A cork is submerged in water by a spring attached to the bottom of a bowl. When the bowl is kept in an elevator moving with acceleration downwards, the length of spring

The correct option is Decreases.

Let's analyze the forces acting on the cork when the bowl is at rest:
1. The buoyant force acting vertically upwards:
$F_B=V·{\rho }_w·g$
where $V$ is the volume of the cork, ${\rho }_w$ is the density of water, and $g$ is the acceleration due to gravity.
2. The gravitational force (weight) acting vertically downwards:
$F_g=V·{\rho }_c·g$
where ${\rho }_c$ is the density of the cork.
3. The spring force acting vertically downwards (since the density of cork is less than that of water, the cork naturally wants to rise to the surface, which stretches the spring downwards):
$F_s=k·x$
where $k$ is the spring constant and $x$ is the elongation of the spring.

In equilibrium at rest, the upward forces balance the downward forces:
$F_B=F_g+F_s$
Substituting the expressions gives:
$V·{\rho }_w·g=V·{\rho }_c·g+k·x$
Solving for the extension $x$:
$k·x=V·({\rho }_w-{\rho }_c)·g$

When the bowl is kept in an elevator accelerating downwards with an acceleration $a$, the system is in an accelerating frame. The effective acceleration due to gravity becomes:
$g_{\mathrm{eff}}=g-a$
Thus, the new elongation $\mathrm{x\text{'}}$ of the spring is:
$k·\mathrm{x\text{'}}=V·({\rho }_w-{\rho }_c)·(g-a)$

Since the term $(g-a)$ is less than $g$, the new elongation $\mathrm{x\text{'}}$ is smaller than the initial elongation $x$:
$\mathrm{x\text{'}}<x$
Because the extension of the spring decreases, the overall length of the spring decreases.