Question 53 of 130

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

A$\dfrac{2}{3}a$B$\dfrac{5}{6}a$C$\dfrac{1}{6}a$D$\dfrac{10}{11}a$

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Answer: (B) $\dfrac{5}{6}a$

Let $\sigma$ be the uniform mass density of disc then
${x}_{COM}=\dfrac{\left(\sigma \pi {a}^{2}\right)a-\sigma \pi \left(\dfrac{{a}^{2}}{4}\right)\times \dfrac{3a}{2}}{\sigma \pi {a}^{2}-\dfrac{\sigma \pi {a}^{2}}{4}}$
$=\dfrac{a-\dfrac{3a}{8}}{1-\dfrac{1}{4}}=\dfrac{5a}{6}$

JEE Main Physics practice question on Center of Mass, Momentum & Collision (Centre of mass of continuous mass distribution), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.