Question

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

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

Answer 1

The correct option is D $\frac{5}{6}a$

Let $\sigma$ be the surface mass density of the disc.

$X_{\text{com}}$ $=$ $\frac{m_1{x}_1-m_2{x}_2}{m_1-m_2}$

Where, $m=\sigma$$\pi$$a^2$

$X_{\text{com}}=\frac{\sigma \pi a^2\times a-\sigma \pi {\left(\frac{a}{2}\right)}^2\times \frac{3a}{2}}{\sigma \pi a^2-\sigma \pi {\left(\frac{a}{2}\right)}^2}$

$X_{\text{com}}$ $=$ $\frac{a-\frac{3a}{8}}{1-\frac{1}{4}}$

$X_{\text{com}}$ $=$ $\frac{\frac{5a}{8}}{\frac{3}{4}}$

$X_{\text{com}}$ $=$ $\frac{5}{6}a$