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

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$\frac{5}{6} a$

$\frac{1}{6} a$

$\frac{2}{3} a$

$\frac{10}{11} a$

answer is D.

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Detailed Solution

Let the surface mass of disc's density be $σ$ ,

$x_{com} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} - m_{2}}$

$m = {σπa}^{2}$

$x_{com} = \frac{{σπa}^{2} \times a - σπ {(\frac{a}{2})}^{2} \times \frac{3 a}{2}}{{σπa}^{2} - σπa {(\frac{a}{2})}^{2}}$

$x_{com} = \frac{a - \frac{3 a}{8}}{1 - \frac{1}{4}} = \frac{\frac{5 a}{8}}{\frac{3}{4}} = \frac{5}{6} a$

Hence the correct answer is $\frac{5}{6} a .$

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