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

Options

A. `(1)/(4)`
B. `(1)/(3)`
C. `(1)/(2)`
D. `(1)/(6)`

Answer

B

Explanation

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)`