Question

From a circular disc of radius 'R' having a centre A, a smaller disc of radius '0.5 R' and centre B, is cut out, such that the point A lies on its circumference, as shown in figure. The centre of gravity of the remaining portion of the larger disc (shaded area) is on the line joining the centres A and B at a distance 'x' from centre A, where 'x' is

• A. 0.5 R
• B. R/6
• C. 3R/4
• D. R/3

Answer 1

The correct option is B R/6

$\overline{x}=\frac{A_1\overline{x_1}+A_2\overline{x_2}}{A_1+A_2}$

$A_1=\pi R^2,A_2=-\pi (0.5R{)}^2$

$x_1=0,x_2=-0.5R$

By taking origin at 'A'

$\overline{x}=\frac{\pi (0.5{)}^2{R}^2\times 0.5R}{\pi (R^2-(0.5R{)}^2)}=\frac{(0.5{)}^{3}R}{1-{0.5}^2}=\frac{R}{6}$