Question

A thin brass sheet at ${10}^{\circ }C$ and a thin steel sheet at ${20}^{\circ }C$ have same surface area. The common temperature at which both would have the same area is :
Coefficient of areal expension for brass and steel are ${10\times {10}^{-6}}^{\circ }{C}^{-1}$ and ${20\times {10}^{-6}}^{\circ }{C}^{-1}$ respectively.

• A. ${10}^{\circ }C$
• B. ${20}^{\circ }C$
• C. ${30}^{\circ }C$
• D. ${35}^{\circ }C$

Answer 1

The correct option is C ${30}^{\circ }C$

We know ,
Final area $A=A_0(1+\beta \Delta T)$
where $A_0$ is initial area.

Given:
$A_0$ is same for both brass and steel sheet at ${10}^{\circ }C$ and ${20}^{\circ }C$ respectively.
Let at temperature $T$, both have same area $A$.
$A_0(1+{\beta }_b\Delta T_1)=A_0(1+{\beta }_s\Delta T_2)$
$A_0(1+{\beta }_b(T-10\left)\right)=A_0(1+{\beta }_s(T-20\left)\right)$
$\frac{{\beta }_b}{{\beta }_s}=\frac{T-20}{T-10}$
$\Rightarrow \frac{{10\times {10}^{-6}}^{\circ }{C}^{-1}}{{20\times {10}^{-6}}^{\circ }{C}^{-1}}=\frac{T-20}{T-10}$
$\Rightarrow T={30}^{\circ }C$