A U-tube of constant cross-sectional area,
open to the atmosphere, is partially filled
with a heavy liquid with density 9.59 g/cm3
A light liquid with density 1.83 g/cm3
is then poured into both arms.
If the equilibrium configuration of the tube
is as shown, with a difference in the height of
the heavy liquid of 0.721 cm, determine the
value of the difference in height of the light
liquid hℓ; Answer in units of cm.

The hydrostatic pressure is equal at the bottom of each arm. This hydrostatic pressure is equal to the product of the density of the liquid layer, acceleration due to gravity, and the depth of the layer.

Define some variables:

ρ₁ is the density of the light liquid,

ρ₂ is the density of the heavy liquid,

h is the height of the light liquid in the right arm,

and hℓ is the height difference between the arms.

Pressure at the bottom of the left tube equals the pressure at the bottom of the right tube.

P = P

ρ₁ g (0.721 + h + hℓ) = ρ₂ g (0.721) + ρ₁ g h

ρ₁ (0.721) + ρ₁ h + ρ₁ hℓ = ρ₂ (0.721) + ρ₁ h

0.721 ρ₁ + ρ₁ hℓ = 0.721 ρ₂

ρ₁ hℓ = 0.721 (ρ₂ − ρ₁)

hℓ = 0.721 (ρ₂ − ρ₁) / ρ₁

Plug in values:

hℓ = 0.721 (9.59 − 1.83) / 1.83

hℓ = 3.06 cm