1. Given data

• Mass of rod: $m = 2\,\text{kg}$
• Length of rod: $L = 30\,\text{cm} = 0.30\,\text{m}$
• Impulse applied at end $B$: $J = 0.2\,\text{N s}$
• Rod is on a smooth horizontal surface, so after the impulse it undergoes:
  • translation of center of mass
  • rotation about center of mass

We need the time taken for the rod to rotate through $90^\circ = \dfrac{\pi}{2}$.

2. Angular impulse about the center of mass

The impulse is applied at the end of the rod, so its perpendicular distance from the center is

$r = \frac{L}{2} = \frac{0.30}{2} = 0.15\,\text{m}$

Angular impulse about center of mass:

$J_{\text{ang}} = rJ = 0.15 \times 0.2 = 0.03\,\text{kg m}^2/\text{s}$

This equals the angular momentum gained:

$I\omega = 0.03$

3. Moment of inertia of rod about its center

For a uniform rod about an axis through its center and perpendicular to its length:

$I = \frac{1}{12}mL^2$

So,

$I = \frac{1}{12}(2)(0.30)^2 = \frac{2}{12}\times 0.09 = 0.015\,\text{kg m}^2$

Thus,

$0.015\,\omega = 0.03$

$\omega = \frac{0.03}{0.015} = 2\,\text{rad/s}$

4. Time to rotate through $90^\circ$

After the impulse, no external torque acts, so angular velocity remains constant at $\omega = 2\,\text{rad/s}$.

Required angular displacement:

$\theta = \frac{\pi}{2}$

Using

$\theta = \omega t$

$t = \frac{\theta}{\omega} = \frac{\pi/2}{2} = \frac{\pi}{4}\,\text{s}$

Given that

$t = \frac{\pi}{x} \text{ s}$

So,

$\frac{\pi}{x} = \frac{\pi}{4} \Rightarrow x = 4$

5. Comparison with stored answer

Derived answer: $4$

Stored correct answer: $4$

They match.