Find out the position of centre of mass of the uniform lamina as shown in figure.from center of cavity

Explanation

We assume that a disc of radius R having mass density `pmsigma` is in the removed section. Now the problem change in following form

When disc of mass density `+sigma` and radius R is include than a complete rectangular plate is make having centre of mass at point O. When consider only disc having mass density `–sigma` and radius R then C.O.M of this disc lie at point O'

Then the position of C.O.M
`=(sigma(4R)^(2).R)/(-sigmapiR^(2)+sigma(4R^(2)))=(sigma16R^(3))/(sigmaR^(2)(16-pi))=(16R)/(16-pi)`
i.e., centre of mass lie in the rightwards side from the cavity.