The ring which can slide along the rod are kept at mid point of a smooth rod of length L. The rod is rotated with constant angular velocity `omega` about vertical axis passing through its one end. The ring is released from mid point. The velocity of the ring, when it just leaves the rod is = `(sqrt(x))/(2)omegaL`. Then x is equal to

Answer

To solve the problem step by step, we need to analyze the motion of the ring as it slides along the rod and determine its velocity when it leaves the rod.

### Step 1: Understand the System
The ring is initially at the midpoint of a smooth rod of length \( L \). The rod is rotated about a vertical axis with a constant angular velocity \( \omega \). When the ring is released, it will experience a centrifugal force due to the rotation.

### Step 2: Identify the Forces
The centrifugal force acting on the ring when it is at a distance \( x \) from the axis of rotation is given by:
\[ F_c = m \omega^2 x \]
where \( m \) is the mass of the ring.

### Step 3: Relate Centrifugal Force to Acceleration
The centrifugal force can be equated to the mass times the acceleration experienced by the ring:
\[ m \omega^2 x = m \frac{dv}{dt} \]
Since the ring is sliding, we can express the acceleration in terms of the change in velocity with respect to the position \( x \):
\[ \omega^2 x = v \frac{dv}{dx} \]

### Step 4: Integrate Both Sides
To find the relationship between velocity and position, we need to integrate both sides. Rearranging gives:
\[ v dv = \omega^2 x dx \]
Integrating from \( x = L/2 \) to \( x = L \) and from \( v = 0 \) to \( v \):
\[ \int_0^v v \, dv = \int_{L/2}^{L} \omega^2 x \, dx \]
This results in:
\[ \frac{v^2}{2} = \omega^2 \left[ \frac{x^2}{2} \right]_{L/2}^{L} \]

### Step 5: Calculate the Integral
Calculating the definite integral:
\[ \frac{v^2}{2} = \omega^2 \left( \frac{L^2}{2} - \frac{(L/2)^2}{2} \right) \]
\[ = \omega^2 \left( \frac{L^2}{2} - \frac{L^2/4}{2} \right) = \omega^2 \left( \frac{L^2}{2} - \frac{L^2}{8} \right) \]
\[ = \omega^2 \left( \frac{4L^2}{8} - \frac{L^2}{8} \right) = \omega^2 \left( \frac{3L^2}{8} \right) \]

### Step 6: Solve for \( v \)
Thus, we have:
\[ \frac{v^2}{2} = \frac{3\omega^2 L^2}{8} \]
Multiplying both sides by 2 gives:
\[ v^2 = \frac{3\omega^2 L^2}{4} \]
Taking the square root:
\[ v = \sqrt{\frac{3}{4}} \omega L = \frac{\sqrt{3}}{2} \omega L \]

### Step 7: Final Comparison
The problem states that the velocity of the ring when it just leaves the rod is given by:
\[ v = \sqrt{\frac{x}{2}} \omega L \]
Comparing both expressions for \( v \):
\[ \frac{\sqrt{3}}{2} \omega L = \sqrt{\frac{x}{2}} \omega L \]
Dividing both sides by \( \omega L \) (assuming \( \omega L \neq 0 \)):
\[ \frac{\sqrt{3}}{2} = \sqrt{\frac{x}{2}} \]
Squaring both sides:
\[ \frac{3}{4} = \frac{x}{2} \]
Multiplying both sides by 2:
\[ x = \frac{3}{2} \cdot 2 = 3 \]

### Conclusion
Thus, the value of \( x \) is:
\[ \boxed{3} \]