Question

A square hole is punched out from a circular laming. The diagonal of the square being equal to radius of the circle, $R$. Find the distance of $CM$ of the remaining portion from centre of disc.

• A. $\frac{R}{4\pi -2}$
• B. $\frac{R}{3\pi -2}$
• C. $\frac{2R}{4\pi -2}$
• D. $\frac{3R}{4\pi +2}$

Answer 1

The correct option is A $\frac{R}{4\pi -2}$

Let mass of circular lamina before removing square $=m$

Mass of the square lamina removed

$=\frac{M}{\pi R^2}=\left(\frac{1}{2}{R}^2\right)=\frac{M}{2\pi }$

As in vertical direction, equal and symmetrical man is removed, cone won't shift in vertical direction

Let cone of remaining lamina$=x$

$\frac{\left(M_{remaininglamina}x\right)+\left(M_{squarelamina}\right)x_{squarelamina}}{Totalmass}=$ cone of circular lamina without removing square.

$=0$

$\frac{(M-\frac{M}{2\pi })x+\left(\frac{M}{2\pi }\right)(-\frac{R}{2})}{M}=0\Rightarrow x(1-\frac{2}{2\pi })+\left(\frac{1}{2\pi }\right)(-\frac{R}{2})=0\Rightarrow x(1-\frac{1}{2\pi })=\frac{R}{4\pi }\Rightarrow x\left(\frac{2\pi -1}{2\pi }\right)=\frac{R}{4\pi }\Rightarrow x=\frac{R}{4\pi -2}$