TS EAMCET · Physics · Rotational Motion
The moment of intertia of a solid cylinder of mass \(M\), length \(2 R\) and radius \(R\) about an axis passing through the centre of mass and perpendicular to the axis of the cylinder is \(I_1\) and about an axis passing through one end of the cylinder and perpendicular to the axis of the cylinder is \(I_2\), then
- A \(I_2 < I_1\)
- B \(I_2-I_1=M R^2\)
- C \(\frac{I_2}{I_1}=\frac{19}{12}\)
- D \(\frac{l_2}{l_1}=\frac{7}{6}\)
Answer & Solution
Correct Answer
(B) \(I_2-I_1=M R^2\)
Step-by-step Solution
Detailed explanation
Moment of inertia of a solid cylinder of mass \(M\), length \(2 R\) and radius \(R\) about an axis passing through the centre of mass and perpendicular to the axis is…
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