A uniform thin rod of length `l` and mass `m` is hinged at a distance `l//4` from one of the end and released from horizontal position as shown in Fig. The angular velocity of the rod as it passes the vertical position is

Options

A. `2sqrt((5g)/(7l))`
B. `2sqrt((6g)/(7l))`
C. `sqrt((3g)/(7l))`
D. `2sqrt(g/l)`

Answer

B

Explanation

`mg(l/4)=1/2[(ml^(2))/12+m(1/4)^(2)]omega^(2)`
`omega=sqrt((24g)/(7l))=2sqrt((6g)/(7l))`