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. becomes \(\dfrac{1}{2}\) of its initial value.
• B. is unchanged.
• C. becomes \(\dfrac{1}{16}\) of initial value.
• D. becomes \(\dfrac{1}{32}\) of initial value.

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 \[ \left(\frac12\right)^5=\frac{1}{32} \]

Answer 1

The Correct Option is D

Step 1: Recall the moment of inertia of a solid sphere.
About a diameter, a solid sphere of mass $M$ and radius $R$ has \[ I = \frac{2}{5}MR^2 \] So $I$ depends on both the mass and the square of the radius, and we must track how each changes.
Step 2: Note that mass scales with volume.
The sphere is made of the same material, so its mass is proportional to its volume, and volume of a sphere is $\dfrac{4}{3}\pi R^3 \propto R^3$. Thus $M \propto R^3$.
Step 3: Find the new mass when the radius is halved.
With $R' = \dfrac{R}{2}$, \[ M' = M\left(\frac{R'}{R}\right)^3 = M\left(\frac{1}{2}\right)^3 = \frac{M}{8} \]
Step 4: Write the new moment of inertia.
\[ I' = \frac{2}{5}M'R'^2 = \frac{2}{5}\left(\frac{M}{8}\right)\left(\frac{R}{2}\right)^2 \]
Step 5: Simplify the expression.
\[ I' = \frac{2}{5}\cdot\frac{M}{8}\cdot\frac{R^2}{4} = \frac{2}{5}MR^2 \cdot \frac{1}{32} = \frac{I}{32} \]
Step 6: State the result.
The new moment of inertia is $\dfrac{1}{32}$ of the original value, matching option (4). \[ \boxed{\dfrac{1}{32}\,I} \]