Q. A metallic solid sphere is routing about its diameter as axis of rotation. If the temperature is increased by 200$℃$, the percentage in its moment of inertia is (Coefficient of linear expansion of the metal=$10^{- 5}$ $℃^{- 1}$)
A) 0.1%
B) 0.2%
C) 0.3%
D) 0.4%
Explanation
(d) The moment of inertia of a solid sphere about the axis along its diameter is
$I = \frac{2}{5}\ mR^{2}\ \Rightarrow I \propto R^{2}$
$\therefore$ $\frac{\mathrm{\Delta}I}{I} \times 100 = 2\left\lbrack \frac{\mathrm{\Delta}R}{R} \right\rbrack 100$
But $\alpha = \frac{\mathrm{\Delta}R}{R \times \mathrm{\Delta}t\ }\ \Rightarrow \frac{\mathrm{\Delta}R}{R} = \ \alpha\mathrm{\Delta}t$
$\therefore$ $\frac{\mathrm{\Delta}I}{I} \times 100 = 2(\alpha)(\mathrm{\Delta}t)100$
= $2\left( 10^{- 5} \right)(200)(100)$
= 0.4%
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.