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
[tex]\rho=\frac{m}{V}[/tex]
where
m is the mass of the substance
V is its volume
For a liquid in a tube, the volume can be written as
[tex]V=AL[/tex]
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
[tex]\rho = \frac{m}{AL}[/tex]
For the light liquid in the right arm of the tube, we have:
m = 92 g (mass)
[tex]\rho=1.7 g/cm^3[/tex] (density)
[tex]A=4.29 cm^2[/tex] (cross-sectional area)
Therefore we can re-arrange the equation to find L, the height of the column of light liquid:
[tex]L=\frac{m}{A\rho}=\frac{92}{(4.29)(1.7)}=12.6 cm[/tex]
2)
The gauge pressure of a column of liquid is given by:
[tex]p=\rho g h[/tex]
where
[tex]\rho[/tex] 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 [tex]A_2 h_2[/tex] is displacement from the right arm to the right arm, and its volume now becomes [tex]A_1 h_1[/tex], where
[tex]A_1 = 10.2 cm^2[/tex] is the cross-sectional area of the left arm
[tex]h_1[/tex] is the change in height of the heavy liquid on the left arm
[tex]A_2 = 4.29 cm^2[/tex] is the cross-sectional area of the right arm
[tex]h_2[/tex] is the initial height of the heavy liquid displaced from the right arm
Since the volume of the liquid must remain the same,
[tex]A_1 h_1 = A_2 h_2\\\rightarrow h_2 = \frac{A_1}{A_2}h_1[/tex] (1)
Moverover, at equilibrium we know that the pressure on the two arms of the U-tube must be the same, so:
[tex]\rho_1 g (h_1 + h_2 )= \rho_2 g L[/tex] (2)
where here we have:
[tex]\rho_1 = 10 g/cm^3[/tex] (density of the heavy liquid)
[tex]\rho_2 = 1.7 g/cm^3[/tex] (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:
[tex]\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[/tex]
Learn more about pressure in a liquid:
brainly.com/question/4868239
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