If a disc of mass m and radius R totates with an angular acceleration a, of the torque acting on the disc is

Options

A. `MR^(2)a `
B. `(MR^(2)a)/(2)`
C. `(2MR^(2)a)/(5)`
D. `(MR^(2) a)/(12)`

Answer

B

Explanation

Moment of inertia of disc, `I = 1/2 MR ^(2)`
Angular acceleration,` alpha = a`(given)
So, torque `tau = I alpha `
`tau = ((1)/(2) MR^(2)) (a) implies tau = (MR^(2) a )/(2)`