Two discs `A` and `B` are mounted coaxially ona vertical axle. The discs have moments of inertia `l` and `2l` respectively about the common axis. Disc A is imparted an initial angular velocity `2 omega` using the centre potential energy of a spring compressed by a distance `x_(1)`. Disc `B` is imparted angular velocity `omega` by a spring having the same spring constant and compressed by a distance `x_(2)`. Both the disc rotate in the clockwise direction.
The loss of kinetic energy the above process is -

Options

A. `(Iomega^(3))/2`
B. `(Iomega^(2))/3`
C. `(Iomega^(2))/4`
D. `(Iomega^(2))/6`

Explanation

To solve the problem of finding the loss of kinetic energy when two discs A and B are brought into contact, we can follow these steps:

### Step 1: Determine the initial kinetic energies of the discs

1. **Disc A** has a moment of inertia \( I_A = I \) and an initial angular velocity \( \omega_A = 2\omega \).
- The kinetic energy \( KE_A \) of disc A is given by:
\[ KE_A = \frac{1}{2} I_A \omega_A^2 = \frac{1}{2} I (2\omega)^2 = \frac{1}{2} I \cdot 4\omega^2 = 2I\omega^2 \]

2. **Disc B** has a moment of inertia \( I_B = 2I \) and an initial angular velocity \( \omega_B = \omega \).
- The kinetic energy \( KE_B \) of disc B is given by:
\[ KE_B = \frac{1}{2} I_B \omega_B^2 = \frac{1}{2} (2I) \omega^2 = I\omega^2 \]

3. **Total initial kinetic energy** \( KE_{initial} \) is:
\[ KE_{initial} = KE_A + KE_B = 2I\omega^2 + I\omega^2 = 3I\omega^2 \]

### Step 2: Calculate the final angular velocity when the discs are brought into contact

Using the conservation of angular momentum:
\[ I_A \omega_A + I_B \omega_B = (I_A + I_B) \omega_{final} \]
Substituting the values:
\[ I \cdot (2\omega) + 2I \cdot \omega = (I + 2I) \omega_{final} \]
This simplifies to:
\[ 2I\omega + 2I\omega = 3I \omega_{final} \]
\[ 4I\omega = 3I \omega_{final} \]
Dividing both sides by \( 3I \):
\[ \omega_{final} = \frac{4\omega}{3} \]

### Step 3: Determine the final kinetic energy of the system

1. The total moment of inertia when both discs are rotating together is:
\[ I_{total} = I_A + I_B = I + 2I = 3I \]

2. The final kinetic energy \( KE_{final} \) is:
\[ KE_{final} = \frac{1}{2} I_{total} \omega_{final}^2 = \frac{1}{2} (3I) \left(\frac{4\omega}{3}\right)^2 \]
Simplifying this:
\[ KE_{final} = \frac{1}{2} (3I) \cdot \frac{16\omega^2}{9} = \frac{48I\omega^2}{18} = \frac{8I\omega^2}{3} \]

### Step 4: Calculate the loss of kinetic energy

The loss of kinetic energy \( \Delta KE \) is given by:
\[ \Delta KE = KE_{initial} - KE_{final} \]
Substituting the values:
\[ \Delta KE = 3I\omega^2 - \frac{8I\omega^2}{3} \]
Finding a common denominator:
\[ \Delta KE = \frac{9I\omega^2}{3} - \frac{8I\omega^2}{3} = \frac{I\omega^2}{3} \]

### Final Answer

The loss of kinetic energy in the above process is:
\[ \Delta KE = \frac{I\omega^2}{3} \]