Question

Fluid Pressure in a U-Tube

A U-tube is filled with water, and the two arms are capped.(Figure 1) The tube is cylindrical, and the right arm has twice the radius of the left arm. The caps have negligible mass, are watertight, and can freely slide up and down the tube.

The locations of the two caps at equilibrium are now as given in this figure. (Figure 4) The dashed line represents the level of the water in the left arm.

What is the mass of the water located in the right arm between the dashed line and the right cap?

Express your answer in kilograms.

Fluid Pressure in a U-Tube A U-tube is filled with water, and the two arms are capped.(Figure 1) The tube is cylindrical, and the right arm has twice the radius of the left arm. The caps have negligible mass, are watertight, and can freely slide up and down the tube. The locations of the two caps at equilibrium are now as given in this figure. (Figure 4) The dashed line represents the level of the water in the left arm. What is the mass of the water located in the right arm between the dashed line and the right cap?

Answer 1

Concepts and reason

The concept required to solve the given problem is pascal’s law.

First calculate the area of both the arms of the U tube.

Then, calculate the force acting on both the arms.

Finally substitute the values of force and area thus obtained into the expression relating the ratio of the force acting on the two arms with the ratio of the areas of the two arms.

Fundamentals

Pressure: It is given by,

$P = \frac{F}{A}$

Here, $F$ is the force and $A$ is the area.

Pascal’s Law: It states that if the pressure in a liquid is changed at a particular point, the change is transmitted to the entire liquid without being diminished in magnitude.

According to Pascal’s law, the pressure in both the arms would be same at the same horizontal level.

Hence,

${P_{left}} = {P_{right}}$ ……(1)

Pressure is given by,

$P = \frac{F}{A}$

Here, $F$ is the force and $A$ is the area.

Thus, equation (1) becomes,

$\frac{{{F_L}}}{{{A_L}}} = \frac{{{F_R}}}{{{A_R}}}$ ……(2)

The force acting on the left arm will be,

${F_L} = mg$

The force acting on the right arm will be,

${F_R} = {M_{water}}g + mg$

Here, $\rho$ is the density of water, $g$ is acceleration due to gravity and $h$ is the height of water above the dashed line in the right arm.

Substitute $mg$ for ${F_L}$ and ${M_{water}}g + mg$ for ${F_R}$ in equation (2).

$\frac{{mg}}{{{A_L}}} = \frac{{{M_{water}}g + mg}}{{{A_R}}}$ ……(3)

The area of cross section of the left arm will be,

${A_L} = \pi {R_L}^2$ ……(4)

The area of cross section of the right arm will be,

${A_R} = \pi {R_R}^2$ ……(5)

Also, it is given that the radius of the right arm has twice the radius of the left arm.

Thus, ${R_R} = 2{R_L}$

Substitute $2{R_L}$ for ${R_R}$ in equation (5).

$\begin{array}{c}\\{A_R} = \pi {\left( {2{R_L}} \right)^2}\\\\ = 4\pi {R_L}^2\\\end{array}$

The area of cross section of the left arm is, ${A_L} = \pi {R_L}^2$.

The area of cross section of the right arm is,${A_R} = 4\pi {R_L}^2$.

Substitute $\pi {R_L}^2$ for ${A_L}$ and $4\pi {R_L}^2$for ${A_R}$ in equation (3).

$\begin{array}{c}\\\frac{{mg}}{{\pi {R_L}^2}} = \frac{{{M_{water}}g + mg}}{{4\pi {R_L}^2}}\\\\{M_{water}} = 3m\\\end{array}$

Substitute $1{\rm{ kg}}$ for $m$ in the above equation.

$\begin{array}{c}\\{M_{water}} = 3\left( {1{\rm{ kg}}} \right)\\\\ = 3{\rm{ kg}}\\\end{array}$

Ans:

The mass of water located in the right arm between the dashed line and right cap is $3{\rm{ kg}}$.