Thermal Expansion for Solid Questions in English

(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.
Taking the logarithmic derivative, we get $\frac{dT}{T} = \frac{1}{2} \frac{dl}{l}$.
Since the linear expansion is given by $\frac{dl}{l} = \alpha \Delta \theta$, we substitute this into the equation:
$\frac{dT}{T} = \frac{1}{2} \alpha \Delta \theta$.
Given $\alpha = 9 \times 10^{-7} /{ }^{\circ} C$ and $\Delta \theta = (30 - 20) = 10^{\circ} C$,
$\frac{dT}{T} = \frac{1}{2} \times 9 \times 10^{-7} \times 10 = 4.5 \times 10^{-6}$.
The loss in time per oscillation is $dT = T \times (4.5 \times 10^{-6})$.
Substituting $T = 0.5 \, s$, we get $dT = 0.5 \times 4.5 \times 10^{-6} = 2.25 \times 10^{-6} \, s$.

(B) The angular momentum $J$ of the rod remains constant because no external torque acts on the system.
$J = I \omega = \text{constant}$.
Initially,the moment of inertia is $I_1 = \frac{1}{3} M L^2$ and angular velocity is $\omega_1 = \omega$.
When the temperature increases by $t^{\circ} C$,the length of the rod becomes $L' = L(1 + \alpha t)$.
The new moment of inertia is $I_2 = \frac{1}{3} M (L')^2 = \frac{1}{3} M L^2 (1 + \alpha t)^2 = I_1 (1 + \alpha t)^2$.
From conservation of angular momentum,$I_1 \omega_1 = I_2 \omega_2$.
$\omega_2 = \omega_1 \left( \frac{I_1}{I_2} \right) = \omega \left( \frac{I_1}{I_1(1 + \alpha t)^2} \right) = \omega (1 + \alpha t)^{-2}$.
Using binomial approximation for small $\alpha t$,$\omega_2 \approx \omega (1 - 2 \alpha t)$.
The change in angular velocity is $\Delta \omega = \omega_2 - \omega_1 = \omega (1 - 2 \alpha t) - \omega = -2 \omega \alpha t$.
Thus,the magnitude of the change in angular velocity $|\Delta \omega|$ is proportional to $\omega$.

(A) Let $A_c$ be the area of the copper ring and $A_s$ be the area of the steel rod at temperature $T_0 = 30^{\circ} C$.
Given: $A_c = 9.98 \ cm^2$,$A_s = 10 \ cm^2$.
Coefficients of linear expansion: $\alpha_c = 17 \times 10^{-6} /{ }^{\circ} C$,$\alpha_s = 11 \times 10^{-6} /{ }^{\circ} C$.
Coefficients of area expansion: $\beta_c = 2\alpha_c = 34 \times 10^{-6} /{ }^{\circ} C$,$\beta_s = 2\alpha_s = 22 \times 10^{-6} /{ }^{\circ} C$.
For the ring to slip onto the rod,the area of the ring must be equal to or greater than the area of the rod at the new temperature $T = T_0 + \Delta T$.
$A_c(1 + \beta_c \Delta T) = A_s(1 + \beta_s \Delta T)$
$9.98(1 + 34 \times 10^{-6} \Delta T) = 10(1 + 22 \times 10^{-6} \Delta T)$
$9.98 + 9.98 \times 34 \times 10^{-6} \Delta T = 10 + 10 \times 22 \times 10^{-6} \Delta T$
$(339.32 - 220) \times 10^{-6} \Delta T = 10 - 9.98$
$119.32 \times 10^{-6} \Delta T = 0.02$
$\Delta T = \frac{0.02}{119.32 \times 10^{-6}} \approx 167.6^{\circ} C$.

(C) The formula for linear expansion is given by $\Delta L = L \alpha \Delta T$,where $\Delta L$ is the change in length,$L$ is the original length,$\alpha$ is the coefficient of linear expansion,and $\Delta T$ is the change in temperature.
Given that the length increases by $0.4 \%$,we have $\frac{\Delta L}{L} = 0.4 \% = \frac{0.4}{100} = 0.004$.
The coefficient of linear expansion is $\alpha = 20 \times 10^{-6} \ {}^{\circ}C^{-1}$.
Substituting these values into the formula $\frac{\Delta L}{L} = \alpha \Delta T$:
$0.004 = (20 \times 10^{-6}) \Delta T$.
Solving for $\Delta T$:
$\Delta T = \frac{0.004}{20 \times 10^{-6}} = \frac{4 \times 10^{-3}}{20 \times 10^{-6}} = \frac{1}{5} \times 10^3 = 0.2 \times 1000 = 200 \ {}^{\circ}C$.
Since the change in temperature in Celsius is equal to the change in temperature in Kelvin,$\Delta T = 200 \ K$.

(C) The length of the metal scale is $L = 1 \ m = 1000 \ mm$. The maximum allowed error in length is $\Delta L = 0.5 \ mm$. The coefficient of linear expansion is $\alpha = 10^{-5} /{ }^{\circ} C$. The change in length due to temperature change $\Delta T$ is given by $\Delta L = L \alpha \Delta T$. Substituting the values: $0.5 = 1000 \times 10^{-5} \times \Delta T$. This simplifies to $0.5 = 10^{-2} \times \Delta T$,which gives $\Delta T = 0.5 / 10^{-2} = 50^{\circ} C$. Since the scale is calibrated at $25^{\circ} C$,the temperature range is $25^{\circ} C \pm 50^{\circ} C$. Thus,the range is from $25 - 50 = -25^{\circ} C$ to $25 + 50 = 75^{\circ} C$.

(D) Change in length of the first rod is given by $\Delta \ell_1 = L(2 \alpha) \Delta T = 2 \,L \alpha \Delta T$.
Change in length of the second rod is given by $\Delta \ell_2 = (2 \,L)(\alpha)(\Delta T) = 2 \,L \alpha \Delta T$.
Total change in length of the composite rod is $\Delta l = \Delta l_1 + \Delta l_2 = 2 \,L \alpha \Delta T + 2 \,L \alpha \Delta T = 4 \,L \alpha \Delta T$.
For the composite rod of length $3 \,L$,the change in length is $\Delta l = (3 \,L) \alpha_C \Delta T$.
Equating the two expressions for $\Delta l$: $4 \,L \alpha \Delta T = 3 \,L \alpha_C \Delta T$.
Solving for $\alpha_C$,we get $\alpha_C = \frac{4 \alpha}{3}$.

(C) The formula for the coefficient of linear expansion $\alpha$ is given by:
$\alpha = \frac{\Delta L}{L \Delta T}$
Given:
Initial length $L = 30 \ cm$
Change in length $\Delta L = 0.027 \ cm$
Initial temperature $T_1 = 30^{\circ} C$
Final temperature $T_2 = 105^{\circ} C$
Change in temperature $\Delta T = T_2 - T_1 = 105^{\circ} C - 30^{\circ} C = 75^{\circ} C$
Substituting the values into the formula:
$\alpha = \frac{0.027}{30 \times 75}$
$\alpha = \frac{0.027}{2250}$
$\alpha = 0.000012 /{ }^{\circ} C$
$\alpha = 12 \times 10^{-6} /{ }^{\circ} C$

(B) The diameter of the iron ring $d_i = 5 \ m$ at $T_1 = 27^{\circ} C$.
The diameter of the wooden wheel $d_w = 5.012 \ m$.
We need the iron ring to expand to a diameter of $5.012 \ m$.
Using the formula for linear expansion: $d_w = d_i(1 + \alpha \Delta T)$.
$\Delta T = \frac{d_w - d_i}{d_i \alpha} = \frac{5.012 - 5}{5 \times 1.2 \times 10^{-5}} = \frac{0.012}{6 \times 10^{-5}} = \frac{1.2 \times 10^{-2}}{6 \times 10^{-5}} = 0.2 \times 10^3 = 200^{\circ} C$.
The final temperature $T_2 = T_1 + \Delta T = 27^{\circ} C + 200^{\circ} C = 227^{\circ} C$.

(A) Let $L_s$ and $L_c$ be the lengths of the steel and copper rods at a certain temperature. The difference in their lengths is given as $L_s - L_c = 4 \ cm$,which remains constant at any temperature.
This implies that the change in length of both rods due to a change in temperature $\Delta T$ must be equal.
Therefore,$\Delta L_s = \Delta L_c$.
Using the formula for linear expansion $\Delta L = \alpha L \Delta T$,we get $\alpha_s L_s \Delta T = \alpha_c L_c \Delta T$.
Canceling $\Delta T$ from both sides,we have $\alpha_s L_s = \alpha_c L_c$.
Substituting the given values: $(1.1 \times 10^{-5}) L_s = (1.7 \times 10^{-5}) L_c$.
Thus,the ratio $\frac{L_s}{L_c} = \frac{1.7 \times 10^{-5}}{1.1 \times 10^{-5}} = \frac{17}{11}$.

(B) The change in temperature is $\Delta T = 230^{\circ} C - 30^{\circ} C = 200^{\circ} C$.
When a metal sheet with a hole is heated,the hole expands just as if it were a solid piece of the same material.
The formula for linear expansion is $L = L_0(1 + \alpha \Delta T)$.
Substituting the given values: $L = 5 \times (1 + 2 \times 10^{-5} \times 200)$.
$L = 5 \times (1 + 400 \times 10^{-5}) = 5 \times (1 + 0.004) = 5 \times 1.004 = 5.02 \ cm$.

(C) The change in length of the combined rod is the sum of the changes in lengths of the individual rods.
$\Delta l = \Delta l_A + \Delta l_B$
Using the formula for linear expansion $\Delta l = l \alpha \Delta T$,where $l = 0.5 \ m$,$\Delta T = 230^{\circ} C - 30^{\circ} C = 200^{\circ} C$,$\alpha_A = 2.0 \times 10^{-5} /^{\circ} C$,and $\alpha_B = 1.0 \times 10^{-5} /^{\circ} C$:
$\Delta l = (l_A \alpha_A \Delta T) + (l_B \alpha_B \Delta T)$
$\Delta l = l \Delta T (\alpha_A + \alpha_B)$
$\Delta l = 0.5 \times 200 \times (2.0 \times 10^{-5} + 1.0 \times 10^{-5})$
$\Delta l = 100 \times (3.0 \times 10^{-5})$
$\Delta l = 3.0 \times 10^{-3} \ m = 3 \ mm$
Therefore,the change in length is $3 \ mm$.

(B) The metal ball tends to expand due to heating,but since its volume is kept constant,thermal stress is developed within the material.
The bulk modulus $B$ is defined as $B = \frac{\Delta P}{\Delta V / V}$,where $\Delta P$ is the change in pressure.
The fractional change in volume due to thermal expansion is given by $\frac{\Delta V}{V} = \gamma \Delta T$,where $\gamma$ is the coefficient of volume expansion and $\Delta T$ is the change in temperature.
For a solid,$\gamma = 3\alpha$,where $\alpha$ is the coefficient of linear expansion.
Given: $\alpha = 10^{-5} \ {^{\circ} C}^{-1}$,$\Delta T = 127^{\circ} C - 20^{\circ} C = 107^{\circ} C$,and $B = 2 \times 10^{11} \ N \ m^{-2}$.
Substituting these into the formula for pressure change: $\Delta P = B \times (3\alpha \Delta T)$.
$\Delta P = (2 \times 10^{11}) \times (3 \times 10^{-5} \times 107) = 6 \times 10^6 \times 107 = 6.42 \times 10^8 \ Pa$.
The final pressure $P_f = P_i + \Delta P = 10^5 \ Pa + 6.42 \times 10^8 \ Pa \approx 6.42 \times 10^8 \ Pa$.
Rounding to the nearest given option,the pressure becomes approximately $6 \times 10^8 \ Pa$.

(B) The initial moment of inertia $I$ of a uniform rod of mass $M$ and length $L$ about its perpendicular bisector is given by:
$I = \frac{1}{12} ML^2$
When the temperature is increased by $\Delta T$,the length of the rod increases to $L' = L + \Delta L$,where $\Delta L = L \alpha \Delta T$. The mass $M$ remains constant.
The new moment of inertia $I' = I + \Delta I$ is:
$I + \Delta I = \frac{1}{12} M(L + \Delta L)^2$
$I + \Delta I = \frac{1}{12} ML^2 \left(1 + \frac{\Delta L}{L}\right)^2$
Since $I = \frac{1}{12} ML^2$,we have:
$I + \Delta I = I \left(1 + \frac{\Delta L}{L}\right)^2$
Using the binomial approximation $(1+x)^n \approx 1+nx$ for $x \ll 1$:
$I + \Delta I \approx I \left(1 + 2 \frac{\Delta L}{L}\right)$
$I + \Delta I \approx I + 2I \frac{\Delta L}{L}$
$\Delta I = 2I \frac{\Delta L}{L}$
Substituting $\frac{\Delta L}{L} = \alpha \Delta T$:
$\Delta I = 2I \alpha \Delta T$
Therefore,$\frac{\Delta I}{I} = 2 \alpha \Delta T$.

(B) Given:
Initial temperature $t_1 = 20^{\circ} C$
Final temperature $t_2 = 60^{\circ} C$
Change in temperature $\Delta T = t_2 - t_1 = 60^{\circ} C - 20^{\circ} C = 40^{\circ} C$
Coefficient of linear expansion $\alpha = 10^{-5} {^{\circ} C}^{-1}$
Initial dimensions are $40 \text{ cm} \times 20 \text{ cm}$.
Initial area $A = 40 \text{ cm} \times 20 \text{ cm} = 800 \text{ cm}^2$.
The coefficient of area expansion is $\beta = 2\alpha$.
The change in area $\Delta A$ is given by the formula:
$\Delta A = A \beta \Delta T = A (2\alpha) \Delta T$
Substituting the values:
$\Delta A = 800 \text{ cm}^2 \times (2 \times 10^{-5} {^{\circ} C}^{-1}) \times 40^{\circ} C$
$\Delta A = 800 \times 2 \times 10^{-5} \times 40 \text{ cm}^2$
$\Delta A = 64000 \times 10^{-5} \text{ cm}^2$
$\Delta A = 0.64 \text{ cm}^2$.

(B) Given,the area of a circular copper coin increases by $0.4 \%$.
This means,$\frac{\Delta A}{A} = 0.004$.
The increase in temperature is $\Delta T = 100^{\circ} C$.
We know that the coefficient of areal expansion $\beta$ is defined as $\beta = \frac{\Delta A}{A \cdot \Delta T}$.
Substituting the values,we get $\beta = \frac{0.004}{100} = 4 \times 10^{-5} /^{\circ} C$.
The relationship between the coefficient of areal expansion $\beta$ and the coefficient of linear expansion $\alpha$ is $\beta = 2\alpha$.
Therefore,$\alpha = \frac{\beta}{2} = \frac{4 \times 10^{-5}}{2} /^{\circ} C = 2 \times 10^{-5} /^{\circ} C$.

$(A)$ Given: Coefficient of linear expansion of steel, $\alpha = 11 \times 10^{-6} /{ }^{\circ} C$.
Required accuracy $\Delta l = 0.06 \,mm = 6 \times 10^{-5} \,m$ for a length $l = 1 \,m$.
We know the formula for linear expansion: $\Delta l = l \alpha \Delta t$.
Rearranging for the temperature change: $\Delta t = \frac{\Delta l}{l \alpha}$.
Substituting the values: $\Delta t = \frac{6 \times 10^{-5}}{1 \times 11 \times 10^{-6}} = \frac{60}{11} \approx 5.45^{\circ} C$.
Since the scale is accurate at $25^{\circ} C$, the allowed temperature range is $25^{\circ} C \pm 5.45^{\circ} C$.
This gives a range from $19.55^{\circ} C$ to $30.45^{\circ} C$.
Rounding to the nearest integer values provided in the options, the range is approximately $19^{\circ} C$ to $31^{\circ} C$.
Thus, option $A$ is correct.

(B) The moment of inertia $I$ of a body is given by $I = Mk^2$,where $M$ is the mass and $k$ is the radius of gyration.
Since $M$ is constant,we have $I \propto k^2$.
Taking the natural logarithm on both sides: $\ln I = \ln M + 2 \ln k$.
Differentiating both sides: $\frac{dI}{I} = 2 \frac{dk}{k}$.
For thermal expansion,the change in length (or any linear dimension like $k$) is given by $\Delta k = k \alpha \Delta T$,which implies $\frac{\Delta k}{k} = \alpha \Delta T$.
Substituting this into the expression for $\frac{\Delta I}{I}$:
$\frac{\Delta I}{I} = 2 \alpha \Delta T$.

(B) Let the initial length of each rod be $l$. In the equilateral triangle $ABC$,the length of the altitude $DC$ is given by:
$DC^2 = AC^2 - AD^2 = l^2 - (l/2)^2 = 3l^2/4$.
After a small change in temperature $\Delta t$,the new lengths are $l' = l(1 + \alpha \Delta t)$.
For rod $AC$,the new length is $l_{AC}' = l(1 + \alpha_2 \Delta t)$.
For segment $AD$,the new length is $l_{AD}' = (l/2)(1 + \alpha_1 \Delta t)$.
Since $DC$ remains constant,$DC^2 = (l_{AC}')^2 - (l_{AD}')^2$.
$3l^2/4 = [l(1 + \alpha_2 \Delta t)]^2 - [(l/2)(1 + \alpha_1 \Delta t)]^2$.
$3l^2/4 = l^2(1 + 2\alpha_2 \Delta t + \alpha_2^2 \Delta t^2) - (l^2/4)(1 + 2\alpha_1 \Delta t + \alpha_1^2 \Delta t^2)$.
Neglecting higher-order terms $\alpha^2 \Delta t^2$ and simplifying:
$3l^2/4 = l^2 + 2l^2 \alpha_2 \Delta t - l^2/4 - (l^2/4)(2\alpha_1 \Delta t)$.
$3l^2/4 = 3l^2/4 + 2l^2 \alpha_2 \Delta t - (l^2/2)\alpha_1 \Delta t$.
$0 = 2l^2 \alpha_2 \Delta t - (l^2/2)\alpha_1 \Delta t$.
$2\alpha_2 = \alpha_1/2 \implies \alpha_1 = 4\alpha_2$.

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

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

(B) Let $L_s$ be the length of the steel rod at $25^{\circ} C$ as measured by the metal scale. The scale is calibrated at $0^{\circ} C$,so at $25^{\circ} C$,its length expands. The length of $1 \ m$ on the scale at $25^{\circ} C$ corresponds to an actual length $L_{actual} = 1(1 + \alpha_{\text{metal}} \Delta T) = 1(1 + 20 \times 10^{-6} \times 25) = 1 + 0.0005 = 1.0005 \ m$.
Now,this is the actual length of the steel rod at $25^{\circ} C$. Let $L_0$ be the length of the steel rod at $0^{\circ} C$.
Using the thermal expansion formula for the steel rod: $L_{actual} = L_0(1 + \alpha_{\text{steel}} \Delta T)$.
$1.0005 = L_0(1 + 12 \times 10^{-6} \times 25)$.
$1.0005 = L_0(1 + 0.0003) = L_0(1.0003)$.
$L_0 = \frac{1.0005}{1.0003} \approx 1.0001999 \ m \approx 1.0002 \ m$.

(B) Let the length of the sides of a square sheet be $L$. The area of the sheet is $A = L^2$.
When the temperature increases by $\Delta T$,the new length becomes $L' = L(1 + \alpha \Delta T)$,where $\alpha$ is the coefficient of linear expansion.
The new area becomes $A' = (L')^2 = L^2(1 + \alpha \Delta T)^2$.
Expanding this,we get $A' = A(1 + 2\alpha \Delta T + \alpha^2 \Delta T^2)$.
Since $\alpha$ is very small,$\alpha^2$ is negligible,so $A' \approx A(1 + 2\alpha \Delta T)$.
The coefficient of superficial expansion $\beta$ is defined by the relation $A' = A(1 + \beta \Delta T)$.
Comparing the two expressions,we find $\beta = 2\alpha$.
Therefore,the ratio of linear expansivity $(\alpha)$ to the coefficient of superficial expansion $(\beta)$ is $\frac{\alpha}{\beta} = \frac{\alpha}{2\alpha} = 0.5$.

(B) Since the rod is rotating freely, the angular momentum $L$ is conserved.
$L = I \omega = \text{constant}$.
Initial moment of inertia $I = \frac{mL^2}{3}$.
After the temperature increase, the new length is $L' = L(1 + \alpha \Delta T)$.
The new moment of inertia is $I' = \frac{m(L')^2}{3} = \frac{m L^2 (1 + \alpha \Delta T)^2}{3} = I(1 + \alpha \Delta T)^2$.
Using conservation of angular momentum: $I \omega = I' \omega'$.
Given $\omega' = \frac{\omega}{2}$, we have $I \omega = I(1 + \alpha \Delta T)^2 \frac{\omega}{2}$.
$1 = \frac{(1 + \alpha \Delta T)^2}{2} \implies (1 + \alpha \Delta T)^2 = 2$.
Taking the square root: $1 + \alpha \Delta T = \sqrt{2}$.
Since $\alpha \ll 1$, we use the binomial approximation $(1 + x)^n \approx 1 + nx$.
$1 + 2 \alpha \Delta T \approx 2$.
$2 \alpha \Delta T = 1$.
$\Delta T = \frac{1}{2 \alpha}$.

(D) Let the initial length be $L_{0}$ and initial breadth be $B_{0}$. The initial area is $S_{0} = L_{0} B_{0}$.
When the temperature increases by $\Delta t$,the new length $L_{t}$ and new breadth $B_{t}$ are given by:
$L_{t} = L_{0}(1 + \alpha_{1} \Delta t)$
$B_{t} = B_{0}(1 + \alpha_{2} \Delta t)$
The new area $S_{t}$ is:
$S_{t} = L_{t} \times B_{t} = L_{0}(1 + \alpha_{1} \Delta t) \times B_{0}(1 + \alpha_{2} \Delta t)$
$S_{t} = L_{0} B_{0} (1 + \alpha_{1} \Delta t + \alpha_{2} \Delta t + \alpha_{1} \alpha_{2} (\Delta t)^{2})$
Since $\alpha_{1} \Delta t \ll 1$ and $\alpha_{2} \Delta t \ll 1$,the term $\alpha_{1} \alpha_{2} (\Delta t)^{2}$ is negligible.
$S_{t} \approx S_{0} (1 + (\alpha_{1} + \alpha_{2}) \Delta t)$
Comparing this with the standard formula for surface expansion $S_{t} = S_{0} (1 + \beta \Delta t)$,where $\beta$ is the coefficient of surface expansion:
$\beta = \alpha_{1} + \alpha_{2}$

(B) Given: Total length $L = 120 \ cm$ at $30^{\circ} C$. Since they have the same length,$\ell_0 = 60 \ cm$ for each rod.
Change in temperature $\Delta T = 100^{\circ} C - 30^{\circ} C = 70^{\circ} C$.
Coefficient of linear expansion for aluminium $\alpha_A = 24 \times 10^{-6} /{ }^{\circ} C$.
Coefficient of linear expansion for steel $\alpha_S = 1.2 \times 10^{-5} /{ }^{\circ} C = 12 \times 10^{-6} /{ }^{\circ} C$.
The final length of the composite rod is the sum of the final lengths of the individual rods:
$\ell_{\text{final}} = \ell_0(1 + \alpha_A \Delta T) + \ell_0(1 + \alpha_S \Delta T)$
$\ell_{\text{final}} = \ell_0 [2 + (\alpha_A + \alpha_S) \Delta T]$
$\ell_{\text{final}} = 60 [2 + (24 \times 10^{-6} + 12 \times 10^{-6}) \times 70]$
$\ell_{\text{final}} = 60 [2 + (36 \times 10^{-6}) \times 70]$
$\ell_{\text{final}} = 60 [2 + 0.00252] = 120 + 0.1512 = 120.1512 \ cm$.
Rounding to two decimal places,the length is $120.15 \ cm$.

(D) Let the initial length at temperature $T_1$ be $L_0$.
For the first temperature change from $T_1$ to $T_2$,the increase in length is $\Delta L_1 = L_0 \alpha (T_2 - T_1) = L_0 \alpha \Delta T$.
The length of the strip at temperature $T_2$ is $L_2 = L_0 + \Delta L_1 = L_0(1 + \alpha \Delta T)$.
For the second temperature change from $T_2$ to $T_3$,the increase in length is $\Delta L_2 = L_2 \alpha (T_3 - T_2)$.
Given $T_3 + T_1 = 2T_2$,we can write $T_3 - T_2 = T_2 - T_1 = \Delta T$.
Substituting $L_2$ and $(T_3 - T_2)$ into the expression for $\Delta L_2$:
$\Delta L_2 = [L_0(1 + \alpha \Delta T)] \alpha \Delta T = (L_0 \alpha \Delta T)(1 + \alpha \Delta T)$.
Since $\Delta L_1 = L_0 \alpha \Delta T$,we get $\Delta L_2 = \Delta L_1(1 + \alpha \Delta T)$.