Question

What is the moment of inertia of a solid sphere of density $\rho $ and radius $R$ about its diameter?
$A)\text{ }\dfrac{105}{176}{{R}^{5}}\rho $
$B)\text{ }\dfrac{105}{176}{{R}^{2}}\rho $
$C)\text{ }\dfrac{176}{105}{{R}^{5}}\rho $
$D)\text{ }\dfrac{176}{105}{{R}^{2}}\rho $

Answer 1

Hint: This problem can be solved by using the direct formula for the moment of inertia of a solid sphere about its diameter in terms of its mass and radius. The mass of the sphere can then be written in terms of its density and volume, thereby getting the required value of the moment of inertia.

Formula used:
$I=\dfrac{2}{5}M{{R}^{2}}$
$M=\rho \times V$
$V=\dfrac{4}{3}\pi {{R}^{3}}$

Complete answer:
We will use the direct formula for the moment of inertia of a solid sphere about its diameter in terms of its mass and radius and then substitute the mass by the radius by using the relation between the mass, volume and density of a body.
The moment of inertia $I$ of a solid sphere of mass $M$ and radius $R$ about its diameter is given by
$I=\dfrac{2}{5}M{{R}^{2}}$ --(1)
The volume $V$ of a solid sphere of radius $R$ is given by
$V=\dfrac{4}{3}\pi {{R}^{3}}$ --(2)
The mass $M$, density $\rho $ and volume $V$ of a body are related as
$M=\rho \times V$ --(3)
Now, let us analyze the question.
Let the mass of the sphere be $M$.
The radius of the sphere is given to be $R$.
The density of the sphere is given to be $\rho $.
Let the volume of the sphere be $V$.
Let the moment of inertia of the solid sphere about its diameter be $I$.
Now, using (1), we get
$I=\dfrac{2}{5}M{{R}^{2}}$ --(4)
Also, using (3), we get
$M=\rho \times V$ --(5)
Now, using (2), we get
$V=\dfrac{4}{3}\pi {{R}^{3}}$ --(6)
Putting (6) in (5), we get
$M=\rho \times \dfrac{4}{3}\pi {{R}^{3}}$ --(7)
Putting (7) in (4), we get
$I=\dfrac{2}{5}\times \rho \times \dfrac{4}{3}\pi {{R}^{3}}\times {{R}^{2}}$
Putting $\pi =\dfrac{22}{7}$ in the above equation, we get
$I=\dfrac{2}{5}\times \rho \times \dfrac{4}{3}\times \dfrac{22}{7}\times {{R}^{3}}\times {{R}^{2}}=\dfrac{2\times 4\times 22}{5\times 3\times 7}\rho {{R}^{5}}=\dfrac{176}{105}\rho {{R}^{5}}$
Therefore, we have got the required value for the moment of inertia of a solid sphere about its diameter.

Therefore, the correct option is $C)\text{ }\dfrac{176}{105}{{R}^{5}}\rho $.

Note:
Students must carefully check about which axis the moment of inertia of the body has to be found out. This is because the moment of inertia of a body generally changes about different axes. For example, in this question, the axis about which the moment of inertia has to be found is the diameter of the sphere. If we had found the moment of inertia about an axis tangent to the sphere, then we would have got a different value for the moment of inertia and arrived at the wrong answer.