Question

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

• 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} $
$\Rightarrow \Delta T=\frac{\Delta L}{L_{0} \times \alpha}=\frac{3}{100 \times 2 \times 10^{-5}}$
$\Delta T =1.5 \times 10^{3}=1500^{\circ} C$