A uniform rod of mass m and length l is suspended by means of two light inextansible strings as shown in the figure. Tension in one string immediately after the other string is cut is

Options

A. `(mg)/(2)`
B. `mg`
C. `2mg`
D. `(mg)/(4)`

Answer

D

Explanation

When one string is cut off, the rod will rotate about the other point A. Let a be the linear acceleration of centre of mass of the rod and `alpha` be the angular acceleration of the rod about A. As it clear from figure.

`mg-T=ma" ...(i)"`
`alpha=(tau)/(l)=(mg(l//2))/(ml^(2)//3)=(3g)/(2l)" ...(ii)"`
`a=l alpha=(l)/(2)alpha=(l)/(2)(3g)/(2l)=(3g)/(4)`
From Eq. (i), we get, `T=mg-ma=mg-(3mg)/(4)=(mg)/(4)`