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)/(3t)`

Answer

B

Explanation

Since, `tau=lalpha`
So, `tau=[(m(2l)^(2))/(12)]((omega)/(t))=(mxx4l^(2)xxomega)/(12xxt)=(4ml^(2)omega)/(12t)=((ml^(2)omega)/(3t))`