Question

Two identical containers are open at the top and are connected at the bottom via a tube of negligible volume and a valve that is closed. Both containers are filled initially to the same height of 1.00 m, one with water, the other with mercury, as the drawing indicates. The valve is then opened. Water and mercury are immiscible. Determine the fluid level in the left container when equilibrium is reestablished.

Answer 1

Given

Initial height of water and mercury h = 1 m

Known

Density of water ρ_w = 1000 Kg/m³

Density of mercury ρ_m = 13,594 Kg/m³

Solution

Before the valve is opened

The pressure on the mercury column

P_m = ρ_mgh

P_m = 13594 x 9.8 x 1

P_m = 133221.2 Pa

The pressure on the water column

P_w = ρ_wgh

P_w = 1000 x 9.8 x 1

P_w = 9800 Pa

Since mercury column has more pressure mercury will flow into water column until equilibrium is obtained

Pressure difference

ΔP = P_m - P_w

Δ P = 133221.2 – 9800

ΔP = 123421.2 Pa

When half of this pressure difference is given to water column, the equilibrium will be obtained

ΔP/2 = 61710.6 Pa

Mercury of pressure 61710.6 Pa will flow into water column, the height of mercury associated with this amount of pressure

ρ_mgh_m = 61710.6

13594 x 9.8 x h_m = 61710.6

h_m = 0.463 m

This is the increased height in water column

Hence the new height in the water column

H = h + h_m

H = 1 + 0.463

H = 1.463 m