A uniform rod of mass M and length L, which is free to rotate about a fixed vertical axis through O, is lying on a frictionless horizontal table. A particle of equal mass strikes the rod with a velocity `V_(0)` and sticks to it. The angular velocity of the combination immediately after the collision is:

Explanation

To solve the problem step by step, we will use the principle of conservation of angular momentum.

### Step-by-Step Solution:

1. **Identify the System**:
We have a uniform rod of mass \( M \) and length \( L \) that is free to rotate about a fixed vertical axis at point \( O \). A particle of mass \( M \) (equal to the mass of the rod) strikes the rod with an initial velocity \( V_0 \) and sticks to it.

2. **Initial Angular Momentum**:
Before the collision, the particle has linear momentum, which can be converted to angular momentum about point \( O \). The distance from point \( O \) to the point of impact (the center of the rod) is \( \frac{L}{2} \). The initial angular momentum \( L_i \) of the system is given by:
\[ L_i = m v_0 \cdot r = M V_0 \cdot \frac{L}{2} \]
where \( m = M \) is the mass of the particle and \( r = \frac{L}{2} \) is the distance from point \( O \) to the center of the rod.

3. **Final Angular Momentum**:
After the collision, the rod and the particle rotate together with an angular velocity \( \omega \). The moment of inertia \( I \) of the system (rod + particle) about point \( O \) needs to be calculated. The moment of inertia of the rod about point \( O \) is given by:
\[ I_{rod} = \frac{1}{3} M L^2 \]
The particle, which is now at a distance \( \frac{L}{2} \) from point \( O \), contributes to the moment of inertia as:
\[ I_{particle} = M \left(\frac{L}{2}\right)^2 = M \frac{L^2}{4} \]
Thus, the total moment of inertia \( I_f \) of the system after the collision is:
\[ I_f = I_{rod} + I_{particle} = \frac{1}{3} M L^2 + M \frac{L^2}{4} \]

4. **Combine the Moments of Inertia**:
To combine the moments of inertia, we need a common denominator:
\[ I_f = \frac{4}{12} M L^2 + \frac{3}{12} M L^2 = \frac{7}{12} M L^2 \]

5. **Conservation of Angular Momentum**:
According to the conservation of angular momentum:
\[ L_i = L_f \]
Therefore:
\[ M V_0 \cdot \frac{L}{2} = I_f \cdot \omega \]
Substituting \( I_f \):
\[ M V_0 \cdot \frac{L}{2} = \frac{7}{12} M L^2 \cdot \omega \]

6. **Solve for Angular Velocity \( \omega \)**:
Cancel \( M \) from both sides (assuming \( M \neq 0 \)):
\[ V_0 \cdot \frac{L}{2} = \frac{7}{12} L^2 \cdot \omega \]
Rearranging gives:
\[ \omega = \frac{V_0 \cdot \frac{L}{2}}{\frac{7}{12} L^2} = \frac{V_0 \cdot 6}{7L} \]
Simplifying:
\[ \omega = \frac{6 V_0}{7 L} \]

### Final Answer:
The angular velocity of the combination immediately after the collision is:
\[ \omega = \frac{6 V_0}{7 L} \]