A small hole is made in a circular disc of mass 'M' and radius ‘R’ at a distance of R/4 from the centre. The disc is supported on a horizontal peg through this hole. The moment of inertia of the disc about the horizontal peg is

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$\frac{{MR}^{2}}{9}$

$\frac{5 {MR}^{2}}{4}$

$\frac{5 {MR}^{2}}{16}$

$\frac{9 {MR}^{2}}{16}$

answer is A.

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Detailed Solution

The disc's moment of inertia around its center is $\frac{1}{2} {MR}^{2}$

By the parallel axis theorem,

$I = I_{CM} + {Md}^{2}$

$I = \frac{1}{2} {MR}^{2} + M {(\frac{R}{4})}^{2}$

$I = \frac{1}{2} {MR}^{2} + \frac{1}{16} {MR}^{2} = \frac{9}{16} {MR}^{2}$

Hence the correct answer is $\frac{9}{16} {MR}^{2} .$

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