A circular hole of radius $(a / 2)$ is cut out of a circular disc of a radius ‘a’ as shown in figure. The centroid of the remaining circular portion with respect to point O will be

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$\frac{10}{11} a$

$\frac{5}{6} a$

$\frac{a}{6}$

$\frac{2}{3} a$

answer is C.

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Detailed Solution

Let M be the mass of circular disk
The density of desk: $a = \frac{M}{π a^{2}}$
Now let $M'$ be the mass of cut out disk

$M' = σ \times π {(\frac{a}{2})}^{2} = \frac{M}{π a^{2}} \times π \frac{a^{2}}{4} = \frac{M}{4} X_{c o m} = \frac{M \times a + \frac{M}{4} \times \frac{3}{2} a}{M = \frac{M}{4}} = \frac{5}{6} a$

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