Question

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

• A. 0.1
• B. 0.2
• C. 0.3
• D. 0.4

Answer 1

Answer: (D)

0.4

Moment of inertia $=m{r}^2$
Now, percentage of change of moment of inertia is given by $\frac{I_2-I_1}{I_1}\times 100$
Now if $r_2=$ increased radius and $r_1=$ initial radius, then
$r_2=r_1(1+\alpha dt)$
Now, $dI=I_2-I_1=m\left(2\alpha dt\right){r_1}^2$, where $m=$ mass of the body
So, $\frac{dI}{I_1}=2\alpha dt$
$\frac{dI}{I_1}\times 100=200\times \alpha \times dt$
$=0.4$