A circular hole a radius $\frac{a}{2}$ is cut of a circular disc of radius a as shown in figure. The centre of mass of the remaining circular portion with respect to the point ’O’ will be

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$\frac{a}{6}$

$\frac{10 a}{11}$

$\frac{5 a}{6}$

$\frac{2 a}{3}$

answer is C.

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Detailed Solution

Let $σ$ the uniform mass density of the disc, then
$x_{c m} = \frac{(π a^{2} σ) a - π {(\frac{a}{2})}^{2} σ x \frac{3 a}{2}}{π a^{2} σ - π {(\frac{a}{2})}^{2} σ} = \frac{5 a}{6}$

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