A circular disc of radius R is removed from a bigger of the discs coincide. The centre of mass of the new disc is ar from the centre of the bigger disc. The value of alpha is

A circular disc of radius R is removed from a bigger of the discs coin

1.	A circular disc of radius R is removed from a bigger of the discs coincide. The centre of mass of the new disc is ar from the centre of the bigger disc. The value of alpha is
Answer» `(1)/(4)` `(1)/(3)` `(1)/(2)` `(1)/(6)` Solution :In figure , O is the CENTRE of circular disc of radius 2 R and MASS M . `C_(1)` is centre of disc of radius R , which is removed . If `sigma` is the mass per unit area of disc , then `M = PI (2R)^(2) sigma` Mass of disc removed , `M_1 = pi R^2 sigma = 1/4 M` Mass of remaining disc , `M_2 = M - M_1 = M - 1/4 M = 3/4 M` Let centre of mass of remaining disc be at `C_2` where `OC_2 = x` As `M_1 xx O C_(1) = M_2 xx O C_(2)` `therefore M/4 R = (3M)/(4) ""x or x = (R)/(3) = alpha R therefore alpha = (1)/(3)`