Two discs `A` and `B` are mounted co-axially one 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 rotation `x_(1)//x_(2)` is.

Options

A. `2`
B. `1/2`
C. `sqrt(2)`
D. `1/(sqrt(2))`

Explanation

To solve the problem, we will use the principle of conservation of energy, which states that the change in rotational kinetic energy of the discs is equal to the change in potential energy stored in the springs.

### Step-by-Step Solution:

1. **Identify the Moments of Inertia**:
- Let the moment of inertia of disc A be \( I \).
- The moment of inertia of disc B is given as \( 2I \).

2. **Determine the Initial Angular Velocities**:
- Disc A is given an initial angular velocity of \( 2\omega \).
- Disc B is given an initial angular velocity of \( \omega \).

3. **Calculate the Rotational Kinetic Energy for Each Disc**:
- For disc A, the rotational kinetic energy \( K_A \) is given by:
\[ K_A = \frac{1}{2} I (2\omega)^2 = \frac{1}{2} I \cdot 4\omega^2 = 2I\omega^2 \]
- For disc B, the rotational kinetic energy \( K_B \) is given by:
\[ K_B = \frac{1}{2} (2I) \omega^2 = I\omega^2 \]

4. **Relate the Kinetic Energy to the Potential Energy of the Springs**:
- The potential energy stored in the spring for disc A when compressed by distance \( x_1 \) is:
\[ PE_A = \frac{1}{2} k x_1^2 \]
- The potential energy stored in the spring for disc B when compressed by distance \( x_2 \) is:
\[ PE_B = \frac{1}{2} k x_2^2 \]

5. **Set Up the Energy Conservation Equations**:
- For disc A:
\[ K_A = PE_A \implies 2I\omega^2 = \frac{1}{2} k x_1^2 \tag{1} \]
- For disc B:
\[ K_B = PE_B \implies I\omega^2 = \frac{1}{2} k x_2^2 \tag{2} \]

6. **Divide Equation (1) by Equation (2)**:
\[ \frac{2I\omega^2}{I\omega^2} = \frac{\frac{1}{2} k x_1^2}{\frac{1}{2} k x_2^2} \]
Simplifying gives:
\[ 2 = \frac{x_1^2}{x_2^2} \]

7. **Solve for the Ratio \( \frac{x_1}{x_2} \)**:
\[ \frac{x_1^2}{x_2^2} = 2 \implies \frac{x_1}{x_2} = \sqrt{2} \]

### Final Answer:
The ratio of the distances the springs are compressed is:
\[ \frac{x_1}{x_2} = \sqrt{2} \]