There is an insulator rod of length L and of negligible mass with two small balls of mass m and electric charge Q attached to its ends. The rod can rotate in the horizontal plane around a vertical axis crossing it ata distance `L//4` from one of its ends.
At first the rod is in unstabele equillbrium in a Horizontal uniform electric field of field strenght E. Then we gently displace it from this position. Determine the maximim velocity attained by the ball taht is closer to the axis in the subsequent motion

Options

A. `sqrt((2QEL)/(m))`
B. `sqrt((2QEL)/(5m))`
C. `sqrt((QEL)/(5m))`
D. `sqrt((4QEL)/(5m))`

Explanation

To solve the problem, we need to analyze the forces and torques acting on the system and use the principles of rotational motion. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the System
We have an insulating rod of length \( L \) with two small balls of mass \( m \) and charge \( Q \) attached at each end. The rod can rotate around a vertical axis located at a distance \( \frac{L}{4} \) from one end. The system is initially in unstable equilibrium in a uniform electric field \( E \).

### Step 2: Identify the Forces Acting on the Balls
When the rod is displaced from its equilibrium position, the electric force acting on each ball due to the electric field \( E \) will be:
- For the ball at distance \( \frac{L}{4} \): \( F_1 = Q \cdot E \)
- For the ball at distance \( \frac{3L}{4} \): \( F_2 = Q \cdot E \)

### Step 3: Calculate the Torque
The torque (\( \tau \)) about the axis of rotation due to the forces acting on the balls can be calculated as follows:
- Torque due to the ball at \( \frac{L}{4} \):
\[ \tau_1 = F_1 \cdot \frac{L}{4} \cdot \sin(\theta) = Q \cdot E \cdot \frac{L}{4} \cdot \sin(\theta) \]
- Torque due to the ball at \( \frac{3L}{4} \):
\[ \tau_2 = F_2 \cdot \frac{3L}{4} \cdot \sin(\theta) = Q \cdot E \cdot \frac{3L}{4} \cdot \sin(\theta) \]

The net torque (\( \tau \)) is:
\[ \tau = \tau_2 - \tau_1 = Q \cdot E \cdot \left(\frac{3L}{4} - \frac{L}{4}\right) \cdot \sin(\theta) = Q \cdot E \cdot \frac{2L}{4} \cdot \sin(\theta) = \frac{Q \cdot E \cdot L}{2} \cdot \sin(\theta) \]

### Step 4: Relate Torque to Angular Acceleration
Using Newton's second law for rotation, we have:
\[ \tau = I \cdot \alpha \]
where \( I \) is the moment of inertia of the system and \( \alpha \) is the angular acceleration. The moment of inertia \( I \) for the two balls about the axis of rotation is:
\[ I = m \left(\frac{L}{4}\right)^2 + m \left(\frac{3L}{4}\right)^2 = m \left(\frac{L^2}{16} + \frac{9L^2}{16}\right) = m \cdot \frac{10L^2}{16} = \frac{5mL^2}{8} \]

Thus, we can write:
\[ \frac{Q \cdot E \cdot L}{2} \cdot \sin(\theta) = \frac{5mL^2}{8} \cdot \alpha \]

### Step 5: Solve for Angular Acceleration
Rearranging gives us:
\[ \alpha = \frac{4Q \cdot E \cdot \sin(\theta)}{5mL} \]

### Step 6: Find Maximum Velocity
The maximum angular velocity \( \omega \) can be related to the maximum linear velocity \( v \) of the ball closer to the axis of rotation (at \( \frac{L}{4} \)):
\[ v = r \cdot \omega \]
where \( r = \frac{L}{4} \).

Using the relationship between angular acceleration and maximum velocity:
\[ \omega = \alpha \cdot t \]
At maximum displacement, we can substitute \( t \) with \( T \) (the period of motion):
\[ v_{\text{max}} = \frac{L}{4} \cdot \omega \]

### Step 7: Substitute and Solve
Substituting \( \omega \) into the equation gives:
\[ v_{\text{max}} = \frac{L}{4} \cdot \left(\frac{4Q \cdot E \cdot \sin(\theta)}{5mL} \cdot T\right) \]

### Final Result
The maximum velocity attained by the ball closer to the axis is:
\[ v_{\text{max}} = \frac{Q \cdot E \cdot T}{5m} \]