Four particles, each of mass m, are lying symmetrically on the rim of a disc of mass M and radius R. M.I. of this system about an axis passing through one of the particles and perpendicular to plane of disc is

Options

A. `16mr^(2)`
B. `(3M+19m)(R^(2))/(2)`
C. `(3M+12m)(R^(2))/(2)`
D. zero

Answer

B

Explanation

According to theorem of parallel axes, moment of inertia of disc about an axis passing through K and perpendicular to plane of disc `=(1)/(2)MR^(2)+MR^(2)=(3)/(2)MR^(2)`
Total moment of inertia of the system
`=(3)/(2)MR^(2)+m(2R)^(2)+m(2sqrtR)^(2)+m(sqrt2R)^(2)`
`=(3M+16m)(R^(2))/(2)`
`OD=R`
`therefore" AC = 2R and AB= AD = "Rsqrt2`