From a circular ring of mass ‘M’ and radius ‘R’ an arc corresponding to a 90° 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 ‘MR²’. Then the value of ‘K’ is :

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1/4

3/4

1/8

7/8

answer is B.

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Detailed Solution

If 90° arc is removed, remaining part is 270° and mass of the remaining part is 3/4 M and moment of inertia is
$\begin{array}{l} \frac{3}{4} {MR}^{2} = {KMR}^{2} \\ K = \frac{3}{4} \end{array}$

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