A unifrom rod of length `l` and mass `m` is free to rotate in a vertical plane about `A`, Fig. The rod initially in horizontal position is released. The initial angular acceleration of the rod is `(MI "of rod about" A "is" (ml^(2))/(3))`

Options

A. `(3g)/(2l)`
B. `(2l)/(3g)`
C. `(3g)/(2l^(2))`
D. `"mg"(l)/(2)`

Answer

a

Explanation

The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is `=(ml^(2))/(3)`
where , m is the mass and l its length .
Torque `(tau=l*alpha)` acting on centre of gravity of rod is given by
`tau=mg(l)/(2)`
also , `(ml^(2))/(3).alpha="mg"(l)/(2)`
`alpha=(3g)/(2l)`