Question

A tube as shown below with constant cross-section throughout is filled with tube liquids for density $\rho \text{ and 2}\rho$ separated by a line-plug. What happens to the height in each of the columns when this plug is removed?

Answer 1

We should understand the pressure difference due to the density difference in the two media separated by the plug. The density difference will change the height of the liquid column on the two sides of the tube which can be found from this information.

Complete answer:
We are given a U-shaped tube as shown in the figure.

The two liquids of densities $\rho \text{ and 2}\rho$ are at an equal height ‘h’ divided by the plug as shown. Let ${{\rho }_{0}}$ be the atmospheric pressure acting as the tube is kept open. The pressure on the left-side can be defined as –
${{P}_{1}}={{\rho }_{0}}-\rho gh$
The right-side can be defined as –
${{P}_{2}}={{\rho }_{0}}-2\rho gh$
The pressures at the same height are equal. This gives –
${{P}_{1}}={{P}_{2}}$
This is the situation when the height is the same.
Now, let us consider the condition when the plug is removed. In this case, there will be a pressure difference as the height attained will vary, such that –

& {{P}_{1}}=\rho g{{h}_{1}} \\\ & \Rightarrow \text{ }{{h}_{1}}=\dfrac{{{P}_{1}}}{\rho g} \\\ & {{P}_{2}}=2\rho g{{h}_{2}} \\\ & \Rightarrow \text{ }{{h}_{2}}=\dfrac{{{P}_{2}}}{2\rho g} \\\ \end{aligned}$$\  From the above relation we can understand that the height of the liquid column is inversely proportional to the density of the liquid. Also, the height has a constraint – $$\Rightarrow 2h={{h}_{2}}+{{h}_{1}}$$ But, $$\rho \propto \dfrac{1}{h}$$ This implies that the heights are related by the ratio – $${{h}_{1}}:{{h}_{2}}\text{ }\Rightarrow \text{ }\dfrac{1}{\rho }:\dfrac{1}{2\rho }=2:1$$ **The heights are related in the ratio 2:1.** **Note:** The height achieved by a liquid in the column is inversely proportional to the density of the medium. The surface tension is neglected in this case or assumed to have equal values. The height of the liquid in left-side will be higher than the initial height.