Q69 / 90 · JEE Main 2021 (24 Feb Shift 2) · Physics

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

Solution

Let $σ$ be the uniform mass density of disc then

$x_{COM} = \frac{(σ π a^{2}) a - σ π (\frac{a^{2}}{4}) \times \frac{3 a}{2}}{σ π a^{2} - \frac{σ π a^{2}}{4}}$

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

Hence, the correct answer is option: 'B'.