A

(i) Maximum compression in the spring:
When the block of mass m is released from rest and falls under the influence of gravity, it acquires velocity until it reaches the maximum compression of the spring. At maximum compression, all the gravitational potential energy will have been converted into elastic potential energy of the spring.

1. The gravitational potential energy (PE) of the block at height h is given by:
$$PE = mgh$$ where g is acceleration due to gravity.
2. The elastic potential energy (EPE) stored in the spring at maximum compression (x) is given by:
$$EPE = \frac{1}{2}kx^2$$
3. At maximum compression, these energies equal each other:
$$mgh = \frac{1}{2}kx^2$$
4. The height h is equal to the compression x (assuming the spring is initially uncompressed and the block falls a distance equal to x).
5. Therefore, we can rewrite the equation as:
$$mgx = \frac{1}{2}kx^2$$
6. Rearranging gives:
$$kx^2 - 2mgx = 0$$
7. Factoring out x gives:
$$x(kx - 2mg) = 0$$
8. Ignoring the solution x = 0 (the spring is indeed compressed), we have:
$$kx - 2mg = 0$$
$$x = \frac{2mg}{k}$$

Answer: The maximum compression in the spring will be \( \frac{2mg}{k} \).

(ii) Compression in the spring at equilibrium:
At equilibrium, the forces on the block must balance, which means that the spring force equals the weight of the block:
1. The spring force is given by Hooke's law as:
$$F_{spring} = kx_{eq}$$
where \( x_{eq} \) is the compression at equilibrium.
2. The weight of the block is:
$$F_{weight} = mg$$
3. Setting these equal gives:
$$kx_{eq} = mg$$
4. Solving for \( x_{eq} \):
$$x_{eq} = \frac{mg}{k}$$

Answer: At equilibrium, the compression in the spring will be \( \frac{mg}{k} \).