A body is attached to the lower end of a vertical spiral spring and it is gradually lowered to its equilibrium position. This stretches the spring by a length x. If the same body attached to the same spring is allowed to fall suddenly, what would be the maximum stretching in this case

The correct option is 2x.

To understand the difference in the spring's stretching in both cases, we can analyze them individually using the laws of mechanics and energy conservation.

Case 1: Gradual lowering to equilibrium
When a body of mass $m$ attached to a spring of spring constant $k$ is lowered gradually, it is kept in equilibrium at every step by supporting it externally until it reaches its final equilibrium position. At this position, the downward gravitational force acting on the body is exactly balanced by the upward restoring force exerted by the spring.
Let $x$ be the stretching in the spring. Applying the equilibrium condition:
$kx=mg$
This gives the equilibrium stretch as:
$x=\frac{mg}{k}$

Case 2: Sudden release (falling suddenly)
When the body is allowed to fall suddenly from the unstretched position, it undergoes oscillation. As the body falls, it accelerates until it passes the equilibrium position, and then starts to decelerate until it momentarily stops at the position of maximum stretching. Let $x_m$ be this maximum stretching of the spring.
Since no non-conservative forces do work, we can apply the law of conservation of mechanical energy between the initial release point (where the spring is unstretched and the body is at rest) and the point of maximum stretching (where the body momentarily comes to rest again).
At both of these positions, the kinetic energy of the body is zero. Therefore:
Loss in gravitational potential energy = Gain in elastic potential energy of the spring
$mg{x}_m=\frac{1}{2}k{x}_m^2$
Since the maximum stretching $x_m$ is non-zero, we can divide both sides by $x_m$:
$mg=\frac{1}{2}k{x}_m$
Solving for $x_m$:
$x_m=\frac{2mg}{k}$

Comparing this with the equilibrium stretch $x=\frac{mg}{k}$ from Case 1:
$x_m=2x$

Thus, the maximum stretching of the spring when the body is allowed to fall suddenly is twice the stretching produced when it is lowered gradually.