Question

If the density of the heavy liquid is 10 g/cm^3

Answer 1

1) Height of the column of light liquid in the right arm: 12.6 cm

2) Height of heavy liquid increases by 0.64 cm

Explanation:

1)

The density of a substance is given by

$\rho=\frac{m}{V}$

where

m is the mass of the substance

V is its volume

For a liquid in a tube, the volume can be written as

$V=AL$

where

A is the cross-sectional area of the tube

L is the length of the column of liquid

Therefore the formula can be rewritten as

$\rho = \frac{m}{AL}$

For the light liquid in the right arm of the tube, we have:

m = 92 g (mass)

$\rho=1.7 g/cm^3$ (density)

$A=4.29 cm^2$ (cross-sectional area)

Therefore we can re-arrange the equation to find L, the height of the column of light liquid:

$L=\frac{m}{A\rho}=\frac{92}{(4.29)(1.7)}=12.6 cm$

2)

The gauge pressure of a column of liquid is given by:

$p=\rho g h$

where

$\rho$ is the density of the liquid

g is the acceleration due to gravity

h is the height of the column of liquid

When the light liquid is added on the right side, a certain volume of heavy liquid $A_2 h_2$ is displacement from the right arm to the right arm, and its volume now becomes $A_1 h_1$, where

$A_1 = 10.2 cm^2$ is the cross-sectional area of the left arm

$h_1$ is the change in height of the heavy liquid on the left arm

$A_2 = 4.29 cm^2$ is the cross-sectional area of the right arm

$h_2$ is the initial height of the heavy liquid displaced from the right arm

Since the volume of the liquid must remain the same,

$A_1 h_1 = A_2 h_2\\\rightarrow h_2 = \frac{A_1}{A_2}h_1$ (1)

Moverover, at equilibrium we know that the pressure on the two arms of the U-tube must be the same, so:

$\rho_1 g (h_1 + h_2 )= \rho_2 g L$ (2)

where here we have:

$\rho_1 = 10 g/cm^3$ (density of the heavy liquid)

$\rho_2 = 1.7 g/cm^3$ (density of the light liquid)

L = 12.6 cm (height of the light liquid in the right arm)

And substituting (1) into (2) and re-arranging for h1, we find by how much the level of heavy liquid increases in the left arm:

$\rho_1 (h_1 + \frac{A_1}{A_2}h_1) = \rho_2 L\\h_1 = \frac{\rho_2 L}{\rho_1 (1+\frac{A_1}{A_2})}=\frac{(1.7)(12.6)}{(10)(1+\frac{10.2}{4.29})}=0.64 cm$

Learn more about pressure in a liquid:

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