Two masses are attached to a rod end to end . If torque is applied they rotate with angular acceleration `alpha`. If their distances are doubled and same torque is applied, then they move with angular acceleration.

Options

A. `4alpha`
B. `alpha`
C. `3alpha`
D. `alpha//4`

Answer

D

Explanation

Suppose the masses `m_(1)` and `m_(2)` are attached to the ends of a massless rod. `C_(1)` is the axis of rotation which is at a distance R from each of the masses.
`therefore ` the moment of inertia of the system,
`I=m_(1)R^(2)+m_(2)R^(2) =(m_(1)+m_(2))^(2)R^(2)`
Now if the distance between the masses is doubled , then the moment of inertia.
`I'=m_(1)(2R)^(2)+m_(2)(2R)^(2)=(m_(1)+m_(2))4R^(2)`
Since the same torque is applied in both the cases,
`tau=Ialpha =(Ialpha)/(I')=((m_(2)+m_(2)R^(2)alpha)/((m_(1)+m_(2)4R^(2))=(alpha)/(4)`