}_{1}$= 10 (1 + 10–5 × 35)
$⇒{L}_{1}$ = 10 + 35 × 10–4
$⇒{L}_{1}$ = 10.0035 m
â€‹So, the length of the slab on summer day when the temperature is 35oC is 10.0035 m.

Given:
Temperature at which the steel metre scale is calibrated, t1 = 20oC
Temperature at which the scale is used, t2 = 10oC
So, the change in temperature, $\Delta$t = (20o$-$10o) C
The distance to be measured by the metre scale, Lo = (51$-$50) = 1 cm = 0.01 m
Coefficient of linear expansion of steel, ${\alpha }_{steel}$1.1 × 10–5 °C1
Let the new length measured by the scale due to expansion of steel be Lâ€‹2, Change in length is given by,

As the temperature is decreasing, therefore length will decrease by $∆L$.
Therefore, â€‹the new length measured by the scale due to expansion of steel (L2) will be,
L2 = 1 cm $-$ 0.00011 cm = 0.99989 cm

#### Question 12:

Given:
Temperature at which the steel metre scale is calibrated, t1 = 20oC
Temperature at which the scale is used, t2 = 10oC
So, the change in temperature, $\Delta$t = (20o$-$10o) C
The distance to be measured by the metre scale, Lo = (51$-$50) = 1 cm = 0.01 m
Coefficient of linear expansion of steel, ${\alpha }_{steel}$1.1 × 10–5 °C1
Let the new length measured by the scale due to expansion of steel be Lâ€‹2, Change in length is given by,

As the temperature is decreasing, therefore length will decrease by $∆L$.
Therefore, â€‹the new length measured by the scale due to expansion of steel (L2) will be,
L2 = 1 cm $-$ 0.00011 cm = 0.99989 cm

Given:
Length of the iron sections when there's no effect of temperature on them, Lo  = 12.0 m
â€‹Temperature at which the iron track is laid in winter, tâ€‹wâ€‹ = 18 oC
Maximum temperature during summers, ts = 48 oC
Coefficient of linear expansion of iron$\alpha$ = 11 × 10–6  °C–1
Let the new lengths attained by each section due to expansion of iron in winter and summer be Lwand Ls,respectively, which can be calculated as follows:

Therefore, the gap ($\Delta$L) that should be left between two iron sections, so that there is no compression during summer, is 0.4 cm.

#### Question 13:

Given:
Length of the iron sections when there's no effect of temperature on them, Lo  = 12.0 m
â€‹Temperature at which the iron track is laid in winter, tâ€‹wâ€‹ = 18 oC
Maximum temperature during summers, ts = 48 oC
Coefficient of linear expansion of iron$\alpha$ = 11 × 10–6  °C–1
Let the new lengths attained by each section due to expansion of iron in winter and summer be Lwand Ls,respectively, which can be calculated as follows:

Therefore, the gap ($\Delta$L) that should be left between two iron sections, so that there is no compression during summer, is 0.4 cm.

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, ($\Delta$t) = 100°C - 0°C = 100°C
The linear expansion coefficient of aluminium, αalâ€‹ = 2.3 × 10–5 °C–1
Let the diameter of the circular hole in the plate at 100oC be d2, which can be written as:

$⇒{d}_{2}$= 2 × 10–2 (1 + 2.3 × 10–5 × 102)
$⇒{d}_{2}$= 2 × 10–2 (1 + 2.3 × 10–3)
$⇒{d}_{2}$= 2 × 10–2 + 2.3 × 2 × 10–5
$⇒{d}_{2}$ = 0.02 + 0.000046
$⇒{d}_{2}$= 0.020046 m
$⇒{d}_{2}$≈ 2.0046 cm
Therefore, the diameter of the circular hole in the aluminium plate at 100oC is â€‹2.0046 cm.

#### Question 14:

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, ($\Delta$t) = 100°C - 0°C = 100°C
The linear expansion coefficient of aluminium, αalâ€‹ = 2.3 × 10–5 °C–1
Let the diameter of the circular hole in the plate at 100oC be d2, which can be written as:

$⇒{d}_{2}$= 2 × 10–2 (1 + 2.3 × 10–5 × 102)
$⇒{d}_{2}$= 2 × 10–2 (1 + 2.3 × 10–3)
$⇒{d}_{2}$= 2 × 10–2 + 2.3 × 2 × 10–5
$⇒{d}_{2}$ = 0.02 + 0.000046
$⇒{d}_{2}$= 0.020046 m
$⇒{d}_{2}$≈ 2.0046 cm
Therefore, the diameter of the circular hole in the aluminium plate at 100oC is â€‹2.0046 cm.

Given:
At 20°C, length of the metre scale made up of steel, Lst= length of the metre scale made up of aluminium, Lalâ€‹
Coefficient of linear expansion for aluminium, αal = 2.3 × 10–5 °C-1
Coefficient of linear expansion for steel, αst = 1.1 × 10–5 °C-1
Let the length of the aluminium scale at 0°C, 40°C and 100°C be L0alâ€‹, Lâ€‹40al and L10â€‹0al.
And let the length of the steel scale at 0°C, 40°C and 100°C be L0stâ€‹, Lâ€‹40st and L10â€‹0st.
(a) So, L0st(1 – αst × 20) = L0al(1 – αal × 20)

(b)

(c)

#### Question 15:

Given:
At 20°C, length of the metre scale made up of steel, Lst= length of the metre scale made up of aluminium, Lalâ€‹
Coefficient of linear expansion for aluminium, αal = 2.3 × 10–5 °C-1
Coefficient of linear expansion for steel, αst = 1.1 × 10–5 °C-1
Let the length of the aluminium scale at 0°C, 40°C and 100°C be L0alâ€‹, Lâ€‹40al and L10â€‹0al.
And let the length of the steel scale at 0°C, 40°C and 100°C be L0stâ€‹, Lâ€‹40st and L10â€‹0st.
(a) So, L0st(1 – αst × 20) = L0al(1 – αal × 20)

(b)

(c)

(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
â€‹So, change in temperature, Δθ = t2 $-$t1 = 30 °C
Coefficient of linear expansion of steel, $\alpha$ = 1.1 × 10–5 °Câ€‹-1
Therefore, change in length,
ΔL = L αΔθ = L × 1.1 × 10–5 × 30

(b) Temperature on a winter day, t2 = 6 °Câ€‹
â€‹So, change in temperature, Δθ = t1$-$t2= 10 °Câ€‹
ΔL = Lâ€‹2 $-$L= L αΔθ = × 1.1 × 10–5 × 10â€‹

#### Question 16:

(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
â€‹So, change in temperature, Δθ = t2 $-$t1 = 30 °C
Coefficient of linear expansion of steel, $\alpha$ = 1.1 × 10–5 °Câ€‹-1
Therefore, change in length,
ΔL = L αΔθ = L × 1.1 × 10–5 × 30

(b) Temperature on a winter day, t2 = 6 °Câ€‹
â€‹So, change in temperature, Δθ = t1$-$t2= 10 °Câ€‹
ΔL = Lâ€‹2 $-$L= L αΔθ = × 1.1 × 10–5 × 10â€‹

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

Hence, the experiment can be performed in the temperature range of 15 °C to 25 °C .

#### Question 17:

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

Hence, the experiment can be performed in the temperature range of 15 °C to 25 °C .

Given:
Density of water at 0°C, ( f0)= 0.998 g cm-3
Density of water at 4°C,  â€‹(f4) = 1.000 g cm-3
Change in temperature, (Δt) = 4oC
Let the average coefficient of volume expansion of water in the temperature range of 0 to 4°C be γ.

As the density decreases,

Therefore,the average coefficient of volume expansion of water in the temperature range of 0 to 4°C will be  $\gamma =-5×{10}^{-4}$oC-1.

#### Question 18:

Given:
Density of water at 0°C, ( f0)= 0.998 g cm-3
Density of water at 4°C,  â€‹(f4) = 1.000 g cm-3
Change in temperature, (Δt) = 4oC
Let the average coefficient of volume expansion of water in the temperature range of 0 to 4°C be γ.

As the density decreases,

Therefore,the average coefficient of volume expansion of water in the temperature range of 0 to 4°C will be  $\gamma =-5×{10}^{-4}$oC-1.

Let the original length of iron rod be LFe and L'â€‹â€‹â€‹Fe be its length when temperature is increased by ΔT.
Let the original length of aluminium rod be LAl and L'â€‹â€‹â€‹Al be its length when temperature is increased by ΔT.

Coefficient of linear expansion of iron, ${\alpha }_{Fe}$ = 12 × 10–6 °Câ€‹$-$1
Coefficient of linear expansion of aluminium, αAl = 23 × 10–6 °Câ€‹â€‹$-$
Since the difference in length is independent of temperature, the difference is always constant.

The ratio of the lengths of the iron to the aluminium rod is 23:12.

#### Question 19:

Let the original length of iron rod be LFe and L'â€‹â€‹â€‹Fe be its length when temperature is increased by ΔT.
Let the original length of aluminium rod be LAl and L'â€‹â€‹â€‹Al be its length when temperature is increased by ΔT.

Coefficient of linear expansion of iron, ${\alpha }_{Fe}$ = 12 × 10–6 °Câ€‹$-$1
Coefficient of linear expansion of aluminium, αAl = 23 × 10–6 °Câ€‹â€‹$-$
Since the difference in length is independent of temperature, the difference is always constant.

The ratio of the lengths of the iron to the aluminium rod is 23:12.

Given:
The temperature at which the pendulum shows the correct time, T1 =
20 °C
Coefficient of linear expansion of steel,  $\alpha$ = 12 × 10–6 °C–1
Let T2 be the temperature at which the value of g is 9.788 ms–2 and $\Delta$T be  the change in temperature.
â€‹
So, the time periods of pendulum at different values of g will be t1 and t2 , such that

Therefore, for a pendulum clock to give correct time, the temperature at which the value of g is 9.788 ms–2 should be $-$ 82oC.

#### Question 20:

Given:
The temperature at which the pendulum shows the correct time, T1 =
20 °C
Coefficient of linear expansion of steel,  $\alpha$ = 12 × 10–6 °C–1
Let T2 be the temperature at which the value of g is 9.788 ms–2 and $\Delta$T be  the change in temperature.
â€‹
So, the time periods of pendulum at different values of g will be t1 and t2 , such that

Therefore, for a pendulum clock to give correct time, the temperature at which the value of g is 9.788 ms–2 should be $-$ 82oC.

Given:
Diameter of the steel sphere at temperature (T110 °C) , dst = 2.005 cm
Diameter of the aluminium sphere, dAl = 2.000 cm
Coefficient of linear expansion of steel, αst  = 11 × 10$-6$  °C$-1$
Coefficient of linear expansion of aluminium, αAl  = 23 × 10$-$6 °C$-$1â€‹
Let the temperature at which the ball will fall be T2 , so that change in temperature be ΔTâ€‹.
d'st = 2.005(1 + αstΔT)

The steel ball will fall when both the diameters become equal.
So, d'st d'Al

Therefore, â€‹the temperature at which the ball will fall is 219 °C.

#### Question 21:

Given:
Diameter of the steel sphere at temperature (T110 °C) , dst = 2.005 cm
Diameter of the aluminium sphere, dAl = 2.000 cm
Coefficient of linear expansion of steel, αst  = 11 × 10$-6$  °C$-1$
Coefficient of linear expansion of aluminium, αAl  = 23 × 10$-$6 °C$-$1â€‹
Let the temperature at which the ball will fall be T2 , so that change in temperature be ΔTâ€‹.
d'st = 2.005(1 + αstΔT)

The steel ball will fall when both the diameters become equal.
So, d'st d'Al

Therefore, â€‹the temperature at which the ball will fall is 219 °C.

Given:
At 40oC, the length and breadth of the glass window are 20 cm and 30 cm, respectively.
Coefficient of linear expansion of glass,  ${\alpha }_{g}$ =  9.0 × 10–6 °C–1
Coefficient of linear expansion for aluminium, ${\alpha }_{Al}$ = 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.
â€‹Change in temperature, $\Delta \theta$ = 40 °C

Let the initial length of aluminium be l.

Let the initial breadth of aluminium be b.

Therefore, the size of the aluminium frame should be 20.012 cm × 30.018 cm.

#### Question 22:

Given:
At 40oC, the length and breadth of the glass window are 20 cm and 30 cm, respectively.
Coefficient of linear expansion of glass,  ${\alpha }_{g}$ =  9.0 × 10–6 °C–1
Coefficient of linear expansion for aluminium, ${\alpha }_{Al}$ = 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.
â€‹Change in temperature, $\Delta \theta$ = 40 °C

Let the initial length of aluminium be l.

Let the initial breadth of aluminium be b.

Therefore, the size of the aluminium frame should be 20.012 cm × 30.018 cm.

At T = 20°C, the volume of the glass vessel,  Vg = 1000 cc.
Let the volume of  mercury be VHg .
Coefficient of cubical expansion of mercury, γHg = 1.8 × 10–4 /°C
Coefficient of cubical expansion of glass, γg = 9 × 10–6 /°C
â€‹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),(Vgâ€‹â€‹ – VHg)
We know:  V'g = Vg (1 + γg ΔT)                       ...(1)
V'Hg = VHg (1 + γ Hg  ΔT)               ...(2)

Subtracting (2) from (1), we get:

Therefore, the volume of mercury that should be poured into the glass vessel is 50 cc.

#### Question 23:

At T = 20°C, the volume of the glass vessel,  Vg = 1000 cc.
Let the volume of  mercury be VHg .
Coefficient of cubical expansion of mercury, γHg = 1.8 × 10–4 /°C
Coefficient of cubical expansion of glass, γg = 9 × 10–6 /°C
â€‹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),(Vgâ€‹â€‹ – VHg)
We know:  V'g = Vg (1 + γg ΔT)                       ...(1)
V'Hg = VHg (1 + γ Hg  ΔT)               ...(2)

Subtracting (2) from (1), we get:

Therefore, the volume of mercury that should be poured into the glass vessel is 50 cc.

Given:
Volume of water contained in the aluminium can, V0 = 500 cm3
Area of inner cross-section of the can, A = 125 cm2
â€‹Coefficient of volume expansion of water, γâ€‹ = 3.2 × 10–4 °C–1
Coefficient of linear expansion of aluminium, ${\alpha }_{AL}$ = 23 × 10
–6 °C–1
If $∆\theta$ is the change in temperature, then final volume of water$\left(V\right)$ 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

#### Question 24:

Given:
Volume of water contained in the aluminium can, V0 = 500 cm3
Area of inner cross-section of the can, A = 125 cm2
â€‹Coefficient of volume expansion of water, γâ€‹ = 3.2 × 10–4 °C–1
Coefficient of linear expansion of aluminium, ${\alpha }_{AL}$ = 23 × 10
–6 °C–1
If $∆\theta$ is the change in temperature, then final volume of water$\left(V\right)$ 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

Given: At 0oâ€‹C, volume of glass vessel, Vg = 10 × 10 × 10 = 1000 cc = volume of mercury, VHg
Let the volume of mercury at 10°C be V'Hg and that of glass be V'g.
At 10â€‹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, αg = 6.5 × 10–6 °C–1
Therefore, the coefficient of volume expansion of glass, γg = 3 × 6.5 × 10–6°C–1â€‹
Let the coefficient of volume expansion of mercury be γHg.
We know that
V'Hg = VHg (1 + γ Hg  ΔT)         ...(1)
V'g= Vg (1 + γg ΔT)              ...(2)
Subtracting (2) from (1) we get,
V'HgV'g= VHgVg + VHg γHg ΔTVg γg ΔT (as VHg = Vg)

Therefore, the coefficient of volume expansion of mercury is 1.8× 10–4 °C–1.

#### Question 25:

Given: At 0oâ€‹C, volume of glass vessel, Vg = 10 × 10 × 10 = 1000 cc = volume of mercury, VHg
Let the volume of mercury at 10°C be V'Hg and that of glass be V'g.
At 10â€‹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, αg = 6.5 × 10–6 °C–1
Therefore, the coefficient of volume expansion of glass, γg = 3 × 6.5 × 10–6°C–1â€‹
Let the coefficient of volume expansion of mercury be γHg.
We know that
V'Hg = VHg (1 + γ Hg  ΔT)         ...(1)
V'g= Vg (1 + γg ΔT)              ...(2)
Subtracting (2) from (1) we get,
V'HgV'g= VHgVg + VHg γHg ΔTVg γg ΔT (as VHg = Vg)

Therefore, the coefficient of volume expansion of mercury is 1.8× 10–4 °C–1.

Given:
Density of wood at 0 °C, fw = 880 kgmâ€‹-3
Density of benzene at 0 °C, fb = 900 kgm-â€‹3
Coefficient of volume expansion for wood, γw = Coefficient of volume expansion for benzene, γb = 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 $\Delta$T = T2$-$T1.
The piece of wood begins to sink when its weight is equal to the weight of the benzene displaced.
Mass = volume $×$density

Therefore, the piece of wood will just sink in benzene at 83 oC.

#### Question 26:

Given:
Density of wood at 0 °C, fw = 880 kgmâ€‹-3
Density of benzene at 0 °C, fb = 900 kgm-â€‹3
Coefficient of volume expansion for wood, γw = Coefficient of volume expansion for benzene, γb = 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 $\Delta$T = T2$-$T1.
The piece of wood begins to sink when its weight is equal to the weight of the benzene displaced.
Mass = volume $×$density

Therefore, the piece of wood will just sink in benzene at 83 oC.

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.

#### Question 27:

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.

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
â€‹So change in temperature,
Coefficient of linear expansion of steel, α = 1.2 × 10–5 °C$-$1â€‹
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

The strain of 3.6 × 10$-$4 will be opposite to the direction of expansion.

#### Question 28:

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
â€‹So change in temperature,
Coefficient of linear expansion of steel, α = 1.2 × 10–5 °C$-$1â€‹
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

The strain of 3.6 × 10$-$4 will be opposite to the direction of expansion.

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
â€‹So, the change in temperature, Δθ = T1$-$T2= 20 °C
Coefficient of linear expansion of steel, α = 1.2 ×10–5 °Câ€‹-1
Young's modulus, γ = 2 ×1011 Nmâ€‹$-2$
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.

Change in length due to tension produced is given by (1) and (2).
So, on equating (1) and (2), we get:

Therefore, the tension produced when the temperature falls to 0°C is 24 N.

#### Question 29:

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
â€‹So, the change in temperature, Δθ = T1$-$T2= 20 °C
Coefficient of linear expansion of steel, α = 1.2 ×10–5 °Câ€‹-1
Young's modulus, γ = 2 ×1011 Nmâ€‹$-2$
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.

Change in length due to tension produced is given by (1) and (2).
So, on equating (1) and (2), we get:

Therefore, the tension produced when the temperature falls to 0°C is 24 N.

Given:
Temperature of the rod at zero tension, T1 = 20 °C
Final temperature, T2 = 100 °C
â€‹Change in temperature, $\Delta \theta$ = 80 °C
Cross-sectional area of the rod, A = 2 mm2 = 2 × 10$-$6 m2
Coefficient of linear expansion of steel, α = 12 ×10–6 °Câ€‹$-$1
Young's modulus of steel, Y = 2 × 1011 Nm$-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

Therefore, the rod will exert a force of 384 N on one of the clamps.

#### Question 30:

Given:
Temperature of the rod at zero tension, T1 = 20 °C
Final temperature, T2 = 100 °C
â€‹Change in temperature, $\Delta \theta$ = 80 °C
Cross-sectional area of the rod, A = 2 mm2 = 2 × 10$-$6 m2
Coefficient of linear expansion of steel, α = 12 ×10–6 °Câ€‹$-$1
Young's modulus of steel, Y = 2 × 1011 Nm$-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

Therefore, the rod will exert a force of 384 N on one of the clamps.

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 αs and αAlâ€‹, respectively.
Temperature is raised by θ.
â€‹So, the change in temperature, $∆\theta$ = θ $-$ 0 °C = θ
Young's modulus of steel and aluminium are γs and γAl, respectively.
If l be the final length of the system at temperature θ,
strain on the system $=\frac{\left(l-{l}_{0}\right)}{{l}_{0}}$ â€‹ ...(1)
Young's modulus = Stress/ Strain
Therefore, the total strain on the system

Now, total stress = stress due to the two steel rods + stress due to the aluminium rod
$\text{S}\mathrm{tress}=\frac{F}{A}=\alpha Y∆\theta$
Total stress = γsαsθ + γsαsθ + γAlαAlθ
= 2γsαsθ + γAlαAlθ   ...(2)
Young's modulus of the system,
Y = γs + γs + γAl = 2γs + γAl      ...(3)
Using (1), (2) and (3), we get:

Therefore, the final length of the system will be , where l0 is its initial length.

#### Question 31:

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 αs and αAlâ€‹, respectively.
Temperature is raised by θ.
â€‹So, the change in temperature, $∆\theta$ = θ $-$ 0 °C = θ
Young's modulus of steel and aluminium are γs and γAl, respectively.
If l be the final length of the system at temperature θ,
strain on the system $=\frac{\left(l-{l}_{0}\right)}{{l}_{0}}$ â€‹ ...(1)
Young's modulus = Stress/ Strain
Therefore, the total strain on the system

Now, total stress = stress due to the two steel rods + stress due to the aluminium rod
$\text{S}\mathrm{tress}=\frac{F}{A}=\alpha Y∆\theta$
Total stress = γsαsθ + γsαsθ + γAlαAlθ
= 2γsαsθ + γAlαAlθ   ...(2)
Young's modulus of the system,
Y = γs + γs + γAl = 2γs + γAl      ...(3)
Using (1), (2) and (3), we get:

Therefore, the final length of the system will be , where l0 is its initial length.

Given:
Initial pressure on the steel ball = 1.0 × 10Pa
The ball is heated from 20 °C to 120 °C.
So, change in temperature, $\Delta \theta$ = 100 °C.
Coefficient of linear expansion of steel, $\alpha$ = 12 × 10–6 °C–1
Bulk modulus of steel, B = 1.6 × 1011 Nm–2
Pressure is given as,

Therefore, the pressure inside the ball is 5.8 × 10Pa.

#### Question 32:

Given:
Initial pressure on the steel ball = 1.0 × 10Pa
The ball is heated from 20 °C to 120 °C.
So, change in temperature, $\Delta \theta$ = 100 °C.
Coefficient of linear expansion of steel, $\alpha$ = 12 × 10–6 °C–1
Bulk modulus of steel, B = 1.6 × 1011 Nm–2
Pressure is given as,

Therefore, the pressure inside the ball is 5.8 × 10Pa.

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, $\left(∆T\right)$ = $\theta$
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
IMR02(1 + 2 αθ)
So, I = I0(1 + 2 αθ)
Hence, proved.

#### Question 33:

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, $\left(∆T\right)$ = $\theta$
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
IMR02(1 + 2 αθ)
So, I = I0(1 + 2 αθ)
Hence, proved.

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 $\left(T\right)$:

Here, I = moment of inertia after change in temperature
When the temperature is changed by $∆\theta$, moment of inertia $\left(I\right)$,
I = I0(1+2$\alpha$$∆\theta$)
On substituting the value of in equation(1), we get:
$T=2\mathrm{\pi }\frac{\sqrt{{I}_{0}\left(1+2\alpha ∆\theta \right)}}{\sqrt{K}}$
In winter, $∆\theta$ = 5 °C
$\therefore$ Time period$\left({T}_{1}\right)$
$=2\mathrm{\pi }\frac{\sqrt{{I}_{0}\left(1+2\alpha ×5\right)}}{\sqrt{K}}$
In summer, $∆\theta$ = 45 °C
Time period $\left({T}_{2}\right)$
$=2\mathrm{\pi }\frac{\sqrt{{\mathrm{I}}_{0}\left(1+2\mathrm{\alpha }×45\right)}}{\sqrt{\mathrm{K}}}$

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

#### Question 34:

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 $\left(T\right)$:

Here, I = moment of inertia after change in temperature
When the temperature is changed by $∆\theta$, moment of inertia $\left(I\right)$,
I = I0(1+2$\alpha$$∆\theta$)
On substituting the value of in equation(1), we get:
$T=2\mathrm{\pi }\frac{\sqrt{{I}_{0}\left(1+2\alpha ∆\theta \right)}}{\sqrt{K}}$
In winter, $∆\theta$ = 5 °C
$\therefore$ Time period$\left({T}_{1}\right)$
$=2\mathrm{\pi }\frac{\sqrt{{I}_{0}\left(1+2\alpha ×5\right)}}{\sqrt{K}}$
In summer, $∆\theta$ = 45 °C
Time period $\left({T}_{2}\right)$
$=2\mathrm{\pi }\frac{\sqrt{{\mathrm{I}}_{0}\left(1+2\mathrm{\alpha }×45\right)}}{\sqrt{\mathrm{K}}}$

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

Let initial radius of the circular disc at 20â‚’C = ${r}_{20}$
Let final radius of the circular disc at 50â‚’C = ${r}_{50}$
Coefficient of linear expansion of iron, $\alpha$ = 1.2 × 10–5 °C–1.
Change in temperature, $∆T$ = 30 â‚€C
Let R' and R be the radius of the paricle at 50â‚’C and 20â‚’C respectively.
If v and v' be the linear speed of the particle at 50â‚’C and 20â‚’C respectively, as the angular velocity remains($\omega$) constant.
Therefore,

Now,
R' = R(1+$\alpha ∆T$)
$⇒$R' = R + R $×$ 1.2 × 10–5 °C–1$×∆T$.
$⇒$R' = 1.00036R
Using equation(1) we have,
$\frac{v}{R}=\frac{v\text{'}}{R\text{'}}\phantom{\rule{0ex}{0ex}}⇒\frac{v}{R}=\frac{v\text{'}}{1.00036R}\phantom{\rule{0ex}{0ex}}⇒v\text{'}=1.00036v$
Percentage change in linear speed will be,
$=\frac{v\text{'}-v}{v}×100\phantom{\rule{0ex}{0ex}}=\frac{1.00036v-v}{v}×100\phantom{\rule{0ex}{0ex}}=3.6×{10}^{-2}\phantom{\rule{0ex}{0ex}}$

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