Thermal Expansion for Solid Questions in English

(B) The formula for linear expansion is $\Delta L = L_0 \alpha \Delta \theta$.
For rod $A$: $0.075 = 20 \times \alpha_A \times 100 \implies \alpha_A = \frac{0.075}{2000} = 3.75 \times 10^{-5} \, ^\circ C^{-1}$.
For rod $B$: $0.045 = 20 \times \alpha_B \times 100 \implies \alpha_B = \frac{0.045}{2000} = 2.25 \times 10^{-5} \, ^\circ C^{-1}$.
Let the length of metal $A$ be $x$ and the length of metal $B$ be $(20 - x)$.
The total expansion of the composite rod is the sum of the expansions of its parts:
$0.060 = (x \alpha_A \Delta \theta) + ((20 - x) \alpha_B \Delta \theta)$
$0.060 = [x(3.75 \times 10^{-5}) + (20 - x)(2.25 \times 10^{-5})] \times 100$
$0.060 = [3.75x + 45 - 2.25x] \times 10^{-3}$
$60 = 1.5x + 45$
$1.5x = 15$
$x = 10 \, cm$.

(B) Let the length of each rod be $\ell$. In an equilateral triangle $PQR$,the height $OR$ is given by $OR^2 = PR^2 - PO^2 = \ell^2 - (\ell/2)^2 = 3\ell^2/4$.
After a temperature change $\Delta t$,the new lengths are $\ell' = \ell(1 + \alpha \Delta t)$.
Let $PR' = \ell(1 + \alpha_2 \Delta t)$ and $PQ' = \ell(1 + \alpha_1 \Delta t)$.
The new midpoint $O'$ divides $PQ'$ into two segments of length $PO' = \frac{\ell(1 + \alpha_1 \Delta t)}{2}$.
Since $OR$ remains constant,$OR^2 = (PR')^2 - (PO')^2$.
$\frac{3\ell^2}{4} = [\ell(1 + \alpha_2 \Delta t)]^2 - [\frac{\ell}{2}(1 + \alpha_1 \Delta t)]^2$.
$\frac{3\ell^2}{4} = \ell^2(1 + 2\alpha_2 \Delta t + \alpha_2^2 \Delta t^2) - \frac{\ell^2}{4}(1 + 2\alpha_1 \Delta t + \alpha_1^2 \Delta t^2)$.
Neglecting higher-order terms like $\Delta t^2$,we get:
$\frac{3\ell^2}{4} = \ell^2 + 2\ell^2 \alpha_2 \Delta t - \frac{\ell^2}{4} - \frac{2\ell^2 \alpha_1 \Delta t}{4}$.
$\frac{3\ell^2}{4} = \frac{3\ell^2}{4} + 2\ell^2 \alpha_2 \Delta t - \frac{\ell^2 \alpha_1 \Delta t}{2}$.
$0 = 2\alpha_2 \Delta t - \frac{\alpha_1 \Delta t}{2}$.
$2\alpha_2 = \frac{\alpha_1}{2} \implies \alpha_1 = 4\alpha_2$.

(C) Let the dimensions of the crystal be $L_0 \times L_0 \times L_0$. The initial volume is $V_0 = L_0^3$.
When the temperature increases by $\Delta \theta$,the new dimensions become $L_1 = L_0(1 + \alpha_1 \Delta \theta)$ and $L_2 = L_0(1 + \alpha_2 \Delta \theta)$ (for the two perpendicular directions).
The new volume $V$ is given by $V = L_1 \times L_2 \times L_2 = L_0(1 + \alpha_1 \Delta \theta) \times L_0(1 + \alpha_2 \Delta \theta) \times L_0(1 + \alpha_2 \Delta \theta)$.
$V = L_0^3(1 + \alpha_1 \Delta \theta)(1 + \alpha_2 \Delta \theta)^2$.
Using the binomial approximation $(1 + x)^n \approx 1 + nx$ for small $\alpha \Delta \theta$,we get:
$V \approx V_0(1 + \alpha_1 \Delta \theta)(1 + 2\alpha_2 \Delta \theta)$.
$V \approx V_0(1 + \alpha_1 \Delta \theta + 2\alpha_2 \Delta \theta + 2\alpha_1 \alpha_2 (\Delta \theta)^2)$.
Neglecting the higher-order term $(\Delta \theta)^2$,we have $V \approx V_0(1 + (\alpha_1 + 2\alpha_2) \Delta \theta)$.
Comparing this with the standard volume expansion formula $V = V_0(1 + \gamma \Delta \theta)$,we find the volume expansion coefficient $\gamma = \alpha_1 + 2\alpha_2$.

(C) Let the fractional change in radius be $\frac{\Delta r}{r} = X$.
The fractional change in area $A = 4 \pi r^2$ is given by $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r} = 2X$.
The fractional change in volume $V = \frac{4}{3} \pi r^3$ is given by $\frac{\Delta V}{V} = 3 \frac{\Delta r}{r} = 3X$.
Comparing the percentage changes,$3X > 2X > X$. Therefore,the maximum percentage change is observed in the volume.

(A) Let the initial length be $l$. The change in length is given as $\Delta l / l = 2\% = 0.02$.
For a square of side $l$,the area is $A = l^2$.
The relative change in area is given by the formula $\Delta A / A = 2(\Delta l / l)$.
Substituting the given value: $\Delta A / A = 2 \times 2\% = 4\%$.
Therefore,the percentage increase in the area of the square copper sheet is $4\%$.

(D) Let the lengths at temperature $\Delta \theta$ be $L_1'$ and $L_2'$.
$L_1' = L_1(1 + \alpha_1 \Delta \theta)$ and $L_2' = L_2(1 + \alpha_2 \Delta \theta)$.
The difference is $D = L_2' - L_1' = (L_2 - L_1) + (L_2 \alpha_2 - L_1 \alpha_1) \Delta \theta$.
For the difference to be constant at any temperature,the coefficient of $\Delta \theta$ must be zero.
Therefore,$L_2 \alpha_2 - L_1 \alpha_1 = 0$.
This implies $L_1 \alpha_1 = L_2 \alpha_2$.

(A) Given that the difference in length is constant at all temperatures,the change in length for both rods must be equal for any change in temperature $\Delta T$.
Let $L_{Fe}$ and $L_{Cu}$ be the lengths of the iron and copper rods respectively.
We are given $L_{Fe} - L_{Cu} = 10 \ cm$ ..... $(i)$
Since the difference is constant,$\Delta L_{Fe} = \Delta L_{Cu}$.
Using the formula for linear expansion $\Delta L = L \alpha \Delta T$,we get:
$L_{Fe} \alpha_{Fe} \Delta T = L_{Cu} \alpha_{Cu} \Delta T$
$L_{Fe} \alpha_{Fe} = L_{Cu} \alpha_{Cu}$
$\frac{L_{Fe}}{L_{Cu}} = \frac{\alpha_{Cu}}{\alpha_{Fe}} = \frac{17 \times 10^{-6}}{11 \times 10^{-6}} = \frac{17}{11}$ ..... $(ii)$
From $(i)$,$L_{Fe} = L_{Cu} + 10$.
Substituting this into $(ii)$:
$\frac{L_{Cu} + 10}{L_{Cu}} = \frac{17}{11}$
$11(L_{Cu} + 10) = 17 L_{Cu}$
$11 L_{Cu} + 110 = 17 L_{Cu}$
$6 L_{Cu} = 110$
$L_{Cu} = \frac{110}{6} \approx 18.33 \ cm$
$L_{Fe} = 18.33 + 10 = 28.33 \ cm$.
Thus,the lengths are approximately $28.3 \ cm$ and $18.3 \ cm$.

(C) The total length of the system at $20^{\circ}C$ is $L = 50 \ cm + 100 \ cm = 150 \ cm$.
Let the temperature change be $\Delta T = 100^{\circ}C - 20^{\circ}C = 80^{\circ}C$.
The new length of the iron rod at $100^{\circ}C$ is $L_{Fe}' = 50(1 + \alpha_{Fe} \Delta T) = 50(1 + 12 \times 10^{-6} \times 80) = 50(1 + 0.00096) = 50.048 \ cm$.
The new length of the aluminum rod at $100^{\circ}C$ is $L_{Al}' = 100(1 + \alpha_{Al} \Delta T) = 100(1 + 24 \times 10^{-6} \times 80) = 100(1 + 0.00192) = 100.192 \ cm$.
The total length at $100^{\circ}C$ is $L' = 50.048 + 100.192 = 150.24 \ cm$.
The change in length is $\Delta L = L' - L = 150.24 - 150 = 0.24 \ cm$.
The effective coefficient of linear expansion $\alpha_{eff}$ is given by $\Delta L = L \alpha_{eff} \Delta T$.
$\alpha_{eff} = \frac{\Delta L}{L \Delta T} = \frac{0.24}{150 \times 80} = \frac{0.24}{12000} = 20 \times 10^{-6} {^{\circ}C}^{-1}$.

(B) The change in length of a rod due to thermal expansion is given by $\Delta L = L \alpha \Delta T$.
For the brass rod:
$\Delta L_{brass} = L \alpha_{brass} \Delta T = 80 \ cm \times (18 \times 10^{-6} \ ^{\circ}C^{-1}) \times (100^{\circ}C - 0^{\circ}C) = 80 \times 18 \times 10^{-4} \ cm = 0.144 \ cm = 1.44 \ mm$.
For the lead rod:
$\Delta L_{lead} = L \alpha_{lead} \Delta T = 80 \ cm \times (28 \times 10^{-6} \ ^{\circ}C^{-1}) \times (100^{\circ}C - 0^{\circ}C) = 80 \times 28 \times 10^{-4} \ cm = 0.224 \ cm = 2.24 \ mm$.
The difference in length between their ends is $\Delta L_{lead} - \Delta L_{brass} = 2.24 \ mm - 1.44 \ mm = 0.8 \ mm$.

(B) Let $V$ be the volume of the sphere and $\rho_L$ be the density of the liquid. Since the sphere floats,the weight of the sphere equals the weight of the displaced liquid.
$Mg = V_{submerged} \rho_L g$
$V_{sphere}(t) \rho_S(t) g = f V_{sphere}(t) \rho_L g$
Since the liquid density $\rho_L$ is constant,$\rho_S(t) = f \rho_L$.
As the sphere expands,its density changes as $\rho_S(t) = \frac{\rho_0}{1 + \gamma t}$,where $\gamma$ is the coefficient of volume expansion.
Thus,$f_1 = \frac{\rho_S(t_1)}{\rho_L} = \frac{\rho_0}{\rho_L(1 + \gamma t_1)}$ and $f_2 = \frac{\rho_S(t_2)}{\rho_L} = \frac{\rho_0}{\rho_L(1 + \gamma t_2)}$.
Taking the ratio: $\frac{f_1}{f_2} = \frac{1 + \gamma t_2}{1 + \gamma t_1}$.
$f_1(1 + \gamma t_1) = f_2(1 + \gamma t_2)$
$f_1 + f_1 \gamma t_1 = f_2 + f_2 \gamma t_2$
$\gamma (f_1 t_1 - f_2 t_2) = f_2 - f_1$
$\gamma = \frac{f_2 - f_1}{f_1 t_1 - f_2 t_2} = \frac{f_1 - f_2}{f_2 t_2 - f_1 t_1}$.
Wait,re-evaluating the ratio: $f_1(1 + \gamma t_2) = f_2(1 + \gamma t_1)$ is incorrect. The correct relation is $f_1(1 + \gamma t_1) = C$ (constant).
Actually,$V(t) = V_0(1 + \gamma t)$. The submerged volume $V_{sub} = f V(t)$.
Since $V_{sub} \rho_L = M$ (constant),$f V_0(1 + \gamma t) \rho_L = M$.
So $f_1(1 + \gamma t_1) = f_2(1 + \gamma t_2)$.
$f_1 + f_1 \gamma t_1 = f_2 + f_2 \gamma t_2$
$\gamma (f_1 t_1 - f_2 t_2) = f_2 - f_1$
$\gamma = \frac{f_2 - f_1}{f_1 t_1 - f_2 t_2} = \frac{f_1 - f_2}{f_2 t_2 - f_1 t_1}$.

(A) The time period of a pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
Taking the derivative,the fractional change in time period is $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta L}{L} = \frac{1}{2} \alpha \Delta \theta$.
Given $\alpha = 12 \times 10^{-6}\,^oC^{-1}$ and $\Delta \theta = (40 - 20) = 20\,^oC$.
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 per day is $\Delta T = \frac{\Delta T}{T} \times T_{day}$,where $T_{day} = 86400\,sec$.
$\Delta T = 1.2 \times 10^{-4} \times 86400 = 10.368\,sec/day \approx 10.3\,sec/day$.

(C) Let the lengths of the brass and steel rods at temperature $T$ be $L_1$ and $L_2$ respectively.
After a change in temperature $\Delta T$,the new lengths are:
$L_1' = l_1(1 + \alpha_1 \Delta T)$
$L_2' = l_2(1 + \alpha_2 \Delta T)$
Given that the difference $(L_2' - L_1')$ remains constant and equal to $(l_2 - l_1)$ at all temperatures:
$L_2' - L_1' = l_2 - l_1$
$l_2(1 + \alpha_2 \Delta T) - l_1(1 + \alpha_1 \Delta T) = l_2 - l_1$
$l_2 + l_2 \alpha_2 \Delta T - l_1 - l_1 \alpha_1 \Delta T = l_2 - l_1$
$(l_2 - l_1) + \Delta T(l_2 \alpha_2 - l_1 \alpha_1) = l_2 - l_1$
For this to hold for any $\Delta T$,the coefficient of $\Delta T$ must be zero:
$l_2 \alpha_2 - l_1 \alpha_1 = 0$
$\alpha_1 l_1 = \alpha_2 l_2$

(D) The thermal force $F$ produced in a wire when it is prevented from expanding due to a temperature change $\Delta \theta$ is given by the formula:
$F = YA\alpha \Delta \theta$
where $Y$ is Young's modulus,$A$ is the area of cross-section,$\alpha$ is the coefficient of linear expansion,and $\Delta \theta$ is the change in temperature.
Given that $Y$,$A$,and $\Delta \theta$ are the same for both wires $A$ and $B$,the force $F$ is directly proportional to the coefficient of linear expansion $\alpha$:
$F \propto \alpha$
Therefore,the ratio of the forces is:
$\frac{F_A}{F_B} = \frac{\alpha_A}{\alpha_B}$
Given that $\alpha_A = \frac{3}{2} \alpha_B$,we have:
$\frac{F_A}{F_B} = \frac{3}{2}$
Thus,the ratio of the forces produced in the two wires is $3/2$.

(B) When a metal sheet is heated,it undergoes thermal expansion. The expansion of a hole in a solid is similar to the expansion of the material itself.
According to the principle of thermal expansion,the distance between any two points on the sheet increases as the temperature increases.
Since the diameter of a hole is defined by the distance between two points on its circumference,this distance also increases.
Therefore,the diameters of both holes,$d_1$ and $d_2$,will increase upon heating. This phenomenon is analogous to a photographic enlargement where every part of the image scales up.

(B) Let $l_1$ be the length of the portion of metal $A$ and $l_2$ be the length of the portion of metal $B$.
Since the total length is $20 \, cm$,we have $l_1 + l_2 = 20 \, cm$.
For a given change in temperature,the change in length $\Delta l$ is proportional to the initial length $l$. Thus,$\Delta l = \alpha \cdot l \cdot \Delta T$.
For metal $A$,$\Delta l_A = 0.075 \, cm$ for $l = 20 \, cm$,so the expansion coefficient factor is $k_A = \frac{0.075}{20} = 0.00375$.
For metal $B$,$\Delta l_B = 0.045 \, cm$ for $l = 20 \, cm$,so the expansion coefficient factor is $k_B = \frac{0.045}{20} = 0.00225$.
The total expansion of the composite rod is $\Delta l_{total} = k_A l_1 + k_B l_2 = 0.06 \, cm$.
Substituting $l_2 = 20 - l_1$:
$0.00375 l_1 + 0.00225(20 - l_1) = 0.06$
$0.00375 l_1 + 0.045 - 0.00225 l_1 = 0.06$
$0.0015 l_1 = 0.015$
$l_1 = \frac{0.015}{0.0015} = 10 \, cm$.

(A) The volume of mercury at $T^\circ C$ is $V_m = V_0(1 + \gamma_m T)$.
The volume of the glass bulb at $T^\circ C$ is $V_g = V_0(1 + 3a_g T)$.
The volume of mercury that overflows into the capillary is $\Delta V = V_m - V_g = V_0(\gamma_m - 3a_g)T$.
The cross-sectional area of the capillary at $T^\circ C$ is $A = A_0(1 + 2a_g T)$.
Let $h$ be the length of the mercury column in the capillary. Then,$\Delta V = A \times h$.
Therefore,$h = \frac{\Delta V}{A} = \frac{V_0 T (\gamma_m - 3a_g)}{A_0 (1 + 2a_g T)}$.

(C) The length of the rod is $L = 2 \ m = 200 \ cm$. The coefficient of linear expansion is $\alpha(x) = (3x + 2) \times 10^{-6} \ ^\circ C^{-1}$.
For a small element $dx$ at distance $x$,the change in length $dL$ for a temperature change $\Delta T = 20^\circ C$ is given by $dL = \alpha(x) \cdot dx \cdot \Delta T$.
Integrating this from $x = 0$ to $x = 200 \ cm$:
$\Delta L = \int_{0}^{200} (3x + 2) \times 10^{-6} \cdot 20 \cdot dx$
$\Delta L = 20 \times 10^{-6} \int_{0}^{200} (3x + 2) \cdot dx$
$\Delta L = 20 \times 10^{-6} [\frac{3x^2}{2} + 2x]_{0}^{200}$
$\Delta L = 20 \times 10^{-6} [\frac{3(200)^2}{2} + 2(200)]$
$\Delta L = 20 \times 10^{-6} [60000 + 400] = 20 \times 10^{-6} \times 60400 = 1.208 \ cm$.
Since $1.208 \ cm = 0.01208 \ m$,the new length is $L' = L + \Delta L = 2 + 0.01208 = 2.01208 \ m \approx 2.01 \ m$.

(C) For an anisotropic material,the coefficient of superficial expansion $\beta$ for a face is the sum of the linear expansion coefficients along the two axes defining that face.
Given: $\alpha_x = 1 \times 10^{-5} /^{\circ}C$,$\alpha_y = 2 \times 10^{-5} /^{\circ}C$,$\alpha_z = 3 \times 10^{-5} /^{\circ}C$.
From the figure,face $CDEH$ lies in the $xy$-plane (assuming $D$ is at origin or aligned with axes as per standard orientation).
Specifically,face $CDEH$ is defined by the $x$ and $y$ axes.
Therefore,$\beta_{CDEH} = \alpha_x + \alpha_y = (1 + 2) \times 10^{-5} /^{\circ}C = 3 \times 10^{-5} /^{\circ}C$.
Thus,option $C$ is correct.

(B) Let the coefficient of linear expansion be $\alpha$. The fractional change in length is given by $\frac{\Delta L}{L} = \alpha \Delta T = 1\% = 0.01$.
The area of the copper plate is $A = 2L \times L = 2L^2$.
The coefficient of area expansion is $\beta = 2\alpha$.
The fractional change in area is given by $\frac{\Delta A}{A} = \beta \Delta T = 2\alpha \Delta T$.
Substituting the value of $\alpha \Delta T = 0.01$:
$\frac{\Delta A}{A} = 2 \times (0.01) = 0.02$.
To find the percentage change in area:
$\frac{\Delta A}{A} \times 100 = 0.02 \times 100 = 2\%$.

(C) The initial length of the combination is $L + 2L = 3L$.
The change in length of the first rod is $\Delta L_1 = L \alpha \Delta t$.
The change in length of the second rod is $\Delta L_2 = (2L)(2\alpha) \Delta t = 4L \alpha \Delta t$.
The total change in length of the composite rod is $\Delta L_{total} = \Delta L_1 + \Delta L_2 = L \alpha \Delta t + 4L \alpha \Delta t = 5L \alpha \Delta t$.
The average coefficient of linear expansion $\alpha_{avg}$ is defined by the relation $\Delta L_{total} = L_{total} \alpha_{avg} \Delta t$.
Substituting the values,we get $5L \alpha \Delta t = (3L) \alpha_{avg} \Delta t$.
Solving for $\alpha_{avg}$,we get $\alpha_{avg} = \frac{5L \alpha \Delta t}{3L \Delta t} = \frac{5}{3}\alpha$.

(C) Let $V_m$ be the volume of mercury and $V_f$ be the volume of the glass flask.
Since the volume of air inside the flask remains constant at different temperatures,the change in the volume of mercury must be equal to the change in the volume of the flask.
$\Delta V_m = \Delta V_f$
$\gamma_m V_m \Delta T = \gamma_f V_f \Delta T$
$\gamma_m V_m = \gamma_f V_f$
We know that the coefficient of volume expansion of the flask $\gamma_f = 3 \alpha_{glass}$.
Given $\gamma_m = 1.8 \times 10^{-4} (^{\circ}C)^{-1}$,$\alpha_{glass} = 9 \times 10^{-6} (^{\circ}C)^{-1}$,and $V_m = 300 \, cm^3$.
Substituting the values:
$1.8 \times 10^{-4} \times 300 = 3 \times (9 \times 10^{-6}) \times V_f$
$5.4 \times 10^{-2} = 2.7 \times 10^{-5} \times V_f$
$V_f = \frac{5.4 \times 10^{-2}}{2.7 \times 10^{-5}} = 2 \times 10^3 = 2000 \, cm^3$.

(C) The time period of a pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$. The fractional change in time period is $\frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta \theta$.
For a day,$T = 24 \times 3600 \; s$.
When the clock loses $12\;s$ at $40^{\circ}C$ (where $\theta$ is the correct temperature): $\frac{12}{T} = \frac{1}{2} \alpha (40 - \theta) \quad ...(1)$
When the clock gains $4\;s$ at $20^{\circ}C$: $\frac{4}{T} = \frac{1}{2} \alpha (\theta - 20) \quad ...(2)$
Dividing equation $(1)$ by $(2)$: $\frac{12}{4} = \frac{40 - \theta}{\theta - 20} \implies 3(\theta - 20) = 40 - \theta \implies 3\theta - 60 = 40 - \theta \implies 4\theta = 100 \implies \theta = 25^{\circ}C$.
Substituting $\theta = 25^{\circ}C$ in equation $(2)$: $\frac{4}{24 \times 3600} = \frac{1}{2} \alpha (25 - 20) \implies \frac{4}{86400} = \frac{1}{2} \alpha (5) \implies \alpha = \frac{8}{86400 \times 5} = \frac{8}{432000} \approx 1.85 \times 10^{-5}/^{\circ}C$.

(C) Let the nominal length of the tape be $L_0 = 30$ $m$ and the expanded length be $L = 30.01$ $m$.
The ratio of the true length to the measured length is equal to the ratio of the actual length of the tape to its nominal length.
Let $d_{true}$ be the true distance and $d_{measured} = 15.52$ $m$ be the measured distance.
Using the principle of proportionality: $\frac{d_{true}}{d_{measured}} = \frac{L}{L_0}$.
$d_{true} = d_{measured} \times \frac{L}{L_0} = 15.52 \times \frac{30.01}{30}$.
$d_{true} = 15.52 \times (1 + \frac{0.01}{30}) = 15.52 + 15.52 \times 0.000333...$
$d_{true} = 15.52 + 0.00517... \approx 15.525$ $m$.

(A) Let $L_S$ and $L_C$ be the lengths of steel and copper rods at $0^o C$.
At any temperature $T$,the lengths are $L_S(1 + \alpha_S T)$ and $L_C(1 + \alpha_C T)$.
The difference in lengths is given as $10\,cm$ at any temperature $T$,so:
$L_S(1 + \alpha_S T) - L_C(1 + \alpha_C T) = 10$
$(L_S - L_C) + (L_S \alpha_S - L_C \alpha_C)T = 10$
For this to be independent of $T$,the coefficient of $T$ must be zero:
$L_S \alpha_S - L_C \alpha_C = 0 \Rightarrow L_S \alpha_S = L_C \alpha_C$
$L_S (1.2 \times 10^{-5}) = L_C (1.8 \times 10^{-5})$
$L_S / L_C = 1.8 / 1.2 = 3 / 2$
Also,from the first equation,$L_S - L_C = 10$.
Substituting $L_S = 1.5 L_C$ into $L_S - L_C = 10$:
$1.5 L_C - L_C = 10 \Rightarrow 0.5 L_C = 10 \Rightarrow L_C = 20\,cm$.
Then $L_S = 1.5 \times 20 = 30\,cm$.

(B) Let the length of each rod be $L$ and mass be $M$. The rods are joined at $B$. The center of mass of rod $AB$ is at a distance $L/2$ from $B$,and the center of mass of rod $BC$ is at a distance $L/2$ from $B$.
When the rod is heated,the length of each rod increases according to the formula $\Delta L = L \alpha \Delta T$.
Since $\alpha_{AB} > \alpha_{BC}$,the rod $AB$ expands more than the rod $BC$.
Let the new lengths be $L_{AB}' = L(1 + \alpha_{AB} \Delta T)$ and $L_{BC}' = L(1 + \alpha_{BC} \Delta T)$.
The new center of mass of rod $AB$ is at $L_{AB}'/2$ from $B$,and the new center of mass of rod $BC$ is at $L_{BC}'/2$ from $B$.
Since $L_{AB}' > L_{BC}'$,the center of mass of rod $AB$ shifts further away from $B$ compared to the center of mass of rod $BC$.
Because the rod is suspended at $B$,the torque due to the weight of rod $AB$ is $\tau_{AB} = M g (L_{AB}'/2)$ and the torque due to the weight of rod $BC$ is $\tau_{BC} = M g (L_{BC}'/2)$.
Since $L_{AB}' > L_{BC}'$,the torque on the side of $AB$ is greater than the torque on the side of $BC$.
Therefore,the rod tilts down on the side of $AB$.

(C) The change in length of a small element $dx$ is given by $d(\Delta L) = \alpha \cdot dx \cdot \Delta T$.
Here,$\Delta T = 10 \, ^oC - 0 \, ^oC = 10 \, ^oC$.
The total change in length $\Delta L$ is the integral of $d(\Delta L)$ from $x = 0$ to $x = 10 \, m$:
$\Delta L = \int_{0}^{10} (2x^2 + 1) \times 10^{-6} \times 10 \, dx$
$\Delta L = 10^{-5} \int_{0}^{10} (2x^2 + 1) \, dx$
$\Delta L = 10^{-5} \left[ \frac{2x^3}{3} + x \right]_{0}^{10}$
$\Delta L = 10^{-5} \left( \frac{2(1000)}{3} + 10 \right) = 10^{-5} \left( \frac{2000}{3} + 10 \right) = 10^{-5} \left( 666.67 + 10 \right) = 10^{-5} \times 676.67 = 0.0067667 \, m$.
The final length $L' = L + \Delta L = 10 + 0.0067667 = 10.0067667 \, m \approx 10.0068 \, m$.

(D) The coefficient of linear expansion,denoted by $\alpha$,is defined by the relation $\Delta L = L_0 \alpha \Delta T$,where $\Delta L$ is the change in length,$L_0$ is the original length,and $\Delta T$ is the change in temperature.
Rearranging this,we get $\alpha = \frac{\Delta L}{L_0 \Delta T}$.
Although the formula involves $L_0$ and $\Delta T$,the coefficient $\alpha$ is a characteristic property of the material itself.
It represents how much a unit length of a specific material expands per unit degree rise in temperature.
Therefore,$\alpha$ depends only on the nature of the material (the metal) and is independent of the dimensions of the rod or the magnitude of the temperature change.

(A) When a material is heated,it expands in length by an amount proportional to $\alpha \Delta T$,where $\alpha$ is the coefficient of linear thermal expansion and $\Delta T$ is the change in temperature.
The metal with the higher coefficient of linear thermal expansion $(\alpha)$ will expand more than the metal with the lower coefficient for the same temperature change.
Because the two metals are bonded together,the one that expands more will form the outer arc of the curve,while the one that expands less will form the inner arc.
Therefore,the bimetallic strip will bend towards the metal with the lower thermal expansion coefficient.

(C) The formula for linear expansion is given by $\Delta L = L_0 \alpha \Delta T$.
Here,$L_0 = 50 \, cm$ is the initial length at the calibration temperature.
$\alpha = 1.0 \times 10^{-5} / ^{\circ}C$ is the coefficient of linear expansion for steel.
$\Delta T = T_f - T_i = 30^{\circ}C - 20^{\circ}C = 10^{\circ}C$ is the change in temperature.
Substituting these values into the formula:
$\Delta L = (50 \, cm) \times (1.0 \times 10^{-5} / ^{\circ}C) \times (10^{\circ}C)$
$\Delta L = 500 \times 10^{-5} \, cm$
$\Delta L = 5 \times 10^{-3} \, cm$.
Therefore,the error in the reading is $5 \times 10^{-3} \, cm$.

(B) The coefficient of volume expansion $\gamma$ is defined by the relation $\frac{\Delta V}{V} = \gamma \Delta T$.
Given $\frac{\Delta V}{V} = 0.12\% = 0.0012$ and $\Delta T = 20^{\circ}C$.
Substituting the values,we get $\gamma = \frac{\Delta V}{V \Delta T} = \frac{0.0012}{20} = 6 \times 10^{-5} /^{\circ}C$.
The relationship between the coefficient of linear expansion $\alpha$ and the coefficient of volume expansion $\gamma$ is $\alpha = \frac{\gamma}{3}$.
Therefore,$\alpha = \frac{6 \times 10^{-5}}{3} = 2 \times 10^{-5} /^{\circ}C$.

(C) The change in length $\Delta L$ for a rod of initial length $L_0$ due to a temperature change $\Delta T$ is given by $\Delta L = L_0 \alpha \Delta T$.
Given that the increase in length for both rods is the same,we have:
$\Delta L_1 = \Delta L_2$
$l_1 \alpha_1 t = l_2 \alpha_2 t$
Dividing both sides by $t$ (assuming $t \neq 0$):
$l_1 \alpha_1 = l_2 \alpha_2$
From this,we can express $l_2$ in terms of $l_1$:
$l_2 = l_1 \frac{\alpha_1}{\alpha_2}$
We want to find the ratio $\frac{l_1}{l_1 + l_2}$. Substituting $l_2$ into the expression:
$\frac{l_1}{l_1 + l_1 \frac{\alpha_1}{\alpha_2}} = \frac{l_1}{l_1 (1 + \frac{\alpha_1}{\alpha_2})} = \frac{1}{\frac{\alpha_2 + \alpha_1}{\alpha_2}} = \frac{\alpha_2}{\alpha_1 + \alpha_2}$

(B) The coefficient of volume expansion $\gamma$ is defined as $\gamma = \frac{1}{V} \frac{\Delta V}{\Delta T}$.
Given,the fractional change in volume $\frac{\Delta V}{V} = 0.12\% = \frac{0.12}{100} = 1.2 \times 10^{-3}$.
The change in temperature $\Delta T = 20\,^{\circ}C$.
Substituting these values,$\gamma = \frac{1.2 \times 10^{-3}}{20} = 0.06 \times 10^{-3} = 6 \times 10^{-5}\,^{\circ}C^{-1}$.
The relationship between the coefficient of linear expansion $\alpha$ and the coefficient of volume expansion $\gamma$ is $\gamma = 3\alpha$.
Therefore,$\alpha = \frac{\gamma}{3} = \frac{6 \times 10^{-5}}{3} = 2 \times 10^{-5}\,^{\circ}C^{-1}$.

(D) 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 $\Delta L = L \alpha \Delta t$,we have $\frac{\Delta L}{L} = \alpha \Delta t$.
Therefore,the fractional change in time is $\frac{\Delta T}{T} = \frac{1}{2} \alpha t$.
This represents the time lost per unit time.
To find the time lost per day,we multiply by the total number of seconds in a day,which is $24 \times 60 \times 60 = 86400 \, s$.
Thus,the time loss per day is $\Delta T_{day} = \frac{1}{2} \alpha t \times 86400$ seconds.

(A) When a bimetallic strip is heated, it bends such that the metal with the higher coefficient of linear expansion $(\alpha)$ forms the outer (convex) side of the curve.
In the given figure, the brass strip $(B)$ is on the outer side of the curve, while the steel strip $(S)$ is on the inner (concave) side.
Therefore, the brass strip has a higher coefficient of linear expansion than the steel strip.

(D) When the temperature increases by $\Delta T$,the rods tend to expand. Since they are fixed between rigid supports,they experience a compressive force $F$. For the total length to remain unchanged,the sum of the thermal expansion and the compression due to the force must be zero for each rod,or more simply,the force $F$ generated in both rods must be equal.
The thermal expansion is $\Delta L_{thermal} = L \alpha \Delta T$.
The compression due to force $F$ is $\Delta L_{comp} = \frac{F L}{A Y}$.
For the total length to remain constant,the net change in length must be zero: $\Delta L_{thermal} - \Delta L_{comp} = 0$,which implies $F = A Y \alpha \Delta T$.
Since the rods are in series,the force $F$ is the same in both rods:
$F_1 = F_2$
$A_1 Y_1 \alpha_1 \Delta T = A_2 Y_2 \alpha_2 \Delta T$
$\frac{A_1}{A_2} = \frac{\alpha_2 Y_2}{\alpha_1 Y_1}$

(B) The change in temperature is $\Delta T = 40\,^{\circ}C - 27\,^{\circ}C = 13\,^{\circ}C$.
The thermal expansion of a rail is given by $\Delta l = l \alpha \Delta T$.
For the gap of width $x$ to close up, the expansion of the rail must equal the gap width, so $\Delta l = x$.
Substituting the values, we get $x = l \alpha (13)$.
Solving for $l$, we find $l = \frac{x}{13\alpha}$.

(D) In the equilateral triangle $PQR,$ the height $OR$ is given by the Pythagorean theorem in $\triangle OPR$ as:
$(OR)^2 = (PR)^2 - (OP)^2$
Since $OP = l/2,$ we have $(OR)^2 = l^2 - (l/2)^2 = 3l^2/4.$
When the temperature changes by $\Delta t,$ the lengths change as $l' = l(1 + \alpha \Delta t).$
Let $l_1 = l(1 + \alpha_1 \Delta t)$ be the new length of $PQ,$ so $OP' = \frac{l}{2}(1 + \alpha_1 \Delta t).$
Let $l_2 = l(1 + \alpha_2 \Delta t)$ be the new length of $PR,$ so $PR' = l(1 + \alpha_2 \Delta t).$
Since $OR$ remains constant,$(OR)^2 = (PR')^2 - (OP')^2.$
Equating the initial and final values of $(OR)^2$:
$l^2 - (l/2)^2 = [l(1 + \alpha_2 \Delta t)]^2 - [\frac{l}{2}(1 + \alpha_1 \Delta t)]^2$
$l^2 - \frac{l^2}{4} = l^2(1 + 2\alpha_2 \Delta t + \alpha_2^2 \Delta t^2) - \frac{l^2}{4}(1 + 2\alpha_1 \Delta t + \alpha_1^2 \Delta t^2)$
Neglecting higher-order terms like $\alpha^2 \Delta t^2$:
$l^2 - \frac{l^2}{4} = l^2 + 2l^2 \alpha_2 \Delta t - \frac{l^2}{4} - \frac{l^2}{2} \alpha_1 \Delta t$
$0 = 2l^2 \alpha_2 \Delta t - \frac{l^2}{2} \alpha_1 \Delta t$
$2 \alpha_2 = \frac{1}{2} \alpha_1 \implies \alpha_1 = 4 \alpha_2.$

(B) When a metal sheet is heated,it undergoes thermal expansion.
For an isotropic solid,the expansion is similar to a photographic enlargement,meaning every part of the object,including the holes,expands in the same proportion.
Therefore,the distance between any two points on the sheet increases,which implies that the diameters of the holes ($d_1$ and $d_2$) will also increase.

(D) Thermal expansion of a solid object is similar to a photographic enlargement. When a solid object is heated,all its linear dimensions (lengths,diameters,radii) increase by the same percentage.
Given that the outer diameter $d_1$ increases by $0.3\%$,the change in length is given by $\Delta d_1 = \alpha d_1 \Delta T$,where $\alpha$ is the coefficient of linear expansion.
Since the inner diameter $d_2$ is also a linear dimension of the same material,it will also expand according to the same coefficient of linear expansion $\alpha$.
Therefore,the percentage change in $d_2$ will be the same as the percentage change in $d_1$.
Thus,$d_2$ will also increase by $0.3\%$.

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
Taking the natural logarithm on both sides: $\ln T = \ln(2\pi) + \frac{1}{2} \ln L - \frac{1}{2} \ln g$.
Differentiating with respect to temperature $\theta$: $\frac{1}{T} \frac{dT}{d\theta} = \frac{1}{2L} \frac{dL}{d\theta}$.
Since $\frac{dL}{d\theta} = L\alpha$,we have $\frac{1}{T} \frac{dT}{d\theta} = \frac{1}{2L} (L\alpha) = \frac{\alpha}{2}$.
Therefore,the slope $S = \frac{dT}{d\theta} = \frac{T\alpha}{2}$.
Given $T = 2 \ s$,we get $S = \frac{2\alpha}{2} = \alpha$.

(A) Let the initial length of both rods be $L_0$ and the initial temperature be $T_i = 30\,^{\circ}C$.
The final length of a rod is given by $L = L_0(1 + \alpha \Delta T)$,where $\alpha$ is the coefficient of linear expansion and $\Delta T$ is the change in temperature.
Since the final lengths are equal,we have:
$L_0(1 + \alpha_A \Delta T_A) = L_0(1 + \alpha_B \Delta T_B)$
This simplifies to:
$\alpha_A \Delta T_A = \alpha_B \Delta T_B$
Given $\frac{\alpha_A}{\alpha_B} = \frac{4}{3}$,$\Delta T_A = 180 - 30 = 150\,^{\circ}C$,and $\Delta T_B = T - 30$.
Substituting the values:
$\frac{4}{3} = \frac{T - 30}{150}$
$T - 30 = \frac{4}{3} \times 150$
$T - 30 = 4 \times 50 = 200$
$T = 200 + 30 = 230\,^{\circ}C$.

(A) The linear expansion formula is $\Delta L = L_0 \alpha \Delta T$.
Given $\Delta L = 0.08 \ mm = 0.008 \ cm$,$L_0 = 10 \ cm$,and $\Delta T = 100 \ ^\circ C - 0 \ ^\circ C = 100 \ ^\circ C$.
Substituting these values: $0.008 = 10 \times \alpha \times 100$.
Thus,$\alpha = \frac{0.008}{1000} = 8 \times 10^{-6} / ^\circ C$.
The coefficient of volume expansion $\gamma$ is related to the linear expansion coefficient $\alpha$ by $\gamma = 3\alpha$.
So,$\gamma = 3 \times 8 \times 10^{-6} = 24 \times 10^{-6} / ^\circ C$.
The volume at temperature $T$ is given by $V_T = V_0(1 + \gamma \Delta T)$.
$V_{100} = 1000(1 + 24 \times 10^{-6} \times 100) = 1000(1 + 24 \times 10^{-4}) = 1000(1 + 0.0024) = 1000(1.0024) = 1002.4 \ cc$.

(B) Let the lengths of the rods at temperature $T$ be $L_1$ and $L_2$.
At a temperature $T + \Delta T$,the new lengths are $L_1' = L_1(1 + \alpha_1 \Delta T)$ and $L_2' = L_2(1 + \alpha_2 \Delta T)$.
The difference between the lengths is independent of temperature,so $L_1' - L_2' = L_1 - L_2$.
Substituting the expressions: $L_1(1 + \alpha_1 \Delta T) - L_2(1 + \alpha_2 \Delta T) = L_1 - L_2$.
Expanding this,we get $L_1 + L_1 \alpha_1 \Delta T - L_2 - L_2 \alpha_2 \Delta T = L_1 - L_2$.
Subtracting $(L_1 - L_2)$ from both sides,we get $L_1 \alpha_1 \Delta T = L_2 \alpha_2 \Delta T$.
Since $\Delta T \neq 0$,we have $L_1 \alpha_1 = L_2 \alpha_2$.
Therefore,$\frac{L_1}{L_2} = \frac{\alpha_2}{\alpha_1}$.

(D) Let $L_0 = 90 \ cm$ be the initial length at $T_1 = 10 \ ^oC$. The change in temperature is $\Delta T = 30 - 10 = 20 \ ^oC$.
The actual length of the copper rod at $30 \ ^oC$ is $L_{cu} = L_0(1 + \alpha_{cu} \Delta T) = 90(1 + 1.7 \times 10^{-5} \times 20) = 90(1 + 3.4 \times 10^{-4}) = 90.0306 \ cm$.
The steel tape also expands. The length of each $1 \ cm$ division on the tape at $30 \ ^oC$ becomes $L'_{st} = 1(1 + \alpha_{st} \Delta T) = 1(1 + 1.2 \times 10^{-5} \times 20) = 1(1 + 2.4 \times 10^{-4}) = 1.00024 \ cm$.
The reading on the tape is the ratio of the actual length of the rod to the length of one division on the tape: $\text{Reading} = \frac{L_{cu}}{L'_{st}} = \frac{90(1 + 3.4 \times 10^{-4})}{1 + 2.4 \times 10^{-4}} \approx 90(1 + 3.4 \times 10^{-4})(1 - 2.4 \times 10^{-4}) \approx 90(1 + 1.0 \times 10^{-4}) = 90.009 \ cm \approx 90.01 \ cm$.

(A) Let the lengths of the copper and iron rods at $0^{\circ}C$ be $L_{c0}$ and $L_{i0}$ respectively.
The lengths at temperature $T$ are given by $l_1 = L_{c0}(1 + \alpha_c T)$ and $l_2 = L_{i0}(1 + \alpha_i T)$.
The difference is $l_2 - l_1 = L_{i0} - L_{c0} + (L_{i0}\alpha_i - L_{c0}\alpha_c)T$.
For the difference to be independent of $T$,the coefficient of $T$ must be zero:
$L_{i0}\alpha_i - L_{c0}\alpha_c = 0 \implies L_{i0}\alpha_i = L_{c0}\alpha_c$.
Given $\alpha_c = 2.0 \times 10^{-6}\,^{\circ}C^{-1}$ and $\alpha_i = 1.0 \times 10^{-6}\,^{\circ}C^{-1}$,we have $L_{i0}(1.0 \times 10^{-6}) = L_{c0}(2.0 \times 10^{-6})$,so $L_{i0} = 2L_{c0}$.
We are given $l_2 - l_1 = 15\,cm$,which implies $L_{i0} - L_{c0} = 15\,cm$.
Substituting $L_{i0} = 2L_{c0}$,we get $2L_{c0} - L_{c0} = 15\,cm$,so $L_{c0} = 15\,cm$ and $L_{i0} = 30\,cm$.
Since the difference is constant,$l_1 = L_{c0} = 15\,cm$ and $l_2 = L_{i0} = 30\,cm$.

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
When the temperature decreases,the length $L$ of the pendulum rod decreases due to thermal contraction.
Since $T \propto \sqrt{L}$,the time period $T$ decreases,which means the clock runs faster and gains time.
The fractional change in time period is given by $\frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta \theta$.
The time gained per day is $\Delta t = \frac{1}{2} \alpha \Delta \theta \times t$,where $t$ is the total number of seconds in a day.
Given: $\alpha = 18 \times 10^{-6}/^{\circ}C$,$\Delta \theta = 20^{\circ}C - 0^{\circ}C = 20^{\circ}C$,and $t = 24 \times 3600 \, s = 86400 \, s$.
Substituting the values: $\Delta t = \frac{1}{2} \times (18 \times 10^{-6}) \times 20 \times 86400$.
$\Delta t = 180 \times 10^{-6} \times 86400 = 0.18 \times 86.4 = 15.552 \, s$.
Thus,the clock gains approximately $15.55 \, s$ per day.

(B) The coefficient of volume expansion $\gamma$ is related to the change in volume by the formula: $\frac{\Delta V}{V} = \gamma \Delta T$.
Given that the volume increases by $0.15\%$,we have $\frac{\Delta V}{V} = \frac{0.15}{100} = 0.0015$.
The change in temperature is $\Delta T = 24\,^{\circ}C$.
Substituting these values,we get $\gamma = \frac{\Delta V}{V \Delta T} = \frac{0.15}{100 \times 24} = \frac{0.0015}{24} = 6.25 \times 10^{-5} /^{\circ}C$.
For a solid,the coefficient of volume expansion $\gamma$ is related to the coefficient of linear expansion $\alpha$ by the relation $\gamma = 3\alpha$.
Therefore,$\alpha = \frac{\gamma}{3} = \frac{6.25 \times 10^{-5}}{3} \approx 2.08 \times 10^{-5} /^{\circ}C$.
Rounding to the nearest provided option,we get $\alpha = 2.0 \times 10^{-5} /^{\circ}C$.

(A) When a solid containing a hole is heated,the material expands in all directions. The expansion of the material follows the same rules as if the hole were filled with the same material. Since the atoms move further apart due to thermal expansion,the linear dimensions of the object,including the diameter of the hole,increase. Therefore,the diameter of the hole will always increase.

(A) When a metal sheet with holes is heated,the material expands as if the entire sheet were solid. The expansion of the material follows the same laws as the expansion of a solid body. Therefore,the distance between any two points on the sheet increases upon heating. Since $A$,$B$,and $C$ are points on the metal sheet,the distances $AB$ (the diameter of the hole) and $BC$ (the distance between the two holes) will both increase due to thermal expansion.