Question

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

• A. ${-3.75}^{o}C$
• B. ${-2.75}^{o}C$
• C. ${2.75}^{o}C$
• D. ${3.75}^{o}C$

Answer 1

The correct option is A ${-3.75}^{o}C$

Area expansion is given as:

$\frac{\Delta A}{A_o}={\alpha }_A\Delta T$

or $A_{t2}=A_{t1}[1+\alpha (t_2-t_1\left)\right]$

Where,$A_{t2}$ = area at temperature $t_2$

$A_{t1}$ = area at temperature $t_1$

$\alpha$ = area expansion coefficient

Given that at $t_2=t$, $A_{t2}$ of steel and brass are equal ($A_{t2}\left(brass\right)=A_{t2}\left(steel\right)$)

and ${\alpha }_{brass}={\alpha }_B=19\times {10}^{-6}{/}^{\circ }C$ and ${\alpha }_{steel}={\alpha }_S=11\times {10}^{-6}{/}^{\circ }C$, for brass $t_1\left(brass\right)=t_{1B}={10}^{\circ }C$ and for steel $t_1\left(steel\right)=t_{1S}={20}^{\circ }C$

Also $A_{t1\left(brass\right)}\left(brass\right)=A_{t1\left(steel\right)}\left(steel\right)=A$

Upon substitution we get

$A[1+{\alpha }_B(t-t_{1B}\left)\right]=A[1+{\alpha }_S(t-t_{1S}\left)\right]$

Or, ${\alpha }_B(t-t_{1B})={\alpha }_S(t-t_{1S})$

$19\times {10}^{-6}(t-10)=11\times {10}^{-6}(t-20)$

$8t=-30$

$t=-{3.75}^{\circ }C$

Therefore option A is correct.