Thermal Expansion for Solid Questions in English

(A) The percentage increase in length is given by $\frac{\Delta l}{l} \times 100 = 2\%$.
For a square sheet of side $l$,the area is $A = l^2$.
Using the concept of relative error,the fractional change in area is $\frac{\Delta A}{A} = 2 \frac{\Delta l}{l}$.
Therefore,the percentage increase in area is $\frac{\Delta A}{A} \times 100 = 2 \times (\frac{\Delta l}{l} \times 100)$.
Substituting the given value: $2 \times 2\% = 4\%$.

(B) The thermal stress developed in a rod when its expansion is prevented is given by $\sigma = Y \alpha \Delta \theta$.
Since stress $\sigma = F/A$,the force $F$ is given by $F = Y A \alpha \Delta \theta$.
Given:
$Y = 10^{11} \, N/m^2$
$A = 1 \, cm^2 = 10^{-4} \, m^2$
$\alpha = 10^{-5} /^\circ C$
$\Delta \theta = 100 - 0 = 100^\circ C = 10^2 \, ^\circ C$
Substituting these values into the formula:
$F = 10^{11} \times 10^{-4} \times 10^{-5} \times 10^2$
$F = 10^{11 - 4 - 5 + 2} = 10^4 \, N$.

(A) The balance wheel in a watch is responsible for regulating the timekeeping mechanism.
For the watch to maintain constant time,the dimensions of the balance wheel must remain unchanged despite fluctuations in ambient temperature.
Invar is a nickel-iron alloy known for its uniquely low coefficient of thermal expansion.
Because the dimensions of Invar do not vary significantly with temperature,it is used to ensure the watch's period of oscillation remains constant.

(D) Let the length of the brass rod at temperature $\Delta \theta$ be ${L_1}$ and the steel rod be ${L_2}$.
${L_1} = {l_1}(1 + {\alpha _1}\Delta \theta )$
${L_2} = {l_2}(1 + {\alpha _2}\Delta \theta )$
The difference in length at temperature $\Delta \theta$ is ${L_2} - {L_1} = {l_2}(1 + {\alpha _2}\Delta \theta ) - {l_1}(1 + {\alpha _1}\Delta \theta )$.
${L_2} - {L_1} = ({l_2} - {l_1}) + \Delta \theta ({l_2}{\alpha _2} - {l_1}{\alpha _1})$.
For the difference in length to remain the same as at $0^{\circ}C$,we require ${L_2} - {L_1} = {l_2} - {l_1}$.
This implies $\Delta \theta ({l_2}{\alpha _2} - {l_1}{\alpha _1}) = 0$.
Since $\Delta \theta \neq 0$,we must have ${l_2}{\alpha _2} - {l_1}{\alpha _1} = 0$,or ${l_1}{\alpha _1} = {l_2}{\alpha _2}$.

(C) When a solid object is heated,it undergoes thermal expansion.
Let the linear expansion coefficient be $\alpha$.
The change in length is $\Delta L = L_0 \alpha \Delta T$,so the fractional change is $\frac{\Delta L}{L_0} = \alpha \Delta T$.
The change in area is $\Delta A = A_0 (2\alpha) \Delta T$,so the fractional change is $\frac{\Delta A}{A_0} = 2\alpha \Delta T$.
The change in volume is $\Delta V = V_0 (3\alpha) \Delta T$,so the fractional change is $\frac{\Delta V}{V_0} = 3\alpha \Delta T$.
Comparing the coefficients,$3\alpha > 2\alpha > \alpha$.
Therefore,the percentage increase is largest for the volume.
Density decreases upon heating because volume increases while mass remains constant.

(D) The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{mgd}}$. For a uniform rod of length $L$,$T \propto \sqrt{L}$.
Taking the derivative,the fractional change in the time period is given by $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta L}{L}$.
Since $\Delta L = L \alpha \Delta \theta$,we have $\frac{\Delta L}{L} = \alpha \Delta \theta$.
Therefore,$\frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta \theta$.
Given $\alpha = 2 \times 10^{-6} /^{\circ}C$ and $\Delta \theta = 10^{\circ}C$:
$\frac{\Delta T}{T} = \frac{1}{2} \times (2 \times 10^{-6}) \times 10 = 10^{-5}$.
The percentage increase is $\frac{\Delta T}{T} \times 100 = 10^{-5} \times 100 = 10^{-3} \%$.

(C) The change in length $\Delta L$ is given by the formula: $\Delta L = L_0 \alpha \Delta T$.
Here,$L_0 = 10 \, cm$,$\alpha = 11 \times 10^{-6} /^{\circ}C$,and $\Delta T = 19^{\circ}C - 20^{\circ}C = -1^{\circ}C$.
Substituting the values:
$\Delta L = 10 \times (11 \times 10^{-6}) \times (-1) \, cm$.
$\Delta L = -110 \times 10^{-6} \, cm = -1.1 \times 10^{-4} \, cm = -11 \times 10^{-5} \, cm$.
The negative sign indicates that the bar becomes shorter.
Therefore,the bar is $11 \times 10^{-5} \, cm$ shorter.

(C) When a material is heated,its particles gain kinetic energy and move further apart,which causes the material to expand.
Since the mass of the material remains constant while the volume increases,the density,defined as $\rho = \frac{m}{V}$,must decrease.
Therefore,thermal expansion leads to a decrease in the density of the material.

(C) bimetallic strip consists of two different metals with different coefficients of linear expansion ($\alpha_A$ and $\alpha_B$).
When the strip is heated, both metals expand, but the metal with the higher coefficient of linear expansion expands more than the other.
As shown in the figure, if $\alpha_A > \alpha_B$, metal $A$ will expand more than metal $B$.
To accommodate this difference in length while remaining bonded together, the strip must bend.
The metal with the higher expansion coefficient $(A)$ forms the outer side of the arc, while the metal with the lower expansion coefficient $(B)$ forms the inner side.
Therefore, the correct option is $C$.

(A) When a solid object with a cavity is heated,the material of the object expands in all directions.
This expansion follows the same laws as if the cavity were filled with the same material.
Since the material expands outward,the inner boundary of the cavity also moves outward,effectively increasing the volume of the cavity.
Therefore,the volume of the cavity increases.

(A) Let the initial diameter of the pin be $D_p$ and the diameter of the hole be $D_h$. We are given $D_p > D_h$ at temperature $T$.
To make them fit,we need the diameters to be equal at some temperature $T + \Delta T$.
Using the linear expansion formula $D' = D(1 + \alpha \Delta T)$,we have $D_p(1 + \alpha_b \Delta T) = D_h(1 + \alpha_s \Delta T)$,where $\alpha_b$ is the coefficient of linear expansion for bronze and $\alpha_s$ is for steel.
Since $\alpha_s > \alpha_b$,heating the block increases the hole diameter $D_h$ faster than the pin diameter $D_p$ increases.
Therefore,heating only the block is the most efficient way to achieve the fit with the minimum temperature change.

(D) The area of the base of a cylinder is given by $A = \pi r^2$.
Since the material is the same,the radius $r$ is proportional to the length $L$ (linear expansion),so $r \propto L$.
Therefore,the area $A$ is proportional to the square of the length: $A \propto L^2$.
Using the error analysis formula for small changes,we have $\frac{\Delta A}{A} \approx 2 \cdot \frac{\Delta L}{L}$.
Given that the percentage increase in length is $\frac{\Delta L}{L} \times 100 = 2\%$,
the percentage increase in area is $\frac{\Delta A}{A} \times 100 = 2 \times (\frac{\Delta L}{L} \times 100) = 2 \times 2\% = 4\%$.
Thus,the area of the base will increase by $4\%$.

(D) The coefficient of superficial expansion (area expansion) is denoted by $\beta$ and the coefficient of linear expansion is denoted by $\alpha$.
These two coefficients are related by the formula $\beta = 2\alpha$.
Given,$\beta = 2 \times 10^{-5} \text{ °C}^{-1}$.
Therefore,$\alpha = \frac{\beta}{2} = \frac{2 \times 10^{-5}}{2} = 1 \times 10^{-5} \text{ °C}^{-1}$.
Thus,the correct option is $D$.

(C) The density $\rho$ is related to temperature $T$ by the relation $\rho = \frac{\rho_0}{1 + \gamma \Delta T} \approx \rho_0 (1 - \gamma \Delta T)$.
Thus,the coefficient of volume expansion $\gamma$ is given by $\gamma = \frac{\rho_0 - \rho}{\rho \Delta T}$.
Given $\rho_0 = 10 \, g/cm^3$,$\rho = 9.7 \, g/cm^3$,and $\Delta T = 100^{\circ}C - 0^{\circ}C = 100^{\circ}C$.
$\gamma = \frac{10 - 9.7}{9.7 \times 100} \approx \frac{0.3}{970} \approx 3.09 \times 10^{-4} \, ^{\circ}C^{-1}$.
Using the approximation $\gamma \approx 3 \times 10^{-4} \, ^{\circ}C^{-1}$.
The coefficient of linear expansion $\alpha$ is related to $\gamma$ by $\alpha = \frac{\gamma}{3}$.
$\alpha = \frac{3 \times 10^{-4}}{3} = 10^{-4} \, ^{\circ}C^{-1}$.

(A) The coefficient of volume expansion $\gamma$ is given by the formula: $\gamma = \frac{\Delta V}{V \cdot \Delta T}$.
Given, $\frac{\Delta V}{V} = 0.24\% = \frac{0.24}{100}$ and $\Delta T = 40^{\circ}C$.
Substituting the values: $\gamma = \frac{0.24}{100 \times 40} = \frac{0.24}{4000} = 0.00006 = 6 \times 10^{-5} \text{ } ^\circ\text{C}^{-1}$.
The relationship between the coefficient of linear expansion $\alpha$ and volume expansion $\gamma$ is $\alpha = \frac{\gamma}{3}$.
Therefore, $\alpha = \frac{6 \times 10^{-5}}{3} = 2 \times 10^{-5} \text{ } ^\circ\text{C}^{-1}$.

(A) The coefficient of linear expansion $(\alpha)$ is defined as the fractional change in length per unit change in temperature.
The coefficient of areal expansion $(\beta)$ is defined as the fractional change in area per unit change in temperature,where $\beta = 2\alpha$.
The coefficient of volume expansion $(\gamma)$ is defined as the fractional change in volume per unit change in temperature,where $\gamma = 3\alpha$.
Therefore,the relationship is $\alpha : \beta : \gamma = \alpha : 2\alpha : 3\alpha$.
Dividing by $\alpha$,we get the ratio $1:2:3$.

(C) When an object made of an isotropic material is heated,it undergoes thermal expansion in all dimensions. This expansion is equivalent to a photographic enlargement of the object.
$1$. The length of the rod,the gap $x$,the radius $r$,and the thickness $d$ are all linear dimensions of the object.
$2$. According to the principle of linear thermal expansion,any linear dimension $L$ changes as $L' = L(1 + \alpha \Delta T)$,where $\alpha$ is the coefficient of linear expansion and $\Delta T$ is the change in temperature.
$3$. Since $\alpha > 0$ for metals and $\Delta T > 0$ during heating,all linear dimensions $x$,$r$,and $d$ will increase.
Therefore,the correct option is $(c)$.

(D) The formula for the coefficient of linear expansion is $\alpha = \frac{\Delta L}{L_0 \times \Delta \theta}$.
Given:
Initial length $L_0 = 5 \, m$
Change in length $\Delta L = 5.01 \, m - 5 \, m = 0.01 \, m$
Change in temperature $\Delta \theta = 100^{\circ}C - 0^{\circ}C = 100^{\circ}C$
Substituting the values:
$\alpha = \frac{0.01}{5 \times 100} = \frac{0.01}{500} = \frac{1}{50000} = 2 \times 10^{-5} \, ^{\circ}C^{-1}$.

(A) Given: Initial length $L_0 = 100\, cm$,Change in temperature $\Delta \theta = 100^{\circ}C - 0^{\circ}C = 100^{\circ}C$,Change in length $\Delta L = 0.19\, cm$.
First,calculate the coefficient of linear expansion $(\alpha)$:
$\alpha = \frac{\Delta L}{L_0 \Delta \theta} = \frac{0.19}{100 \times 100} = \frac{0.19}{10000} = 1.9 \times 10^{-5} {^{\circ}C^{-1}}$.
Now,calculate the coefficient of cubical expansion $(\gamma)$:
$\gamma = 3\alpha = 3 \times 1.9 \times 10^{-5} {^{\circ}C^{-1}} = 5.7 \times 10^{-5} {^{\circ}C^{-1}}$.
Therefore,the correct option is $A$.

(D) The coefficient of linear expansion of brass $(\alpha_{brass})$ is greater than that of steel $(\alpha_{steel})$.
When the system is cooled, both the brass disc and the steel plate contract.
Since $\alpha_{brass} > \alpha_{steel}$, the brass disc contracts more than the steel plate.
As a result, the gap between the disc and the hole increases, and the disc becomes loose.

(B) The formula for the increase in length due to thermal expansion is given by $\Delta L = L_0 \alpha \Delta \theta$.
Here,the initial length $L_0 = 10\, m$,the coefficient of linear expansion $\alpha = 10 \times 10^{-6} {^{\circ}C^{-1}}$,and the change in temperature $\Delta \theta = 100^{\circ}C - 0^{\circ}C = 100^{\circ}C$.
Substituting these values into the formula:
$\Delta L = 10 \times (10 \times 10^{-6}) \times 100$
$\Delta L = 10 \times 10^{-5} \times 100 = 10^{-2}\, m$.
Since $1\, m = 100\, cm$,we have $\Delta L = 10^{-2} \times 100 = 1\, cm$.
Therefore,the increase in the length of the bar is $1.0\, cm$.

(A) The diameter of the hole $D_0 = 0.9997 \, cm$ must increase to $D = 1.0 \, cm$ to allow the cylinder to fit.
Using the formula for linear expansion: $\Delta D = D_0 \alpha \Delta T$.
Here,$\Delta D = D - D_0 = 1.0 - 0.9997 = 0.0003 \, cm$.
Substituting the values: $0.0003 = 0.9997 \times (12 \times 10^{-6}) \times \Delta T$.
$\Delta T = \frac{0.0003}{0.9997 \times 12 \times 10^{-6}} \approx \frac{0.0003}{0.0000119964} \approx 25.007^{\circ}C$.
Rounding to the nearest integer,the minimum required rise in temperature is $25^{\circ}C$.

(C) The change in length of a rod due to thermal expansion is given by $\Delta l = l \alpha \Delta T$.
Given that the change in length for both rods is the same,we have $\Delta l_1 = \Delta l_2$.
Substituting the formula,we get $l_1 \alpha_a t = l_2 \alpha_s t$.
Canceling $t$ from both sides,we get $l_1 \alpha_a = l_2 \alpha_s$,which implies $\frac{l_1}{l_2} = \frac{\alpha_s}{\alpha_a}$.
To find the ratio $\frac{l_1}{l_1 + l_2}$,we use the property of ratios: if $\frac{a}{b} = \frac{c}{d}$,then $\frac{a}{a+b} = \frac{c}{c+d}$.
Applying this to our equation,we get $\frac{l_1}{l_1 + l_2} = \frac{\alpha_s}{\alpha_a + \alpha_s}$.

(A) The coefficient of real expansion of mercury $(\gamma_{real})$ is constant regardless of the vessel used.
We know the relation: $\gamma_{real} = \gamma_{app} + \gamma_{vessel}$,where $\gamma_{vessel} = 3\alpha$.
For the glass vessel: $\gamma_{real} = 153 \times 10^{-6} + \gamma_{glass}$.
For the steel vessel: $\gamma_{real} = 144 \times 10^{-6} + 3\alpha_{steel}$.
Given $\alpha_{steel} = 12 \times 10^{-6} \, ^{\circ}C^{-1}$,so $\gamma_{steel} = 3 \times 12 \times 10^{-6} = 36 \times 10^{-6} \, ^{\circ}C^{-1}$.
Equating the two expressions for $\gamma_{real}$:
$153 \times 10^{-6} + \gamma_{glass} = 144 \times 10^{-6} + 36 \times 10^{-6}$.
$153 \times 10^{-6} + \gamma_{glass} = 180 \times 10^{-6}$.
$\gamma_{glass} = (180 - 153) \times 10^{-6} = 27 \times 10^{-6} \, ^{\circ}C^{-1}$.
Since $\gamma_{glass} = 3\alpha_{glass}$,we have $3\alpha_{glass} = 27 \times 10^{-6} \, ^{\circ}C^{-1}$.
Therefore,$\alpha_{glass} = 9 \times 10^{-6} \, ^{\circ}C^{-1}$.

(D) The expansion of solids can be well understood by the potential energy curve for two adjacent atoms in a crystalline solid as a function of their internuclear separation $(r)$.
At ordinary temperature: Each molecule of the solid vibrates about its equilibrium position $P_1$ between $A$ and $B$,where $r_0$ is the equilibrium distance from another molecule.
At high temperature: The amplitude of vibration increases (e.g.,$C \leftrightarrow D$ and $E \leftrightarrow F$). Due to the asymmetry of the potential energy curve,the average equilibrium positions ($P_2$ and $P_3$) of the molecules are displaced. Hence,the average distance from other molecules increases $(r_2 > r_1 > r_0)$.
Thus,on raising the temperature,the average equilibrium distance between the molecules increases and the solid as a whole expands.

(C) Initial diameter of the wheel $D_w = 1000 \, mm$.
Initial diameter of the tyre $D_t = 1000 - 6 = 994 \, mm$.
Initial radius of the tyre $R = 497 \, mm$.
Change in radius required $\Delta R = 3 \, mm$.
The coefficient of linear expansion $\alpha$ is related to the coefficient of cubical expansion $\gamma$ by $\alpha = \frac{\gamma}{3}$.
Given $\gamma = 3.6 \times 10^{-5} \, ^oC^{-1}$,so $\alpha = \frac{3.6 \times 10^{-5}}{3} = 1.2 \times 10^{-5} \, ^oC^{-1}$.
The change in circumference $\Delta L$ is given by $\Delta L = 2\pi \Delta R = L \alpha \Delta \theta = (2\pi R) \alpha \Delta \theta$.
Thus,$\Delta R = R \alpha \Delta \theta$.
$\Delta \theta = \frac{\Delta R}{R \alpha} = \frac{3}{497 \times 1.2 \times 10^{-5}} \approx \frac{3}{0.005964} \approx 503 \, ^oC$.
Rounding to the nearest provided option,the required temperature increase is $500 \, ^oC$.

(A) Let the initial length be $L_0 = 80.0\,cm$ at $T_1 = 20^\circ C$. The temperature change is $\Delta T = 40^\circ C - 20^\circ C = 20^\circ C$.
The actual length of the copper wire at $40^\circ C$ is $L_{Cu} = L_0(1 + \alpha_{Cu} \Delta T)$.
The length marked on the steel scale at $40^\circ C$ is $L_{steel} = L_0(1 + \alpha_{steel} \Delta T)$.
The scale reading is the ratio of the actual length of the wire to the length of a unit division on the scale at $40^\circ C$. Since the scale itself expands,the reading $L'$ is given by:
$L' = \frac{L_{Cu}}{1 + \alpha_{steel} \Delta T} = \frac{L_0(1 + \alpha_{Cu} \Delta T)}{1 + \alpha_{steel} \Delta T}$.
Using the binomial approximation $(1+x)^{-1} \approx 1-x$ for small $\alpha \Delta T$:
$L' \approx L_0(1 + \alpha_{Cu} \Delta T)(1 - \alpha_{steel} \Delta T) \approx L_0(1 + (\alpha_{Cu} - \alpha_{steel}) \Delta T)$.
Substituting the values:
$L' = 80.0 \times (1 + (17 \times 10^{-6} - 11 \times 10^{-6}) \times 20)$
$L' = 80.0 \times (1 + 6 \times 10^{-6} \times 20) = 80.0 \times (1 + 120 \times 10^{-6})$
$L' = 80.0 + 80.0 \times 0.00012 = 80.0 + 0.0096 = 80.0096\,cm$.

(D) Let $L_0$ be the initial length of each strip before heating.
Let $d$ be the total thickness of the bimetallic strip.
After heating,the lengths of the brass and copper strips are:
$L_B = L_0(1 + \alpha_B \Delta T) = (R + d)\theta$
$L_C = L_0(1 + \alpha_C \Delta T) = R\theta$
Dividing the two equations:
$\frac{R + d}{R} = \frac{1 + \alpha_B \Delta T}{1 + \alpha_C \Delta T}$
$1 + \frac{d}{R} = \frac{1 + \alpha_B \Delta T}{1 + \alpha_C \Delta T}$
$\frac{d}{R} = \frac{1 + \alpha_B \Delta T}{1 + \alpha_C \Delta T} - 1 = \frac{1 + \alpha_B \Delta T - 1 - \alpha_C \Delta T}{1 + \alpha_C \Delta T} = \frac{(\alpha_B - \alpha_C)\Delta T}{1 + \alpha_C \Delta T}$
Since $\alpha \Delta T \ll 1$,we can approximate $1 + \alpha_C \Delta T \approx 1$.
Thus,$R = \frac{d}{(\alpha_B - \alpha_C)\Delta T}$.
Therefore,$R \propto \frac{1}{\Delta T}$ and $R \propto \frac{1}{|\alpha_B - \alpha_C|}$.
Hence,both $(B)$ and $(C)$ are correct.

(D) thermostat is a device used in electric appliances like refrigerators and irons for automatic temperature control.
It consists of a bimetallic strip made of two different metals joined together.
When the temperature changes, the two metals expand by different amounts because they have different coefficients of linear expansion $(\alpha)$.
This difference in expansion causes the strip to bend, which triggers the switching mechanism to turn the appliance on or off.
Therefore, the two metal strips must necessarily differ in their coefficient of linear expansion.

(C) Let the dimensions of the crystal be $L_0, L_0, L_0$. After a temperature change $\Delta \theta$,the new dimensions are $L_0(1 + \alpha_1 \Delta \theta)$,$L_0(1 + \alpha_2 \Delta \theta)$,and $L_0(1 + \alpha_2 \Delta \theta)$.
The new volume $V$ is given by $V = L_0(1 + \alpha_1 \Delta \theta) \times L_0(1 + \alpha_2 \Delta \theta) \times L_0(1 + \alpha_2 \Delta \theta)$.
$V = L_0^3(1 + \alpha_1 \Delta \theta)(1 + \alpha_2 \Delta \theta)^2$.
Since $V_0 = L_0^3$ and $V = V_0(1 + \gamma \Delta \theta)$,we have:
$1 + \gamma \Delta \theta = (1 + \alpha_1 \Delta \theta)(1 + 2\alpha_2 \Delta \theta + \alpha_2^2 \Delta \theta^2)$.
Neglecting higher-order terms of $\Delta \theta$,we get:
$1 + \gamma \Delta \theta \approx 1 + \alpha_1 \Delta \theta + 2\alpha_2 \Delta \theta$.
Comparing the coefficients of $\Delta \theta$,we find $\gamma = \alpha_1 + 2\alpha_2$.

(C) The volume of air in the flask remains constant at different temperatures. This implies that the increase in the volume of the glass flask must be exactly equal to the increase in the volume of the mercury contained within it.
Let $V_g$ be the total volume of the glass flask $(1000 \, cc)$ and $V_L$ be the volume of mercury.
The condition for constant air volume is $\Delta V_g = \Delta V_L$.
Using the formula for volume expansion,$\Delta V = V \gamma \Delta \theta$,we have:
$V_g \gamma_g \Delta \theta = V_L \gamma_L \Delta \theta$
Given that the coefficient of linear expansion of glass is $\alpha_g = 9 \times 10^{-6} /^{\circ}C$,the coefficient of volume expansion of glass is $\gamma_g = 3 \alpha_g = 3 \times 9 \times 10^{-6} = 27 \times 10^{-6} /^{\circ}C$.
The coefficient of volume expansion of mercury is $\gamma_L = 1.8 \times 10^{-4} /^{\circ}C$.
Substituting the values:
$V_L = \frac{V_g \gamma_g}{\gamma_L} = \frac{1000 \times 27 \times 10^{-6}}{1.8 \times 10^{-4}}$
$V_L = \frac{27 \times 10^{-3}}{1.8 \times 10^{-4}} = \frac{270}{1.8} = 150 \, cc$.

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
Taking the derivative,the fractional change in time period is $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l}$.
Since $\Delta l = l \alpha \Delta \theta$,we have $\frac{\Delta l}{l} = \alpha \Delta \theta$.
Therefore,$\frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta \theta$.
Given $\alpha = 12 \times 10^{-6} /^\circ C$ and $\Delta \theta = 40^\circ C - 20^\circ C = 20^\circ C$.
$\frac{\Delta T}{T} = \frac{1}{2} \times 12 \times 10^{-6} \times 20 = 120 \times 10^{-6} = 1.2 \times 10^{-4}$.
The time lost or gained in a day ($86400$ seconds) is $\Delta T = \frac{\Delta T}{T} \times 86400$.
$\Delta T = 1.2 \times 10^{-4} \times 86400 = 10.368$ seconds.
Since the temperature increases,the length of the pendulum increases,the time period increases,and the clock loses time. Thus,it loses approximately $10.3$ seconds per day.

(C) The Young's modulus of the rod is given by $Y = \frac{\text{Stress}}{\text{Linear Strain}}$.
Therefore,$\text{Stress} = Y \times \text{Linear Strain} = Y \times \left( \frac{\Delta l}{l} \right)$.
Since the linear expansion is prevented,the thermal strain is $\frac{\Delta l}{l} = \alpha \Delta \theta$,where $\alpha$ is the coefficient of linear expansion and $\Delta \theta$ is the change in temperature.
Substituting this,we get $\text{Stress} = Y \alpha \Delta \theta$.
Thus,the stress developed in the rod is independent of the initial length $l$ of the rod.

(C) The change in volume is given by the formula $\Delta V = \gamma V \Delta T$,where $V = \pi r^2 l$.
Since both rods are made of the same material (brass),the coefficient of volume expansion $\gamma$ is the same for both.
For rod $A$: $V_A = \pi (2r)^2 l = 4\pi r^2 l$.
For rod $B$: $V_B = \pi (r)^2 (2l) = 2\pi r^2 l$.
Since the change in temperature $\Delta T$ is the same for both,the ratio of the increase in volume is:
$\frac{\Delta V_A}{\Delta V_B} = \frac{\gamma V_A \Delta T}{\gamma V_B \Delta T} = \frac{V_A}{V_B} = \frac{4\pi r^2 l}{2\pi r^2 l} = \frac{2}{1}$.
Therefore,the ratio is $2 : 1$.

(D) The time period of a pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
When the temperature increases,the length $L$ of the pendulum increases as $L' = L(1 + \alpha \Delta \theta)$.
The change in time $\Delta t$ for a total time $t$ is given by the formula $\Delta t = \frac{1}{2} \alpha \Delta \theta \times t$.
Here,$\alpha = 10^{-5} {^{\circ}C}^{-1}$,$\Delta \theta = (40 - 20) = 20^{\circ}C$,and $t = 86400 \, s$ (number of seconds in a day).
Substituting the values:
$\Delta t = \frac{1}{2} \times 10^{-5} \times 20 \times 86400$
$\Delta t = 10^{-5} \times 10 \times 86400$
$\Delta t = 10^{-4} \times 86400 = 8.64 \, s$.

(D) The relationship between the coefficient of area expansion $(\beta)$ and the coefficient of linear expansion $(\alpha)$ is given by $\beta = 2\alpha$.
Given,$\beta = 2 \times 10^{-5} \, ^{\circ}C^{-1}$.
Therefore,$\alpha = \frac{\beta}{2}$.
$\alpha = \frac{2 \times 10^{-5}}{2} = 1 \times 10^{-5} \, ^{\circ}C^{-1}$.

(C) The density $\rho$ at temperature $\theta$ is given by $\rho = \frac{\rho_0}{1 + \gamma \Delta \theta}$,where $\gamma$ is the coefficient of volume expansion.
For small changes,$\gamma \approx \frac{\rho_0 - \rho}{\rho \Delta \theta}$.
Given $\rho_0 = 10 \, g/cm^3$,$\rho = 9.7 \, g/cm^3$,and $\Delta \theta = 100 \, ^\circ C$.
$\gamma = \frac{10 - 9.7}{9.7 \times 100} \approx \frac{0.3}{970} \approx 3.09 \times 10^{-4} \, ^\circ C^{-1}$.
Using the approximation $\gamma \approx \frac{\rho_0 - \rho}{\rho_0 \Delta \theta}$ (often used in textbooks for small expansion):
$\gamma = \frac{10 - 9.7}{10 \times 100} = \frac{0.3}{1000} = 3 \times 10^{-4} \, ^\circ C^{-1}$.
The coefficient of linear expansion $\alpha = \frac{\gamma}{3}$.
$\alpha = \frac{3 \times 10^{-4}}{3} = 10^{-4} \, ^\circ C^{-1}$.

(D) The surface area $A$ of a cylinder is given by $A = 2\pi r^2 + 2\pi rL$. For a thin cylinder or considering the dominant change in length,the surface area is proportional to the square of the length $(A \propto L^2)$.
Taking the derivative,we get $\Delta A = 2L \Delta L$.
Dividing by $A$ (where $A = L^2$ for proportionality),we get $\frac{\Delta A}{A} = \frac{2L \Delta L}{L^2} = 2 \frac{\Delta L}{L}$.
Given $\frac{\Delta L}{L} = 2\%$,the percentage increase in surface area is $\frac{\Delta A}{A} \times 100 = 2 \times 2\% = 4\%$.

(B) The formula for linear expansion is given by $\Delta l = \alpha l \Delta T$.
Here,the initial length $l = 10 \ m$,the coefficient of linear expansion $\alpha = 10 \times 10^{-6} \ ^{\circ}C^{-1}$,and the change in temperature $\Delta T = 100 \ ^{\circ}C - 0 \ ^{\circ}C = 100 \ ^{\circ}C$.
Substituting these values into the formula:
$\Delta l = (10 \times 10^{-6} \ ^{\circ}C^{-1}) \times (10 \ \text{m}) \times (100 \ ^{\circ}C)$
$\Delta l = 10 \times 10^{-6} \times 10 \times 100 \ \text{m}$
$\Delta l = 10^{-2} \ \text{m}$
Since $1 \ \text{m} = 100 \ \text{cm}$,we have $\Delta l = 10^{-2} \times 100 \ \text{cm} = 1.0 \ \text{cm}$.
Thus,the increase in length is $1.0 \ \text{cm}$.

(C) The change in length for a rod is given by $\Delta l = l \alpha \Delta T$.
Given that the increase in length for both rods is the same,we have $\Delta l_a = \Delta l_s$.
Substituting the formula,we get $l_1 \alpha_a \Delta T = l_2 \alpha_s \Delta T$.
Since $\Delta T$ is the same for both,we can cancel it out: $l_1 \alpha_a = l_2 \alpha_s$.
This implies $\frac{l_1}{l_2} = \frac{\alpha_s}{\alpha_a}$.
To find $\frac{l_1}{l_1 + l_2}$,we use the property of ratios: $\frac{l_1}{l_1 + l_2} = \frac{\alpha_s}{\alpha_s + \alpha_a}$.
Thus,the correct option is $C$.

(C) The change in length due to thermal expansion is given by $\Delta l = l_0 \alpha \Delta T$.
Here,$l_0 = 10 \ cm$,$\alpha = 11 \times 10^{-6} \ ^{\circ}C^{-1}$,and $\Delta T = 20^{\circ}C - 19^{\circ}C = 1^{\circ}C$.
Substituting the values:
$\Delta l = 10 \ cm \times (11 \times 10^{-6} \ ^{\circ}C^{-1}) \times 1^{\circ}C$.
$\Delta l = 110 \times 10^{-6} \ cm = 11 \times 10^{-5} \ cm$.
Since the temperature decreases,the length of the rod will decrease by $11 \times 10^{-5} \ cm$.

(C) bimetallic strip consists of two different metals with different coefficients of linear expansion $(\alpha)$.
When the strip is heated, both metals expand, but the metal with the higher coefficient of linear expansion $(\alpha)$ expands more than the other.
Because the two metals are bonded together, the one that expands more must form the outer part of the curve to accommodate its greater length.
Therefore, the strip bends such that the metal with the higher coefficient of linear expansion is on the outside of the arc.

(C) The coefficient of volume expansion $(\gamma)$ is related to the coefficient of area expansion $(\beta)$ and the coefficient of linear expansion $(\alpha)$ by the relations $\gamma = 3\alpha$ and $\gamma = 1.5\beta$.
$1$. The percentage increase in length (diameter) is proportional to $\alpha \Delta T$.
$2$. The percentage increase in surface area is proportional to $\beta \Delta T = 2\alpha \Delta T$.
$3$. The percentage increase in volume is proportional to $\gamma \Delta T = 3\alpha \Delta T$.
Since $3\alpha > 2\alpha > \alpha$,the percentage increase in volume is the largest. Density decreases upon heating,so it does not show a percentage increase.

(A) Let the linear expansion coefficient be $\alpha$.
Then,the area expansion coefficient $\beta = 2\alpha$.
And the volume expansion coefficient $\gamma = 3\alpha$.
Therefore,the ratio $\alpha : \beta : \gamma = \alpha : 2\alpha : 3\alpha = 1 : 2 : 3$.

(D) The coefficient of linear expansion of brass $(\alpha_b)$ is greater than that of steel $(\alpha_s)$.
When the system is cooled, both materials contract.
Since $\alpha_b > \alpha_s$, the brass disc contracts more than the steel plate hole.
As a result, the disc becomes loose in the hole.

(A) When a solid object with a cavity is heated,the material expands as if the cavity were also made of the same material.
According to the principle of thermal expansion,the distance between any two points in the solid increases upon heating.
Since the cavity is defined by the surrounding metal,as the metal expands outward,the boundary of the cavity moves outward as well.
Therefore,the radius of the spherical cavity increases,which leads to an increase in the volume of the cavity $(V = \frac{4}{3} \pi r^3)$.

(A) The change in volume $\Delta V$ is given by the formula $\Delta V = \gamma \cdot V \cdot \Delta T$,where $\gamma$ is the coefficient of volume expansion.
For a solid,$\gamma = 3\alpha$,where $\alpha$ is the coefficient of linear expansion.
Given: Diameter $d = 20 \; cm$,so radius $r = 10 \; cm$.
Initial volume $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (10)^3 = \frac{4000}{3} \pi \; cm^3$.
Change in temperature $\Delta T = 100^{\circ} C - 0^{\circ} C = 100^{\circ} C$.
Coefficient of linear expansion $\alpha = 23 \times 10^{-6} \; /^{\circ} C$.
Substituting the values:
$\Delta V = 3 \times (23 \times 10^{-6}) \times (\frac{4}{3} \pi \times 10^3) \times 100$
$\Delta V = 3 \times 23 \times 10^{-6} \times \frac{4}{3} \times 3.14159 \times 1000 \times 100$
$\Delta V = 23 \times 4 \times 3.14159 \times 10^{-1} = 92 \times 3.14159 \times 0.1 \approx 28.9 \; cm^3$.

(C) The change in length for a rod is given by $\Delta \ell = \ell \alpha \Delta T$.
Given that the increase in length for both rods is the same: $\Delta \ell_1 = \Delta \ell_2$.
Therefore,$\ell_1 \alpha_a t = \ell_2 \alpha_s t$.
This simplifies to $\ell_1 \alpha_a = \ell_2 \alpha_s$,which means $\frac{\ell_1}{\ell_2} = \frac{\alpha_s}{\alpha_a}$.
To find the ratio $\frac{\ell_1}{\ell_1 + \ell_2}$,we use the property of ratios:
$\frac{\ell_1}{\ell_1 + \ell_2} = \frac{\alpha_s}{\alpha_a + \alpha_s}$.

(D) The volume of a cylinder is given by $V = A \times L$,where $A$ is the base area and $L$ is the length.
Assuming the volume of the cylinder remains constant during heating (or considering the relationship between dimensions),if the length $L$ increases by $2\%$,the cross-sectional area $A$ must decrease to keep the volume constant.
However,in the context of thermal expansion,if we consider the linear expansion of the radius $r$ due to temperature change,the area $A = \pi r^2$ changes as $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.
If the question implies that the radius also expands by the same coefficient as the length (isotropic expansion),then $\frac{\Delta r}{r} = \frac{\Delta L}{L} = 2\%$.
Therefore,the percentage increase in area is $\frac{\Delta A}{A} \times 100 = 2 \times (2\%) = 4\%$.

(A) The formula for volume expansion is given by $\frac{\Delta V}{V} = \gamma \Delta \theta$,where $\gamma$ is the coefficient of volume expansion and $\Delta \theta$ is the change in temperature.
Given $\frac{\Delta V}{V} = 0.24\% = \frac{0.24}{100} = 2.4 \times 10^{-3}$ and $\Delta \theta = 40^{\circ}C$.
Substituting the values: $2.4 \times 10^{-3} = \gamma \times 40$.
Solving for $\gamma$: $\gamma = \frac{2.4 \times 10^{-3}}{40} = 0.06 \times 10^{-3} = 6 \times 10^{-5} \, ^{\circ}C^{-1}$.
The relationship between the coefficient of linear expansion $(\alpha)$ and the coefficient of volume expansion $(\gamma)$ is $\alpha = \frac{\gamma}{3}$.
Therefore,$\alpha = \frac{6 \times 10^{-5}}{3} = 2 \times 10^{-5} \, ^{\circ}C^{-1}$.