The density of water is approximately 1/13.5 the density of mercury. What height would be maintained by a column of water inverted in a dish of water at sea level?

10,300 mm or 10.3 meter

Explanation:

1) The hydrostatic pressure (pressure of a column of fluid) is equal to:

P = gρh

Where:

P is the hydrostatic pressure ρ is the density of the fluid h is the height of the column of fluid. g is gravity acceleration (about 9.81 m/s²)

2) Using subscript 1 for water and subscript 2 for mercury, the hydrostatic equations for both water and mercury will be:

P₁ = gρ₁h₁ P₂ = gρ₂h₂

3) For water and mercury columns be in equilibrium P₁ = P₂; and you get:

gρ₁h₁ = gρ₂h₂

⇒ ρ₁h₁ = ρ₂h₂

⇒ h₁ = ρ₂h₂ / ρ₁

⇒ h₁ = h₂ (ρ₂ / ρ₁)

it is given that the density of water is approximately 1/13.5 the density of mercury, which is ρ₁ / ρ₂ = 1 / 13.5. Hence, ρ₂ / ρ₁ = 13.5 / 1

⇒ h₂ = h₁ (13.5 / 1)

4) At sea level, the atmospheric pressure is 1 atm or 760 mmHg, which means that the height of a colum of mercury is 760 mmHg.

So, you calculate the height maintained by a column of water inverted in a dish of water at sea level would be:

h₂ = h₁ (ρ₂ / ρ₁) = 760 (13.5 / 1) mm = 10,260 mm.

Using 3 significant figures that is 10,300 mm or 10.3 meter.

Then, you can tell that the atmospheric pressure (760 mmHg) would be capable of mantaining a column of water 10.3 meters in height.