Question

From a circular disc of radius R, a square is cut out with a radius as its diagonal. The centre of mass of remainder is at a distance (from the centre)

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

Answer 1

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

$A_1\left(C{C}_1\right)=A_2\left(C{C}_2\right)$

Side of square will be$\frac{R}{\sqrt{2}}$
$\therefore C{C}_2=\frac{A_1}{A_2}\left(C{C}_1\right)$
$=\frac{{\left(\frac{R}{\sqrt{2}}\right)}^2}{\pi R^2-{\left(\frac{R}{\sqrt{2}}\right)}^2}\left(\frac{R}{2}\right)$
$=\left(\frac{R}{4\pi -2}\right)$