Variation of pressure with depth

Dive into a swimming pool and by the time you are a couple of metres down, your ears already ache. Nobody piled extra water on top of you on purpose — you simply went deeper, and the water above you got heavier. That ache is a direct, bodily measurement of one of the cleanest results in all of physics: pressure in a fluid grows in a straight line with depth, and the constant of proportionality is nothing more exotic than the density of the fluid times $g$ . Fishermen, submarine engineers, dam builders and doctors reading a patient's blood pressure all lean on this one idea. A dam wall is thickest at the bottom not by tradition but because $P = \rho g h$ says the pressure there is enormous compared with the pressure near the top. A deep-sea diver's suit must resist crushing forces that a snorkeller never feels, even though both are in the "same" ocean. And the humble mercury column in an old-style blood-pressure cuff works precisely because a column of liquid at rest converts a pressure into a height you can read off a scale. This section builds that single formula from a bone-simple force balance on a thin slab of fluid, shows why it depends on depth alone and never on the shape of the container — the so-called hydrostatic paradox — and works it into board-style and competitive-style numbers you can actually be examined on. By the end you should be able to look at any static fluid problem, liquid or (approximately) gas, and know instantly which pressure to add, which to ignore, and why.

Key takeaways

Pressure in a static fluid increases linearly with depth: $P = P_0 + \rho g h$ , where $P_0$ is the pressure at the reference surface and $h$ is measured straight down from it.
The formula comes from a simple force (weight) balance on a thin horizontal slab of fluid — no calculus beyond adding up thin layers.
Pressure at a point depends only on $\rho$ , $g$ , and depth $h$ — never on the shape, cross-section, or total volume of the container. This is the hydrostatic paradox.
All points at the same horizontal level in one connected, static fluid are at the same pressure — the reason liquid "finds its own level" and manometers work.
Absolute (total) pressure = atmospheric pressure + $\rho g h$ ; gauge pressure is just the $\rho g h$ part, the excess over atmospheric.
The linear formula is exact for incompressible liquids but only an approximation for gases such as air, whose own density falls with height.
Key idea: in a fluid at rest, pressure rises linearly with depth as P = P0 + rho g h because each layer must support the weight of everything above it, and this depends only on depth and density, never on the container's shape.

The core insight

In a fluid at rest, pressure increases linearly with depth according to $P = P_0 + \rho g h$ , purely because each horizontal layer must support the weight of the fluid column above it, so pressure depends only on depth and density — not on the container's shape or volume.

Where it fits in the syllabus

This topic sits inside the "Mechanical Properties of Fluids" unit of CBSE Class 11 Physics, coming right after the basic definition of pressure and feeding directly into Pascal's law, hydraulic machines, and atmospheric/gauge pressure. You are expected to derive $P = P_0 + \rho g h$ from a force balance on a thin fluid slab, apply it to find pressure at any depth in a liquid, and correctly combine it with atmospheric pressure to get absolute pressure — all firmly Class 11 board-and-competitive material, with no calculus and no fluid-dynamics machinery beyond this static picture.

The physical picture

Picture a liquid of density $\rho$ sitting quietly in a container, in equilibrium under gravity. Imagine slicing out, mentally, a thin horizontal disc of that liquid — cross-sectional area $A$ , thickness $dh$ — sitting at some depth. This disc does not accelerate, so every force on it must balance. It is pushed down by the fluid above it, pushed up by the fluid below it, and pulled down by its own weight; horizontally the pressure from the surrounding fluid cancels by symmetry. Working through this balance (done in full in the derivation below) gives the central result of this section:

P = P_0 + \rho g h

where: $P$ is the absolute pressure at a point at depth $h$ ; $P_0$ is the pressure at the reference level directly above that point (usually the free surface of the liquid, where $P_0$ is atmospheric pressure if the liquid is open to the air); $\rho$ is the (constant) density of the liquid; $g = 9.8\ \text{m/s}^2$ is the acceleration due to gravity; and $h$ is the vertical depth below the reference level, measured straight down. The physical reading is simple: the deeper you go, the more liquid sits on top of you, so the more weight per unit area you must support — and because $\rho$ and $g$ are fixed, that extra load grows in exact proportion to $h$ . Three consequences follow immediately from this one equation, and each is a favourite exam trap in disguise. First, pressure grows linearly, not with the square of depth or any other power — double $h$ and the extra ("gauge") pressure due to the liquid exactly doubles. Second, and most striking, the formula contains no reference at all to the shape, cross-section, or total volume of the container — only $\rho$ , $g$ , and $h$ . So a narrow test tube and a huge, oddly shaped tank filled with the same liquid to the same height produce identical pressure at the bottom, even though the tank might hold a thousand times more liquid by weight. This is the hydrostatic paradox, and it is resolved by realising that the container's slanted or curved walls bear the extra horizontal and vertical components of force that the liquid's own weight would otherwise transmit downward — the walls, not the base alone, carry the rest of the load in a wide vessel. Third, because pressure depends only on $h$ , every point at the same horizontal level inside one connected body of static liquid has exactly the same pressure — even if the liquid snakes through bends, wide bulbs, and narrow necks to get there. This is why liquid poured into a U-tube always settles to the same height in both arms, and why a manometer can be used to measure an unknown pressure by comparing liquid levels.

Figure

A container with a wide left arm and a narrow right arm connected at the bottom, both filled with the same liquid to the same height, with horizontal dashed lines at three depths showing equal pressure at each shared level despite the very different widths

The derivation

Derivation ADeriving P = P0 + rho g h from a force balance

The goal here is to derive the linear pressure-depth relation for a liquid at rest, starting from nothing more than Newton's first law applied to a thin slab of fluid and the definition of pressure as force per unit area. There is no calculus needed beyond treating a very thin layer as having essentially constant pressure across its own tiny thickness; the result nonetheless holds exactly, because we can imagine building up any depth from arbitrarily many such thin layers stacked on top of each other, each one simply adding its own small weight to the load carried by the ones below.

Step 1. Set up the slab. Consider a horizontal disc of liquid of cross-sectional area $A$ and small thickness $dh$ , located at depth $h$ below the free surface of the liquid, entirely surrounded by more of the same liquid. Because the liquid is at rest and the whole system is in equilibrium, this slab has zero net force acting on it in every direction.

Step 2. Identify the vertical forces. Liquid above the slab pushes down on its top face with pressure $P$ , giving a downward force $P\,A$ . Liquid below the slab pushes up on its bottom face with the slightly larger pressure $P + dP$ , giving an upward force $(P+dP)A$ . The slab also has its own weight acting downward, equal to its mass times $g$ : since the slab's volume is $A\,dh$ and its density is $\rho$ , the weight is $\rho\,A\,dh\,g$ .

Step 3. Balance the vertical forces. Since the slab does not accelerate, the upward force must exactly equal the sum of the downward forces: $(P+dP)A = P\,A + \rho\,A\,dh\,g$ . The area $A$ appears in every term and cancels directly, leaving $dP = \rho\,g\,dh$ . This says the pressure increases by a small amount $dP$ for every small increase $dh$ in depth, and the rate of increase is the constant $\rho g$ .

Step 4. Add up all the thin slabs from the surface down to depth $h$ . Since $\rho$ and $g$ do not change with depth for an incompressible liquid, the total increase in pressure from the surface down to depth $h$ is simply $\rho g$ multiplied by the total depth: the pressure increase is $\rho g h$ . Adding this to whatever pressure $P_0$ already exists at the surface (atmospheric pressure, if the liquid is open to air) gives the full result: $P = P_0 + \rho g h$ .

Step 5. Read off the two headline consequences. Because $A$ cancelled completely in Step 3, the final formula never involves the container's shape, cross-section, or total liquid volume at all — only $\rho$ , $g$ , and $h$ survive, which is exactly the hydrostatic paradox. And because $dP/dh = \rho g$ is a constant rather than something that itself depends on $h$ , the growth of pressure with depth is strictly linear — a straight-line graph, not a curve.

Worked examples

Board level - pressure at the bottom of a lake

Find the pressure at a depth of 10 m below the surface of a lake. Take the density of water $\rho = 1000\ \text{kg/m}^3$ , atmospheric pressure $P_0 = 1.013\times10^5\ \text{Pa}$ , and $g = 9.8\ \text{m/s}^2$ .

Step 1. Write down the depth-pressure relation to be used: $P = P_0 + \rho g h$ .

Step 2. Compute the extra pressure contributed by the 10 m water column alone (the gauge pressure): $\rho g h = 1000 \times 9.8 \times 10 = 9.8\times10^{4}\ \text{Pa}$ .

Step 3. Add the atmospheric pressure pressing down on the free surface to get the total, absolute pressure: $P = 1.013\times10^5 + 9.8\times10^4 = 1.993\times10^5\ \text{Pa}$ . The absolute pressure at 10 m depth is about $1.99\times10^5\ \text{Pa}$ , nearly double atmospheric pressure, while the gauge pressure due to the water alone is $9.8\times10^4\ \text{Pa}$ — always keep these two numbers distinct, since exam questions frequently ask for one or the other by name.

Competitive level - the hydrostatic paradox in two different vessels

Two vessels, one narrow and cylindrical and one wide and conical, are filled with the same liquid of density $\rho = 800\ \text{kg/m}^3$ to the same height of 0.5 m. Find the pressure at the bottom of each vessel and comment on whether they are equal, even though the total weight of liquid in each is very different. Take $g = 9.8\ \text{m/s}^2$ and ignore atmospheric pressure (find gauge pressure only).

Step 1. Recall the key structural fact proved in the derivation above: pressure at depth $h$ in a static liquid depends only on $\rho$ , $g$ , and $h$ , since the cross-sectional area $A$ cancelled out of the force balance entirely — it never depends on the shape or total volume of the container.

Step 2. Apply $P = \rho g h$ to both vessels using the same depth $h = 0.5\ \text{m}$ for each, since both are filled to the same height with the same liquid: $P = 800 \times 9.8 \times 0.5 = 3920\ \text{Pa}$ .

Step 3. Compare the two results. Both the narrow cylindrical vessel and the wide conical vessel give exactly the same pressure of 3920 Pa at the bottom, even though the wide conical vessel contains far more liquid, and hence far more total weight, than the narrow one. The extra weight in the wide vessel is carried by the slanted container walls rather than being transmitted straight down onto the base, which is precisely the resolution of the hydrostatic paradox. Timing note: this kind of "same depth, different shape" question is a fast mark in competitive exams once you recognise it — do not waste time trying to compute or compare the volumes or weights of liquid in each vessel; the shape is a deliberate distractor, and the depth alone decides the answer.

Common traps

Assuming pressure at the bottom depends on the total volume or weight of liquid in the container — it depends only on depth, density, and $g$ , never on the container's shape.
Forgetting to add atmospheric pressure $P_0$ when a question asks for absolute pressure rather than gauge pressure (or, just as commonly, adding it when the question explicitly asks only for gauge pressure).
Measuring "depth" from the wrong reference point — always measure $h$ straight down from the free surface of the liquid, never up from the bottom of the container or from some other arbitrary point.
Applying $P = \rho g h$ to a gas such as air over a large height range as though its density were constant — the formula assumes constant $\rho$ , which holds well for incompressible liquids but is only a rough approximation for a compressible gas whose own density changes with altitude.
Mixing up horizontal and vertical distances in a tilted or irregular container — only the vertical depth below the free surface matters, not distance measured along a slanted wall or pipe.

You should be able to…

Derive $P = P_0 + \rho g h$ from a force balance on a thin horizontal fluid slab.
Explain why pressure at a point depends only on depth and density, never on the shape or volume of the vessel (the hydrostatic paradox).
State why all points at the same horizontal level in a connected, static fluid have equal pressure.
Calculate absolute pressure and gauge pressure at a given depth in a liquid, and know which one a question is actually asking for.
Explain why this linear formula is exact for liquids but only an approximation for gases such as air.

Ask the tutor: Want to see how this same P = P0 + rho g h formula turns into Pascal's law and a working hydraulic lift, or try a few more depth-pressure numericals with different liquids — ask the AI tutor for extra practice.

Why does pressure depend on depth but not on how much liquid is around at that depth?

Because the force balance in the derivation is done on a thin horizontal slab of liquid, and the area of that slab cancels out completely when you write down Newton's first law for it. What survives is only the density of the liquid, g, and how many such thin slabs are stacked above the point in question, which is exactly the depth h. A huge lake and a narrow well of the same liquid, at the same depth, genuinely do exert the same pressure on a diver — the extra liquid in the lake pushes sideways on its own banks, not downward onto that one point.

Do I always have to add atmospheric pressure, or is that only sometimes needed?

It depends entirely on what the question asks for. If a question asks for absolute (total) pressure, you must add atmospheric pressure P0 to rho g h. If it asks for gauge pressure — the pressure "due to the liquid" or "in excess of atmospheric" — you report rho g h alone and leave P0 out. Read the question wording carefully, because both versions look almost identical numerically once you are ten or more metres deep, and examiners exploit exactly that similarity to test whether you understood the difference.

Why doesn't P = rho g h work well for air the same way it works for water?

The derivation assumes rho stays constant as you add up thin slabs from the surface downward. Water is essentially incompressible, so its density barely changes even under large pressure differences, and the linear formula is essentially exact. Air, however, is a gas whose density itself depends on pressure — as you go higher, the air thins out and rho drops, so the same formula with a fixed rho only works as a rough approximation over modest height changes near the ground; over the full height of the atmosphere it is not accurate, though the details of that correction belong to a later stage of study.

Is the hydrostatic paradox actually a paradox, or is something being missed?

It only feels paradoxical if you expect pressure to track the total weight of liquid in a container, the way you might expect a heavier backpack to press harder on your shoulders. In a fluid at rest, though, a container with slanted or curved walls does not let all of the liquid's weight act straight down onto the base — some of that weight pushes outward or upward on the walls themselves, and the walls push back. Once you trace where every bit of weight actually goes, the numbers agree perfectly with P = rho g h, and there is no contradiction with Newton's laws at all.