Question

A ring of mass m and radius r is melted and then molded into a sphere. Then the moment of inertia of the sphere will be
A. More than that of the ring
B. Less than that of the ring
C. Equal to that of the ring
D. None of these.

Answer 1

Recall the moment of inertia of ring and sphere. After converting the ring into a sphere, the volume and mass of the sphere will be the same as that of the ring. Take the ratio of moment of inertia of sphere and ring to answer the question.

Complete step by step answer: We have given that the ring of mass m and r is converted into a sphere. Therefore, the volume and mass of the sphere will be the same as that of the ring.

We know the moment of inertia of the ring of radius r is,
${I_{ring}} = m{r^2}$ …… (1)

Here, m is the mass of the ring.

The moment of inertia of the sphere of same mass as that of the ring is,
${I_{sphere}} = \dfrac{2}{5}mr_1^2$ …… (2)

Here, ${r_1}$ is the radius of the sphere.

Divide equation (1) by equation (2).
$\dfrac{{{I_{sphere}}}}{{{I_{ring}}}} = \dfrac{{\dfrac{2}{5}mr_1^2}}{{m{r^2}}}$
$\Rightarrow {I_{sphere}} = \left( {\dfrac{2}{5}} \right)\dfrac{{r_1^2}}{{{r^2}}}{I_{ring}}$
After converting the same volume of ring into a sphere, the radius of the sphere will be very small than that of the ring. $r > > {r_1}$.
Therefore, the moment of inertia of the sphere will be very small as compared to the moment of inertia of the ring for the same volume.

So, the correct answer is option (B).

Note: Students should remember the moment of inertia for objects like ring, disk, rod, sphere etc. This will ease in answering such questions in no time. Since the moment of inertia for both ring and sphere is proportional to square of the radius, one could have directly answered the question after knowing the fact that the volume of the ring and sphere is the same.