Question

How much should the temperature of a brass rod be increased so as to increase its length $3\%$ ( $\alpha$ for brass is 2 $\times 10^{-5}/{}^{\circ }{C}^{-1})?$

• A. $1500^{\circ }C$
• B. $150^{\circ }C$
• C. $2000^{\circ }C$
• D. $2500^{\circ }C$

Answer 1

Answer: (A)

$\alpha =\frac{\Delta L}{L_0\Delta T}$
$\Rightarrow \Delta T=\frac{\Delta L}{L_0\times \alpha }=\frac{3}{100\times 2\times 10^{-5}}$
$\Delta T=1.5\times 10^3=1500^{\circ }C$