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 center )

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

Answer 1

Answer: (A)

$\frac{R}{(4\pi -2)}$

We know that density is mass per unit area

mass of disc is $\rho \pi R^2$

Diagonal of square is given by

$d=R=\sqrt{2}aa=\frac{\sqrt{2}}{R}$

mass of cut square from disc is

$\rho a^2=\rho \times \frac{R^2}{2}=\frac{\rho R^2}{2}$

Point of centre of mass is $(0,0)$ and centre of mass of square is $(R/2,0)$

Centre of remaining portion is

$M=\frac{\rho \pi R^2\left(0\right)+\left(\frac{\rho R^2}{2}\right)\left(\frac{R}{2}\right)}{\rho \pi R^2-\rho \frac{R^2}{2}}M=\frac{R}{4\pi -2}$

Hence, correct option is $\left(A\right)$