Question:
If the moment of inertia of a solid sphere of mass M and radius R about its diameter is I, then that of another sphere of mass 2M and radius 2R about its diameter is
If the moment of inertia of a solid sphere of mass M and radius R about its diameter is I, then that of another sphere of mass 2M and radius 2R about its diameter is
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\(I \propto MR^2\). If M doubles and R doubles, \(I\) becomes \(2 \times 4 = 8\) times.
Updated On: Apr 24, 2026
- \(2I\)
- \(4I\)
- \(8I\)
- \(16I\)
- \(I\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Moment of inertia of solid sphere about diameter: \(I = \frac{2}{5}MR^2\).
Step 2: Detailed Explanation:
For first sphere: \(I = \frac{2}{5}MR^2\)
For second sphere: \(M' = 2M\), \(R' = 2R\)
\(I' = \frac{2}{5}M'R'^2 = \frac{2}{5}(2M)(2R)^2 = \frac{2}{5} \cdot 2M \cdot 4R^2 = \frac{2}{5} \cdot 8MR^2 = 8 \cdot \frac{2}{5}MR^2 = 8I\)
Step 3: Final Answer:
Moment of inertia = \(8I\).
Moment of inertia of solid sphere about diameter: \(I = \frac{2}{5}MR^2\).
Step 2: Detailed Explanation:
For first sphere: \(I = \frac{2}{5}MR^2\)
For second sphere: \(M' = 2M\), \(R' = 2R\)
\(I' = \frac{2}{5}M'R'^2 = \frac{2}{5}(2M)(2R)^2 = \frac{2}{5} \cdot 2M \cdot 4R^2 = \frac{2}{5} \cdot 8MR^2 = 8 \cdot \frac{2}{5}MR^2 = 8I\)
Step 3: Final Answer:
Moment of inertia = \(8I\).
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