Question

A solid sphere of mass $ M $ and radius $ R $ cannot have moment of inertia
(A) $ \dfrac{2}{3}M{R^2} $
(B) $ \dfrac{1}{2}M{R^2} $
(C) $ \dfrac{2}{5}M{R^2} $
(D) $ \dfrac{2}{7}M{R^2} $

Answer 1

Hint: Moment of Inertia of a solid sphere along its diameter, which is the ais passing through its center of mass, is $ I = \dfrac{2}{5}M{R^2} $. Use the parallel axis theorem to get a relationship between the moment of inertia of a body about the axis passing through the center of mass and other axes.

Complete step by step solution:
We here have a solid sphere with mass $ M $ and radius $ R $. The moment of inertia of such a sphere about the axis passing through the center can be easily found to be $ {I_{CM}} = \dfrac{2}{5}M{R^2} $ .
Now according to the parallel axis theorem, for an axis through any other point $ P $ which is at a distance of $ d $ from the center of mass of the body, is given by $ I = {I_{CM}} + M{d^2} $.
We can now observe that the quantity $ M{d^2} $ is always positive and so we have the inequality $ I \geqslant {I_{CM}} $ for an axis through any point $ P $.
In other words, $ {I_{CM}} $ is the least possible inertia for a body. So, any possible inertia of the body should be greater than or equal to $ {I_{CM}} $ . As already mentioned, for a solid sphere $ {I_{CM}} = \dfrac{2}{5}M{R^2} $. So, we will now check this for each and every option and find the inertia that the body cannot possibly have.
 $ \dfrac{2}{3}M{R^2} > \dfrac{2}{5}M{R^2} = {I_{CM}} $. So, it can be a moment of inertia of the sphere.
 $ \dfrac{1}{2}M{R^2} > \dfrac{2}{5}M{R^2} = {I_{CM}} $. So, this value is also a possible value.
 $ \dfrac{2}{5}M{R^2} = {I_{CM}} $ . Here this value is the least possible moment of inertia for the body and so it surely can have such a value as its moment of inertia.
But here, $ \dfrac{2}{7}M{R^2} < \dfrac{2}{5}M{R^2} = {I_{CM}} $. So, this value cannot be a possible value for the moment of inertia of the body.
Hence, the correct option is D. $ \dfrac{2}{7}M{R^2} $ .

Note:
Remember that it is easy to confuse the moment of inertia of a solid sphere along its diameter $ I = \dfrac{2}{5}M{R^2} $ with that of a hollow sphere $ I = \dfrac{2}{3}M{R^2} $ . Always do not forget to take a brief look at the axis mentioned before proceeding with solving the problem.