A thin rod of mass `m` and length `2L` is made to rotate about an axis passing through its center and perpendicular to it. If its angular velocity changes from `O` to `omega` in time `t`, the torque acting on it is

Options

A. `(mL^(2)omega)/(12t)`
B. `(mL^(2)omega)/(3t)`
C. `(mL^(2)omega)/(t)`
D. `(4mL^(2)omega)/(6t)`

Explanation

`I=(m(2L)^(2))/(12)=(mL^(2))/(3)`
`tau Delta t=Delta L= I Delta omega`
`tau t=(mL^(2))/(3)(omega-0)implies tau=(mL^(2)omega)/(3t)`