Hydrostatics — pressure = ρgh, derivation: Worked examples

The scenario matrix

Before working problems, let us list every distinct kind of situation the formula $P = P_{0} + ρ g h$ can produce. Each row is a "case class." The worked examples below are tagged with the cell they cover, and together they hit every row.

#	Case class	What is special about it	Covered by
A	Plain gauge pressure	Just $ρ g h$ , one liquid, ask "extra" pressure	Ex 1
B	Absolute pressure	Must remember to add $P_{0}$	Ex 2
C	Solve backwards for $h$	Pressure given, find depth	Ex 3
D	Solve backwards for $ρ$	Height + pressure given, find density	Ex 4
E	Two stacked liquids	Add each layer's $ρ g h$ separately	Ex 5
F	Slanted / bent tube	Only vertical height counts, not path length	Ex 6
G	Zero / degenerate input	$h = 0$ (at surface), or a vacuum top ( $P_{0} = 0$ )	Ex 7
H	Limiting / real-world	Very deep ocean; compare $ρ g h$ to $P_{0}$	Ex 8
I	Exam twist	Same-depth trick / hydrostatic paradox with numbers	Ex 9

Recall Quick self-test before you start

A tube is bent so the water travels 5 m along a slanted path but only rises 3 m vertically. Which number goes into $ρ g h$ ? ::: The vertical rise, 3 m. Path length is irrelevant.

Example 1 — Plain gauge pressure (Cell A)

Worked example The extra squeeze at 3 m

A diver reaches a depth of $h = 3 m$ in fresh water, $ρ = 1000 kg/m^{3}$ , $g = 9.8 m/s^{2}$ . Find the gauge pressure (pressure due to the water alone).

Forecast: Roughly a third of an atmosphere? A full atmosphere? Guess a power of ten before reading on.

Step 1 — Pick the right piece of the formula. Gauge pressure means "above atmosphere," which is exactly $ρ g h$ — no $P_{0}$ . Why this step? The word gauge tells us to drop $P_{0}$ ; only the water column matters.

Step 2 — Plug in SI numbers. $P_{gauge} = ρ g h = 1000 \times 9.8 \times 3 = 2.94 \times 1 0^{4} Pa$ Why this step? Every quantity is already in SI, so the product comes out in pascals directly.

Verify: Units: $\frac{kg}{m ^{3}} \cdot \frac{m}{s ^{2}} \cdot m = \frac{kg}{m \cdot s ^{2}} = \frac{N}{m ^{2}} = Pa$ ✓. And $2.94 \times 1 0^{4}$ is about $0.29$ of an atmosphere — sensible for only 3 m (recall $\sim 10 m$ = 1 atm).

Example 2 — Absolute pressure (Cell B)

Example 3 — Solve backwards for depth (Cell C)

Example 4 — Solve backwards for density (Cell D)

Worked example Mystery liquid in a tube

A liquid column of height $h = 0.758 m$ balances atmospheric pressure $P_{0} = 1.01 \times 1 0^{5} Pa$ (a barometer with vacuum on top). What is its density? Which liquid is it?

Forecast: Denser than water or lighter? (Hint: the column is short.)

Step 1 — Write the balance. Vacuum above means the top pressure is $0$ , so the column's $ρ g h$ must supply the full atmosphere: $ρ g h = P_{0}$ Why this step? This is the barometer condition — see Atmospheric Pressure & Barometer.

Step 2 — Solve for $ρ$ . $ρ = \frac{P _{0}}{g h} = \frac{1.01 \times 1 0 ^{5}}{9.8 \times 0.758}$ Why this step? Now $ρ$ is the lone unknown; divide it out.

Step 3 — Compute. $ρ = \frac{1.01 \times 1 0 ^{5}}{7.43} \approx 1.36 \times 1 0^{4} kg/m^{3}$

Verify: $1.36 \times 1 0^{4} = 13600 kg/m^{3}$ = mercury ✓ (see Density and Specific Gravity). A short heavy column — consistent with our "denser than water" expectation.

Example 5 — Two stacked liquids (Cell E)

Figure — Hydrostatics — pressure = ρgh, derivation

Worked example Oil floating on water

A tank holds $h_{1} = 2 m$ of oil ( $ρ_{1} = 800 kg/m^{3}$ ) floating on top of $h_{2} = 3 m$ of water ( $ρ_{2} = 1000 kg/m^{3}$ ). Find the gauge pressure at the very bottom.

Forecast: More or less than 5 m of pure water would give? Guess first.

Step 1 — Treat each layer separately. Pressure adds up as you descend through each layer. Look at the figure: the blue oil sits on the green water. $P_{bottom} = through oil ρ_{1} g h_{1} + through water ρ_{2} g h_{2}$ Why this step? Each layer contributes its own $ρ g h$ . The pressure at the oil–water boundary becomes the "surface pressure" for the water layer below it, so the contributions simply add.

Step 2 — Compute each piece. $ρ_{1} g h_{1} = 800 \times 9.8 \times 2 = 1.568 \times 1 0^{4} Pa$ $ρ_{2} g h_{2} = 1000 \times 9.8 \times 3 = 2.94 \times 1 0^{4} Pa$ Why this step? Plug SI numbers into each layer independently.

Step 3 — Add. $P_{bottom} = 1.568 \times 1 0^{4} + 2.94 \times 1 0^{4} = 4.508 \times 1 0^{4} Pa$

Verify: If all 5 m were water, we'd get $1000 \times 9.8 \times 5 = 4.9 \times 1 0^{4} Pa$ . Our answer $4.508 \times 1 0^{4}$ is a bit less — correct, because the top 2 m is lighter oil ✓.

Example 6 — Slanted tube (Cell F)

Worked example The tilted straw trap

Water fills a straight tube that runs slanted at $3 0^{\circ}$ below horizontal. The water travels a path length $L = 4 m$ along the tube. Find the gauge pressure at the lower end.

Forecast: Do we use $4 m$ , or something smaller?

Step 1 — Identify the true vertical depth. Only the vertical drop enters $ρ g h$ . From the figure, the vertical height is $h = L sin 3 0^{\circ} = 4 \times 0.5 = 2 m$ Why this step? $h$ in $ρ g h$ was defined as vertical depth below the surface. A slanted path of length $L$ that descends at angle $θ$ drops by $L sin θ$ — the red vertical arrow in the figure, not the blue slanted one.

Step 2 — Apply the formula. $P_{gauge} = ρ g h = 1000 \times 9.8 \times 2 = 1.96 \times 1 0^{4} Pa$ Why this step? With the correct vertical $h$ , it's the ordinary one-liquid formula.

Verify: If we (wrongly) used $L = 4 m$ , we'd get $3.92 \times 1 0^{4} Pa$ — exactly double, the classic mistake. The physical answer must use $h = 2 m$ , giving $1.96 \times 1 0^{4} Pa$ ✓.

Example 7 — Degenerate inputs (Cell G)

Worked example The two "nothing" cases

(a) What is the gauge pressure right at the free surface, $h = 0$ ? (b) In a mercury barometer the space above the column is a vacuum. What is $P_{0}$ (top pressure) there?

Forecast: Both should be a specific tidy number — what?

Step 1 — Case (a): set $h = 0$ . $P_{gauge} = ρ g \times 0 = 0 Pa$ Why this step? At the surface no fluid sits above you, so there is zero extra squeeze. This is the sanity anchor of the whole topic.

Step 2 — Case (b): a vacuum has no pressure. $P_{0} = 0 Pa$ Why this step? "Vacuum" literally means no gas, no molecules pushing, hence zero pressure. That's why in Example 4 the whole atmosphere had to be balanced by $ρ g h$ alone.

Verify: (a) The absolute pressure at the surface is then $P_{0} + 0 = P_{0}$ = atmosphere ✓, matching everyday life (air pressure at a pool's surface). (b) Setting $P_{0} = 0$ turns $P = P_{0} + ρ g h$ into $P = ρ g h$ , exactly the barometer equation we used ✓.

Example 8 — Limiting / real-world (Cell H)

Worked example Deep ocean crush

At the bottom of the Mariana Trench, $h \approx 11000 m$ , seawater $ρ \approx 1030 kg/m^{3}$ . Find the gauge pressure, and compare it to one atmosphere.

Forecast: Tens of atmospheres? Hundreds? Thousands?

Step 1 — Compute $ρ g h$ . $P_{gauge} = 1030 \times 9.8 \times 11000 = 1.110 \times 1 0^{8} Pa$ Why this step? Same one-liquid formula — depth is just enormous. Seawater is slightly denser than fresh water, hence $1030$ .

Step 2 — Express in atmospheres. $\frac{P _{gauge}}{P _{0}} = \frac{1.110 \times 1 0 ^{8}}{1.01 \times 1 0 ^{5}} \approx 1099$ Why this step? Dividing by $P_{0}$ tells us how many atmospheres of squeeze — the intuitive scale.

Verify: About 1100 atmospheres. Using the 10 m ≈ 1 atm benchmark: $11000/10 = 1100$ atm ✓. The benchmark and the exact formula agree, and $P_{0}$ ( $1 0^{5}$ ) is utterly negligible next to $1 0^{8}$ — which is why deep-sea problems ignore the atmosphere.

Example 9 — Exam twist: the hydrostatic paradox with numbers (Cell I)

Worked example Two containers, one trick

Container X is a wide barrel holding $500 L$ of water, filled to depth $h = 1.2 m$ . Container Y is a thin pipe holding only $0.5 L$ of water, also filled to depth $h = 1.2 m$ . Compare the gauge pressure at the bottom of each.

Forecast: Surely the 500 L barrel presses harder... right?

Step 1 — Compute X. $P_{X} = ρ g h = 1000 \times 9.8 \times 1.2 = 1.176 \times 1 0^{4} Pa$ Why this step? Only depth enters the formula, so we use $h = 1.2 m$ .

Step 2 — Compute Y. $P_{Y} = ρ g h = 1000 \times 9.8 \times 1.2 = 1.176 \times 1 0^{4} Pa$ Why this step? Same depth, same $ρ$ , same $g$ → same pressure. Volume never entered.

Step 3 — State the resolution. $P_{X} = P_{Y}$ . The extra $499.5 L$ in the barrel is held up by the wide base and walls, not stacked above one bottom point (see the figure: only the narrow red column above a point presses on it). Why this step? This is the hydrostatic paradox — the exam's favourite "gotcha."

Verify: $P_{X} / P_{Y} = 1$ exactly ✓. Despite a $1000 \times$ volume difference, pressure is identical because both share $h = 1.2 m$ . See Manometers and Buoyancy & Archimedes' Principle for where equal-depth reasoning pays off.

Connections

Parent: Hydrostatics derivation — where $P = P_{0} + ρ g h$ was proven.
Pascal's Law — why surface pressure $P_{0}$ reaches every point below.
Atmospheric Pressure & Barometer — Examples 4 & 7 in action.
Manometers — reads pressure as a height difference (multi-liquid, like Example 5).
Buoyancy & Archimedes' Principle — uses the difference in pressure with depth.
Bernoulli's Equation — reduces to $ρ g h$ when the fluid is static.
Density and Specific Gravity — supplies the $ρ$ used everywhere here.