# Liquid Pressure Numericals

Question 1) A swimming pool contains water with a density of 10^{3}kg m^{–3}. Calculate the pressure due to the water at a depth of 4.5 m. Ignore the effect of the atmosphere.

**Solution**

Pressure P = ρgh = 1000 × 9.81 × 4.5 = 44145 Pa = 44.14 kPa (Answer)

**Note for students**: If we include the pressure of the atmosphere pressing on the water’s surface, then we would need to add another 101 kPa to this answer.

Question 2) a) Calculate, in Pa, the pressure difference that will cause a U-tube manometer to have a height difference of 3.0 cm of mercury pressure between the arms. The density of mercury is 1.36 × 10^{4}kg m^{–3}.

b) State this pressure difference as a percentage of atmospheric pressure. Atmospheric pressure is 101 kPa.

**Solution**

a) The height difference is 0.030 m. So Δp = ρgΔh = 1.36 × 10^{4} × 9.81 × 3 × 10^{–2} = 4.00 × 10^{3} Pa.

This should be stated as 4.0 kPa.

b) Pressure difference as a percentage of atmospheric pressure = 4/101× 100% = 4%