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The moment of inertia of a disc of mass M and radius R about any of its diameters is (MR²)/4. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be x/2MR². The value of x is ______.

Asked in JEE Main 1st Feb 2nd Shift 2023 · Standard bodies and axis theorems

Answer: 3

Step-by-step solution

First get the central perpendicular value. By the perpendicular-axis theorem, the two diameters add to it: I_z=2×(MR²)/4=(MR²)/2.

Now shift the axis out to the edge, a distance R from the centre.

I=(MR²)/2+MR²=(3MR²)/2.

Comparing with x/2MR² gives x=3.

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