JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A uniform solid cylinder with radius \(R\) and length L has moment of inertia \(I_1\), about the axis of cylinder. A concentric solid cylinder of radius \(R^{\prime}=\frac{R}{2}\) and length \(L^{\prime}=\frac{L}{2}\) is caned out of the original cylinder. If \(I_2\) is the moment of inertia of the carved out portion ot the cylinder then \(\frac{I_1}{I_2}=..........\) (Both \(I_1\) and \(I_2\) are about the axis of the cylinder)
- A \(30\)
- B \(31\)
- C \(32\)
- D \(33\)
Answer & Solution
Correct Answer
(C) \(32\)
Step-by-step Solution
Detailed explanation
\(I_1=\frac{ m _1 R ^2}{2} \quad I_2=\frac{ m _2( R / 2)^2}{2}\) \(\frac{I_1}{I_2}=\frac{4 m_1}{m_2}=\frac{4 \cdot \rho \pi R^2 \ell}{\rho \cdot \frac{\pi R^2}{4} \times \frac{\ell}{2}} \Rightarrow \frac{I_1}{I_2}=32\)
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