Two communicating vessels contain mercury. The diameter of one vessel is n times larger than the diameter of the other. A column of water of height h is poured into the left vessel. The mercury level will rise in the right-hand vessel (s = relative density of mercury and ρ = density of water) by

Let the left vessel have a diameter $d_1$ and the right vessel have a diameter $d_2$. According to the problem, the diameter of one vessel is $n$ times larger than the diameter of the other. Let the right vessel be the wider one, so that its diameter is $n$ times the diameter of the left vessel:
$d_2=n{d}_1$

The cross-sectional area of a vessel is proportional to the square of its diameter. Thus, the cross-sectional area of the left vessel $A_1$ and the right vessel $A_2$ are related by:
$A_2=n^2{A}_1$

Initially, the mercury levels in both communicating vessels are at the same horizontal level. When a column of water of height $h$ is poured into the left vessel, it depresses the mercury surface in the left vessel by some height $x_1$.

Because mercury is incompressible, the volume of mercury depressed in the left vessel must be equal to the volume of mercury that rises in the right vessel. If the mercury level in the right vessel rises by a height $x_2$, we have:
$A_1{x}_1=A_2{x}_2$

Substituting $A_2=n^2{A}_1$ into the volume conservation equation gives:
$A_1{x}_1=n^2{A}_1{x}_2\Rightarrow x_1=n^2{x}_2$

Now, let us consider the pressure at the new interface between water and mercury in the left vessel. This interface is at a depth $x_1$ below the initial mercury level.
The total pressure at this interface in the left column is due to the water column of height $h$:
$P_{\text{left}}=P_0+\rho gh$
where $P_0$ is the atmospheric pressure, $\rho$ is the density of water, and $g$ is the acceleration due to gravity.

In the right vessel, the mercury level has risen by $x_2$ above the initial level. Therefore, the height of the mercury column above the interface level in the left vessel is:
$H=x_1+x_2$

The pressure at the horizontal level of the interface in the right column is:
$P_{\text{right}}=P_0+{\rho }_{m}g(x_1+x_2)$
where ${\rho }_m$ is the density of mercury.

Equating the pressures at the same horizontal level ($P_{\text{left}}=P_{\text{right}}$):
$P_0+\rho gh=P_0+{\rho }_{m}g(x_1+x_2)$
Subtracting $P_0$ and dividing by $g$ on both sides gives:
$\rho h={\rho }_m(x_1+x_2)$

Using the relative density of mercury $s=\frac{{\rho }_m}{\rho }$, we can rewrite ${\rho }_m$ as $s\rho$:
$\rho h=s\rho (x_1+x_2)$
Dividing both sides by $\rho$ yields:
$h=s(x_1+x_2)$

Substitute $x_1=n^2{x}_2$ into the equation:
$h=s(n^2{x}_2+x_2)$
$h=s{x}_2(n^2+1)$

Solving for the rise in the mercury level in the right-hand vessel, $x_2$:
$x_2=\frac{h}{(n^2+1)s}$

Thus, the mercury level in the right-hand vessel will rise by $\frac{h}{(n^2+1)s}$.