A solid cylinder of mass m and volume v is suspended from ceiling by a spring of spring constant k. It has cross-section area A. It is submerged in a liquid of density $ρ$ upto half its length. If a small block of mass $M_{0}$ is kept at the centre of the top, the amplitude of small oscillation will be :-

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Before the added block, spring force and buoyancy balance the cylinder weight.
An extra downward displacement x makes the spring force rise by:
$k x$
and increases the displaced liquid volume by Ax, adding buoyant force:
$ρ A gx$
The added block contributes downward force:
$M_{0} g$
Hence the displacement from the old equilibrium is:
$(k + ρ A g) x = M_{0} g$
$x = \frac{M _{0} g}{k + ρ A g}$
This is the oscillation amplitude about the new equilibrium: B.

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JEE Main 2027 (Class 12+)