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Physics Question on System of Particles & Rotational Motion

From a circular ring of mass ‘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 ‘MR2 ’. Then the value of ‘K’ is

A

18\frac{1}{8}

B

34\frac{3}{4}

C

78\frac{7}{8}

D

14\frac{1}{4}

Answer

34\frac{3}{4}

Explanation

Solution

The correct answer is option (B): 34\frac{3}{4}