Two discs A and B are mounted coaxiallay on a vertical axle. The discs have moments of inertia I and 2 I respectively about the common axis. Disc A is imparted an initial angular velocity `2 omega` using the entire potential energy of a spring compressed by a distance `x_1` Disc B is imparted an angular velocity `omega` by a spring having the same spring constant and compressed by a distance `x_2` Both the discs rotate in the clockwise direction.
When disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is

Options

A. `(2Iomega)/(3t)`
B. `(9Iomega)/(2t)`
C. `(9Iomega)/(4t)`
D. `(3Iomega)/(2t)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the Moments of Inertia and Initial Angular Velocities
- Disc A has a moment of inertia \( I_A = I \) and an initial angular velocity \( \omega_A = 2\omega \).
- Disc B has a moment of inertia \( I_B = 2I \) and an initial angular velocity \( \omega_B = \omega \).

### Step 2: Apply Conservation of Angular Momentum
When the discs come into contact, they will rotate together with a common angular velocity \( \omega' \). According to the conservation of angular momentum:
\[ I_A \omega_A + I_B \omega_B = (I_A + I_B) \omega' \]
Substituting the values:
\[ I(2\omega) + (2I)(\omega) = (I + 2I) \omega' \]
This simplifies to:
\[ 2I\omega + 2I\omega = 3I\omega' \]
Thus, we have:
\[ 4I\omega = 3I\omega' \]

### Step 3: Solve for the Common Angular Velocity \( \omega' \)
Dividing both sides by \( 3I \):
\[ \omega' = \frac{4}{3}\omega \]

### Step 4: Calculate the Change in Angular Momentum for Disc B
The change in angular momentum for disc B can be calculated as follows:
- Initial angular momentum of disc B:
\[ L_{B_{initial}} = I_B \omega_B = 2I \cdot \omega \]
- Final angular momentum of disc B:
\[ L_{B_{final}} = I_B \omega' = 2I \cdot \frac{4}{3}\omega = \frac{8}{3}I\omega \]
- Change in angular momentum:
\[ \Delta L_B = L_{B_{final}} - L_{B_{initial}} = \frac{8}{3}I\omega - 2I\omega \]
This simplifies to:
\[ \Delta L_B = \frac{8}{3}I\omega - \frac{6}{3}I\omega = \frac{2}{3}I\omega \]

### Step 5: Relate Change in Angular Momentum to Average Frictional Torque
The average frictional torque \( \tau \) exerted by disc A on disc B over time \( t \) is given by:
\[ \tau \cdot t = \Delta L_B \]
Thus:
\[ \tau = \frac{\Delta L_B}{t} = \frac{\frac{2}{3}I\omega}{t} = \frac{2I\omega}{3t} \]

### Final Answer
The average frictional torque on one disc by the other during this period is:
\[ \tau = \frac{2I\omega}{3t} \]