Question

Moment of inertia of solid sphere about its diameter is I. If that sphere is recast into 8 identical small spheres, then the moment of inertia of a small sphere about its diameter is:
(A) $\dfrac{I}{8}$
(B) $\dfrac{I}{{16}}$
(C) $\dfrac{I}{{24}}$
(D) $\dfrac{I}{{32}}$

Answer 1

Moment of Inertia (M.I.) of the solid sphere along its diameter is $I = \dfrac{{2M{R^2}}}{5}$.As this sphere is recast into 8 smaller spheres hence the mass of smaller spheres is $\dfrac{M}{8}$. As the material of both the materials is the same thus density remains the same.
Formula used
Moment of Inertia (M.I.) of the solid sphere along its diameter is $I = \dfrac{{2M{R^2}}}{5}$
$\rho = \dfrac{M}{V}$ where$\rho$ is the density, $M$ is the mass, $V$ is the volume.
$V = \dfrac{{4\pi {R^3}}}{3}$ Where $R$is the radius and $V$ is the volume.

Complete step by step solution:
Let Mass and radius of the bigger sphere be $M$ and$R$.
So the moment of inertia is $I = \dfrac{{2M{R^2}}}{5}$
As this sphere is recast into 8 smaller spheres hence the mass of smaller spheres is $\dfrac{M}{8}$ and let radius be $r$ .
As the material of both the materials is the same thus density remains the same from this we can calculate the radius r of the new smaller sphere formed.
From,
($\rho$ is the density, $M$ is the mass, $V$ is the volume)
$\rho = \dfrac{M}{V}$
Volume of a sphere of radius R is ${V_R} = \dfrac{{4\pi {R^3}}}{3}$
For a sphere of mass $M$ and radius$R$,
$\rho = \dfrac{M}{{\dfrac{{4\pi {R^3}}}{3}}}$
Volume of a sphere of radius r is ${V_r}= \dfrac{{4\pi {r^3}}}{3}$
For a sphere of mass $\dfrac{M}{8}$ and radius $r$
$\rho = \dfrac{{\dfrac{M}{8}}}{{\dfrac{{4\pi {r^3}}}{3}}}$
From the above two equation and as both spheres have density we can assert that,
$\rho = \dfrac{M}{{\dfrac{{4\pi {R^3}}}{3}}} = \dfrac{{\dfrac{M}{8}}}{{\dfrac{{4\pi {r^3}}}{3}}}$
$\Rightarrow \dfrac{M}{{\dfrac{{4\pi {R^3}}}{3}}} = \dfrac{{\dfrac{M}{8}}}{{\dfrac{{4\pi {r^3}}}{3}}}$
$\Rightarrow {r^3} = \dfrac{{{R^3}}}{8}$
$\Rightarrow r = \dfrac{R}{2}$

As the moment of inertia of solid sphere along its diameter is $I = \dfrac{{2M{R^2}}}{5}$

So the moment of inertia of the smaller sphere whose mass (M) is $\dfrac{M}{8}$ and radius(R) is$\dfrac{R}{2}$
${I_{smaller-sphere}}= \dfrac{{2 \times \dfrac{M}{8} \times {{(\dfrac{R}{2})}^2}}}{5}$
$\Rightarrow {I_{smaller-sphere}}= \dfrac{{2M{R^2}}}{{5 \times 32}}$
As $I = \dfrac{{2M{R^2}}}{5}$
$\Rightarrow {I_{smaller-sphere}} = \dfrac{I}{{32}}$

Hence the answer to this question is (D) $\dfrac{I}{{32}}$

Note:
Always remember that $I = \dfrac{{2M{R^2}}}{5}$is the moment of inertia of solid sphere along its diameter and not $I = \dfrac{{2M{R^2}}}{3}$ which is the moment of inertia of hollow sphere along its diameter also be careful about the mentioned axis about which the moment of inertia is being written these small checks while attempting a question can save you from silly mistakes in the exam.