A circular hole of radius (a/2) is cut out of a circular disc of… | JEE Main 2021 (24 Feb, Shift 2)

Q6JEE Main 2021 · 24 FEB · SHIFT 2Medium

Composite & Cut-Out Bodies

A circular hole of radius $\left(\frac{a}{2}\right)$ 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:

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Answer: (c)

Full disc: area $\pi a^2$ , centre at $x = a$ . Hole: area $\frac{\pi a^2}{4}$ , centre at $x = \frac{3a}{2}$ .
$x_{cm} = \frac{\pi a^2 (a) - \frac{\pi a^2}{4}\cdot\frac{3a}{2}}{\pi a^2 - \frac{\pi a^2}{4}} = \frac{a - \frac{3a}{8}}{\frac{3}{4}}$
$x_{cm} = \frac{5a}{8}\cdot\frac{4}{3} = \frac{5a}{6}$
So the answer is (c) $\frac{5}{6}a$ .

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Composite & Cut-Out Bodies question from JEE Main 2021 (24 Feb, Shift 2)

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