Question

A thin brass sheet at $10^{\circ} C$, and a thin steel sheet at $20^{\circ} C$ have the same surface area. The common temperature at which both would have the same area is (Coefficient of linear expansion for brass and steel are $19 \times 10^{-60} C ^{-1}$ and $11 \times 10^{-6 \circ} C ^{-1}$ respectively)

• A. $-3.75^{\circ} C$
• B. $-2.75^{\circ} C$
• C. $+2.75^{\circ} C$
• D. $+3.75^{\circ} C$

Answer 1

Answer: (A)

The area of brass sheet at $10^{\circ} C =l_{1} b_{1}$
The area of steel sheet at $20^{\circ} C =l_{2} b_{2}$
Given that $l_{1} b_{1}=l_{2} b_{2}$
Let at temperature $T^{\circ} C$ the areas of brass and steel sheets be same.
Length of brass sheet at $T^{\circ} C =l_{1}\left[1+\alpha_{b} \Delta T_{b}\right]$
Length of steel sheet at $T^{\circ} C =l_{2}\left[1+\alpha_{s} \Delta T_{s}\right]$
Breadth of brass sheet at $T^{\circ} C =b_{1}\left[1+\alpha_{b} \Delta T_{b}\right]$
Breadth of steel sheet at $T^{\circ} C =b_{2}\left[1+\alpha_{s} \Delta T_{s}\right]$
According to problem,
$ l_{1}\left[1+\alpha_{b} \Delta T_{b}\right] \times b_{1}\left[1+\alpha_{b} \Delta T_{b}\right]$
$=l_{2}\left[1+\alpha_{s} \Delta T_{s}\right] \times b_{2}\left[1+\alpha_{s} \Delta T_{s}\right] $
or $ l_{1} b_{1}\left[1+\alpha_{b} \Delta T_{b}\right]^{2}=l_{2} b_{2}\left[1+\alpha_{s} \Delta T_{s}\right]^{2} $
or $ l_{1} b_{1}\left[1+19 \times 10^{-6} \times(10-T)\right]^{2}=l_{2} b_{2}\left[1+11 \times 10^{-6}(20-T)\right]^{2} $
or $ 19 \times 10^{-6}(10-T)=11 \times 10^{-6}(20-T) $
$ \therefore T=-\frac{30}{8}=-3.75^{\circ} C $