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A quarter sector is cut from a unifrom circular disc of radius R. This sector has mass M. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is

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$\frac{{MR}^{2}}{2}$

$\frac{{MR}^{2}}{4}$

$\frac{{MR}^{2}}{8}$

$\sqrt{2} {MR}^{2}$

answer is A.

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Detailed Solution

Moment of inertia for a full disc with mass ′ 4M ′

along a given axis, the formula is $(4 M) (\frac{R^{2}}{2}) = 2 {MR}^{2}$

Therefore, according to symmetry, for the specified quarter of the disc

Moment of inertia $= \frac{2 {MR}^{2}}{4} = \frac{1}{2} {MR}^{2}$

Hence the correct answer is $\frac{1}{2} {MR}^{2} .$

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