NEET · Physics · STD 11 - 6. system of particles and rotational motion
From a circular ring of mass \('\mathrm{M}'\) and radius \('R'\) an arc corresponding to a \(90^{\circ}\) sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is \('K'\) times \(\mathrm{MR}^{2}.\) Then the value of \('\mathrm{K}'\) is
- A \(\frac{3}{4}\)
- B \(\frac{7}{8}\)
- C \(\frac{1}{4}\)
- D \(\frac{1}{8}\)
Answer & Solution
Correct Answer
(A) \(\frac{3}{4}\)
Step-by-step Solution
Detailed explanation
\(I=M R^{2}\) \(I^{\prime}=\frac{3 M}{4} R^{2}\)
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