A uniform rod AB  of mass m nd length l is kept at rest on a smooth horizontal surface.

An impulse is applied to the one end B perpendicular to the length of the rod.

The time taken by the rod to turn through a right angle is

A)pie*ml/12p

B)pie*ml/6p

C)ml/6p

D)none of these

Here we apply the conservation of momentum as follows

Angular momentum imparted is given by

I x w = L x P

where P = impulse imparted

I = moment of inertia of the rod across one end  or I = mL²/3

w = angular velocity

L = length of the rod

Thus w = LP/I = 3LP/mL²

or w = 3P/mL  ...(1)

Now let time taken by the rod to become perpendicular =  t (here we assume that another end is fixed i.e rod is movinf about it)

thus w t = pi /2   ...(2)

From (1) and (2) we obtain the following relation

or t = pi/ 2w =  pi * mL/6P

Thus correct option is B) pi *mL/6p

However if opposite end of the rod is not fixed and free.

Then impuse will make rotate the rod about its center.

The approach will exactly be like the one suggested above, but moment of inertia will be calulated around center of the rod.

hey guys thanks a lot !!!