Question

How much should the temperature of a brass rod be increased so as to increase its length by $1 \%$ ? (Given, $\alpha$ for brass is $0.00002^{\circ} C ^{-1}$ ).

• A. $300^{\circ} C$
• B. $400^{\circ} C$
• C. $500^{\circ} C$
• D. $550^{\circ} C$

Answer 1

Answer: (C)

Here, $\Delta T=?, \frac{\Delta L}{L}=\frac{1}{100}, \alpha=0.00002^{\circ} C ^{-1}$
As, $ \Delta L=\alpha L \Delta T$
$\therefore \alpha \Delta T =\frac{\Delta L}{L}$
or $\Delta T=\frac{\Delta L}{L \alpha}=\frac{1}{100 \times 0.00002} $
$\Delta T =\frac{10^{5}}{2 \times 10^{2}}=500^{\circ} C$
So, $500^{\circ} C$ of temperature of a brass rod should be increased, so as to increase its length by $1 \%$.