Question

A solid sphere of mass $M$ and radius $R$ is divided into two unequal parts. The first part has a mass $\dfrac{7M}{8}$ and is converted into a uniform disc of radius $2R$. The second part is converted into a uniform solid sphere. Let ${{I}_{1}}$ be the moment of inertia of the disc about its axis and ${{I}_{2}}$ be the moment of inertia of the new sphere about its axis. The ratio ${{I}_{1}}/{{I}_{2}}$ is given by:
$A)\text{ }185$
$B)\text{ 6}5$
$C)\text{ 2}85$
$D)\text{ }140$

Answer 1

Hint: This problem can be solved by using the direct formula for the moment of inertia of a solid disc about the axis passing through its centre and perpendicular to its surface and the moment of inertia of a solid sphere about an axis passing through its centre. Since, the solid sphere of mass $M$ is divided to make the disc and the smaller sphere, the sum of their masses will be equal to the mass of the larger sphere.

Formula used:
${{I}_{disc}}=\dfrac{M{{R}^{2}}}{2}$
${{I}_{sphere}}=\dfrac{2}{5}M{{R}^{2}}$
${{V}_{sphere}}=\dfrac{4}{3}\pi {{R}^{3}}$

Complete step by step answer:
We will use the direct formula for the moments of inertia about the required axes for the sphere and the disc.
Now let us analyze the question.
The given mass of the bigger sphere is $M$.
The radius of the bigger sphere is $R$.
Let the volume of this sphere be $V$

Let the mass of the disc be ${{m}_{disc}}=\dfrac{7M}{8}$
The radius of the disc is ${{r}_{disc}}=2R$.
The moment of inertia of the disc about the axis passing through its centre and perpendicular to its surface is given to be ${{I}_{1}}$.

Let the mass of the smaller sphere be ${{m}_{sphere}}$.
Let the radius of the smaller sphere be ${{r}_{sphere}}$.
Let the volume of the smaller sphere be ${{V}_{sphere}}$.
The moment of inertia of the smaller sphere about an axis passing through its centre is given to be ${{I}_{2}}$.
Let the density of the material of which all of the bodies are made be $\rho $.
Now since the smaller sphere and the disc are made by dividing the bigger sphere, the sum of their masses must equal the mass of the bigger sphere.
$\therefore M={{m}_{disc}}+{{m}_{sphere}}$
$\therefore M=\dfrac{7M}{8}+{{m}_{sphere}}$
$\therefore {{m}_{sphere}}=M-\dfrac{7M}{8}=\dfrac{8-7}{8}M=\dfrac{M}{8}$ --(1)
Now, the volume $V$ of a sphere of radius $R$ is
$V=\dfrac{4}{3}\pi {{R}^{3}}$ --(2)
Using (2), we get
$V=\dfrac{4}{3}\pi {{R}^{3}}$ --(3)
${{V}_{sphere}}=\dfrac{4}{3}\pi {{r}_{sphere}}^{3}$ --(4)
Now, the density $\rho $ of an object is defined as the ratio of the total mass $M$ to the total volume $V$.
$\therefore \rho =\dfrac{M}{V}$ --(5)
Using (5), we get
$\therefore \rho =\dfrac{M}{V}$ --(6)
Also,
$\rho =\dfrac{{{m}_{sphere}}}{{{V}_{sphere}}}$ --(7)
Equating (6) and (7), we get
$\dfrac{M}{V}=\dfrac{{{m}_{sphere}}}{{{V}_{sphere}}}$ --(8)
Using (1), (3) and (4) in (8), we get,
$\dfrac{M}{\dfrac{4}{3}\pi {{R}^{3}}}=\dfrac{\dfrac{M}{8}}{\dfrac{4}{3}\pi {{r}_{sphere}}^{3}}$
$\therefore \dfrac{1}{{{R}^{3}}}=\dfrac{1}{8{{r}_{sphere}}^{3}}$
$\therefore {{r}_{sphere}}^{3}=\dfrac{1}{8}{{R}^{3}}$
Cube rooting both sides, we get
$\sqrt[3]{{{r}_{sphere}}^{3}}=\sqrt[3]{\dfrac{1}{8}{{R}^{3}}}$
$\therefore {{r}_{sphere}}=\dfrac{R}{2}$ --(9)
Now, the moment of inertia ${{I}_{disc}}$ of a disc of mass $M$ and radius $R$ about an axis passing through its centre and perpendicular to its surface is given by
${{I}_{disc}}=\dfrac{M{{R}^{2}}}{2}$ --(10)
Using (10), we get
${{I}_{1}}=\dfrac{\dfrac{7M}{8}{{\left( 2R \right)}^{2}}}{2}=\dfrac{\dfrac{7M}{8}4{{R}^{2}}}{2}=\dfrac{7}{4}M{{R}^{2}}$ --(11)
Now, the moment of inertia ${{I}_{sphere}}$ of a sphere of mass $M$ and radius $R$ about an axis passing through its centre is given by
${{I}_{sphere}}=\dfrac{2}{5}M{{R}^{2}}$ --(12)
Using (12), we get
${{I}_{2}}=\dfrac{2}{5}{{m}_{sphere}}{{r}_{sphere}}^{2}$
Using (1) and (9) in the above equation, we get
${{I}_{2}}=\dfrac{2}{5}\dfrac{M}{8}{{\left( \dfrac{R}{2} \right)}^{2}}=\dfrac{2}{5}\dfrac{M}{8}\left( \dfrac{{{R}^{2}}}{4} \right)=\dfrac{1}{80}M{{R}^{2}}$ --(13)
Using (11) and (13), we get
$\dfrac{{{I}_{1}}}{{{I}_{2}}}=\dfrac{\dfrac{7}{4}M{{R}^{2}}}{\dfrac{1}{80}M{{R}^{2}}}=20\left( 7 \right)=140$
Hence, the required ratio is $140:1$.
Therefore, the correct option is $D)\text{ }140$.

Note: Students must be careful of the axis about which the moment of inertia has to be found out. This is because the moment of inertia of a body generally changes about different axes of rotation. In this question also, normally the axis of a disc means the axis that passes through its centre and is perpendicular to the plane. If the student had found out the moment of inertia about some other axis, for example an axis passing through the centre and lying in the plane of the disc, then he or she would have arrived at the wrong answer.