# How to calculate the height of liquid in a communicating vessel after another liquid is added [closed]

A u-pipe is filled with liquid $A$ (density: $1.2\ \mathrm{g/cm^3}$. When the left leg of the pipe is filled with liquid $B$ $3.75\ \mathrm{cm}$ high, the liquid’s level in the left leg falls $2.5\ \mathrm{cm}$.

So is the height of liquid $A$ $2.5\ \mathrm{cm}$ or $5\ \mathrm{cm}$ ($2.5 \times 2$)?

Let $h_0$ be the initial level in both arms. Now we add the $3.75\ \mathrm{cm}$ (call it $h_1$) of $B$ to the right hand side and the level of $A$ in the right arm drops by $h$ ($2.5\ \mathrm{cm}$).

As $A$ is incompressible the level in the left arm must also rise by the same amount $h$.

We can now calculate the hydrostatic pressure using Pascal's law at the very bottom of the U tube, with $P_0$ the atmospheric pressure:

From the Left hand side:

$$P=P_0+\rho_Ag(h_0+h)$$

From the Right hand side:

$$P=P_0+\rho_Ag(h_0-h)+\rho_Bgh_1$$

Equate and the $P_0$ and $g$ drop out: $$\rho_A(h_0+h)=\rho_A(h_0-h)+\rho_Bh_1$$

From which $\rho_B$ can easily be calculated:

$$\rho_B=\rho_A\frac{2h}{h_1}$$

• Thanks. I only used mental visualization to get (2h). I wasn't sure with my conclusion. Now your equation supports the conclusion of my imagination. Nov 20, 2017 at 21:57
• I made an edit. It's better to consider hydrostatic pressure than column weight, even though they are very closely related. The result remains the same.
– Gert
Nov 20, 2017 at 22:40