In the above question, the velocity of the image of `Q` in plane mirror of block `P` with respect to ground when `Q` is at the lowest point first time is:

Options

A. `sqrt((2mg^(2))/(K))`
B. `sqrt((4mg^(2))/(K))`
C. `sqrt((3mg^(2))/(K))`
D. 0

Explanation

To solve the problem, we need to analyze the motion of the masses P and Q attached to their respective springs and determine the velocity of the image of Q in the plane mirror of block P when Q is at its lowest point for the first time.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have two masses, P and Q, attached to springs with spring constants 4K and K, respectively.
- Both masses can oscillate independently, executing simple harmonic motion (SHM).

2. **Finding the Time Periods**:
- The time period (T) for a mass-spring system is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
- For mass P (attached to spring constant 4K):
\[ T_P = 2\pi \sqrt{\frac{M}{4K}} = \pi \sqrt{\frac{M}{K}} \]
- For mass Q (attached to spring constant K):
\[ T_Q = 2\pi \sqrt{\frac{M}{K}} \]
- From this, we can see that:
\[ T_Q = 2 T_P \]
- This means the time period of Q is twice that of P.

3. **Understanding the Motion**:
- When Q is at its lowest point for the first time, it has completed half of its oscillation (downward motion).
- Since T_Q is twice T_P, it implies that while Q is at its lowest point, mass P would have completed its full oscillation (up and down).

4. **Velocity at the Lowest Point**:
- At the lowest point of its motion, the velocity of mass Q is zero (as it momentarily stops before changing direction).
- The velocity of mass P at this moment can be determined by its motion characteristics. Since P oscillates faster, it would be at its mean position when Q is at its lowest point.

5. **Velocity of P**:
- The velocity of mass P at its mean position is at its maximum.
- The maximum speed \( v_{max} \) for a mass in SHM is given by:
\[ v_{max} = \omega A \]
- Where \( \omega = \frac{2\pi}{T} \) and \( A \) is the amplitude of oscillation.

6. **Relative Velocity**:
- Since Q is at its lowest point (velocity = 0) and P is at its mean position (velocity = maximum), the velocity of the image of Q in the mirror of block P with respect to the ground is equal to the velocity of P.
- Therefore, the velocity of the image of Q is equal to the velocity of P.

### Final Answer:
The velocity of the image of Q in the plane mirror of block P with respect to the ground when Q is at the lowest point for the first time is equal to the maximum velocity of mass P.