Question

A sphere made of iron is rotating about its diameter as axis, $\alpha=1 \times 10^{-5}{ }^{\circ} C ^{-1}$. If the temperature rises by $100^{\circ} C ,$ the percentage increase in its moment of inertia is

• A. $0.1 \%$
• B. $0.2 \%$
• C. $0.5 \%$
• D. $0.002 \%$

Answer 1

Answer: (B)

For a solid sphere $I=\frac{2}{5} M R^{2}$
$\frac{\Delta R}{R}=\alpha \Delta T=\left(1 \times 10^{-5}{ }^{\circ} C ^{-1}\right)\left(100^{\circ} C \right)=10^{-3}{ }^{\circ} C ^{-1}$
As $I \propto R^{2}$
$\frac{\Delta I}{I}=\frac{2 \Delta R}{R}$
$=2 \times 10^{-3}=2 \times 10^{-3} \times 100 \%=0.2 \%$