Question

A U-tube having horizontal arm length $20cm$, has uniform cross-sectional area=$1{cm}^2$. It is filled with water of volume of $60cc$. What minimum volume of a liquid of density $4g/cc$ should be poured from one side into the U-tube so that no water is left in the horizontal arm of the tube?

• A. $60cc$
• B. $45cc$
• C. $50cc$
• D. $35cc$

Answer 1

The correct option is D $35cc$

Given that the U-tube is filled with water of volume $60cc$ and the horizontal length is given as $20cm$. So $20cc$ of water will be in the horizontal part and the remaining $40cc$ will be in the two vertical tubes. For water to be equilibrium the $40cc$ is distributed as $20cc$ in each of the vertical tubes. That is depicted in Case 1 of the diagram.
Now another liquid of higher density is added so that it displaces the water in the horizontal arm. This is shown in case 2 of the diagram. Now the pressure at points A and B is equal and that is atmospheric pressure $P_o$.
Pressure at point X$=P_o+{\rho }_1\times g\times h.$
Pressure at point Y $=P_o+\rho \times g\times 60.$
where $g$= gravitational constant
$\rho$= density of water
${\rho }_1$ =density of liquid added
But X and Y points lie on the same horizontal plane with respect to ground.
So pressure at X = pressure at Y
$\Rightarrow P_o+{\rho }_1\times g\times h=P_o+\rho \times g\times 60$
$\Rightarrow {\rho }_1\times h=\rho \times 60$
$\Rightarrow 4g/cc\times h=1g/cc\times 60cm$
$\Rightarrow h=15cm.$
Therefore volume of liquid in vertical tube is $15cc$ and volume of liquid of in horizontal arm is $20cc$. Therefore total volume of liquid that needs to be added so that no water remains in horizontal arm is $20+15=35cc.$