Question:
A solid sphere of radius \(R\) has its outer half removed, so that its radius becomes \(\left(\dfrac{R}{2}\right)\). Then its moment of inertia about the diameter is
A solid sphere of radius \(R\) has its outer half removed, so that its radius becomes \(\left(\dfrac{R}{2}\right)\). Then its moment of inertia about the diameter is
Show Hint
For a solid sphere,
\[
I\propto MR^2
\]
and since \(M\propto R^3\), overall
\[
I\propto R^5
\]
So halving the radius reduces moment of inertia by
\[
\left(\frac12\right)^5=\frac{1}{32}
\]
Updated On: Jun 22, 2026
- becomes \(\dfrac{1}{2}\) of its initial value.
- is unchanged.
- becomes \(\dfrac{1}{16}\) of initial value.
- becomes \(\dfrac{1}{32}\) of initial value.
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The Correct Option is D
Solution and Explanation
Step 1: Write the formula for moment of inertia of a solid sphere.
The moment of inertia of a solid sphere about its diameter is \[ I=\frac{2}{5}MR^2 \] where \(M\) is the mass of the sphere and \(R\) is its radius.
Step 2: Find how the mass changes when radius becomes \(\dfrac{R}{2}\).
Mass of a sphere is proportional to its volume.
Since volume of a sphere is \[ V=\frac{4}{3}\pi R^3, \] mass is proportional to \[ R^3 \] If the new radius becomes \[ R'=\frac{R}{2}, \] then the new mass becomes \[ M'=M\left(\frac{R'}{R}\right)^3 \] \[ =M\left(\frac{1}{2}\right)^3 \] \[ =\frac{M}{8} \]
Step 3: Find the new moment of inertia.
New moment of inertia is \[ I'=\frac{2}{5}M'R'^2 \] Substitute \[ M'=\frac{M}{8} \] and \[ R'=\frac{R}{2} \] \[ I' = \frac{2}{5}\left(\frac{M}{8}\right)\left(\frac{R}{2}\right)^2 \] \[ = \frac{2}{5}\cdot \frac{M}{8}\cdot \frac{R^2}{4} \] \[ = \frac{2MR^2}{160} \] \[ = \frac{1}{32}\left(\frac{2}{5}MR^2\right) \] Since \[ I=\frac{2}{5}MR^2, \] we get \[ I'=\frac{I}{32} \]
Step 4: Final conclusion.
Hence, the moment of inertia becomes \[ \boxed{\frac{1}{32}} \] of its initial value.
The moment of inertia of a solid sphere about its diameter is \[ I=\frac{2}{5}MR^2 \] where \(M\) is the mass of the sphere and \(R\) is its radius.
Step 2: Find how the mass changes when radius becomes \(\dfrac{R}{2}\).
Mass of a sphere is proportional to its volume.
Since volume of a sphere is \[ V=\frac{4}{3}\pi R^3, \] mass is proportional to \[ R^3 \] If the new radius becomes \[ R'=\frac{R}{2}, \] then the new mass becomes \[ M'=M\left(\frac{R'}{R}\right)^3 \] \[ =M\left(\frac{1}{2}\right)^3 \] \[ =\frac{M}{8} \]
Step 3: Find the new moment of inertia.
New moment of inertia is \[ I'=\frac{2}{5}M'R'^2 \] Substitute \[ M'=\frac{M}{8} \] and \[ R'=\frac{R}{2} \] \[ I' = \frac{2}{5}\left(\frac{M}{8}\right)\left(\frac{R}{2}\right)^2 \] \[ = \frac{2}{5}\cdot \frac{M}{8}\cdot \frac{R^2}{4} \] \[ = \frac{2MR^2}{160} \] \[ = \frac{1}{32}\left(\frac{2}{5}MR^2\right) \] Since \[ I=\frac{2}{5}MR^2, \] we get \[ I'=\frac{I}{32} \]
Step 4: Final conclusion.
Hence, the moment of inertia becomes \[ \boxed{\frac{1}{32}} \] of its initial value.
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