A rod of mass m and length l in placed on a smooth table. An another particle of same mass m strikes the rod with velocity `v_(0)` in a direction perpendicular to the rod at distance `x( lt l//2)` from its centre . Particle sticks to the end. Let `omega` be the angular speed of system after collision , then
As x is increased from 0 to l/2 , the angular speed `omega`.

Options

A. will continuously increase
B. will continuously decrease
C. will first increase and then decrease
D. will first increase and then decrease

Answer

To solve the problem, we will analyze the situation using the principles of conservation of angular momentum and the moment of inertia.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a rod of mass \( m \) and length \( l \) placed on a smooth table.
- A particle of mass \( m \) strikes the rod perpendicularly at a distance \( x \) from the center of the rod.
- The particle sticks to the end of the rod after the collision.

2. **Initial Angular Momentum**:
- The initial linear momentum of the particle before the collision is \( mv_0 \).
- The distance from the center of the rod to the point of impact is \( x \).
- Therefore, the initial angular momentum \( L_i \) about the center of the rod is given by:
\[ L_i = mv_0 \cdot x \]

3. **Final Angular Momentum**:
- After the collision, the system (rod + particle) will rotate about the center of mass.
- Let \( \omega \) be the angular speed of the system after the collision.
- The moment of inertia \( I \) of the rod about its center is:
\[ I_{\text{rod}} = \frac{1}{3} ml^2 \]
- The moment of inertia of the particle (considered as a point mass at a distance \( \frac{l}{2} \) from the center) is:
\[ I_{\text{particle}} = m \left(\frac{l}{2}\right)^2 = \frac{ml^2}{4} \]
- The total moment of inertia \( I_f \) of the system after the collision is:
\[ I_f = I_{\text{rod}} + I_{\text{particle}} = \frac{1}{3} ml^2 + \frac{ml^2}{4} \]
- To combine these, we find a common denominator (12):
\[ I_f = \frac{4ml^2}{12} + \frac{3ml^2}{12} = \frac{7ml^2}{12} \]

4. **Applying Conservation of Angular Momentum**:
- According to the conservation of angular momentum:
\[ L_i = L_f \]
- Therefore:
\[ mv_0 \cdot x = I_f \cdot \omega \]
- Substituting \( I_f \):
\[ mv_0 \cdot x = \frac{7ml^2}{12} \cdot \omega \]
- Canceling \( m \) from both sides:
\[ v_0 \cdot x = \frac{7l^2}{12} \cdot \omega \]
- Solving for \( \omega \):
\[ \omega = \frac{12v_0 \cdot x}{7l^2} \]

5. **Analyzing \( \omega \) as \( x \) varies**:
- As \( x \) increases from 0 to \( \frac{l}{2} \):
- When \( x = 0 \):
\[ \omega = 0 \]
- When \( x = \frac{l}{4} \):
\[ \omega = \frac{12v_0 \cdot \frac{l}{4}}{7l^2} = \frac{3v_0}{7l} \]
- When \( x = \frac{l}{2} \):
\[ \omega = \frac{12v_0 \cdot \frac{l}{2}}{7l^2} = \frac{6v_0}{7l} \]
- Since \( \frac{3v_0}{7l} < \frac{6v_0}{7l} \), we conclude that \( \omega \) increases as \( x \) increases from 0 to \( \frac{l}{2} \).

### Conclusion:
As \( x \) increases from 0 to \( \frac{l}{2} \), the angular speed \( \omega \) of the system increases.