Exemplar MCQs

Hint

Bi-metallic strip- Two strips of equal lengths but of different materials (different coefficient of linear expansion) when join together, it is called “bi-metallic strip”, and can be used in thermostat to break or make electrical contact. This strip has the characteristic property of bending on heating due to unequal linear expansion of the two metal. The strip will bend with metal of greater a on outer side, i.e. convex side.

On heating, the metallic strip with higher coefficient of linear expansion \(\left(\alpha_{\mathrm{Al}}\right)\) will expand more. According to the question, \(\propto_{\mathrm{Al}}>\alpha_{\text {steel, }}\) so aluminum will expand more. So, it should have larger radius of curvature. Hence, aluminium will be on convex side. The metal of smaller \(\propto\) (i.e., steel) bends on the inner side, i.e., concave side.

Hint

(b) When the rod is heated uniformly to raise its temperature slightly, it expands. So, moment of inertia of the rod will increase.Moment of inertia of a uniform rod about its perpendicular bisector,
\(
I=\frac{1}{12} M L^2
\)
\(\Delta T=\) Increase in the temperature of the rod.
\(\therefore \quad\) Changed length, \(L^{\prime}=L(1+\alpha \Delta T) \dots(i)\)
\(\therefore\) New moment of inertia of rod,
\(
\begin{aligned}
I^{\prime} & =\frac{M L^{\prime 2}}{12}=\frac{M}{12} L[1+\alpha \Delta T]^2 \\
& =\frac{M L^2}{12}\left[1+2 \alpha \Delta T+\alpha^2(\Delta T)^2\right] \\
\therefore \quad I^{\prime} & =I[1+2 \alpha \Delta T]
\end{aligned}
\)
[Using (i)]
\(\left(\because \alpha^2(\Delta T)^2\right.\) is very small)
If the temperature increases, moment of inertia will increase.
No external torque is acting on the system, so angular momentum should be conserved.
\(\quad L=\) Angular momentum \(=I \omega=\) constant
\(\Rightarrow \quad I^{\prime} \omega^{\prime}=I \omega\)
where, \(I=\) initial M.I. of rod before heating
\(\omega=\) initial angular velocity
\(I^{\prime}=\) final M.I after heating
\(\omega^{\prime}=\) final angular velocity after heating
Due to expansion of the rod \(I^{\prime}>I\).
\(\Rightarrow \quad \frac{\omega^{\prime}}{\omega^{\prime}}=\frac{I}{I^{\prime}}<1\)
\(\Rightarrow \quad \omega^{\prime}<\omega\)
So, angular velocity (speed of rotation) decreases.

Hint

The temperature on one scale can be converted into other scales by using the following identity.
Reading on any scale – LFP/UFP – LFP = Constant for all scales
Where LFP → Lower fixed point
UFP \(\rightarrow\) Upper fixed point
From the graph, it is clear that the lowest point for scale \(\mathrm{A}\) is \(30^{\circ}\) and the highest point for scale \(\mathrm{A}\) is \(180^{\circ}\).
The lowest point for scale \(B\) is \(0^{\circ}\) and the highest point for scale \(B\) is \(100^{\circ}\). Hence, the relation between the two scales \(A\) and \(B\) is given by
\(
\begin{aligned}
& \frac{T_A-(L F P)_A}{(U F P)_A-(L F P)_A}=\frac{T_B-(L F P)_B}{(U F P)_B-(L F P)_B} \\
& \Rightarrow \frac{T_A-30}{180-30}=\frac{T_B-0}{100-0} \\
& \Rightarrow \frac{t_A-30}{150}=\frac{t_B}{100}
\end{aligned}
\)

Hint

Key concept: Liquids generally increase in volume with increasing temperature but in case of water, it expands on heating if its temperature is greater than \(4^{\circ} \mathrm{C}\). The density of water reaches a maximum value of \(1.000 \mathrm{~g} / \mathrm{cm} 3\) at \(4^{\circ} \mathrm{C}\)
This behaviour of water in the range from \(0^{\circ} \mathrm{C}\) to \(4^{\circ} \mathrm{C}\) is called anomalous expansion.
Let volume of the sphere be \(V\) and \(\rho\) be its density, then we can write buoyant force at \(0^{\circ} \mathrm{C}\).
\(F_{0^{\circ} \mathrm{C}}=V \rho_{0^{\circ}} \mathrm{g} \quad\left(\rho_{0^{\circ}}=\right.\) density at \(\left.0^{\circ} \mathrm{C}\right)\)
Buoyancy at \(4^{\circ} \mathrm{C}\)
\(
\begin{aligned}
& F_{4^{\circ} \mathrm{C}}=V \rho_{4^{\circ} \mathrm{C}} \\
\Rightarrow \quad & \frac{F_{4^{\circ} \mathrm{C}}}{F_{0^{\circ} \mathrm{C}}}=\frac{\rho_{4^{\circ} \mathrm{C}}}{\rho_{0^{\circ} \mathrm{C}}}>1 \\
\Rightarrow \quad & F_{4^{\circ} \mathrm{C}}>F_{0^{\circ} \mathrm{C}}
\end{aligned} \quad\left(\because \rho_{4^{\circ} \mathrm{C}}>\rho_{0^{\circ} \mathrm{C}}\right)
\)
Hence, buoyancy will be less in water at \(0^{\circ} \mathrm{C}\) than that in water at \(4^{\circ} \mathrm{C}\).

Hint

(a) A pendulum clock keeps proper time at temperature \(\theta_0\). If temperature is increased to \(\theta\left(>\theta_0\right)\), then due to linear expansion, length of pendulum increases and hence its time period will increase
Let \(T=2 \pi \sqrt{\frac{L_0}{g}}\) at temperature \(\theta_0\) and \(T^{\prime}=2 \pi \sqrt{\frac{L}{g}}\) at temperature \(\theta\).
\(
\frac{T^{\prime}}{T}=\sqrt{\frac{L^{\prime}}{L}}=\sqrt{\frac{L[1+\alpha \Delta \theta]}{L}}=1+\frac{1}{2} \alpha \Delta \theta
\)
Therefore change (loss or gain) in time per unit time lapsed is \(\frac{T^{\prime}-T}{T}=\frac{1}{2} \alpha \Delta \theta\) Fractional change in time period \(\frac{\Delta T}{T}=\frac{1}{2} \alpha \Delta \theta\)
So, as the temperature increases, length of pendulum increases and hence time period of pendulum increases. Due to increment in its time period, a pendulum clock becomes slow in summer and will lose time.

Hint

(a) When a body is heated its temperature rises and in liquids and gases vibration of molecules about their mean position increases, hence kinetic energy associated with random motion of molecules increases. So, thermal energy or heat associated with the random and translatory motions of molecules.

Hint

Key concept: The co-efficient \(\alpha, \beta\) and \(\gamma\) for a solid are related to each other as follows:
\(
\alpha=\frac{\beta}{2}=\frac{\gamma}{3} \Rightarrow \alpha: \beta: \gamma=1: 2: 3
\). The values of \(\alpha, \beta\) and \(\gamma\) are independent of the units of length, area and volume respectively. For anisotropic solids \(\gamma=\alpha_x+\alpha_y+\alpha_z\) where \(\alpha_x, \alpha_y\) and \(\alpha_z\) represent the mean coefficients of linear expansion along three mutually perpendicular directions.
As the temperature increases radius of the sphere increases as shown. So, the volume of the sphere increases.
Original volume of the sphere \(V_0=\frac{4}{3} \pi R^3\)
Coefficient of linear expansion \(=\alpha\)
Hence, coefficient of volume expansion \(=3 \alpha\)
\(\therefore \quad\) Change in volume is \(\Delta V=V_0 \gamma \Delta T\)
\(
\Delta V=\frac{4}{3} \pi R^3(3 \alpha) \Delta T
\)
Increase in volume \(=4 \pi R^3 \alpha \Delta T\)

Hint

\(
\rho=\text { mass } / \text { volume }
\)

So, the volume of all object will also be same. Here the cooling will be done in the form of radiations that is according to Stefan’s law. Since, emissive power is directly proportional to the surface. Here, for given volume, sphere has least surface area and circular plate of greatest surface area.

As thickness of the plate is least, hence surface area of the plate is maximum. According to Stefan’s law, heat loss (cooling) is directly proportional to the surface area.
\(H_{\text {sphere }}: H_{\text {cube }}: H_{\text {plate }}=A_{\text {sphere }}: A_{\text {cube }}: A_{\text {plate }}\)
As \(\mathrm{A}_{\text {plate }}\) is maximum, hence the plate will cool fastest.
As the sphere is having minimum surface area, hence the sphere cools slowest.

Hint

Key concept:Two bodies are said to be in thermal equilibrium with each other, when no heat flows from one body to the other. That is when both the bodies are at the same temperature.

According to the problem,
(b) If two systems \(X\) and \(Y\) are in thermal equilibrium, i.e., \(T_x=T_y\) and \(x\) is not in thermal equilibrium with \(z\),
i.e., \(T_x \neq T_z\)
then clearly, \(T_y \neq T_z\)
Hence, \(y\) and \(z\) are not in thermal equilibrium.
(d) if \(X\) is not in thermal equilibrium with \(Y\), i.e, \(T_x \neq T_y\) and also \(X\) is not in thermal equilibrium with \(Y\), i.e., \(T_x \neq T_z\)
Then we cannot say about equilibrium of \(Y\) and \(Z\), they may or may not be in equilibrium.

Hint

(b, c) Between these four which has the least surface area will be heated first because of less heat radiation. So, smaller gulab jamuns are having least surface area, hence they will be heated first.
Similarly, smaller pizzas are heated before bigger ones because they are of small surface areas.

Hint

\((a, d)\) During phase change process, the temperature of the system remains constant.
(a) In region AB, a phase change takes place, heat is supplied and ice melts but the temperature of the system is \(0^{\circ} \mathrm{C}\). it remains constant during process. The heat supplied is used to break bonding between molecules.
(d) In region \(C D\), again a phase change takes place from a liquid to a vapour state during which temperature remains constant. It shows water and steam are in equilibrium at boiling point.

Hint

(b, c, d) When hot milk spread on the table heat is transferred to the surroundings by conduction, convection and radiation. Because the surface area of poured milk on a table is more than the surface area of milk filled in a glass. Hence, its temperature falls off exponentially according to Newton’s law of cooling. Heat also will be transferred from surroundings to the milk but will be lesser than that of transferred from milk to the surroundings. So, option (b), (c) and (d) are correct.

Hint

The bulb of a thermometer is made up of diathermic wall because diathermic walls allow the exchange of heat energy between two systems but adiabatic walls do not. So it receives heat from the body to measure the temperature of the body.

Hint

If the first observation is correct, hence from the 1 st observation we get the coefficient of linear expansion
\(
\alpha=\frac{\Delta l}{l \times \Delta T}=\frac{4 \times 10^{-4}}{2 \times 10}=2 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}
\)
For 2nd observation, \(\Delta l=\alpha l \Delta T\)
\(
\begin{aligned}
& =2 \times 10^{-5} \times 1 \times 10 \\
& =2 \times 10^{-4} \mathrm{~m} \neq 4 \times 10^{-4} \mathrm{~m}
\end{aligned}
\)
which is incorrect.
For 3rd observation, \(\Delta l=\alpha l \Delta T\)
\(
\begin{aligned}
& =2 \times 10^{-5} \times 2 \times 20 \\
& =8 \times 10^{-4} \mathrm{~m} \neq 2 \times 10^{-4} \mathrm{~m}
\end{aligned}
\)
which is incorrect.

For 4th observation, \(\Delta l=\alpha l \Delta T\)
\(=2 \times 10^{-5} \times 3 \times 10=6 \times 10^{-4} \mathrm{~m}\) [i.e., observed value is correct]

Hint

The rate of transferring heat in metal is larger than that in wood, as the conductivity of the metal bar is extremely high as compared to wood. Also, metal requires exceedingly small quantities of heat than wood to change temperature as the specific heat of metal is exceptionally low as compared to the wood. Thus, due to larger conductivity and smaller specific heat metal becomes colder and hotter than wood when kept in cold and hot regions, respectively.

Thus if \(a_{\text {practical }}\) and \(E_{\text {practical }}\) represent the absorptive and emissive power of a given surface, while \(a_{\text {black }}\) and \(E_{\text {black }}\) for a perfectly black body, then according to law \(\frac{E_{\text {practical }}}{a_{\text {practical }}}=\frac{E_{\text {biack }}}{a_{\text {black }}}\)
But for a perfectly black body \(a_{\text {black }}=1\), so \(\frac{E_{\text {practical }}}{a_{\text {practical }}}=E_{\text {black }}\)
If emissive and absorptive powers are considered for a particular wavelength \(\lambda,\left(\frac{E_\lambda}{a_\lambda}\right)_{\text {practical }}=\left(E_\lambda\right)_{\text {black }}\)

Now since \(\left(E_\lambda\right)_{\text {Black }}\) is constant at a given temperature, according to this law if a surface is a good absorber of a particular wavelength it is also a good emitter of that wavelength. This in turn implies that a good absorber is a good emitter (or radiator). The conductivities of metals are very high compared to wood. Due to the difference in conductivity, if one touch the hot metal with a finger, heat from the surrounding flows faster to the finger from metals and so one feels the heat. Similarly, when one touches a cold metal the heat from the finger flows away to the surroundings faster. So we can say that a good radiator can be a good absorber.

Hint

To construct a scale of temperature, two fixed points are taken. First fixed point is the freezing point of water, it is called lower fixed point. The second fixed point is the boiling point of water, it is called upper fixed point. Temperature on one scale can be converted into other scale by using the following identity.

\(\frac{\text { Reading on any scale }-\text { (LFP) }}{(UFP) – (LFP)}=\) Constant for all scales

\(
\frac{C-0}{100}=\frac{F-32}{212-32}
\)
Let \(T\) be the value of temperature having same value on Celsius and Fahrenheit scale, i.e. \(F=C=T\)
\(
\begin{array}{ll}
& \frac{T-32}{180}=\frac{T}{100} \\
\Rightarrow \quad & T-32=\frac{9}{5} T \Rightarrow \frac{4}{5} T=-32 \\
\Rightarrow \quad & T=-40^{\circ} \mathrm{C}=-40^{\circ} \mathrm{F}
\end{array}
\)

Hint

The copper bottom of the steel utensil gets heated quickly.
Because of the reason that copper is a good conductor of heat as compared to steel. But steel does not conduct as quickly, thereby allowing food inside to get heated uniformly.

Hint

\(
\text { Moment of inertia of a uniform rod of mass } \mathrm{M} \text { and length } {l} \text { about its perpendicular bisector }
\)
\(
I=\frac{1}{12} M l^2
\)
Increase in length of the rod when temperature is increased by \(\Delta t\), is given by \(L^{\prime}=L(1+\alpha \Delta T)\)
\(\therefore\) New moment of inertia of the rod
\(
\begin{aligned}
I^{\prime} & =\frac{M L^{\prime 2}}{12}=\frac{M}{12} L^2[1+\alpha \Delta T]^2 \\
& =\frac{M L^2}{12}\left[1+2 \alpha \Delta T+\alpha^2(\Delta T)^2\right]
\end{aligned}
\)
\(\therefore\) As change in length \(\Delta l\) is very small, therefore, neglecting \((\Delta l)^2\) \(\left(\because \alpha^2(\Delta T)^2\right.\) is very small) we get
\(
I^{\prime}=I[1+2 \alpha \Delta T]
\)
\(\therefore\) Increase in moment of inertia
\(
=I^{\prime}-I=I[1+2 \alpha \Delta T]-I=2 \alpha I \Delta T
\)

Hint

According to the problem, mass of water \((m)=100 \mathrm{~g}\)
Change in temperature \(\Delta T=0-(-10)=10^{\circ} \mathrm{C}\)
Specific heat of water \(\left(S_{\mathrm{n}}\right)=1 \mathrm{cal} / \mathrm{g} /{ }^{\circ} \mathrm{C}\)
Latent heat of fusion of water \(L^w\) fusion \(=80 \mathrm{cal} / \mathrm{g}\)
Amount of heat required to change the temperature of \(100 \mathrm{~g}\) of water at \(-10^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\)
\(
\begin{aligned}
Q & =m S_w \Delta T \\
& =100 \times 1 \times[0-(-10)]=1000 \mathrm{cal}
\end{aligned}
\)
Let \(m\) gram of ice be melted.
\(
\begin{array}{ll}
\therefore & Q=m L \\
\text { or } & m=\frac{Q}{L}=\frac{1000}{80}=12.5 \mathrm{~g}
\end{array}
\)
As small mass of ice is melted, thus the temperature of the resultant mixture will remain \(0^{\circ} \mathrm{C}\) which contains \(12.5 \mathrm{~g}\) of ice and \(87.5 \mathrm{~g}\) of water.

Hint

According to the Newton’s law of cooling, the rate of loss of heat is directly proportional to the difference of temperature. Or we can say which gives a consequence about rate of fall of temperature of a body with respect to the difference of temperature of body and surroundings.
The first option would have kept water warmer because by adding hot water to cold water, the temperature of the mixture decreases. Due to this temperature difference between the mixed water in the bucket and the surrounding decreases, thereby the decrease in the rate of loss of the heat by the water.
In second option, the hot water in the bucket will lose heat quickly. So if he first attend to the work and fill the remaining bucket with cold water which already lose much heat in 5-10 minutes then the water become more colder as compared with first case.

Hint

According to the problem, \(L_I-L_b=10 \mathrm{~cm}\) where,
\(L_I=\) length of iron scale
\(L_b=\) Length of brass scale
This condition is possible if change in length both the rods is remain same at all temperatures.
Change in length of iron rod,
\(
\Delta L=\alpha_I L_I \Delta T
\)
Change in length of brass rod,
\(
\Delta L=\alpha_B L_B \Delta T
\)
As the change will equal in both the rods, so
\(
\begin{array}{ll}
& \alpha_I L_I \Delta T=\alpha_B L_B \Delta T \\
\Rightarrow & \alpha_I L_I=\alpha_B L_B \Rightarrow \frac{L_I}{L_B}=\frac{\alpha_B}{\alpha_I} \\
\text { Here, } & \alpha_B=1.8 \times 10^{-5} \mathrm{~K}^{-1}, \alpha_I=1.2 \times 10^{-5} \mathrm{~K}^{-1} \\
\therefore & \frac{L_I}{L_B}=\frac{1.8 \times 10^{-5}}{1.2 \times 10^{-5}}=\frac{3}{2} \\
& L_I=\frac{3}{2} L_B \\
\text { As, } & L_I-L_B=10 \mathrm{~cm}
\end{array}
\)
\(
\frac{3}{2} L_B-L_B=10 \Rightarrow \frac{1}{2} L_B=10 \Rightarrow L_B=20 \mathrm{~cm}
\)

Hint

Here we are making a vessel whose Brass volume does not change with temperature.
To make the desired vessel, we should have an iron vessel with a brass rod inside as shown in the diagram.

Volume of vessel, \(V_0=100 \mathrm{~cm}^3\) \(=10^{-4}=\) constant
Volume of iron vessel \(\left(V_I\right)\) – Volume of brass \(\operatorname{rod}\left(V_B\right)=10^{-4} \mathrm{~m}^3\)
\(
\Rightarrow \quad V_I-V_B=10^{-4} \mathrm{~m}^3 \dots(i)
\)
This condition is possible if
\(
\begin{array}{ll}
& \beta_I V_I \Delta T=\beta_B V_B \Delta T \\
\therefore & V_I=\left(\frac{\beta_B}{\beta_I}\right) V_B=\left(\frac{6 \times 10^{-5}}{3.55 \times 10^{-5}}\right) V_B=1.69 V_B
\end{array}
\)
From equation (i),
\(
\begin{array}{ll}
& 1.69 V_B-V_B=10^{-4} \\
\Rightarrow \quad & V_B=\frac{10^{-4}}{0.69}=1.449 \times 10^{-4} \mathrm{~m}^3 \\
\Rightarrow \quad & V_B=144.9 \mathrm{~cm}^3 \\
& V_I=1.69 V_B=1.69 \times 144.9=244.9 \mathrm{~cm}^3
\end{array}
\)
Therefore, an iron vessel with a volume of \(249.9 \mathrm{~cm}^3\) fitted with a brass rod of volume \(144.9 \mathrm{~cm}^3\) will serve as a vessel of volume \(100 \mathrm{~cm}^3\), which will not change with temperature.
Important points:
– Solids can expand in one dimension (linear expansion), two dimensions (superficial expansion) and three dimensions (volume expansion) while liquids and gases usually suffers change in volume only.
– Thermal expansion is minimum in case of solids but maximum in case of gases because intermoleeular force is maximum in solids but minimum in gases.

Hint

According to the problem, decrease in temperature
\(
(\Delta t)=57-37=20^{\circ} \mathrm{C}
\)
Coefficient of linear expansion
\(
(\alpha)=1.7 \times 10^{-5} /{ }^{\circ} \mathrm{C}
\)
Bulk modulus for copper \((B)=140 \times 10^9 \mathrm{~N} / \mathrm{m}^2\)
Coefficient of cubical expansion
\(
(\gamma)=3 \alpha=5.1 \times 10^{-5} /{ }^{\circ} \mathrm{C}
\)
Let initial volume of the cavity be \(V\) and its volume increases by \(\Delta V\) due to increase in temperature.
\(
\therefore \quad \Delta V=\gamma V \Delta t \Rightarrow \frac{\Delta V}{V}=\gamma \Delta t
\)
We know, \(B=\frac{\text { stress }}{\text { volume strain }}\)
\(
\begin{aligned}
\therefore \quad \text { Thermal stress }=B \times\left(\frac{\Delta V}{V}\right) & =B(\gamma \Delta T) \\
& =B(3 \alpha \Delta T) \quad(\because \gamma=3 \alpha) \\
& =140 \times 10^9 \times 3 \times 1.7 \times 10^{-5} \times 20 \\
& =1.428 \times 10^8 \mathrm{Nm}^{-2}
\end{aligned}
\)
This is about \(10^3\) times of atmospheric pressure.

Hint

\(
\text { Diagram shows the deformation of a railway track due to rise in temperature. }
\)
\(
\alpha_{\text {steel }}=1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}
\)
Applying Pythagoras theorem in right angled triangle,
\(
\begin{aligned}
x^2 & =\left(\frac{L+\Delta L}{2}\right)^2-\left(\frac{L}{2}\right)^2 \\
\Rightarrow \quad x & =\sqrt{\left(\frac{L+\Delta L}{2}\right)^2-\left(\frac{L}{2}\right)^2} \\
& =\sqrt{\left(\frac{L}{2}\right)^2+\frac{2 L \Delta L}{4}+\left(\frac{\Delta L}{2}\right)^2-\left(\frac{L}{2}\right)^2} \\
& =\frac{1}{2} \sqrt{\left(L^2+\Delta L^2+2 L \Delta L\right)-L^2}=\frac{1}{2} \sqrt{\left(\Delta L^2+2 L \Delta L\right)}
\end{aligned}
\)
As increase in length \(\Delta L\) is very small, therefore, neglecting \((\Delta L)^2\), we get
\(
x=\frac{\sqrt{2 L \Delta L}}{2} \dots(i)
\)
But \(\quad \Delta L=L \alpha \Delta t \dots(ii)\)
According to the problem, \(L=10 \mathrm{~m}\),
\(
\alpha=1.2 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}, \Delta T=20^{\circ} \mathrm{C}
\)
Substituting value of \(\Delta L\) in Eq. (i) from Eq. (ii)
\(
\begin{aligned}
x & =\frac{1}{2} \sqrt{2 L \times L \alpha \Delta t}=\frac{1}{2} L \sqrt{2 \alpha \Delta t} \\
& =\frac{10}{2} \times \sqrt{2 \times 1.2 \times 10^{-5} \times 20} \\
& =5 \times \sqrt{4 \times 1.2 \times 10^{-4}}=5 \times 2 \times 1.1 \times 10^{-2}=0.11 \mathrm{~m}=11 \mathrm{~cm}
\end{aligned}
\)

Hint

When temperature of a rod varies linearly, then average temperature of the middle point of the rod can be taken as mean of temperatures at the two ends. According to the diagram,
According to the diagram,
\(
\theta=\frac{\theta_1+\theta_2}{2}
\)
Let temperature varies linearly in the rod from its one end to other end from \(\theta_1\) to \(\theta_2\). Let \(\theta\) be the temperature of the mid-point of the rod. Therefore, average temperature of the mid-point of the rod is
\(
\Rightarrow \quad \theta=\frac{\theta_1+\theta_2}{2}
\)
Using relation, \(L=L_0(1+\alpha \theta)\)
or
\(
L=L_0\left[1+\alpha\left(\frac{\theta_1+\theta_2}{2}\right)\right]
\)

Hint

\(
\begin{aligned}
& P=\sigma A T^4=\sigma\left(4 \pi R^2\right) \mathrm{T}^4 \\
& =5.67 \times 10^{-8} \times 4 \times 3.14 \times\left(0.5^{2} \times\left(10^6\right)^4\right. \\
& =1.78 \times 10^{17} \mathrm{~J} / \mathrm{s} \\
& \approx 1.8 \times 10^{17} \mathrm{~J} / \mathrm{s}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& P=\sigma A T^4=\sigma\left(4 \pi R^2\right) \mathrm{T}^4 \\
& =5.67 \times 10^{-8} \times 4 \times 3.14 \times\left(0.5^{2} \times\left(10^6\right)^4\right. \\
& =1.78 \times 10^{17} \mathrm{~J} / \mathrm{s} \\
& \approx 1.8 \times 10^{17} \mathrm{~J} / \mathrm{s}
\end{aligned}
\)

The energy available per second,
\(
\begin{aligned}
& E=1.8 \times 10^{17} \mathrm{~J} / \mathrm{s} \\
\end{aligned}
\)
The energy required to evaporate water
\(
=10 \% \text { of } E=1.8 \times 10^{16} \mathrm{~J} / \mathrm{s}
\)
Energy used (per second) to raise temperature \(30^{\circ} \mathrm{C} \rightarrow 100^{\circ} \mathrm{C}\) and then into vapour at \(\left(100^{\circ} \mathrm{C}\right)\)
\(
\begin{aligned}
& =\mathrm{mS}_{\mathrm{w}} \Delta \theta+\mathrm{ml}_{\mathrm{v}} \\
& =\mathrm{m} \times 4186 \times(100-30)+\mathrm{m} \times 22.6 \times 10^5 \\
& =2.93 \times 10^5 \mathrm{~m}+22.6 \times 10^5 \mathrm{~m} \\
& =\mathrm{m}\left(25.53 \times 10^5 \mathrm{~J} / \mathrm{s}\right)
\end{aligned}
\)
From question, \(\mathrm{m}\left(25.53 \times 10^5\right)=1.8 \times 10^{16}\)
\(
\begin{aligned}
& \text { or } \mathrm{m}=\frac{1.8 \times 10^{16}}{25.53 \times 10^5} \\
& =7.0 \times 10^9 \mathrm{~kg}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& P=\sigma A T^4=\sigma\left(4 \pi R^2\right) \mathrm{T}^4 \\
& =5.67 \times 10^{-8} \times 4 \times 3.14 \times\left(0.5^{2} \times\left(10^6\right)^4\right. \\
& =1.78 \times 10^{17} \mathrm{~J} / \mathrm{s} \\
& \approx 1.8 \times 10^{17} \mathrm{~J} / \mathrm{s}
\end{aligned}
\)

Momentum per unit time,
\(
\begin{aligned}
& \mathrm{P}=\frac{E}{c} \\
& =\frac{1.8 \times 10^{17}}{3 \times 10^8} \\
& =6 \times 10^9 \mathrm{~kg} .
\end{aligned}
\)
Momentum per unit time per unit area,
\(
\begin{aligned}
& P_1=\frac{P}{4 \pi R^2}=\frac{6 \times 10^8}{4 \times 3.14 \times\left(10^3\right)^2} \\
& P_1=47.7 \mathrm{~N} / \mathrm{m}^2
\end{aligned}
\)

Hint

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 the temperature of one body due to change in frame, the same change will be acquired by the other body.

Hint

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 the temperature of the system.

Hint

The ideal gas thermometer is based on the ideal gas equation, \(P V=n R T\), where \(P\) is pressure of the gas at constant volume \(\mathrm{V}\) with \(\mathrm{n}\) number of moles at temperature \(\mathrm{T}\). Therefore, \(\mathrm{P}=\) constant \(\times \mathrm{T}\). 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 the kinetic theory of gases or any experimental results.

Hint

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.

Hint

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.

Hint

At sea level, the pressure is around 1 atmosphere and at high altitude, the density of air reduces.
Pressure of liquid,
\(
P=h \rho g \text {, }
\)
where \(\rho=\) density of fluid
The above equation shows that pressure depends on density. Therefore at \(4^{\circ} \mathrm{C}\), the density of water will be less at high altitude, compared to the density at sea level.

Hint

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.

Hint

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.

Hint

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.

Hint

When a spherical shell is heated, its volume changes according to the equation, \(V_\theta=V_0(1+\gamma \Delta \theta)\). 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, \(\gamma\).

Hint

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).

Hint

Celsius and Fahrenheit temperatures are related in the following way:
\(
C=\frac{5}{9} \mathrm{~F}-\frac{160}{9}
\)
Here, \(\mathrm{F}=\) 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 \(5 / 9\).

Hint

If we consider \(T_F\) and \(T_K\) as the temperatures measured in Fahrenheit and Kelvin scales respectively, then \(T_F=T_K=\theta=574.59^{\circ}\) where, \(\theta\) is the temperature in Fahrenheit and Kelvin scales. But in case of Celsius and Kelvin scales, such \(\theta\) does not exist.
Also, in Kelvin and Platinum scales, they depend on the properties of the thermometric substance used to define the scale. Hence, such value does not exist in Kelvin and Platinum.

Let \(\theta\) be the temperature in Fahrenheit and Kelvin scales.
We know that the relation between the temperature in Fahrenheit and Kelvin scales is given by
\(
\begin{aligned}
& \frac{T_F-32}{180}=\frac{T_K-273.15}{100} \\
& T_F=T_K=\theta
\end{aligned}
\)
Therefore,
\(
\begin{aligned}
& \frac{\theta-32}{180}=\frac{\theta-273.15}{100} \\
& 5 \theta-160=9 \theta-2458.5 \\
& 4 \theta=2298.35 \\
& \theta=574.59^{\circ}
\end{aligned}
\)

Hint

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.

Hint

The coefficient of linear expansion,
\(
\begin{aligned}
& \propto=\frac{1}{L} \frac{\Delta L}{\Delta T} \\
& =\frac{|L|}{|L T|}=K^{-1}
\end{aligned}
\)
Here, \(L=\) initial length
\(\Delta \mathrm{L}=\) change in length
\(\Delta T=\) change in temperature
On the other hand, the coefficient of volume expansion,
\(
\gamma=\frac{1}{V} \frac{\Delta V}{\Delta T}=\frac{\left|L^3\right|}{\left|L^3 T\right|}=K^{-1}
\)
Here, V = initial volume
\(\Delta \mathrm{V}=\) change in volume
\(\Delta \mathrm{T}=\) change in temperature
\(K=\) kelvin, the S.I. unit of temperature

Hint

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.

Hint

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.

Hint

The change in moment of inertia of uniform rod with change in temperature is given by, \(I^{\prime}=I(1+2 \propto \Delta t)\)
Here, \(I\) = initial moment of inertia
\(\mathrm{I}^{\prime}=\) new moment of inertia due to change in temperature
\(\alpha=\) expansion coefficient
\(\Delta \mathrm{t}=\) change in temperature
So, \(I^{\prime}-I=2 \alpha I \Delta t\)

Hint

The change in moment of inertia of uniform rod with change in temperature is given by, \(I^{\prime}=I(1+2 \propto \Delta t)\)
Here, \(I\) = initial moment of inertia
\(\mathrm{I}^{\prime}=\) new moment of inertia due to change in temperature
\(\alpha=\) expansion coefficient
\(\Delta \mathrm{t}=\) change in temperature
So, \(I^{\prime}-I=2 \alpha I \Delta t\)

Hint

The density of water is maximum at \(4^{\circ} \mathrm{C}\), 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 \(4^{\circ} \mathrm{C}\).

Hint

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.

Hint

(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.

Hint

(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.

Hint

(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 whether body B’s internal energy will be or will not be greater than that of body A.

Hint

(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.

Hint

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.

Hint

In general, the time period of a pendulum, \(t\), is given by
\(
\mathrm{t}=\frac{1}{2 \pi} \sqrt{\frac{1}{g}}
\)
When the temperature \((T)\) is increased, the length of the pendulum \((l)\) is given by,
\(
l=l_0(1+\alpha \mathrm{T}) \text {, }
\)
where \(l_0=\) length at \(0^{\circ} \mathrm{C}\)
\(\alpha=\) linear coefficient of expansion.
Therefore, the time period of a pendulum will be
\(
\mathrm{t}=\frac{1}{2 \pi} \sqrt{\frac{l_0(1+\propto T)}{g}}
\)
Hence, time period of a pendulum will increase with increase in temperature.

Hint

Given:
Ice point of a mercury thermometer, \(T_0=20^{\circ} \mathrm{C}\)
Steam point of a mercury thermometer, \(T_{100}=80^{\circ} \mathrm{C}\)
Temperature on thermometer that is to be calculated in centigrade scale, \(T_1=32^{\circ} \mathrm{C}\)
Temperature on a centigrade mercury scale, \(\mathrm{T}\), is given as:
\(
\begin{aligned}
& \mathrm{T}=\frac{T_1-T_0}{T_{100}-T_0} \times 100 \\
& \Rightarrow T=\frac{32-20}{80-20} \times 100 \\
& \Rightarrow T=\frac{12}{60} \times 100 \\
& \Rightarrow T=\frac{120}{6} \\
& \Rightarrow \mathrm{T}=20^{\circ} \mathrm{C}
\end{aligned}
\)
Therefore, the temperature on a centigrade mercury scale will be \(20^{\circ} \mathrm{C}\).

Hint

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

Hint

Given:
In a gas thermometer, the pressure measured at the melting point of lead, \(P=2.20 \times\) Pressure at triple point \(\left(P_{t r}\right)\)
So the melting point of lead, \((T)\) is given as:
\(
\begin{aligned}
& T=\frac{P}{P_{t r}} \times 273.16 \mathrm{~K} \\
& \Rightarrow T=\frac{2.20 \times P_{t r}}{P_{t r}} \times 273.16 \mathrm{~K} \\
& \Rightarrow \mathrm{T}=2.20 \times 273.16 \mathrm{~K} \\
& \Rightarrow \mathrm{T}=600.952 \mathrm{~K} \\
& \Rightarrow \mathrm{T} \cong 601 \mathrm{~K}
\end{aligned}
\)
Therefore, the melting point of lead is \(601 \mathrm{~K}\).

Hint

Given:
Pressure measured by a constant volume gas thermometer at the triple point of water, \(P_{t r}=40 \mathrm{kPa}=40 \times\) \(10^3 \mathrm{~Pa}\)
Boiling point of water, \(T=100^{\circ} \mathrm{C}=373.16 \mathrm{~K}\)
Let the pressure measured at the boiling point of water be \(P\).
For a constant volume gas thermometer, temperature-pressure relation is given below:
\(
\begin{aligned}
& T=\frac{P}{P_{t r}} \times 273.16 K \\
& \Rightarrow P=\frac{T \times P_{t r}}{273.16} \\
& \Rightarrow P=\frac{373.16 \times 40 \times 10^3}{273.16} \\
& \Rightarrow P=54643 P a \\
& \Rightarrow P=54.6 \times 10^3 \mathrm{~Pa} \\
& \Rightarrow P \simeq 55 \mathrm{kPa}
\end{aligned}
\)
Therefore, the pressure measured at the boiling point of water is \(55 \mathrm{kPa}\).

Hint

Given:
Temperature of ice point, \(T_1=273.15 \mathrm{~K}\)
Temperature of steam point, \(T_2=373.15 \mathrm{~K}\)
Pressure of the gas in a constant volume thermometer at the ice point, \(P_1=70 \mathrm{kPa}\),
Let \(P_{t r}\) be the pressure at the triple point and \(P_2\) be the pressure at the steam point.
The temperature-pressure relations for ice point and steam point are given below:
For ice point,
\(
\begin{aligned}
& T_1=\frac{P_1}{P_{t r}} \times 273.16 K \\
& \Rightarrow 273.15=\frac{70}{P_{t r}} \times 10^3 \times 273.16 \\
& \Rightarrow P_{t r}=\frac{70 \times 273.16 \times 10^3}{273.15} P a
\end{aligned}
\)
For steam point,
\(
T_2=\frac{P_2 \times 273.16}{P_{t r}} K
\)
On substituting the value of \(P_{\text {tr }}\), we get:
\(
\begin{aligned}
& 373.15=\frac{P_2 \times 273.15 \times 273.16}{70 \times 273.16 \times 10^3} \\
& \Rightarrow P_2=\frac{373.15 \times 70 \times 10^3}{273.15} \\
& \Rightarrow P_2=95.626 \times 10^3 \mathrm{~Pa} \\
& \Rightarrow P_2 \simeq 96 \mathrm{kPa}
\end{aligned}
\)
Therefore, the pressure at steam point is \(96 \mathrm{kPa}\).

Hint

Given:
In a constant-volume gas thermometer,
Pressure of the gas at the ice point, \(P_0=80 \mathrm{~cm}\) of \(\mathrm{Hg}\)
Pressure of the gas at the steam point, \(P_{100}=90 \mathrm{~cm}\) of \(\mathrm{Hg}\)
Pressure of the gas in a heated wax bath, \(P=100 \mathrm{~cm}\) of \(\mathrm{Hg}\)
The temperature of the wax bath
\((T)\) is given by:
\(
\begin{aligned}
& T=\frac{P-P_0}{P_{100}-P_0} \times 100^{\circ} \mathrm{C} \\
& \Rightarrow T=\frac{100-80}{90-80} \times 100 \\
& \Rightarrow T=\frac{20}{10} \times 100 \\
& \Rightarrow T=200^{\circ} C
\end{aligned}
\)
Therefore, the temperature of the wax bath is \(200^{\circ} \mathrm{C}\).

Hint

Given:
Volume of the bulb in a Callender’s compensated constant pressure air thermometer, \((\mathrm{V})=\) \(1800 \mathrm{cc}\)
Volume of mercury that has to be poured out, \(V^{\prime}=200 \mathrm{cc}\)
Temperature of ice bath, \(T_0=273.15 \mathrm{~K}\)
So the temperature of the vessel \(\left(T^{\prime}\right)\) is given by:
\(
\begin{aligned}
& T^{\prime}=\frac{V}{V-V^{\prime}} \times T_0 \\
& \Rightarrow T^{\prime}=\frac{1800}{1600} \times 273.15 K \\
& \Rightarrow T^{\prime}=307.293 \\
& \Rightarrow T^{\prime} \simeq 307 K
\end{aligned}
\)
Therefore, the temperature of the vessel is \(307 \mathrm{~K}\).

Hint

Given:
\(
\text { Resistance at } 0^{\circ} \mathrm{C}, R_0=80 \Omega
\)
Resistance at \(100^{\circ} \mathrm{C}, R_{100}=90 \Omega\)
Let \(t\) be the temperature at which the resistance \(\left(R_t\right)\) is 86 \(\Omega\)
\(
\begin{aligned}
& t=\frac{R_t-R_0}{R_{100}-R_0} \times 100 \\
& \Rightarrow t=\frac{86-80}{90-80} \times 100 \\
& \Rightarrow t=\frac{6}{10} \times 100 \\
& \Rightarrow t=60^{\circ}
\end{aligned}
\)
Therefore, the resistance is 86 \(\Omega\) at \(60^{\circ} \mathrm{C}\)

Hint

Given:
Reading on resistance thermometer at ice point, \(R_0=20 \Omega\)
Reading on resistance thermometer at steam point, \(R_{100}=27.5 \Omega\)
Reading on resistance thermometer at zinc point, \(R_{420}=50 \Omega\)
The variation of resistance with temperature in Celsius scale, \(\theta\), is given as:
\(
\begin{aligned}
& R_{100}=R_0\left(1+\alpha \theta+\beta \theta^2\right) \\
& \Rightarrow R_{100}=R_0+R_0 \alpha \theta+R_0 \beta \theta^2 \\
& \Rightarrow R_{100}=R_0+R_0 \alpha \theta+R_0 \beta \theta^2 \\
& \Rightarrow \frac{\left(R_{100}-R_0\right)}{R_0}=\alpha \theta+\beta \theta^2 \\
& \Rightarrow \frac{27.5-20}{20}=\alpha \theta+\beta \theta^2 \\
& \Rightarrow \frac{7.5}{20}=\alpha \times 100+\beta \times 10000 \ldots(i)
\end{aligned}
\)
\(
\begin{aligned}
& \text { Also, } R_{420}=R_0\left(1+\alpha \theta+\beta \theta^2\right) \\
& \Rightarrow R_{420}=R_0+R_0 \alpha \theta+R_0 \beta \theta^2 \\
& \Rightarrow \frac{R_{420}-R_0}{R_0}=\alpha \theta+\beta \theta^2 \\
& \Rightarrow \frac{50-20}{20}=420 \alpha+176400 \beta \\
& \Rightarrow \frac{3}{2}=420 \alpha+176400 \beta \ldots(i i)
\end{aligned}
\)
Solving (i) and (ii), we get:
\(
\begin{aligned}
& \alpha=3.8 \times 10^{-3 \circ} \mathrm{C}^{-1} \\
& \beta=-5.6 \times 10^{-7 \circ} \mathrm{C}^{-1}
\end{aligned}
\)
Therefore, resistance \(R_0\) is \(20 \Omega\) and the value of \(\alpha\) is \(3.8 \times 10^{-3 \circ} \mathrm{C}^{-1}\) and that of \(\beta\) is \(-5.6 \times 10^{-7 \circ} \mathrm{C}^{-2}\).

Hint

Given:
Length of the slab when the temperature is \(0^{\circ} \mathrm{C}, L_0=10 \mathrm{~m}\)
Temperature on the summer day, \(t=35^{\circ} \mathrm{C}\)
Let \(L_1\) be the length of the slab on a summer day when the temperature is \(35^{\circ} \mathrm{C}\).
The coefficient of linear expansion of concrete, \(\alpha=1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\)
\(
\begin{aligned}
& L_1=L_0(1+\propto t) \\
& \Rightarrow L_1=10\left(1+10^{-5} \times 35\right) \\
& \Rightarrow L_1=10+35 \times 10^{-4} \\
& \Rightarrow L_1=10.0035 \mathrm{~m}
\end{aligned}
\)
So, the length of the slab on summer day when the temperature is \(35^{\circ} \mathrm{C}\) is \(10.0035 \mathrm{~m}\).

Hint

Given:
Temperature at which the steel metre scale is calibrated, \(t_1=20^{\circ} \mathrm{C}\)
Temperature at which the scale is used, \(\mathrm{t}_2=10^{\circ} \mathrm{C}\)
So, the change in temperature,
\(
\begin{aligned}
& \Delta \mathrm{t}=(20^{\circ} & \left.-10^{\circ}\right) \mathrm{C} \\
\end{aligned}
\)
The distance to be measured by the metre scale, \(L_0=51\)
\(
-50)=1 \mathrm{~cm}=0.01 \mathrm{~m}
\)
Coefficient of linear expansion of steel,
\(
\alpha_{\text {steel }}=1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}
\)
Let the new length measured by the scale due to expansion of steel be \(L_2\), Change in length is given by,
\(
\begin{aligned}
& \Delta \mathrm{L}=\mathrm{L}_1 \propto_{\text {steel }} \Delta \mathrm{t} \\
& \Rightarrow \Delta L=1 \times 1.1 \times 10^{-5} \times 10 \\
& \Rightarrow \Delta L=0.00011 \mathrm{~cm}
\end{aligned}
\)
As the temperature is decreasing, therefore length will decrease by
\(\Delta L\)
Therefore, the new length measured by the scale due to expansion of steel \(\left(L_2\right)\) will be,
\(
\mathrm{L}_2=1 \mathrm{~cm}
\)
\(-0.00011 \mathrm{~cm}=0.99989 \mathrm{~cm}\)

Hint

Given, Length of the iron sections when there’s no effect of temperature on them, \(L_0=12.0 \mathrm{~m}\)
Temperature at which the iron track is laid in winter, \(t_w=18^{\circ} \mathrm{C}\)
Maximum temperature during summers, \(\mathrm{t}_{\mathrm{s}}=48^{\circ} \mathrm{C}\)
Coefficient of linear expansion of iron,
\(
\alpha=11 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}
\)
Let the new lengths attained by each section due to expansion of iron in winter and summer be \(L_w\) and \(L_s\), respectively, which can be calculated as follows:
\(
\begin{aligned}
& L_w=L_0\left(1+\alpha t_w\right) \\
& \Rightarrow L_w=12\left(1+11 \times 10^{-6} \times 18\right) \\
& \Rightarrow L_w=12.00237 \mathrm{~m} \\
& L_s=L_0\left(1+\alpha t_s\right) \\
& \Rightarrow L_s=12\left(1+11 \times 10^{-6} \times 48\right) \\
& \Rightarrow L_s=12.006336 \mathrm{~m} \\
& \therefore \Delta L=L_s-L_w \\
& \Rightarrow \Delta L=12.006336-12.002376 \\
& \Rightarrow \Delta L=0.00396 \mathrm{~m} \\
& \Rightarrow \Delta L \approx 0.4 \mathrm{~cm}
\end{aligned}
\)
Therefore, the gap \((\Delta L)\) that should be left between two iron sections, so that there is no compression during summer, is \(0.4 \mathrm{~cm}\).

Hint

Given:
Diameter of a circular hole in an aluminium plate at \(0^{\circ} \mathrm{C}, \mathrm{d}_1=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}\) Initial temperature, \(\mathrm{t}_1=0^{\circ} \mathrm{C}\)
Final temperature, \(t_2=100^{\circ} \mathrm{C}\)
So, the change in temperature, \((\Delta t)=100^{\circ} \mathrm{C}-0^{\circ} \mathrm{C}=100^{\circ} \mathrm{C}\)
The linear expansion coefficient of aluminium, \(\alpha_{a l}=2.3 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\)
Let the diameter of the circular hole in the plate at \(100^{\circ} \mathrm{C}\) be \(\mathrm{d}_2\), which can be written as
\(
\begin{aligned}
& d_2=d_1(1+\alpha \Delta t) \\
& \Rightarrow d_2=2 \times 10^{-2}\left(1+2.3 \times 10^{-5} \times 10^2\right) \\
& \Rightarrow d_2=2 \times 10^{-2}\left(1+2.3 \times 10^{-3}\right) \\
& \Rightarrow d_2=2 \times 10^{-2}+2.3 \times 2 \times 10^{-5} \\
& \Rightarrow d^2=0.02+0.000046 \\
& \Rightarrow d^2=0.020046 \mathrm{~m} \\
& \Rightarrow d^2 \approx 2.0046 \mathrm{~cm}
\end{aligned}
\)
Therefore, the diameter of the circular hole in the aluminium plate at \(100^{\circ} \mathrm{C}\) is \(2.0046 \mathrm{~cm}\).

Hint

Given:
At \(20^{\circ} \mathrm{C}\), length of the metre scale made up of steel, \(L_{s t}=\) length of the metre scale made up of aluminium, \(L_{\text {al }}\) Coefficient of linear expansion for aluminium, \(\alpha_{a l}=2.3 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\)
Coefficient of linear expansion for steel, \(\alpha_{\text {st }}=1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\)
Let the length of the aluminium scale at \(0^{\circ} \mathrm{C}, 40^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\) be \(L_{\text {0al }}, L_{40 \text { al }}\) and \(L_{100 a l}\).
And let the length of the steel scale at \(0^{\circ} \mathrm{C}, 40^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\) be \(L_{0 s t}, L_{40 \text { st }}\) and \(L_{100 s t}\).
(a) So, \(L_{0 s t}\left(1-\alpha_{\text {st }} \times 20\right)=L_{0 a l}\left(1-\alpha_{a l} \times 20\right)\)
\(
\begin{aligned}
& \frac{L_{0 s t}}{L_{0 a l}}=\frac{\left(1-\alpha_{a l} \times 20\right)}{\left(1-\alpha_{s t} \times 20\right)} \\
& \Rightarrow \frac{L_{0 s t}}{L_{0 a l}}=\frac{1-2.3 \times 10^{-5} \times 20}{1-1.1 \times 10^{-5} \times 20} \\
& \Rightarrow \frac{L_{0 s t}}{L_{0 a l}}=\frac{0.99954}{0.99978} \\
& \Rightarrow \frac{L_{0 s t}}{L_{0 a l}}=0.999759
\end{aligned}
\)
\(
\begin{aligned}
& \text { (b) } \frac{L_{40 a l}}{L_{40 s t}}=\frac{L_{0 a l}\left(1+\alpha_{a l} \times 40\right)}{L_{0 s t}\left(1+\alpha_{s t} \times 40\right)} \\
& \Rightarrow \frac{L_{40 a l}}{L_{40 s t}}=\frac{L_{0 a l}}{L_{0 s t}} \times \frac{\left(1+2.3 \times 10^{-5} \times 40\right)}{\left(1+1.1 \times 10^{-5} \times 40\right)} \\
& \Rightarrow \frac{L_{40 a l}}{L_{40 s t}}=\frac{0.99977 \times 1.00092}{1.00044} \\
& \Rightarrow \frac{L_{40 a l}}{L_{40 s t}}=1.0002496 \\
& \text { (c) } \frac{L_{100 a l}}{L_{100 s t}}=\frac{L_{0 a l}\left(1+\alpha_{a l} \times 100\right)}{L_{0 s t}\left(1+\alpha_{s t} \times 100\right)} \\
& \Rightarrow \frac{L_{100 a l}}{L_{100 s t}}=\frac{0.99977 \times 1.0023}{1.0023} \\
& \Rightarrow \frac{L_{100 a l}}{L_{100 s t}}=1.00096
\end{aligned}
\)

Hint

The change in length is given as,
\(
\begin{aligned}
& \Delta \mathrm{L}=\mathrm{L} \alpha \Delta \mathrm{T} \\
& \frac{\Delta \mathrm{L}}{\mathrm{L}}=11 \times 10^{-6} \times -30
\end{aligned}
\)
The error percentage is given as,
\(
\begin{aligned}
& \% \mathrm{e}=\left(\frac{\Delta \mathrm{L}}{\mathrm{L}} \times 100\right) \% \\
& \% \mathrm{e}=\left(11 \times 10^{-6} \times -30 \times 100\right) \% \\
& =-3.3 \times 10^{-2} \%
\end{aligned}
\)
(b)
The change in length is given as,
\(
\begin{aligned}
& \Delta \mathrm{L}=\mathrm{L} \alpha \Delta \mathrm{T} \\
& \frac{\Delta \mathrm{L}}{\mathrm{L}}=11 \times 10^{-6} \times 10
\end{aligned}
\)
The error percentage is given as,
\(
\begin{aligned}
& \% \mathrm{e}=\left(\frac{\Delta \mathrm{L}}{\mathrm{L}} \times 100\right) \% \\
& \% \mathrm{e}=\left(11 \times 10^{-6} \times 10 \times 100\right) \% \\
& =1.1 \times 10^{-2} \%
\end{aligned}
\)
Thus, the error percentages are \(-3.3 \times 10^{-2} \%\) and \(1.1 \times 10^{-2} \%\).

Hint

Given:
Temperature at which a metre scale gives an accurate reading, \(T_1=20^{\circ} \mathrm{C}\)
The value of variation admissible, \(\Delta \mathrm{L}=0.055 \mathrm{~mm}=0.055 \times 10^{-3} \mathrm{~m}\), in the length, \(\mathrm{L}_0=1 \mathrm{~m}\)
Coefficient of linear expansion of steel, \(\alpha=11 \times 10^{-6} \mathrm{C}^{-1}\)
Let the range of temperature in which the experiment can be performed be \(T_2\).
We know: \(\Delta \mathrm{L}=\mathrm{L}_0 \alpha \Delta \mathrm{T}\)
\(
\begin{aligned}
& \Rightarrow 0.055 \times 10^{-3}=1 \times 11 \times 10^{-6} \times\left(T_1 \pm T_2\right) \\
& \Rightarrow 5 \times 10^{-3}=\left(20 \pm T_2\right) \times 10^{-3} \\
& \Rightarrow 20 \pm T_2=5 \\
& \text { Either } T_2=20+5=25^{\circ} \mathrm{C} \\
& \text { or } T_2=20-5=15^{\circ} \mathrm{C}
\end{aligned}
\)
Hence, the experiment can be performed in the temperature range of \(15^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\)

Hint

Given:
Density of water at \(0^{\circ} \mathrm{C},\left(f_0\right)=0.998 \mathrm{~g} \mathrm{~cm}^{-3}\)
Density of water at \(4^{\circ} \mathrm{C},\left(f_4\right)=1.000 \mathrm{~g} \mathrm{~cm}-3\)
Change in temperature, \((\Delta t)=4^{\circ} \mathrm{C}\)
Let the average coefficient of volume expansion of water in the temperature range of 0 to \(4^{\circ} \mathrm{C}\) be \(\gamma\).
Weknow \(: f_4=f_0(1+\gamma \Delta t)\)
\(
\begin{aligned}
& \Rightarrow f_0=\frac{f_4}{1+\gamma \Delta t} \\
& \Rightarrow 0.998=\frac{1}{1+\gamma .4} \\
& \Rightarrow 1+4 \gamma=\frac{1}{0.998} \\
& \Rightarrow 4 \gamma=\left(\frac{1}{0.998}\right)-1 \\
& \Rightarrow \gamma=0.0005=5 \times 10^{-4 o} \mathrm{C}^{-1}
\end{aligned}
\)
As the density decreases,
\(
\gamma=-5 \times 10^{-4 o} \mathrm{C}^{-1}
\)
Therefore,the average coefficient of volume expansion of water in the temperature range of 0 to \(4^{\circ} \mathrm{C}\) will be
\(
\gamma=-5 \times 10^{-4}
\)

Hint

Let the original length of iron rod be \(L_{F e}\) and \(L_{F e}\) be its length when temperature is increased by \(\Delta T\).
Let the original length of aluminium rod be \(L_{A l}\) and \(L_{A l}^{\prime}\) be its length when temperature is increased by \(\Delta T\).
Coefficient of linear expansion of iron,
\(
\alpha_{F e}=12 \times 10^{-6}{ }^{\circ} \mathrm{C}-1\)
Coefficient of linear expansion of aluminium, \(\alpha_{\mathrm{Al}}=23 \times 10^{-6}{ }^{\circ} \mathrm{C}-1\)
Since the difference in length is independent of temperature, the difference is always constant.
\(
\begin{aligned}
& L_{F e}^{\prime}=L_{F e}\left(1+\alpha_{F e} \times \Delta T\right) \\
& \text { and } L_{A l}^{\prime}=L_{A l}\left(1+\alpha_{A l} \times \Delta T\right) \\
& \Rightarrow L_{F e}^{\prime}-L_{A l}^{\prime}=L_{F e}-L_{A l}+L_{F e} \times \alpha_{F e} \Delta T-L_{A l} \times \alpha_{A l} \times \Delta T \quad-(1)
\end{aligned}
\)
Given :
\(
L_{F e}^{\prime}-L_{A l}^{\prime}=L_{F e}-L_{A l}
\)
Hence, \(L_{F e} \alpha_{F e}=L_{A l} \alpha_{A l}\)[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 \(23: 12\).

Hint

Given: The temperature at which the pendulum shows the correct time, \(T_1=20^{\circ} \mathrm{C}\) Coefficient of linear expansion of steel,
\(
\alpha=12 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}
\)
Let \(T_2\) be the temperature at which the value of \(\mathrm{g}\) is \(9.788 \mathrm{~ms}^{-2}\) and
\(\Delta \mathrm{T}\) be the change in temperature.
So, the time periods of pendulum at different values of \(g\) will be \(t_1\) and \(t_2\), such that
\(
\begin{aligned}
& t_1=2 \pi \sqrt{\frac{l_1}{g_1}} \\
& t_2=2 \pi \sqrt{\frac{l_2}{g_2}} \\
& =\frac{2 \pi \sqrt{l_1(1+\alpha \Delta T)}}{\sqrt{g_2}}\left(\because l_2=l_1(1+\alpha \Delta T)\right)
\end{aligned}
\)
Given, \(t_1=t_2\)
\(
\begin{aligned}
& \Rightarrow \frac{2 \pi \sqrt{l_1}}{\sqrt{g_1}}=\frac{2 \pi \sqrt{l_1(1+\alpha \Delta T)}}{\sqrt{g_2}} \\
& \Rightarrow \sqrt{\left(\frac{l_1}{g_1}\right)}=\frac{\sqrt{l_1(1+\alpha \Delta T)}}{\sqrt{g_2}} \\
& \Rightarrow \frac{1}{9.8}=\frac{1+12 \times 10^{-6} \times \Delta T}{9.788} \\
& \Rightarrow \frac{9.788}{9.8}=1+12 \times 10^{-6} \times \Delta T \\
& \Rightarrow \frac{9.788}{9.8}-1=12 \times 10^{-6} \times \Delta T \\
& \Rightarrow \Delta T=\frac{-0.00122}{12 \times 10^{-6}} \\
& \Rightarrow T_2-20=-102.4 \\
& \Rightarrow T_2=-102.4+20 \\
& =-82^{\circ} 4 \\
& \Rightarrow T_2 \approx-82^{\circ} \mathrm{C}
\end{aligned}
\)
Therefore, for a pendulum clock to give correct time, the temperature at which the value of \(\mathrm{g}\) is \(9.788 \mathrm{~ms}^{-2}\) should be \(-82^{\circ} \mathrm{C}\)

Hint

Given:
Diameter of the steel sphere at temperature \(\left(T_1=10^{\circ} \mathrm{C}\right), d_{s t}=2.005 \mathrm{~cm}\)
Diameter of the aluminium sphere, \(\mathrm{d}_{\mathrm{Al}}=2.000 \mathrm{~cm}\)
Coefficient of linear expansion of steel, \(\alpha_{\text {st }}=11 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\)
Coefficient of linear expansion of aluminium, \(\alpha_{\mathrm{Al}}=23 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\)
Let the temperature at which the ball will fall be \(T_2\) so that change in temperature be \(\Delta T\).
\(
\begin{aligned}
& d_{s t}^{\prime}=2.005\left(1+\alpha_{s t} \Delta T\right) \\
& \Rightarrow d_{s t}^{\prime}=2.005+2.005 \times 11 \times 10^{-6} \times \Delta T \\
& d_{A l}^{\prime}=2\left(1+\alpha_{A l} \times \Delta T\right) \\
& \Rightarrow d_{A l}^{\prime}=2+2 \times 23 \times 10^{-6} \times \Delta T
\end{aligned}
\)
The steel ball will fall when both the diameters become equal.
So, \(d_{\mathrm{st}}^{\prime}=d_{\mathrm{Al}}^{\prime}\)
\(
\begin{aligned}
& \Rightarrow 2.005+2.005 \times 11 \times 10^{-6} \Delta T=2+2 \times 23 \times 10^{-6} \Delta T \\
& \Rightarrow(46-22.055) \times 10^{-6} \Delta T=0.005 \\
& \Rightarrow \Delta T=\frac{0.005 \times 10^6}{23.945}=208.81 \\
& \text { Now, } \Delta T=T_2-T_1=T_2-10^{\circ} \mathrm{C} \\
& \Rightarrow T_2=\Delta T+T_1=208.81+10 \\
& \Rightarrow T_2=218.8 \cong 219^{\circ} \mathrm{C}
\end{aligned}
\)
Therefore, the temperature at which the ball will fall is \(219^{\circ} \mathrm{C}\).

Hint

Given:
At \(40^{\circ} \mathrm{C}\), the length and breadth of the glass window are \(20 \mathrm{~cm}\) and \(30 \mathrm{~cm}\), respectively. Coefficient of linear expansion of glass,
\(
\alpha_g=9.0 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}
\)
Coefficient of linear expansion for aluminium,
\(
\alpha_{A l}=24 \times 100^{-6}{ }^{\circ} \mathrm{C}^{-1}
\)
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 \(0^{\circ} \mathrm{C}\).
Change in temperature,
\(
\Delta \theta=40^{\circ} \mathrm{C}
\)
Let the initial length of aluminium be I.
\(
\begin{aligned}
& l\left(1-\alpha_{A l} \Delta \theta\right)=20\left(1-\alpha_g \Delta \theta\right) \\
& \Rightarrow l\left(1-24 \times 10^{-6} \times 40\right)=20\left(1-9 \times 10^{-6} \times 40\right) \\
& \Rightarrow l(1-0.00096)=20(1-0.00036) \\
& \Rightarrow l=\frac{20 \times 0.99964}{1-0.00096} \\
& =\frac{20 \times 0.99964}{0.99904} \\
& \Rightarrow l=20.012 \mathrm{~cm}
\end{aligned}
\)
Let the initial breadth of aluminium be \(b\).
\(
\begin{aligned}
& b\left(1-\alpha_{A l} \Delta \theta\right)=30\left(1-\alpha_g \Delta \theta\right) \\
& \Rightarrow b=\frac{30 \times\left(1-9 \times 10^{-6} \times 40\right)}{\left(1-24 \times 10^{-6} \times 40\right)} \\
& =\frac{30 \times 0.99964}{0.99904} \\
& \Rightarrow b=30.018 \mathrm{~cm}
\end{aligned}
\)
Therefore, the size of the aluminium frame should be \(20.012 \mathrm{~cm} \times 30.018 \mathrm{~cm}\).

Hint

At \(\mathrm{T}=20^{\circ} \mathrm{C}\), the volume of the glass vessel, \(\mathrm{V}_{\mathrm{g}}=1000 \mathrm{cc}\).
Let the volume of mercury be \(\mathrm{V}_{\mathrm{Hg}}\).
Coefficient of cubical expansion of mercury, \(\gamma_{\mathrm{Hg}}=1.8 \times 10^{-4} /{ }^{\circ} \mathrm{C}\)
Coefficient of cubical expansion of glass, \(\gamma_g=9 \times 10^{-6} /{ }^{\circ} \mathrm{C}\)
Change in temperature, \(\Delta T\), is same for glass and mercury.
Let the volume of glass and mercury after rise in temperature be \(\mathrm{V}^{\prime} \mathrm{g}\) and \(\mathrm{V}^{\prime} \mathrm{Hg}\) respectively.[/latex]
Volume of remaining space after change in temperature, \(
\left(V_g^{\prime}-V^{\prime}{ }_{H g}\right)=
\)Volume of the remaining space (initial), \(\left(\mathrm{V}_{\mathrm{g}}-\mathrm{V}_{\mathrm{Hg}}\right)\)
We know: \(V_g^{\prime}=V_g\left(1+V_g \Delta T\right)\)
\(
V^{\prime}{ }_{H g}=V_{H g}\left(1+Y_{H g} \Delta T\right)
\)
Subtracting (2) from (1), we get:
\(
\begin{aligned}
& V_g^{\prime}-V_{H g}^{\prime}=V_g-V_{H g}+V_g \gamma_g \Delta T-V_{H g} \gamma_{H g} \Delta T \\
& \Rightarrow V_g \gamma_g \Delta T-V_{H g} \gamma_{H g} \Delta T=0 \\
& \Rightarrow \frac{V_g}{V_{H g}}=\frac{\gamma_{H g}}{\gamma_g} \\
& \Rightarrow \frac{1000}{V_{H g}}=\frac{1.8 \times 10^{-4}}{9 \times 10^{-6}} \\
& \Rightarrow V_{H g}=\frac{9 \times 10^{-3}}{1.8 \times 10^{-4}} \\
& \Rightarrow V_{H g}=50 c c
\end{aligned}
\)
Therefore, the volume of mercury that should be poured into the glass vessel is \(50 \mathrm{cc}\).

Hint

Given:
Volume of water contained in the aluminium can, \(V_0=500 \mathrm{~cm}^3\)
Area of inner cross-section of the can, \(A=125 \mathrm{~cm}^2\)
Coefficient of volume expansion of water, \(\gamma=3.2 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\)
Coefficient of linear expansion of aluminium,
\(
\alpha_{A L}=23 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}
\)
If \(\Delta \theta\) is the change in temperature, then final volume of water
\((V)\) due to expansion,
\(
\begin{aligned}
V & =V_0(1+\gamma \Delta \theta) \\
& =500\left[1+3.2 \times 10^{-4} \times(80-10)\right] \\
& =500\left[1+3.2 \times 10^{-4} \times 70\right] \\
& =511.2 \mathrm{~cm}^3
\end{aligned}
\)
The aluminium vessel expands in its length only.
So, area of expansion of the base can be neglected.
Increase in volume of water \(=11.2 \mathrm{~cm}^3\)
Consider a cylinder of volume \(11.2 \mathrm{~cm}^3\)
\(\therefore\) Increase in height of the water
\(
\begin{aligned}
& =\frac{11.2}{125}=0.0896 \\
& =0.089 \mathrm{~cm}
\end{aligned}
\)

Hint

Given: At \(0^{\circ} \mathrm{C}\), volume of glass vessel, \(\mathrm{V}_{\mathrm{g}}=10 \times 10 \times 10=1000 \mathrm{cc}=\) volume of mercury, \(\mathrm{V}_{\mathrm{Hg}}\) Let the volume of mercury at \(10^{\circ} \mathrm{C}\) be \(\mathrm{V}^{\prime} \mathrm{Hg}\) and that of glass be \(\mathrm{V}^{\prime} \mathrm{g}\).

At \(10^{\circ} \mathrm{C}\), the additional volume of mercury than glass, due to heating, \(\mathrm{V}^{\prime} \mathrm{Hg}-\mathrm{V}^{\prime} \mathrm{g}=1.6 \mathrm{~cm}^3\)
So change in temperature, \(\Delta \mathrm{T}=10^{\circ} \mathrm{C}\)
Coefficient of linear expansion of glass, \(\alpha_g=6.5 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\)

\(
\text { Therefore, the coefficient of volume expansion of glass, } \gamma_g=3 \times 6.5 \times 10^{-6{ }^{\circ}} \mathrm{C}^{-1}
\)
Let the coefficient of volume expansion of mercury be \(\gamma_{\mathrm{Hg}}\).
We know that
\(
\begin{aligned}
& \mathrm{V}^{\prime}{ }_{\mathrm{Hg}}=\mathrm{V}_{\mathrm{Hg}}\left(1+\gamma_{\mathrm{Hg}} \Delta \mathrm{T}\right) \dots(1)\\
& \mathrm{V}_{\mathrm{g}}^{\prime}=\mathrm{V}_{\mathrm{g}}\left(1+\gamma_{\mathrm{g}} \Delta \mathrm{T}\right) \dots(2)
\end{aligned}
\)
Subtracting (2) from (1) we get,
\(
\begin{aligned}
& \mathrm{V}^{\prime} \mathrm{Hg}-\mathrm{V}_{\mathrm{g}}^{\prime}=\mathrm{V}_{\mathrm{Hg}}-\mathrm{V}_{\mathrm{g}}+\mathrm{V}_{\mathrm{Hg}} \gamma_{\mathrm{Hg}} \Delta \mathrm{T}-\mathrm{V}_{\mathrm{g}} \gamma_{\mathrm{g}} \Delta \mathrm{T} \text { (as } \mathrm{V}_{\mathrm{Hg}}=\mathrm{V}_{\mathrm{g}} \text { ) } \\
& \Rightarrow 1.6=1000 \times \gamma_{H g} \times 10-1000 \times 6.5 \times 3 \times 10^{-6} \times 10 \\
& \Rightarrow \gamma_{H g}=\frac{1.6+19.5 \times 10^{-2}}{10000} \\
& \Rightarrow \gamma_{H g}=\frac{1.6+0.195}{10000} \\
& \Rightarrow \gamma_{H g}=\frac{1.795}{10000} \\
& \Rightarrow \gamma_{H g}=1.795 \times 10^{-4} \\
& \Rightarrow \gamma_{\mathrm{Hg}} \cong 1.8 \times 10^{-4{ }^{\circ}} \mathrm{C}^{-1}
\end{aligned}
\)
Therefore, the coefficient of volume expansion of mercury is \(1.8 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\).

Hint

\(
\begin{aligned}
& f_{\mathrm{w}}=880 \mathrm{Kg} / \mathrm{m}^3, f_{\mathrm{b}}=900 \mathrm{Kg} / \mathrm{m}^3 \\
& \mathrm{~T}_1=0^{\circ} \mathrm{C}, \\
& \gamma_\omega=1.2 \times 10^{-3} /{ }^{\circ} \mathrm{C} \\
& \gamma_{\mathrm{b}}=1.5 \times 10^{-3} /{ }^{\circ} \mathrm{C} \\
& \text { The sphere begins } \mathrm{t} \text { sink when, } \\
& (\mathrm{mg})_{\text {sphere }}=\text { displaced water } \\
& \Rightarrow \mathrm{V} f_\omega^{\prime} \mathrm{g}=\mathrm{V} f_b^{\prime} \mathrm{g} \\
& \Rightarrow \frac{f_\omega}{1+\gamma_\omega \Delta \theta}=\frac{f_{\mathrm{b}}}{1+\gamma_{\mathrm{b}} \Delta \theta} \\
& \Rightarrow \frac{880}{1+1.2 \times 10^{-3} / \theta}={ }^{\circ} \mathrm{C} \\
& \Rightarrow 880+880 \times 1.5 \times 10^{-3}(\Delta \theta)=900+900 \times 1.2 \times 10^{-3}(\Delta \theta) \\
& \Rightarrow\left(880 \times 1.5 \times 10^{-3}-900 \times 1.2 \times 10^{-3}\right)(\Delta \theta)=20 \\
& \Rightarrow(1320-1080) \times 10^{-3}(\Delta \theta)=20 \\
& \Rightarrow \Delta \theta=83.3^{\circ} \mathrm{C} \approx 83^{\circ} \mathrm{C}
\end{aligned}
\)

Hint

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

Hint

Given, \(\theta_1=20^{\circ} \mathrm{C}, \quad \theta_2=50^{\circ} \mathrm{C}\)
\(
a_{\text {steel }}=1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}
\)
Longitudinal strain \(=\)?
Strain \(=\frac{\Delta L}{L}-\frac{L a \Delta \theta}{L}=a \Delta \theta\)
\(=1.2 \times 10^{-5} \times(50-20)\)
\(=1.2 \times 10^{-5} \times 30\)
\(=36 \times 10^{-5}=3.6 \times 10^{-4}\)
The strain is opposite to the direction of expansion.

Hint

\(
\begin{aligned}
& \mathrm{A}=0.5 \mathrm{~mm}^2=0.5 \times 10^{-6} \mathrm{~m}^2 \\
& \mathrm{~T}_1=20^{\circ} \mathrm{C}, \mathrm{T}_2=0^{\circ} \mathrm{C} \\
& \alpha_{\mathrm{s}}=1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}, \quad \gamma=2 \times 2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2
\end{aligned}
\)
Decrease in length due to compression \(=L \alpha \Delta \theta\)
\(
\gamma=\frac{\text { Stress }}{\text { Strain }}=\frac{\mathrm{F}}{\mathrm{A}} \times \frac{\mathrm{L}}{\Delta \mathrm{L}} \Rightarrow \Delta \mathrm{L}=\frac{\mathrm{FL}}{\mathrm{A\gamma}}
\)
Tension is developed due to (1) & (2) Equating them,
\(
\begin{aligned}
& L \alpha \Delta \theta=\frac{F L}{A \gamma} \Rightarrow F=\alpha \Delta \theta A \gamma \\
& =1.2 \times 10^{-5} \times(20-0) \times 0.5 \times 10^{-5} 2 \times 10^{11}=24 \mathrm{~N}
\end{aligned}
\)

Hint

Given:
Temperature of the rod at zero tension, \(T_1=20^{\circ} \mathrm{C}\)
Final temperature, \(T_2=100^{\circ} \mathrm{C}\)
Change in temperature, \(\Delta \theta=80^{\circ} \mathrm{C}\)
Cross-sectional area of the rod, \(A=2 \mathrm{~mm}^2=2 \times 10^{-6} \mathrm{~m}^2\)
Coefficient of linear expansion of steel, \(\alpha=12 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\)
Young’s modulus of steel, \(Y=2 \times 10^{11} \mathrm{Nm}^{-2}\)
Let \(L\) be the length of the steel rod at \(20^{\circ} \mathrm{C}\) and \(L^{\prime}\) be the length of steel rod at \(100^{\circ} \mathrm{C}\).
Change of length of the rod, \(\Delta L=L^{\prime}-L\)
If \(F\) be the force exerted by the rod on one of the clamps due to rise in temperature, then
\(
\begin{aligned}
& \mathrm{Y}=\frac{\text { stress }}{\text { strain }}=\frac{\mathrm{F}}{\mathrm{A}} / \frac{\Delta \mathrm{L}}{\mathrm{L}} \\
& \Rightarrow \mathrm{F}=\frac{\mathrm{Y} \times \Delta \mathrm{L}}{\mathrm{L}} \times \mathrm{A} \\
& \Delta \mathrm{L}=\mathrm{L} \alpha \Delta \theta \\
& \Rightarrow \mathrm{F}=\frac{\mathrm{YL} \alpha \Delta \theta \mathrm{A}}{\mathrm{L}} \\
& \mathrm{F}=\mathrm{YA} \alpha \Delta \theta \\
& =2 \times 10^{11} \times 2 \times 10^{-6} \times 12 \times 10^{-6} \times 80 \\
& =48 \times 80 \times 10^{-1}
\end{aligned}
\)
So, \(F=384 \mathrm{~N}\)
Therefore, the rod will exert a force of \(384 \mathrm{~N}\) on one of the clamps.

Hint

Given:
Initial pressure on the steel ball \(=1.0 \times 10^5 \mathrm{~Pa}\)
The ball is heated from \(20^{\circ} \mathrm{C}\) to \(120^{\circ} \mathrm{C}\).
So, change in temperature, \(\Delta \theta=100^{\circ} \mathrm{C}\).
Coefficient of linear expansion of steel, \(\alpha=12 \times 10^{-6 \circ} \mathrm{C}^{-1}\)
Bulk modulus of steel,\(B=1.6 \times 10^{11} \mathrm{Nm}^{-2}\)
Pressure is given as,
\(
\begin{aligned}
& \Rightarrow P=B \times \gamma \Delta \theta \\
& \Rightarrow P=B \times 3 \alpha \Delta \theta(\because \gamma=3 \alpha) \\
& \Rightarrow P=1.6 \times 10^{11} \times 3 \times 12 \times 10^{-6} \times(120-20) \\
& =1.6 \times 3 \times 12 \times 10^{11} \times 10^{-6} \times 10^2 \\
& =57.6 \times 10^7 \\
& \Rightarrow P=5.8 \times 10^8 \mathrm{~Pa}
\end{aligned}
\)
Therefore, the pressure inside the ball is \(5.8 \times 10^8 \mathrm{~Pa}\).

Hint

Given:
Coefficient of linear expansion of solid \(=\alpha\)
Moment of inertia at \(0{ }^{\circ} \mathrm{C}=I_0\)
If temperature changes to \(\theta\) from \(0^{\circ} \mathrm{C}\), then change in temperature, \((\Delta \mathrm{T})=\theta\)
Let I be the new moment of inertia attained due to rise in temperature.
Let \(R_0\) be the radius of gyration at \(0{ }^{\circ} \mathrm{C}\).
We know that on heating, radius of gyration will change as
\(
R=R_0(1+\alpha \theta)
\)
Here, \(R\) is the radius of gyration after heating.
\(\mathrm{I}_0=\mathrm{MR}_0{ }^2\), where \(\mathrm{M}=\) mass of the body
Now, I \(=\mathrm{MR}^2=\mathrm{MR}_0^2(1+\alpha \theta)^2\)
Expanding binomially and neglecting the higher terms of \(\operatorname{order}(\alpha \theta)\) that will be very small, we get \(\mathrm{I}=\mathrm{MR}_0{ }^2(1+2 \alpha \theta)\)
So, \(I=I_0(1+2 \alpha \theta)\)
Hence, proved.

Hint

Let the initial M.I.a at \(0^{\circ} \mathrm{C}\) be \(\mathrm{I}_0\)
\(
\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{K}}} \\
& \mathrm{I}=\mathrm{I}_0(1+2 \alpha \Delta \theta) \quad \text { (from above question) }
\end{aligned}
\)
At \(5^{\circ} \mathrm{C}, \quad T_1=2 \pi \sqrt{\frac{I_0(1+2 \alpha \Delta \theta)}{K}}=2 \pi \sqrt{\frac{I_0(1+2 \alpha 5)}{K}}=2 \pi \sqrt{\frac{I_0(1+10 \alpha)}{\mathrm{K}}}\)
At \(45^{\circ} \mathrm{C}, \quad T_2=2 \pi \sqrt{\frac{\mathrm{I}_0(1+2 \alpha 45)}{\mathrm{K}}}=2 \pi \sqrt{\frac{\mathrm{I}_0(1+90 \alpha)}{\mathrm{K}}}\)
\(
\frac{T_2}{T_1}=\sqrt{\frac{1+90 \alpha}{1+10 \alpha}}=\sqrt{\frac{1+90 \times 2.4 \times 10^{-5}}{1+10 \times 2.4 \times 10^{-5}}} \sqrt{\frac{1.00216}{1.00024}}
\)
\(\%\) change \(=\left(\frac{T_2}{T_1}-1\right) \times 100=0.0959 \%=9.6 \times 10^{-2}\)

Hint

\(
\begin{aligned}
& \mathrm{T}_1=20^{\circ} \mathrm{C}, \mathrm{T}_2=50^{\circ} \mathrm{C}, \Delta \mathrm{T}=30^{\circ} \mathrm{C} \\
& \alpha=1.2 \times 10^5 /{ }^{\circ} \mathrm{C}
\end{aligned}
\)
\(\omega\) remains constant
(I) \(\omega=\frac{V}{R}\)
(II) \(\omega=\frac{V^{\prime}}{R^{\prime}}\)
Now, \(R^{\prime}=R(1+\alpha \Delta \theta)=R+R \times 1.2 \times 10^{-5} \times 30=1.00036 R\)
From (I) and (II)
\(
\begin{aligned}
& \frac{V}{R}=\frac{V^{\prime}}{R^{\prime}}=\frac{V^{\prime}}{1.00036 R} \\
& \Rightarrow V^{\prime}=1.00036 \mathrm{~V} \\
& \% \text { change }=\frac{(1.00036 \mathrm{~V}-\mathrm{V})}{\mathrm{V}} \times 100=0.00036 \times 100=3.6 \times 10^{-2}
\end{aligned}
\)