A U-shaped tube, open to the air on both ends, contains mercury. Water is poured into the left arm until the water column is $10.0cm$deep. How far upward from its initial position does the mercury in the right arm rise?

We know that water and mercury are liquids that are both incompressible and immiscible.

The diagram which can support the fact is given below :

As we know that the water floats on top of the mercury in the left arm, pushing it down from its initial position.

This will happen because the mercury levels are identical at the point$1$and $2$in the figure, the pressure at the point $1$and $2$in the figure must be equal.

Hence, we can say that if the pressures at the sites and are not equal, the fluid will flow in a way that violates the static equilibrium requirement.

In the figure, the pressure at the point $1$is due to water having depth of $10cm$.

$P_1=P_{atm}+{\rho }_{water}gd............\left(1\right)$

Here, $p_{atm}$is the atmospheric pressure $P_1$is the pressure at point $1$, ${\rho }_{water}$is the density of the water, $g$$g$is the acceleration,

due to gravity, and $d_{water}$is the depth of water.

Because mercury is an incompressible fluid, it will travel down a distance $x$in the left arm while travelling up a distance $x$in the right arm.

This implies that the pressure at point $2$is due to the mercury having a depth of $2x$.

Now, we can see that the pressure at point $2$is $P_2=P_{atm+}{\rho }_{Hg}g{d}_{Hg}$

In the above expression here, $P_{atm}$is the atmospheric pressure $P_2$is the pressure at point $2$, ${\rho }_{Hg}$is the density of the mercury. $g$is the acceleration due to gravity, and $d_{Hg}$is the depth of mercury.

Substitute $2x$for $d_{Hg}$$2x$

$P_2=P_{atm}+2{\rho }_{\mathrm{Hg}}gx$

Equate the equation $\left(1\right)$and $\left(2\right)$

$P_{atm}+{\rho }_{water}g{d}_{water}=P_{atm}+2{\rho }_{Hg}gx$

Rearrange the above equation in terms of $x$

$x=\frac{1}{2}\frac{{\rho }_{water}}{{\rho }_{\mathrm{Hg}}}{d}_{water}$

Substitute $1000kg/m^3$for ${\rho }_{water}$,$13,600kg/m^3$for ${\rho }_{Hg}$,$10cm$for $d_{water}$

$x=\frac{1}{2}\frac{\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)}{\left(13,600\mathrm{kg}/{\mathrm{m}}^3\right)}(10\mathrm{cm})$

$=3.68mm$

The temperature in the right arm has increased $3.68mm$from its original

position.