Find the position of center of mass of the uniform lamina shown in figure, if small disc of radius `alpha/2` is cut from disc of radius a.

Options

A. `(pib^(2))/(a^(2) - c^(2))`
B. `(cb^(2))/(a^(2)-b^(2))`
C. `(pic^(2))/(a^(2)-b^(2))`
D. `(ca^(2))/((c^(2)-b^(2)))`

Answer

B

Explanation

Center of mass of whole system was at point O. Hence `x_(2)(pia^(2) - pib^(2)) = c(pib^(2)) rArr x_92) = (cb^(2))/(a^(2)-b^(2))`