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 $\left.\alpha_{1} t< < 1, \alpha_{2} t< < 1\right)$

• A

  $\frac{\alpha_{1}+\alpha_{2}}{2}$
• B

  $2\left(\alpha_{1}+\alpha_{2}\right)$
• C

  $\frac{4 \alpha_{1} \alpha_{2}}{\alpha_{1}+\alpha_{2}}$
• Correct answer
  D

  $\alpha_{1}+\alpha_{2}$

Solution

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

Hence, the correct answer is option: 'D'.