A uniform rod of mass `m`, length `l` rests on a smooth horizontal surface. Rod is given a sharp horizontal impulse `p` perpendicular to the rod at a distance `l//4` from the centre. The angular velocity of the rod will be

Options

A. a)`(3p)/(ml)`
B. b)`p/(ml)`
C. c)`p/(2ml)`
D. d)`(2p)/(ml)`

Explanation

To find the angular velocity of a uniform rod of mass `m` and length `l` after it is given a sharp horizontal impulse `p` at a distance `l/4` from its center, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a uniform rod of mass `m` and length `l` resting on a smooth horizontal surface.
- An impulse `p` is applied perpendicular to the rod at a distance of `l/4` from the center.

2. **Identify the Point of Application of Impulse**:
- The impulse is applied at a distance `l/4` from the center of the rod. This means the distance from the center to the point of application is `d = l/4`.

3. **Calculate the Torque due to the Impulse**:
- The torque (τ) caused by the impulse about the center of mass is given by:
\[ \tau = p \cdot d = p \cdot \frac{l}{4} \]

4. **Angular Impulse-Momentum Theorem**:
- According to the angular impulse-momentum theorem, the change in angular momentum (ΔL) is equal to the torque multiplied by the time duration (Δt) of the impulse:
\[ \tau = \Delta L \]
- Since the impulse is instantaneous, we can consider the angular momentum just after the impulse is applied.

5. **Moment of Inertia of the Rod**:
- The moment of inertia (I) of a uniform rod about its center is given by:
\[ I = \frac{1}{12} m l^2 \]

6. **Relate Torque to Angular Velocity**:
- The angular momentum (L) just after the impulse is:
\[ L = I \cdot \omega \]
- Setting the torque equal to the change in angular momentum gives:
\[ p \cdot \frac{l}{4} = I \cdot \omega \]

7. **Substituting the Moment of Inertia**:
- Substitute the expression for moment of inertia into the equation:
\[ p \cdot \frac{l}{4} = \frac{1}{12} m l^2 \cdot \omega \]

8. **Solving for Angular Velocity (ω)**:
- Rearranging the equation to solve for ω:
\[ \omega = \frac{p \cdot \frac{l}{4}}{\frac{1}{12} m l^2} \]
- Simplifying this expression:
\[ \omega = \frac{p \cdot 12}{m l} \cdot \frac{1}{4} = \frac{3p}{ml} \]

### Final Result:
The angular velocity of the rod after the impulse is given by:
\[ \omega = \frac{3p}{ml} \]