Question

The coefficient of linear expansion of an inhomogeneous rod changes linearly from $\alpha_1$ to $\alpha_2$ from one end to other end of the rod. The effective coefficient of linear expansion of rod is :-

• A. $\alpha_1 + \alpha_2$
• B. $\frac{\alpha_1 + \alpha_2}{2}$
• C. $\sqrt{\alpha_1 \alpha_2}$
• D. $\alpha_1 - \alpha_2$

Answer 1

Answer: (B)

$\frac{\alpha_1 + \alpha_2}{2}$

Answer 2

The coefficient of linear expansion $\alpha$ varies linearly along the length $L$ of the rod from $\alpha_1$ to $\alpha_2$. Let $\alpha(x)$ be the coefficient at a distance $x$ from one end. We can express $\alpha(x)$ as: $\alpha(x) = \alpha_1 + \frac{\alpha_2 - \alpha_1}{L} x$

The change in length $d(\Delta L)$ of a small segment $dx$ due to a temperature change $\Delta T$ is given by $d(\Delta L) = \alpha(x) dx \Delta T$. The total change in length of the rod, $\Delta L$, is the integral of $d(\Delta L)$ over the entire length of the rod: $\Delta L = \int_0^L \alpha(x) dx \Delta T$

The effective coefficient of linear expansion, $\alpha_{eff}$, is defined by the relation $\Delta L = \alpha_{eff} L \Delta T$. Equating the two expressions for $\Delta L$: $\alpha_{eff} L \Delta T = \int_0^L \alpha(x) dx \Delta T$ $\alpha_{eff} = \frac{1}{L} \int_0^L \alpha(x) dx$

Substituting the expression for $\alpha(x)$: $\alpha_{eff} = \frac{1}{L} \int_0^L \left( \alpha_1 + \frac{\alpha_2 - \alpha_1}{L} x \right) dx$ $\alpha_{eff} = \frac{1}{L} \left[ \alpha_1 x + \frac{\alpha_2 - \alpha_1}{L} \frac{x^2}{2} \right]_0^L$ $\alpha_{eff} = \frac{1}{L} \left( \alpha_1 L + \frac{\alpha_2 - \alpha_1}{L} \frac{L^2}{2} \right)$ $\alpha_{eff} = \frac{1}{L} \left( \alpha_1 L + \frac{(\alpha_2 - \alpha_1)L}{2} \right)$ $\alpha_{eff} = \alpha_1 + \frac{\alpha_2 - \alpha_1}{2}$ $\alpha_{eff} = \frac{2\alpha_1 + \alpha_2 - \alpha_1}{2}$ $\alpha_{eff} = \frac{\alpha_1 + \alpha_2}{2}$

Alternatively, for a quantity that varies linearly over an interval, its average value is the arithmetic mean of its values at the endpoints. Since $\alpha(x)$ varies linearly from $\alpha_1$ to $\alpha_2$, its average value over the length $L$ is $\frac{\alpha_1 + \alpha_2}{2}$. This average value is the effective coefficient of linear expansion.