Thermal Expansion for Solid Questions in English

(C) When an isotropic solid is heated,it undergoes thermal expansion in all dimensions. This expansion is similar to a photographic enlargement,where every linear dimension of the object increases.
$1$. The radius $r$ of the ring increases due to linear expansion.
$2$. The gap width $x$ also increases because the entire circumference of the ring expands,and the gap behaves like a segment of the circumference.
$3$. The thickness $d$ of the rod increases because the material itself expands in all directions.
Therefore,$x$,$r$,and $d$ all increase.

(A) The moment of inertia of a rod about an axis passing through its center and perpendicular to its length is given by:
$I = \frac{1}{12} ML^2$ --- $(i)$
where $M$ is the mass and $L$ is the length of the rod.
Since the mass $M$ remains constant during thermal expansion,we differentiate the expression with respect to $L$:
$\Delta I = \frac{1}{12} M (2L \Delta L) = 2 (\frac{1}{12} ML^2) \frac{\Delta L}{L} = 2I \frac{\Delta L}{L}$ --- $(ii)$
Dividing equation $(ii)$ by equation $(i)$,we get:
$\frac{\Delta I}{I} = 2 \frac{\Delta L}{L}$ --- $(iii)$
We know that the change in length due to thermal expansion is given by $\Delta L = L \alpha \Delta t$,which implies:
$\frac{\Delta L}{L} = \alpha \Delta t$
Substituting this into equation $(iii)$,we get:
$\frac{\Delta I}{I} = 2 \alpha \Delta t$

(A) bimetallic strip consists of two metal strips with different coefficients of linear thermal expansion $(\alpha)$ fastened together.
When the strip is heated, the metal with the higher coefficient of linear thermal expansion $(\alpha)$ expands more than the metal with the lower coefficient.
Because the two metals are bonded together, the strip must bend to accommodate the difference in lengths.
The side with the higher expansion coefficient becomes the outer arc, and the side with the lower expansion coefficient becomes the inner arc.
Therefore, the strip bends towards the metal with the lower linear thermal expansion coefficient.

(B) The coefficient of linear expansion $\alpha$ determines how much a material changes its length with temperature change,given by $\Delta L = L_0 \alpha \Delta T$.
Since the coefficient of linear expansion of metal $X$ is greater than that of metal $Y$ $(\alpha_X > \alpha_Y)$,metal $X$ will contract more than metal $Y$ when the temperature decreases (cooling).
Because metal $X$ is on the left side and it contracts more than metal $Y$,the strip will bend towards the side that contracts more.
Therefore,the bimetallic strip will bend towards the left.

(A) The coefficient of cubical expansion $\gamma$ for an anisotropic solid is given by the sum of the coefficients of linear expansion along three mutually perpendicular axes.
Given:
$\alpha_1 = 13 \times 10^{-7} \ K^{-1}$
$\alpha_2 = 231 \times 10^{-7} \ K^{-1}$
$\alpha_3 = 231 \times 10^{-7} \ K^{-1}$
Formula:
$\gamma = \alpha_1 + \alpha_2 + \alpha_3$
Calculation:
$\gamma = (13 + 231 + 231) \times 10^{-7} \ K^{-1}$
$\gamma = 475 \times 10^{-7} \ K^{-1}$

(D) The problem states that the increase in length for both rods is independent of the increase in temperature,which implies that the change in length $(\Delta \ell)$ for both rods must be equal for any given change in temperature $(\Delta T)$.
Given:
$\ell_{Cu} = 88\; cm$
$\alpha_{Cu} = 1.7 \times 10^{-5}\; K^{-1}$
$\alpha_{Al} = 2.2 \times 10^{-5}\; K^{-1}$
The formula for thermal expansion is $\Delta \ell = \ell \alpha \Delta T$.
Since $(\Delta \ell)_{Cu} = (\Delta \ell)_{Al}$,we have:
$\ell_{Cu} \alpha_{Cu} \Delta T = \ell_{Al} \alpha_{Al} \Delta T$
Canceling $\Delta T$ from both sides:
$\ell_{Cu} \alpha_{Cu} = \ell_{Al} \alpha_{Al}$
Substituting the values:
$88 \times 1.7 \times 10^{-5} = \ell_{Al} \times 2.2 \times 10^{-5}$
Solving for $\ell_{Al}$:
$\ell_{Al} = \frac{88 \times 1.7}{2.2}$
$\ell_{Al} = 40 \times 1.7 = 68\; cm$.

(D) For a non-isotropic solid,the coefficient of volume expansion $\gamma$ is the sum of the coefficients of linear expansion along the three mutually perpendicular axes:
$\gamma = \alpha_{x} + \alpha_{y} + \alpha_{z}$
Given:
$\alpha_{x} = 5 \times 10^{-5} /^{\circ} C = 50 \times 10^{-6} /^{\circ} C$
$\alpha_{y} = 5 \times 10^{-6} /^{\circ} C$
$\alpha_{z} = 5 \times 10^{-6} /^{\circ} C$
Substituting these values into the formula:
$\gamma = (50 \times 10^{-6} + 5 \times 10^{-6} + 5 \times 10^{-6}) /^{\circ} C$
$\gamma = (50 + 5 + 5) \times 10^{-6} /^{\circ} C$
$\gamma = 60 \times 10^{-6} /^{\circ} C$
Comparing this with the given form $C \times 10^{-6} /^{\circ} C$,we get $C = 60$.

(N/A) Consider a rectangular sheet of a solid material of length $a$ and breadth $b$. When the temperature increases by $\Delta T$,the length $a$ increases by $\Delta a = \alpha_{l} a \Delta T$ and the breadth $b$ increases by $\Delta b = \alpha_{l} b \Delta T$.
The increase in area $\Delta A$ is given by the sum of the three shaded regions in the diagram:
$\Delta A = \Delta A_{1} + \Delta A_{2} + \Delta A_{3}$
$\Delta A = a \Delta b + b \Delta a + (\Delta a)(\Delta b)$
Substituting the expressions for $\Delta a$ and $\Delta b$:
$\Delta A = a(\alpha_{l} b \Delta T) + b(\alpha_{l} a \Delta T) + (\alpha_{l} a \Delta T)(\alpha_{l} b \Delta T)$
$\Delta A = 2 \alpha_{l} ab \Delta T + \alpha_{l}^{2} ab (\Delta T)^{2}$
Since $A = ab$,we have:
$\Delta A = 2 \alpha_{l} A \Delta T + \alpha_{l}^{2} A (\Delta T)^{2}$
$\Delta A = \alpha_{l} A \Delta T (2 + \alpha_{l} \Delta T)$
Dividing by $A \Delta T$:
$\frac{\Delta A}{A \Delta T} = \alpha_{l} (2 + \alpha_{l} \Delta T)$
Since $\alpha_{l}$ is very small (typically $\approx 10^{-5} \text{ K}^{-1}$),the term $\alpha_{l} \Delta T$ is negligible compared to $2$. Therefore:
$\frac{\Delta A}{A \Delta T} \approx 2 \alpha_{l}$

(B) Given:
Initial temperature $T_{1} = 27^{\circ}C$
Initial length (diameter) of the iron ring $L_{T1} = 5.231\; m$
Final length (diameter) required $L_{T2} = 5.243\; m$
Coefficient of linear expansion $\alpha = 1.20 \times 10^{-5} K^{-1}$
Using the formula for linear expansion:
$L_{T2} = L_{T1} [1 + \alpha(T_{2} - T_{1})]$
Substituting the values:
$5.243 = 5.231 [1 + 1.20 \times 10^{-5} (T_{2} - 27)]$
Divide by $5.231$:
$1.002294 = 1 + 1.20 \times 10^{-5} (T_{2} - 27)$
$0.002294 = 1.20 \times 10^{-5} (T_{2} - 27)$
$T_{2} - 27 = \frac{0.002294}{1.20 \times 10^{-5}}$
$T_{2} - 27 \approx 191.17$
$T_{2} \approx 191.17 + 27 = 218.17^{\circ}C$
Rounding to the nearest integer,the temperature is $218^{\circ}C$.

(N/A) The length of the steel tape at $T = 27.0 \; ^{\circ}C$ is $l = 100 \; cm$.
At $T_1 = 45.0 \; ^{\circ}C$,the length of the tape $l'$ increases due to thermal expansion:
$l' = l(1 + \alpha \Delta T) = 100 \times [1 + 1.20 \times 10^{-5} \times (45.0 - 27.0)]$
$l' = 100 \times [1 + 1.20 \times 10^{-5} \times 18] = 100 \times [1 + 0.000216] = 100.0216 \; cm$.
The measured length of the rod is $63.0 \; cm$ using the expanded tape. The actual length $l_2$ at $45.0 \; ^{\circ}C$ is:
$l_2 = \frac{l'}{l} \times 63.0 = \frac{100.0216}{100} \times 63.0 = 63.0136 \; cm$.
To find the length at $27.0 \; ^{\circ}C$,we use $l_2 = l_0(1 + \alpha \Delta T)$:
$63.0136 = l_0(1 + 1.20 \times 10^{-5} \times 18) = l_0(1.000216)$
$l_0 = \frac{63.0136}{1.000216} \approx 63.0 \; cm$.

(C) Initial temperature $T = 27^{\circ}C = 300\;K$.
Initial diameter of the shaft $d_1 = 8.70\;cm$.
Diameter of the hole in the wheel $d_2 = 8.69\;cm$.
Coefficient of linear expansion $\alpha = 1.20 \times 10^{-5}\;K^{-1}$.
The wheel will slip on the shaft when the diameter of the shaft becomes equal to the diameter of the hole,i.e.,$d_2 = 8.69\;cm$.
The change in diameter is $\Delta d = d_2 - d_1 = 8.69 - 8.70 = -0.01\;cm$.
Using the formula for linear expansion $\Delta d = d_1 \alpha \Delta T$,where $\Delta T = T_f - T_i$:
$-0.01 = 8.70 \times (1.20 \times 10^{-5}) \times (T_f - 300)$.
$T_f - 300 = \frac{-0.01}{8.70 \times 1.20 \times 10^{-5}} = \frac{-0.01}{0.0001044} \approx -95.78\;K$.
$T_f = 300 - 95.78 = 204.22\;K$.
Converting to Celsius: $T_f(^{\circ}C) = 204.22 - 273.15 \approx -68.93^{\circ}C$.
Rounding to the nearest integer,the temperature is $-69^{\circ}C$.

(D) Initial temperature $T_{1} = 27.0 \; ^{\circ}C$.
Initial diameter $d_{1} = 4.24 \; cm$.
Final temperature $T_{2} = 227 \; ^{\circ}C$.
Change in temperature $\Delta T = T_{2} - T_{1} = 227 - 27 = 200 \; K$.
Coefficient of linear expansion $\alpha = 1.70 \times 10^{-5} \; K^{-1}$.
Thermal expansion of a hole in a material follows the same rule as the expansion of a solid disc of the same material.
The change in diameter $\Delta d$ is given by the formula $\Delta d = d_{1} \alpha \Delta T$.
Substituting the values:
$\Delta d = 4.24 \times (1.70 \times 10^{-5}) \times 200$.
$\Delta d = 4.24 \times 3.40 \times 10^{-3}$.
$\Delta d = 14.416 \times 10^{-3} \; cm$.
$\Delta d = 1.4416 \times 10^{-2} \; cm \approx 1.44 \times 10^{-2} \; cm$.

(N/A) Initial temperature,$T_{1} = 40.0\; ^{\circ}C$
Final temperature,$T_{2} = 250\; ^{\circ}C$
Change in temperature,$\Delta T = T_{2} - T_{1} = 210\; ^{\circ}C$
Length of each rod,$l = 50\; cm$
Coefficient of linear expansion of brass,$\alpha_{b} = 2.0 \times 10^{-5}\; K^{-1}$
Coefficient of linear expansion of steel,$\alpha_{s} = 1.2 \times 10^{-5}\; K^{-1}$
Change in length of brass rod,$\Delta l_{b} = l \alpha_{b} \Delta T = 50 \times (2.0 \times 10^{-5}) \times 210 = 0.21\; cm$
Change in length of steel rod,$\Delta l_{s} = l \alpha_{s} \Delta T = 50 \times (1.2 \times 10^{-5}) \times 210 = 0.126\; cm$
Total change in length of the combined rod,$\Delta l = \Delta l_{b} + \Delta l_{s} = 0.21 + 0.126 = 0.336\; cm$
Since the ends of the rod are free to expand,there is no constraint on the expansion,and therefore no 'thermal stress' is developed at the junction.

(N/A) Let the lengths of the sides be $AB = l_1$,$BC = l_3$,and $AC = l_2$. Initially,$l_1 = l_2 = l_3 = l$.
Using the law of cosines for $\angle ABC = \theta$:
$\cos \theta = \frac{l_1^2 + l_3^2 - l_2^2}{2 l_1 l_3}$
$2 l_1 l_3 \cos \theta = l_1^2 + l_3^2 - l_2^2$
Taking the differential of both sides:
$2(l_3 dl_1 + l_1 dl_3) \cos \theta - 2 l_1 l_3 \sin \theta d\theta = 2 l_1 dl_1 + 2 l_3 dl_3 - 2 l_2 dl_2$
Since $AB$ and $BC$ are copper $(\alpha_1)$ and $AC$ is aluminium $(\alpha_2)$:
$dl_1 = l_1 \alpha_1 \Delta T$,$dl_3 = l_3 \alpha_1 \Delta T$,$dl_2 = l_2 \alpha_2 \Delta T$
Substituting these into the equation and using $l_1 = l_2 = l_3 = l$ and $\theta = 60^\circ$:
$2(l^2 \alpha_1 \Delta T + l^2 \alpha_1 \Delta T) \cos 60^\circ - 2 l^2 \sin 60^\circ d\theta = 2 l^2 \alpha_1 \Delta T + 2 l^2 \alpha_1 \Delta T - 2 l^2 \alpha_2 \Delta T$
$4 l^2 \alpha_1 \Delta T (1/2) - 2 l^2 (\sqrt{3}/2) d\theta = 4 l^2 \alpha_1 \Delta T - 2 l^2 \alpha_2 \Delta T$
$2 l^2 \alpha_1 \Delta T - \sqrt{3} l^2 d\theta = 4 l^2 \alpha_1 \Delta T - 2 l^2 \alpha_2 \Delta T$
$-\sqrt{3} d\theta = 2 l^2 \alpha_1 \Delta T - 2 l^2 \alpha_2 \Delta T$
$d\theta = \frac{2(\alpha_2 - \alpha_1) \Delta T}{\sqrt{3}}$

(N/A) Thermal expansion is the phenomenon where the dimensions of a body increase due to an increase in its temperature. When a substance is heated,its particles gain kinetic energy and vibrate more vigorously,leading to an increase in the average separation between them,which results in macroscopic expansion.
There are three types of thermal expansion:
$(a)$ Linear expansion: The expansion in the length of a solid is called linear expansion.
$(b)$ Area expansion: The expansion in the surface area of a solid is called area expansion.
$(c)$ Volume expansion: The expansion in the volume of a solid or fluid is called volume expansion.
These are illustrated as follows:
$(a)$ Linear expansion: $\frac{\Delta l}{l} = \alpha_l \Delta T$
$(b)$ Area expansion: $\frac{\Delta A}{A} = 2\alpha_l \Delta T$
$(c)$ Volume expansion: $\frac{\Delta V}{V} = 3\alpha_l \Delta T$

Linear expansion refers to the increase in the length of a solid material when its temperature is increased.
The increase in length $(\Delta l)$ is directly proportional to the original length $(l)$ and the change in temperature $(\Delta T)$.
$\Delta l \propto l$ and $\Delta l \propto \Delta T$
Combining these,we get $\Delta l \propto l \Delta T$.
$\frac{\Delta l}{l} = \alpha_{l} \Delta T$,where $\alpha_{l}$ is the coefficient of linear expansion.
Rearranging,$\Delta l = \alpha_{l} l \Delta T$.
The coefficient $\alpha_{l}$ is a characteristic property of the material and depends on the type of material.
The unit of $\alpha_{l}$ is $(^{\circ}C)^{-1}$ or $K^{-1}$.
If $l_{1}$ is the initial length at temperature $T_{1}$ and $l_{2}$ is the final length at temperature $T_{2}$,then:
$l_{2} - l_{1} = \alpha_{l} l_{1} (T_{2} - T_{1})$
$l_{2} = l_{1} [1 + \alpha_{l} (T_{2} - T_{1})]$

(A) Consider a cube of side length $l$. When its temperature is increased by $\Delta T$,it expands equally in all dimensions.
From the volume formula,$V = l^3$.
The change in volume is given by $\Delta V = (l + \Delta l)^3 - l^3$.
Expanding this,we get $\Delta V = l^3 + 3l^2 \Delta l + 3l(\Delta l)^2 + (\Delta l)^3 - l^3$.
Since $\Delta l$ is very small,$(\Delta l)^2$ and $(\Delta l)^3$ are negligible. Thus,$\Delta V \approx 3l^2 \Delta l$ ... $(1)$.
From the definition of linear expansion,$\Delta l = \alpha_l l \Delta T$ ... $(2)$.
Substituting equation $(2)$ into equation $(1)$:
$\Delta V = 3l^2 (\alpha_l l \Delta T) = 3l^3 \alpha_l \Delta T$.
Since $V = l^3$,we have $\Delta V = 3 V \alpha_l \Delta T$ ... $(3)$.
Comparing equation $(3)$ with the general equation for volume expansion,$\Delta V = \alpha_V V \Delta T$,we get:
$\alpha_V = 3 \alpha_l$.

(B) When a rod is fixed rigidly at both ends,it cannot expand freely upon heating.
As the temperature increases,the rod attempts to expand by $\Delta l = l \alpha \Delta T$. Since the supports are rigid,they exert an inward force on the rod,preventing this expansion.
This constraint results in the development of compressive strain and,consequently,thermal stress within the material. The thermal stress is given by $\sigma = Y \alpha \Delta T$.
If the internal compressive force exceeds the structural limit of the rod or its supports,the rod may buckle or bend.
For example,if a steel rod of cross-sectional area $40 \text{ cm}^2$ is constrained,a temperature change of $10^{\circ}C$ produces a compressive strain of $\epsilon = \alpha \Delta T = 1.2 \times 10^{-5} \times 10 = 1.2 \times 10^{-4}$.
The resulting force $F = Y A \epsilon = (2 \times 10^{11} \text{ Pa}) \times (40 \times 10^{-4} \text{ m}^2) \times (1.2 \times 10^{-4}) = 96,000 \text{ N}$,which is sufficient to cause bending or buckling in many structures.

(N/A) Areal expansion (or superficial expansion) is the increase in the surface area of a solid body due to an increase in its temperature.
Definition of coefficient of areal expansion $(\beta)$: It is defined as the fractional change in area per unit change in temperature.
Mathematically,$\beta = \frac{\Delta A}{A_0 \Delta T}$,where $\Delta A$ is the change in area,$A_0$ is the initial area,and $\Delta T$ is the change in temperature.
Unit: The $SI$ unit of the coefficient of areal expansion is $\text{K}^{-1}$ or $^\circ\text{C}^{-1}$.

(A) The coefficient of linear expansion $\alpha_l$ is defined by the relation $\Delta L = L_0 \alpha_l \Delta T$,where $\Delta L$ is the change in length,$L_0$ is the original length,and $\Delta T$ is the change in temperature.
From this,$\alpha_l = \frac{\Delta L}{L_0 \Delta T}$.
$\alpha_l$ depends solely on the nature of the material of the substance.
Its $SI$ unit is per Kelvin $(K^{-1})$ or per degree Celsius $(^{\circ}C^{-1})$.

(N/A) The coefficient of linear expansion $(\alpha_l)$ is defined as the fractional change in length per unit change in temperature.
The coefficient of area expansion $(\alpha_A)$ is defined as the fractional change in area per unit change in temperature,which is approximately $2\alpha_l$.
The coefficient of volume expansion $(\alpha_V)$ is defined as the fractional change in volume per unit change in temperature,which is approximately $3\alpha_l$.
Therefore,the relationship between them is given by the ratio: $\alpha_l : \alpha_A : \alpha_V = 1 : 2 : 3$.

(C) The coefficient of volume expansion $(\alpha_V)$ for a solid is defined as the fractional change in volume per unit change in temperature,i.e.,$\alpha_V = \frac{1}{V} \frac{dV}{dT}$.
For most solids,$\alpha_V$ is not constant but depends on temperature.
At very low temperatures (near $0 \ K$),$\alpha_V$ is very small and increases with temperature.
As the temperature increases,$\alpha_V$ rises and eventually approaches a constant value at high temperatures.
Therefore,the graph of $\alpha_V$ versus $T$ is a curve that starts from a low value and increases,eventually becoming nearly constant at high temperatures.

(N/A) Let a solid cube have a side length $L$ at temperature $T$. The initial volume is $V = L^3$.
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 volume $V'$ is given by $V' = (L')^3 = [L(1 + \alpha \Delta T)]^3$.
Using the binomial expansion $(1 + x)^n \approx 1 + nx$ for small $x$,we get $V' \approx L^3(1 + 3\alpha \Delta T) = V(1 + 3\alpha \Delta T)$.
The coefficient of volume expansion $\gamma$ is defined by the relation $V' = V(1 + \gamma \Delta T)$.
Comparing the two expressions,we find that $\gamma = 3\alpha$.

(B) The density $\rho$ of a substance is defined as the ratio of its mass $m$ to its volume $V$,given by the formula $\rho = \frac{m}{V}$.
When the temperature of a solid substance increases,its atoms vibrate with greater amplitude,causing the average distance between them to increase. This phenomenon is known as thermal expansion.
As a result of thermal expansion,the volume $V$ of the solid increases.
Since the mass $m$ of the substance remains constant during this process,the increase in volume $V$ leads to a decrease in density $\rho$ because $\rho$ is inversely proportional to $V$.

(C) The coefficient of linear expansion,denoted by $\alpha$,is a property that characterizes how a material changes its length with temperature.
It depends primarily on the nature of the material (the atomic structure and bonding) and the temperature range over which the expansion is measured.
While it is often treated as a constant for small temperature changes,it is technically a function of temperature.

(N/A) The formula for linear thermal expansion is $\Delta l = \alpha l \Delta T$,where $\alpha$ is the coefficient of linear expansion.
From the first observation:
$\alpha = \frac{\Delta l}{l \Delta T} = \frac{4 \times 10^{-4}}{2 \times 10} = 2 \times 10^{-5} \, ^{\circ}C^{-1}$.
Now,we check the other observations using $\alpha = 2 \times 10^{-5} \, ^{\circ}C^{-1}$:
For observation $(2)$:
$\Delta l = \alpha l \Delta T = (2 \times 10^{-5}) \times 1 \times 10 = 2 \times 10^{-4} \, m$.
Since the recorded value is $4 \times 10^{-4} \, m$,observation $(2)$ is incorrect.
For observation $(3)$:
$\Delta l = \alpha l \Delta T = (2 \times 10^{-5}) \times 2 \times 20 = 8 \times 10^{-4} \, m$.
Since the recorded value is $2 \times 10^{-4} \, m$,observation $(3)$ is incorrect.
For observation $(4)$:
$\Delta l = \alpha l \Delta T = (2 \times 10^{-5}) \times 3 \times 10 = 6 \times 10^{-4} \, m$.
Since the recorded value is $6 \times 10^{-4} \, m$,observation $(4)$ is correct.

(N/A) Let $l_{\text{iron}}$ and $l_{\text{brass}}$ be the lengths of the iron and brass strips respectively.
According to the problem,the difference in their lengths must remain constant at $10 \ cm$ for any temperature change $\Delta T$.
So,$l_{\text{iron}} - l_{\text{brass}} = 10 \ cm \ldots (1)$
When temperature changes by $\Delta T$,the new lengths are $l_{\text{iron}}' = l_{\text{iron}}(1 + \alpha_{\text{iron}} \Delta T)$ and $l_{\text{brass}}' = l_{\text{brass}}(1 + \alpha_{\text{brass}} \Delta T)$.
For the difference to be constant,$l_{\text{iron}}' - l_{\text{brass}}' = l_{\text{iron}} - l_{\text{brass}}$.
This implies $l_{\text{iron}} \alpha_{\text{iron}} \Delta T = l_{\text{brass}} \alpha_{\text{brass}} \Delta T$.
Therefore,$l_{\text{iron}} \alpha_{\text{iron}} = l_{\text{brass}} \alpha_{\text{brass}}$.
$\frac{l_{\text{iron}}}{l_{\text{brass}}} = \frac{\alpha_{\text{brass}}}{\alpha_{\text{iron}}} = \frac{1.8 \times 10^{-5}}{1.2 \times 10^{-5}} = \frac{3}{2}$.
So,$l_{\text{iron}} = 1.5 \ l_{\text{brass}}$.
Substituting this into equation $(1)$:
$1.5 \ l_{\text{brass}} - l_{\text{brass}} = 10 \ cm$.
$0.5 \ l_{\text{brass}} = 10 \ cm \implies l_{\text{brass}} = 20 \ cm$.
Then,$l_{\text{iron}} = 1.5 \times 20 \ cm = 30 \ cm$.
Thus,the lengths should be $30 \ cm$ for iron and $20 \ cm$ for brass.

(N/A) Let $V_{i,0}$ and $V_{b,0}$ be the volumes of iron and brass at $0^{\circ} C$ respectively.
Let $V_i$ and $V_b$ be the volumes of iron and brass at temperature $\Delta T^{\circ} C$ respectively.
Let $\gamma_i$ and $\gamma_b$ be the coefficients of volume expansion for iron and brass respectively.
The net volume of the vessel is $V_c = V_i - V_b = 100 \, cc$.
For the volume to remain constant with temperature,the change in volume must be zero:
$\Delta V_c = \Delta V_i - \Delta V_b = 0$
$V_{i,0} \gamma_i \Delta T - V_{b,0} \gamma_b \Delta T = 0$
$V_{i,0} \gamma_i = V_{b,0} \gamma_b$
$\frac{V_{i,0}}{V_{b,0}} = \frac{\gamma_b}{\gamma_i} = \frac{6 \times 10^{-5}}{3.55 \times 10^{-5}} = \frac{6}{3.55} \approx 1.69$
Given $V_{i,0} - V_{b,0} = 100 \, cc$,we substitute $V_{i,0} = \frac{6}{3.55} V_{b,0}$:
$\left( \frac{6}{3.55} - 1 \right) V_{b,0} = 100$
$\left( \frac{6 - 3.55}{3.55} \right) V_{b,0} = 100$
$\frac{2.45}{3.55} V_{b,0} = 100$
$V_{b,0} = \frac{355}{2.45} \approx 144.9 \, cc$
$V_{i,0} = 100 + 144.9 = 244.9 \, cc$.

(N/A) By using the Pythagoras theorem for the given right-angled triangle:
$\left(\frac{L+\Delta L}{2}\right)^{2} = \left(\frac{L}{2}\right)^{2} + x^{2}$
$x = \sqrt{\left(\frac{L+\Delta L}{2}\right)^{2} - \left(\frac{L}{2}\right)^{2}}$
$x = \frac{1}{2} \sqrt{(L+\Delta L)^{2} - L^{2}}$
$x = \frac{1}{2} \sqrt{L^{2} + 2L\Delta L + (\Delta L)^{2} - L^{2}}$
$x = \frac{1}{2} \sqrt{2L\Delta L + (\Delta L)^{2}}$
Since $\Delta L$ is very small,$(\Delta L)^{2}$ can be neglected.
$x \approx \frac{1}{2} \sqrt{2L\Delta L}$
Given $\Delta L = L \alpha \Delta T$,we substitute this into the equation:
$x = \frac{1}{2} \sqrt{2L(L \alpha \Delta T)} = \frac{L}{2} \sqrt{2 \alpha \Delta T}$
Substituting the values $L = 10\,m$,$\alpha = 1.2 \times 10^{-5} \,^oC^{-1}$,and $\Delta T = 20\,^oC$:
$x = \frac{10}{2} \sqrt{2 \times 1.2 \times 10^{-5} \times 20}$
$x = 5 \sqrt{48 \times 10^{-5}} = 5 \sqrt{4.8 \times 10^{-4}}$
$x = 5 \times 2.19 \times 10^{-2} \approx 0.1095\,m \approx 11\,cm$.

(N/A) In thermal steady state,the temperature gradient along the rod is constant. The temperature varies linearly from $\theta_1$ to $\theta_2$. The average temperature of the rod is $\theta_{avg} = \frac{\theta_1 + \theta_2}{2}$.
The change in length $\Delta L$ is given by $\Delta L = L_0 \alpha \Delta T$,where $\Delta T$ is the change in temperature from the reference temperature $(0\,^{\circ}C)$.
Thus,the new length $L$ is:
$L = L_0(1 + \alpha \theta_{avg})$
$L = L_0 \left( 1 + \alpha \left( \frac{\theta_1 + \theta_2}{2} \right) \right)$

(B) Given that the fractional change in length is $\frac{\Delta L}{L} = 0.02 \% = 2 \times 10^{-4}$.
Since $\Delta L = L \alpha \Delta T$,we have $\alpha \Delta T = 2 \times 10^{-4}$.
The volume $V$ of the wire is $V = A \times L$,where $A$ is the cross-sectional area.
The density $\rho$ is given by $\rho = \frac{M}{V} = \frac{M}{AL}$.
Taking the logarithmic derivative,we get $\frac{\Delta \rho}{\rho} = - (\frac{\Delta A}{A} + \frac{\Delta L}{L})$.
Since $\frac{\Delta A}{A} = 2 \alpha \Delta T$ and $\frac{\Delta L}{L} = \alpha \Delta T$,the fractional change in density is $\frac{\Delta \rho}{\rho} = -(2 \alpha \Delta T + \alpha \Delta T) = -3 \alpha \Delta T$.
Substituting $\alpha \Delta T = 0.02 \%$,we get $\frac{\Delta \rho}{\rho} = -3 \times 0.02 \% = -0.06 \%$.
The magnitude of the percentage change is $0.06 \%$.

(D) At temperature $T$,the total length is $L = L_{1} + L_{2}$.
When the temperature increases to $T + \Delta T$,the new lengths are:
$L_{1}' = L_{1}(1 + \alpha_{1} \Delta T)$
$L_{2}' = L_{2}(1 + \alpha_{2} \Delta T)$
The total new length $L_{eq}'$ is the sum of the individual new lengths:
$L_{eq}' = L_{1}' + L_{2}' = L_{1}(1 + \alpha_{1} \Delta T) + L_{2}(1 + \alpha_{2} \Delta T)$
$L_{eq}' = L_{1} + L_{1} \alpha_{1} \Delta T + L_{2} + L_{2} \alpha_{2} \Delta T$
$L_{eq}' = (L_{1} + L_{2}) + (L_{1} \alpha_{1} + L_{2} \alpha_{2}) \Delta T$
For the equivalent system with effective coefficient $\alpha_{avg}$,the new length is:
$L_{eq}' = (L_{1} + L_{2})(1 + \alpha_{avg} \Delta T) = (L_{1} + L_{2}) + (L_{1} + L_{2}) \alpha_{avg} \Delta T$
Equating the two expressions for $L_{eq}'$:
$(L_{1} + L_{2}) \alpha_{avg} \Delta T = (L_{1} \alpha_{1} + L_{2} \alpha_{2}) \Delta T$
$\alpha_{avg} = \frac{L_{1} \alpha_{1} + L_{2} \alpha_{2}}{L_{1} + L_{2}}$

(A) The real coefficient of volume expansion of mercury $(\gamma_{\text{real}})$ is constant regardless of the vessel used.
The relationship between real expansion, apparent expansion, and vessel expansion is given by:
$\gamma_{\text{real}} = \gamma_{\text{app}} + \gamma_{\text{vessel}}$
Since $\gamma_{\text{real}}$ is the same for both cases:
$(\gamma_{\text{app}} + \gamma_{\text{vessel}})_{\text{glass}} = (\gamma_{\text{app}} + \gamma_{\text{vessel}})_{\text{steel}}$
We know that the volume expansion coefficient of a solid is $\gamma_{\text{vessel}} = 3\alpha$.
For steel:
$\gamma_{\text{vessel, steel}} = 3 \times (12 \times 10^{-6} /{ }^{\circ} C) = 36 \times 10^{-6} /{ }^{\circ} C$
Substituting the values into the equation:
$153 \times 10^{-6} + \gamma_{\text{vessel, glass}} = 144 \times 10^{-6} + 36 \times 10^{-6}$
$153 \times 10^{-6} + \gamma_{\text{vessel, glass}} = 180 \times 10^{-6}$
$\gamma_{\text{vessel, glass}} = (180 - 153) \times 10^{-6} = 27 \times 10^{-6} /{ }^{\circ} C$
Since $\gamma_{\text{vessel, glass}} = 3\alpha_{\text{glass}}$:
$3\alpha_{\text{glass}} = 27 \times 10^{-6} /{ }^{\circ} C$
$\alpha_{\text{glass}} = 9 \times 10^{-6} /{ }^{\circ} C$

(A) First,we calculate the coefficient of linear expansion $(\alpha)$:
$\alpha = \frac{\Delta L}{L_{0} \Delta \theta}$
$\alpha = \frac{0.19 \; cm}{100 \; cm \times (100^{\circ} C - 0^{\circ} C)}$
$\alpha = \frac{0.19}{100 \times 100} = \frac{0.19}{10000} = 1.9 \times 10^{-5} /{ }^{\circ} C$
The coefficient of cubical expansion $(\gamma)$ is related to the coefficient of linear expansion by the formula $\gamma = 3\alpha$.
$\gamma = 3 \times (1.9 \times 10^{-5} /{ }^{\circ} C)$
$\gamma = 5.7 \times 10^{-5} /{ }^{\circ} C$

(D) The change in length of a metal strip is given by $\Delta L = L \alpha \Delta T$,where $\alpha$ is the coefficient of linear expansion and $\Delta T$ is the change in temperature.
Since the strip is placed in a cold bath,the temperature decreases,so $\Delta T$ is negative,meaning both metals will contract.
Given that $\alpha_{A} > \alpha_{B}$,the contraction in metal $A$ will be greater than the contraction in metal $B$ (i.e.,$|\Delta L_{A}| > |\Delta L_{B}|$).
Since metal $A$ is on the left and it contracts more than metal $B$,the strip will bend towards the side that contracts more,which is the left side.

(A) The initial volume of the cubic box is $V = a^{3}$.
The coefficient of volume expansion $\gamma$ is related to the coefficient of linear expansion $\alpha$ by the relation $\gamma = 3\alpha$.
The change in volume $\Delta V$ for a temperature change $\Delta T$ is given by the formula $\Delta V = V \gamma \Delta T$.
Substituting the values of $V$ and $\gamma$ into the formula,we get:
$\Delta V = a^{3} \times (3\alpha) \times \Delta T$.
Therefore,the increase in the volume of the metal box is $\Delta V = 3 a^{3} \alpha \Delta T$.

(A) Assertion $A$ is true: When a rod is heated freely,it expands without any external constraint. Since there is no constraint to prevent expansion,no internal restoring force or thermal stress is generated.
Reason $R$ is true: Thermal expansion causes the length of the rod to increase upon heating,which is a physical fact.
However,Reason $R$ does not explain why no thermal stress is developed. The absence of thermal stress is due to the lack of external constraints (freedom to expand),not simply because the length increases. Therefore,both $A$ and $R$ are true,but $R$ is not the correct explanation of $A$.

(B) The elastic potential energy stored per unit volume is given by $u = \frac{1}{2} \times \text{Young's Modulus} \times (\text{strain})^2$.
Since the track is constrained,the thermal strain is $\text{strain} = \alpha \Delta T$.
Given $\alpha = 10^{-5} /^{\circ}C$ and $\Delta T = 10^{\circ}C$,the strain is $\text{strain} = 10^{-5} \times 10 = 10^{-4}$.
The energy stored per unit length is $U = u \times \text{Area} = \frac{1}{2} Y (\text{strain})^2 \times A$.
Substituting the values: $U = \frac{1}{2} \times 10^{11} \times (10^{-4})^2 \times 0.01$.
$U = 0.5 \times 10^{11} \times 10^{-8} \times 10^{-2} = 0.5 \times 10^1 = 5\, J/m$.

(B) The increase in volume is given by $\Delta V = \gamma V_{0} \Delta T$.
Since $\gamma = 3\alpha$,we have $\Delta V = (3\alpha) V_{0} \Delta T$.
The total surface area of a cube is $6a^{2}$,where $a$ is the side length.
Given $6a^{2} = 24 \; m^{2}$,so $a^{2} = 4 \; m^{2}$,which means $a = 2 \; m$.
The initial volume $V_{0} = a^{3} = (2)^{3} = 8 \; m^{3}$.
Substituting the values: $\Delta V = (3 \times 5.0 \times 10^{-4} \; ^{\circ}C^{-1}) \times (8 \; m^{3}) \times (10 \; ^{\circ}C)$.
$\Delta V = 15 \times 10^{-4} \times 80 = 1200 \times 10^{-4} = 0.12 \; m^{3}$.
Since $1 \; m^{3} = 10^{6} \; cm^{3}$,then $\Delta V = 0.12 \times 10^{6} \; cm^{3} = 1.2 \times 10^{5} \; cm^{3}$.

(D) The change in length (diameter) required is $\Delta L = L_2 - L_1 = 6.241 \,cm - 6.230 \,cm = 0.011 \,cm$.
The formula for linear thermal expansion is $\Delta L = L_1 \alpha_L \Delta T$,where $\Delta T = T_f - T_i$.
Substituting the given values: $0.011 = 6.230 \times (1.4 \times 10^{-5}) \times (T_f - 27)$.
Solving for $\Delta T$: $\Delta T = \frac{0.011}{6.230 \times 1.4 \times 10^{-5}} = \frac{0.011 \times 10^5}{8.722} \approx 126.11^{\circ} C$.
Therefore,the final temperature $T_f = 27 + 126.11 = 153.11^{\circ} C$.
Comparing with the given options,the closest value is $152.7^{\circ} C$.

(C) Let $\ell_{B}$ be the length of brass and $\ell_{i}$ be the length of iron.
The condition is that the difference between their lengths remains constant with temperature change $\Delta T$:
$\ell_{B}(1 + \alpha_{B} \Delta T) - \ell_{i}(1 + \alpha_{i} \Delta T) = \ell_{B} - \ell_{i}$
Expanding the terms:
$\ell_{B} + \ell_{B} \alpha_{B} \Delta T - \ell_{i} - \ell_{i} \alpha_{i} \Delta T = \ell_{B} - \ell_{i}$
This simplifies to:
$\ell_{B} \alpha_{B} \Delta T = \ell_{i} \alpha_{i} \Delta T$
Dividing by $\Delta T$:
$\ell_{B} \alpha_{B} = \ell_{i} \alpha_{i}$
Given $\ell_{B} = 40\,cm$,$\alpha_{B} = 1.8 \times 10^{-5} K^{-1}$,and $\alpha_{i} = 1.2 \times 10^{-5} K^{-1}$:
$40 \times 1.8 \times 10^{-5} = \ell_{i} \times 1.2 \times 10^{-5}$
Solving for $\ell_{i}$:
$\ell_{i} = \frac{40 \times 1.8}{1.2} = \frac{40 \times 3}{2} = 60\,cm$.

(B) The potential energy $U$ of two atoms in a solid as a function of their separation $r$ is represented by an asymmetric curve.
As the temperature of the solid increases,the vibrational energy of the atoms increases.
Due to the asymmetry of the potential energy curve,the atoms vibrate with larger amplitudes at higher energies.
Because the curve is steeper for $r < r_0$ (where $r_0$ is the equilibrium separation) and shallower for $r > r_0$,the mean separation between the atoms increases as the total energy increases.
This increase in the mean separation of atoms manifests as the macroscopic thermal expansion of the solid.

(A) The correct answer is $(A)$.
The potential energy $U$ versus interatomic separation $r$ plot for two atoms in a solid is asymmetric about the equilibrium position.
As the temperature increases,the total energy of the atoms increases $(E_3 > E_2 > E_1)$.
Due to the asymmetry of the potential energy curve,the mean separation between the atoms increases as the energy increases $(r_3 > r_2 > r_1)$.
Consequently,crystalline solids generally expand upon heating.

(A) The thermal expansion of a solid with a hole or a gap behaves as if the hole or gap were made of the same material as the solid.
When the temperature of the ring increases by $\Delta T$,every linear dimension of the ring,including the gap width $d$,expands according to the linear expansion formula $\Delta L = L \alpha \Delta T$.
Therefore,the change in the width of the gap is given by $\Delta d = d \alpha \Delta T$.
Since $\Delta T > 0$,the width of the gap increases by $d \alpha \Delta T$.

(B) The area $A$ of a circular coin is given by $A = \pi r^2$,where $r$ is the radius.
Since the diameter $D = 2r$,the percentage change in diameter is the same as the percentage change in radius: $\frac{\Delta D}{D} \times 100 = \frac{\Delta r}{r} \times 100 = 0.15 \%$.
The area is $A = \pi r^2$. Using the concept of relative error (or differentials),the fractional change in area is given by $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.
To find the percentage increase in area:
$\frac{\Delta A}{A} \times 100 = 2 \times (\frac{\Delta r}{r} \times 100)$
$= 2 \times 0.15 \% = 0.30 \%$.

(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 (from $C \leftrightarrow D$ to $E \leftrightarrow F$).
Due to the asymmetry of the potential energy curve,the average equilibrium positions ($P_2$ and $P_3$) of the molecules are displaced outward. 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) When a solid containing a hole is heated,the material expands in all directions,including the material surrounding the hole.
This results in the hole expanding as if the entire sheet were made of the same material.
Mathematically,the change in diameter is given by $\Delta D = D \alpha \Delta T$,where $\alpha$ is the coefficient of linear expansion for copper.
Since $\Delta T = 35^{\circ} C - 27.0^{\circ} C = 8.0^{\circ} C$ is positive,the diameter $D$ will increase.
Therefore,the new diameter will be more than $4.24 \,cm$.

(D) Let the initial length be $L$ and the initial radius be $r$. The initial area of the cross-section is $A = \pi r^2$.
When the cylinder is heated, the length increases by $3 \%$, so the new length is $L' = L(1 + 0.03)$.
Since the material is uniform, the linear expansion coefficient $\alpha$ is constant. The change in length is $\Delta L = L \alpha \Delta T$, where $\Delta L / L = 0.03$.
The area of the cross-section depends on the radius $r$. The change in radius is $\Delta r = r \alpha \Delta T$.
The new area is $A' = \pi (r + \Delta r)^2 = \pi (r^2 + 2r \Delta r + (\Delta r)^2)$.
Neglecting the higher-order term $(\Delta r)^2$, we get $A' \approx \pi r^2 + 2 \pi r \Delta r = A + 2A (\Delta r / r)$.
The fractional change in area is $\Delta A / A = 2 (\Delta r / r) = 2 \alpha \Delta T$.
Since $\Delta L / L = \alpha \Delta T = 0.03$, we have $\Delta A / A = 2 \times 0.03 = 0.06$.
Thus, the area of the cross-section increases by $6 \%$.

(C) When a metallic object is heated,it undergoes thermal expansion. The expansion of a solid is similar to a photographic enlargement,meaning that all linear dimensions of the object increase proportionally.
For a circular disc with a cavity,the material expands in all directions. The outer radius $R$ increases because the circumference expands. Similarly,the inner radius $r$ also increases because the material surrounding the cavity expands outward,effectively pushing the boundary of the cavity further away from the center.
Therefore,both $R$ and $r$ increase upon heating.

(D) The rod is placed on a frictionless horizontal surface and is not constrained or fixed at its ends.
When the temperature of the rod is increased,it undergoes thermal expansion.
Since there are no external constraints (like walls) to prevent this expansion,the rod expands freely.
Because the rod expands freely without any resistance,no thermal stress is developed within the material.
Consequently,the tension (or compressive force) developed in the copper rod is $0 \,N$.