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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 $\left.\alpha_{1} t< < 1, \alpha_{2} t< < 1\right)$
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Verified Answer
The correct answer is:
$\alpha_{1}+\alpha_{2}$
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,
$$
\begin{aligned}
S_{t} &=L_{r} \times B_{r} \\
&=L_{0} b_{0}\left(1+\alpha_{1} \Delta t\right)\left(1+\alpha_{2} \Delta t\right) \\
&=S_{0}\left(1+\alpha_{1} \Delta t+\alpha_{2} \Delta t+\ldots\right) \\
S_{t} &=S_{0}(1+\beta \Delta t) \\
\therefore S_{0}(1+\beta \Delta t) &=S_{0}\left(1+\alpha_{1} \Delta t+\alpha_{2} \Delta t+\ldots\right) \\
\beta \cdot \Delta t &=\alpha_{1} \Delta t+\alpha_{2} \Delta t \\
\beta &=\alpha_{1}+\alpha_{2}
\end{aligned}
$$
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,
$$
\begin{aligned}
S_{t} &=L_{r} \times B_{r} \\
&=L_{0} b_{0}\left(1+\alpha_{1} \Delta t\right)\left(1+\alpha_{2} \Delta t\right) \\
&=S_{0}\left(1+\alpha_{1} \Delta t+\alpha_{2} \Delta t+\ldots\right) \\
S_{t} &=S_{0}(1+\beta \Delta t) \\
\therefore S_{0}(1+\beta \Delta t) &=S_{0}\left(1+\alpha_{1} \Delta t+\alpha_{2} \Delta t+\ldots\right) \\
\beta \cdot \Delta t &=\alpha_{1} \Delta t+\alpha_{2} \Delta t \\
\beta &=\alpha_{1}+\alpha_{2}
\end{aligned}
$$
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