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(D) The moment of inertia of a solid sphere about its diameter is given by $I = \frac{2}{5} m R^2$.Since mass $m$ remains constant, $I \propto R^2$.Th

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$0.4$

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(D) The moment of inertia of a solid sphere about its diameter is given by $I = \frac{2}{5} m R^2$.Since mass $m$ remains constant, $I \propto R^2$.The fractional change in moment of inertia is given by $\frac{\Delta I}{I} = 2 \frac{\Delta R}{R}$.We know that the linear expansion is given by $\frac{\Delta R}{R} = \alpha \Delta t$, where $\alpha = 10^{-5} /^{\circ} C$ and $\Delta t = 200^{\circ} C$.Substituting these values, we get $\frac{\Delta I}{I} = 2 \alpha \Delta t$.To find the percentage increase, we multiply by $100$:$	ext{Percentage increase} = 2 	imes 10^{-5} 	imes 200 	imes 100 = 0.4 \%$.

$A$ metallic solid sphere is rotating about its diameter as an axis of rotation. If the temperature is increased by $200^{\circ} C$, the percentage increase in its moment of inertia is (Coefficient of linear expansion of the metal $= 10^{-5} /^{\circ} C$) (in $\%$)

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