JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
Moment of inertia \((M.I.)\) of four bodies, having same mass and radius, are reported as; \(I _{1}= M\). of thin circular ring about its diameter. \(I _{2}=\) \(M.I.\) of circular disc about an axis perpendicular to the disc and going through the centre, \(I _{3}=\) \(M.I.\) of solid cylinder about its axis and \(I _{4}= M . I.\) of solid sphere about its diameter. Then
- A \(I _{1}+ I _{3}< I _{2}+ I _{4}\)
- B \(I _{1}+ I _{2}= I _{3}+\frac{5}{2} I _{4}\)
- C \(I _{1}= I _{2}= I _{3}> I _{4}\)
- D \(I _{1}= I _{2}= I _{3}< I _{4}\)
Answer & Solution
Correct Answer
(C) \(I _{1}= I _{2}= I _{3}> I _{4}\)
Step-by-step Solution
Detailed explanation
Ring \(I _{1}=\frac{ MR ^{2}}{2}\) about diameter Disc \(I_{2}=\frac{M R^{2}}{2}\) Solid cylinder \(I _{3}=\frac{ MR ^{2}}{2}\) Solid sphere \(I _{4}=\frac{2}{5} MR ^{2}\) \(I _{1}= I _{2}= I _{3}> I _{4}\)
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