Question

To increase the length of brass rod by $2\%$ its temperature should increase by $\left(\alpha =0.00002{}^{\circ }{C}^{-1}\right)$

• A. $800{}^{\circ }C$
• B. $900{}^{\circ }C$
• C. $1000{}^{\circ }C$
• D. $1100{}^{\circ }C$

Answer 1

Answer: (C)

$\frac{\Delta L}{L}=2\%=\frac{2}{100}$
$\alpha =0.00002{}^{\circ }{C}^{-1}$
As $\Delta L=\alpha L\Delta T$
or $\Delta T=\frac{\Delta L}{\alpha L}=\frac{2}{100\times 0.00002}=\frac{1}{0.001}$
$=10^3=1000{}^{\circ }C$