IGCSE Physics 0625: Calculating Liquid Pressure with Δp = ρgΔh

Two swimmers dive into different pools. Take the density of the fresh water in the first pool as $1000\text{ kg/m}^3$ , the density of the salt water in the second pool as $1030\text{ kg/m}^3$ , the gravitational field strength as $g = 10\text{ N/kg}$ , and atmospheric pressure at the surface of both pools as $1.0\times10^{5}\text{ Pa}$ .

(a) The first swimmer dives to a depth of $2.5\text{ m}$ below the surface of the fresh-water pool. Calculate the increase in pressure, due to the water alone, at this depth. [3]

(b) Calculate the total pressure (due to the water and the atmosphere together) acting on the first swimmer at this depth. [2]

(c) The second swimmer dives to a depth of only $1.0\text{ m}$ , but in the salt-water pool. Calculate the increase in pressure, due to the water alone, at this depth. [2]

(d) Using your answers to (a) and (c), state which swimmer experiences the greater increase in pressure due to the water above them. [1]

Part (a): Pressure increase for the first swimmer

Use $\Delta p = \rho g \Delta h$ with $\rho = 1000\text{ kg/m}^3$ , $g = 10\text{ N/kg}$ and $\Delta h = 2.5\text{ m}$ :

$\Delta p = 1000 \times 10 \times 2.5 = 25\,000\text{ Pa} = 2.5\times10^{4}\text{ Pa}$

Part (b): Total pressure on the first swimmer

The total pressure is the pressure due to the water plus the atmospheric pressure already pushing down at the surface:

$p_{\text{total}} = p_{\text{atmosphere}} + \Delta p = 1.0\times10^{5}\text{ Pa} + 2.5\times10^{4}\text{ Pa} = 1.25\times10^{5}\text{ Pa}$

Part (c): Pressure increase for the second swimmer

Use $\Delta p = \rho g \Delta h$ with $\rho = 1030\text{ kg/m}^3$ , $g = 10\text{ N/kg}$ and $\Delta h = 1.0\text{ m}$ :

$\Delta p = 1030 \times 10 \times 1.0 = 10\,300\text{ Pa} = 1.03\times10^{4}\text{ Pa}$

Part (d): Comparing the two swimmers

Comparing the two results:

$25\,000\text{ Pa} \; (\text{first swimmer}) \quad \text{vs} \quad 10\,300\text{ Pa} \; (\text{second swimmer})$

Even though the second swimmer is in the denser salt water, the first swimmer is over twice as deep, and depth has the larger effect here. So the first swimmer experiences the greater increase in pressure due to the water above them.

Final answers

(a) Pressure increase (first swimmer) $=$ $2.5\times10^{4}\text{ Pa}$
(b) Total pressure (first swimmer) $=$ $1.25\times10^{5}\text{ Pa}$
(c) Pressure increase (second swimmer) $=$ $1.03\times10^{4}\text{ Pa}$
(d) The first swimmer experiences the greater increase in pressure due to the water