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

Let  $\sigma$  be surface mass density. Then   $m_{disc}=\left(\pi R^2\right)\sigma$

$m_{cavity}=-l^2\sigma$

Where l  is length of square. now,  $l\cos \left({45}^o\right)=\frac{R}{2}$

$\Rightarrow l=\frac{R}{\sqrt{2}} \Rightarrow m_{cavity}=-\frac{R^2\sigma }{2}$

Now assume origin to be centre of disc, then

$\Rightarrow r_{CMremainder}=\frac{\left(\pi R^2\sigma \right)0-\left(\frac{R^2\sigma }{2}\right)\left(\frac{R}{2}\right)}{\pi R^2\sigma -\frac{R^2\sigma }{2}} \Rightarrow r_{CMremainder}=-\frac{R}{2(2\pi -1)}$

Negative sign indicates that CM of remained is to the left centre of disc i.e, origin.
Hence, the correct answer is (C)