A wire of radius r, Young’s modulus Y and length l is hung from a fixed point and supports a heavy metal cylinder of volume V at its lower end. The change in length of wire when cylinder is immersed in a liquid of density ρ is in fact

The correct option is: Decrease by Vlρg/Yπr²

Let us derive the solution step-by-step:
Initially, the wire of length $l$, radius $r$, and Young's modulus $Y$ supports a heavy metal cylinder of volume $V$.
Let the mass of the cylinder be $M$. The downward gravitational force acting on the cylinder is:
$F_1=Mg$

When the cylinder is completely immersed in a liquid of density $\rho$, it experiences an upward buoyant force (Archimedes' principle) equal to the weight of the liquid displaced by the cylinder.
The buoyant force is given by:
$F_b=V\rho g$

Therefore, the net downward tension (force) acting on the wire when the cylinder is immersed in the liquid becomes:
$F_2=Mg-F_b=Mg-V\rho g$

The change in the force stretching the wire is:
$\Delta F=F_2-F_1=-V\rho g$
The negative sign indicates that the stretching force on the wire decreases by $V\rho g$.

According to Hooke's law, Young's modulus $Y$ is defined as:
$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\Delta l/l}$
where $A=\pi r^2$ is the cross-sectional area of the wire.

Rearranging the formula for the change in length $\Delta l$:
$\Delta l=\frac{Fl}{YA}=\frac{Fl}{Y\pi r^2}$

Since the stretching force decreases by $V\rho g$, the corresponding decrease in the extension (change in length) of the wire is:
$\mathrm{Decrease\; in\; length}=\frac{\left(V\rho g\right)l}{Y\pi r^2}=\frac{Vl\rho g}{Y\pi r^2}$

Thus, the change in length of the wire is a decrease by $\frac{Vl\rho g}{Y\pi r^2}$.