Question:
The moment of inertia of a solid sphere of radius $20cm$ about its diameter is same as that of a solid cylinder of same mass about its axis, then the radius of the cylinder in $cm$ is
The moment of inertia of a solid sphere of radius $20cm$ about its diameter is same as that of a solid cylinder of same mass about its axis, then the radius of the cylinder in $cm$ is
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Equate moments carefully when masses are same.
Updated On: Apr 24, 2026
- $3\sqrt{5}$
- $5\sqrt{5}$
- $2\sqrt{5}$
- $8\sqrt{5}$
- $7\sqrt{5}$
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The Correct Option is D
Solution and Explanation
Concept:
• Sphere: $I = \frac{2}{5}MR^2$
• Cylinder: $I = \frac{1}{2}MR^2$
Step 1: Equate moments
\[ \frac{2}{5}M(20)^2 = \frac{1}{2}M r^2 \]
Step 2: Simplify
\[ \frac{2}{5}\cdot 400 = \frac{1}{2}r^2 \] \[ 160 = \frac{1}{2}r^2 \Rightarrow r^2 = 320 \]
Step 3: Find radius
\[ r = \sqrt{320} = 8\sqrt{5} \] Final Conclusion:
Option (D)
• Sphere: $I = \frac{2}{5}MR^2$
• Cylinder: $I = \frac{1}{2}MR^2$
Step 1: Equate moments
\[ \frac{2}{5}M(20)^2 = \frac{1}{2}M r^2 \]
Step 2: Simplify
\[ \frac{2}{5}\cdot 400 = \frac{1}{2}r^2 \] \[ 160 = \frac{1}{2}r^2 \Rightarrow r^2 = 320 \]
Step 3: Find radius
\[ r = \sqrt{320} = 8\sqrt{5} \] Final Conclusion:
Option (D)
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