Question

A solid rectangular sheet has two different coefficients of linear expansion $\alpha_{1}$ and $\alpha_{2}$ along its length and breadth respectively. The coefficient of surface expansion is (for $\alpha_{1} t < < 1, \alpha_{2} t < < 1$)

• A. $\frac{\alpha_{1}+\alpha_{2}}{2}$
• B. $2\left(\alpha_{1}+\alpha_{2}\right)$
• C. $\frac{4\alpha_{1}\alpha_{2}}{\alpha _{1}+\alpha _{2}}$
• D. $\alpha_{1}+\alpha_{2}$

Answer 1

The Correct Option is D

The coefficient of linear expansion along its length $= \alpha_1$

The coefficient of linear expansion along its breadth $=\alpha_{2}$
Increase in length,
$L_{t}=l_{0}\left(1+\alpha_{1} \Delta t\right)$
Increase in breadth,
$B_{ t }=b_{0}\left(1+\alpha_{2} \Delta t_{2}\right)$
Let coefficient of surface expansion is $\beta$
Area = length $ \times$ breadth
$=l_{0}\left(1+\alpha_{1} \Delta t\right) \times b_{0}\left(1+\alpha_{2} \Delta t\right) $
$=l_{0} b_{0}\left(1+\alpha_{1} \Delta t\right)\left(1+\alpha_{2} \Delta t\right) $
$=S_{0}\left(1+\alpha_{1} \Delta t+\alpha_{2} \Delta t+\ldots\right)$
where, $S_{0}=l_{0} \cdot b_{0}$
$=$ Initial area of surface
In state of expansion,
$S_{t} =L_{t} \times B_{t} $
$=l_{0} b_{0}\left(1+\alpha_{1} \Delta t\right)\left(1+\alpha_{2} \Delta t\right) $
$=S_{0}\left(1+\alpha_{1} \Delta t+\alpha_{2} \Delta t+\ldots\right) $
$S_{t} =S_{0}(1+\beta \Delta t) $
$\therefore S_{0}(1+\beta \Delta t) =S_{0}\left(1+\alpha_{1} \Delta t+\alpha_{2} \Delta t+\ldots\right)$
$\beta \cdot \Delta t =\alpha_{1} \Delta t+\alpha_{2} \Delta t$
$\beta =\alpha_{1}+\alpha_{2}$