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. `2 sqrt ((5g)/(71))`
B. `2 sqrt ((6g)/( 71))`
C. `sqrt (( 3g)/( 71))`
D. `2 sqrt ((2)/(1))`

Answer

B

Explanation

`mg ((1)/(4)) = (1)/(2) [ ( m 1 ^(2))/( 12) + m ((1)/(4)) ^(2) ] omega^(2)`
`omega = sqrt ((24 g )/( 71)) = 2 sqrt (( 6 g )/( 71))`