Question:
A uniform solid cylinder with radius R and length L has moment of inertia \(I_1\) about the axis of the cylinder. A concentric solid cylinder of radius \(R/2\) and length \(L/2\) is carved out of the original cylinder. If \(I_2\) is the moment of inertia of the carved out portion of the cylinder then \(I_1/I_2\) is (Both \(I_1\) and \(I_2\) are about the axis of the cylinder)
A uniform solid cylinder with radius R and length L has moment of inertia \(I_1\) about the axis of the cylinder. A concentric solid cylinder of radius \(R/2\) and length \(L/2\) is carved out of the original cylinder. If \(I_2\) is the moment of inertia of the carved out portion of the cylinder then \(I_1/I_2\) is (Both \(I_1\) and \(I_2\) are about the axis of the cylinder)
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Mass scales with volume, so find the mass of the small cylinder first.
Updated On: Oct 1, 2026
- \(4:1\)
- \(8:1\)
- \(16:1\)
- \(32:1\)
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The Correct Option is D
Solution and Explanation
Step 1: Mass of carved cylinder
Volume scales as \(r^2 L\). The small cylinder has radius \(\frac R2\) and length \(\frac L2\), so its volume is \(\frac14\times\frac12 = \frac18\) of the original. Its mass is \(\frac M8\).
Step 2: Moments of inertia
\(I_1 = \frac12MR^2\). \(I_2 = \frac12\left(\frac M8\right)\left(\frac R2\right)^2 = \frac{MR^2}{64}\).
Step 3: Ratio
\(\frac{I_1}{I_2} = \frac{1/2}{1/64} = 32\). So \(32:1\), option (D).
Final Answer:
The ratio is 32:1. \[ \boxed{\text{(D)}\ 32:1} \]
Volume scales as \(r^2 L\). The small cylinder has radius \(\frac R2\) and length \(\frac L2\), so its volume is \(\frac14\times\frac12 = \frac18\) of the original. Its mass is \(\frac M8\).
Step 2: Moments of inertia
\(I_1 = \frac12MR^2\). \(I_2 = \frac12\left(\frac M8\right)\left(\frac R2\right)^2 = \frac{MR^2}{64}\).
Step 3: Ratio
\(\frac{I_1}{I_2} = \frac{1/2}{1/64} = 32\). So \(32:1\), option (D).
Final Answer:
The ratio is 32:1. \[ \boxed{\text{(D)}\ 32:1} \]
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