Question

A circular disc of radius R and thickness $\dfrac{R}{6}$ has moment of inertia I about an axis passing through its Centre perpendicular to its plane. It is melted and recast into a solid sphere. The moment of inertia of the sphere about its diameter is
A. $I$
B. $\dfrac{{2I}}{8}$
C. $\dfrac{I}{5}$
D. $\dfrac{I}{{10}}$

Answer 1

Hint: Moment of inertia is a resisting property of a rigid body, when the body tends to rotate. It is defined as a summation of mass of particles multiplied by square of its distance from the axis.

Complete step by step solution:
As we know that if an object of any shape is melted and recast into another shape, then its volume remains the same(or say conserved),.
For Circular Disc,
\[Radius{\text{ }}of{\text{ }}circular{\text{ }}disc\; = \;R\] ,
\[Thickness\;of\;circular{\text{ }}disc{\text{ }} = \dfrac{R}{6}\] ,
$Volume$ \[of{\text{ }}circular{\text{ }}disc{\text{ }} = \pi {R^2} \times \dfrac{R}{6}\] .
For Solid Sphere,
Let radius of solid sphere be r,
\[Volume{\text{ }}of{\text{ }}solid{\text{ }}sphere{\text{ }} = \;\dfrac{{4\pi {r^3}}}{3}\] ,
\[Mass{\text{ }}of{\text{ }}circular{\text{ }}disc{\text{ }} = {\text{ }}Mass{\text{ }}of{\text{ }}solid{\text{ }}sphere\]

As volume remains conserved,
\[Volume{\text{ }}of{\text{ }}circular{\text{ }}disc{\text{ }} = {\text{ }}Volume{\text{ }}of{\text{ }}solid{\text{ }}sphere\]
$\pi {R^2} \times \dfrac{R}{6} = \dfrac{{4\pi {r^3}}}{3}$
After solving it, you will get
$r = \dfrac{R}{2}\left( {equation \to 1} \right)$
Till now, we have find the radius of solid sphere,
Let’s move to our last step, that is to find a moment of inertia of the solid sphere.
\[Moment{\text{ }}of{\text{ }}inertia{\text{ }}of{\text{ }}circular{\text{ }}disc = I\left( {given} \right)\]
$\dfrac{{M{R^2}}}{2} = I$
\[M = \dfrac{{2 \times I}}{{{R^2}}}\left( {equation \to 2} \right)\]
Moment of inertia of solid sphere about its diameter =$\dfrac{{2M{r^2}}}{5}$
Now, put the values of M and r from equation 1 and equation 2.
$ = \dfrac{{2 \times 2 \times I}}{{5 \times {R^2}}} \times \dfrac{{{R^2}}}{4}$
$ = \dfrac{I}{5}$

$\therefore $ Option(C) is correct.

Note: Moment of inertia of different rigid bodies is different.
Volume before melting of an object remains equal to Volume after melting.