In short

Pressure is force per unit area, P = F/A, a scalar measured in pascals (Pa = N/m²). At a point inside a fluid at rest, pressure is the same in every direction — this is Pascal's insight and follows from a force balance on a tiny wedge of fluid. Pressure grows linearly with depth: \boxed{P = P_0 + \rho g h}, where P_0 is the pressure at the surface and \rho g h is the weight per unit area of the fluid column above the point. Gauge pressure is the excess above atmospheric; absolute pressure is gauge plus atmospheric. The SI unit is the pascal, but atmospheric-scale pressures are quoted in atm, bar, mmHg, or torr: 1 \text{ atm} = 1.013 \times 10^5 \text{ Pa} = 1.013 \text{ bar} = 760 \text{ mmHg} = 760 \text{ torr}.

Dive three metres to the bottom of the deep end of a swimming pool and your ears hurt. Dive ten metres at the Andaman coast and they scream. The sea is not heavier there than at the surface; the water is not pressing against your ears with a different kind of force. What has changed is the weight of water stacked above you — a column of water one ear-drum wide and ten metres tall, pressing down through every layer beneath it until it reaches your face. That stack of weight, divided by the area it presses on, is pressure. It grows with depth because the stack does.

This article builds pressure from that one picture. You will derive the formula P = P_0 + \rho g h from Newton's second law applied to a slice of fluid, convince yourself (with a wedge argument that is much older than calculus) that pressure at a point has no direction, sort out when a physicist writes gauge and when they write absolute, and finish by fluently converting between the five pressure units a JEE problem might throw at you. By the end, you will know why a dam is thicker at its base than its top, why your ears pop in a lift descending the Burj Khalifa, and why a 33-foot column of water weighs the same — on any square centimetre of floor — as the entire atmosphere above Chennai.

Pressure as force per unit area

Stand on one foot in your chappals: the sole of one chappal carries all your weight. Stand on one foot in a pair of heeled shoes: the same weight lands on a much smaller area, and the ground feels correspondingly sharper. Nothing about your weight has changed — only the area over which it is spread.

Pressure captures exactly this. If a force F is distributed uniformly over an area A perpendicular to the force, the pressure is

P = \frac{F}{A}.

The SI unit is the pascal (Pa), defined as 1 newton per square metre. A pascal is a small unit — the pressure at the bottom of a 10-centimetre-tall glass of water is already about 980 Pa, and atmospheric pressure is more than 10⁵ Pa — so in the lab you will see kilopascals (kPa), bars, and atmospheres more often than bare pascals.

Pressure is a scalar, not a vector

This is the part that always surprises. Force is a vector: you can push a block left, right, up, or down, and those are four different forces. But pressure is a scalar. When a physicist says "the pressure at this point in the fluid is 2 × 10⁵ Pa", they are not committing to a direction. The pressure acts equally outward on any tiny surface placed at that point, no matter how the surface is oriented.

Why does the vector nature of force disappear when you divide by area? Because both the force and the orientation of the surface rotate together. Take a tiny flat surface of area A floating inside the fluid. The fluid pushes perpendicular to this surface with a force F. Rotate the surface 90° and the fluid now pushes perpendicular to the new orientation, with the same magnitude F (as we will prove in the wedge argument below). The ratio F/A — which is what pressure is — is the same before and after the rotation. The scalar P captures the strength of the push; the normal to the surface captures the direction. You do not need to pack the direction into P itself.

This is why we can say "the pressure 10 m below the surface of the ocean is about 2 atm" without specifying a direction. It is the same in every direction at that point.

The wedge argument — why pressure at a point has no direction

A tiny wedge of fluid used to prove pressure is isotropic A small right-angled wedge of fluid is drawn in equilibrium. Three pressures — P_x on the vertical face, P_y on the horizontal face, and P_n on the slanted face — push perpendicular to each face. The three forces must balance when the wedge is shrunk to a point, forcing P_x = P_y = P_n. horizontal face (area $A_x$) vertical face ($A_y$) slanted face ($A_n$) $P_y$ $P_x$ $P_n$ $\theta$
A small wedge of fluid. The surrounding fluid exerts pressures $P_x$, $P_y$, $P_n$ perpendicular to the three faces. Force balance (in the limit the wedge shrinks to a point) forces the three pressures to be equal.

Isolate a tiny right-angled wedge of fluid, with legs of length dx and dy and hypotenuse of length ds. Let the wedge have thickness b into the page. The three faces have areas A_y = dx \cdot b (horizontal), A_x = dy \cdot b (vertical), and A_n = ds \cdot b (slanted). Let P_x, P_y, P_n denote the pressures on the three faces, and let the angle at the bottom-right corner be \theta, so dx = ds \cos\theta and dy = ds \sin\theta.

Step 1. Write horizontal force balance.

P_x \cdot A_x = P_n \cdot A_n \cdot \sin\theta

Why: only the horizontal component of the pressure force on the slanted face contributes; that component is P_n A_n \sin\theta pointing left. P_x A_x pushes right. The wedge is at rest, so these balance.

Substitute A_x = dy \cdot b = ds \sin\theta \cdot b and A_n = ds \cdot b:

P_x \cdot ds \sin\theta \cdot b = P_n \cdot ds \cdot b \cdot \sin\theta

The \sin\theta, ds, and b cancel, leaving

P_x = P_n.

Step 2. Write vertical force balance.

P_y \cdot A_y = P_n \cdot A_n \cdot \cos\theta + W_{\text{wedge}}

The vertical-component of the slanted-face force is P_n A_n \cos\theta pushing up (into the wedge from below), and P_y A_y pushes down from the top face. The wedge also has its own weight W_{\text{wedge}} = \rho g \cdot \tfrac{1}{2} dx \, dy \, b pulling down.

Why: the weight of the wedge scales as dx \cdot dy \cdot b — a third-order small quantity — while the pressure-area forces scale as ds \cdot b — only second-order small. As the wedge shrinks to a point (dx, dy \to 0), the weight term becomes negligible compared to the others.

Substitute and shrink to a point (drop the weight term):

P_y \cdot dx \cdot b = P_n \cdot ds \cdot b \cdot \cos\theta.

Using dx = ds \cos\theta, everything cancels to give

P_y = P_n.

Combining both balances:

\boxed{\; P_x = P_y = P_n. \;}

The pressure at a point is the same whether you look at it from the left, from below, or along the slant. The direction of the tiny surface does not matter. This is what a physicist means when they say "the pressure at a point in a fluid is isotropic."

Pressure varies with depth

A fluid at rest has one direction in which pressure clearly does depend on position: the vertical. Shallower water is under less stacked weight than deeper water, and the deeper fluid has to support everything above it. This is the key equation of hydrostatics, and you are about to derive it from Newton's second law.

A slab of fluid used to derive P = P₀ + ρgh A rectangular slab of fluid of area A and height h sits inside a larger tank of water. The top face is at depth 0 with pressure P₀. The bottom face is at depth h with pressure P. The slab is in equilibrium under three vertical forces: P₀ A pushing down from above, weight ρ g A h pulling down, and P A pushing up from below. top face: area $A$, depth 0 bottom face: depth $h$ $P_0 A$ (down) weight $\rho g A h$ (down) $P A$ (up) $h$
A vertical cylinder of fluid inside a larger container. The three vertical forces — pressure on top, pressure on bottom, and weight — must add to zero because the cylinder is at rest.

Imagine a vertical cylindrical slab of the fluid itself — not a solid object, just a column of the liquid that is already there. Let it have cross-sectional area A and height h. Its top face sits at depth 0 (where the pressure is P_0), and its bottom face sits at depth h (where the pressure is P, the quantity you want).

Three vertical forces act on this slab:

  1. The fluid above pushes down on the top face: F_{\text{top}} = P_0 A.
  2. Gravity pulls the slab down: W = m g = (\rho A h) g, using m = \rho V = \rho A h.
  3. The fluid below pushes up on the bottom face: F_{\text{bot}} = P A.

The slab is at rest — it is not accelerating — so Newton's second law demands that these three forces sum to zero. Taking up as positive:

P A - P_0 A - \rho g A h = 0.

Why: PA up minus P_0 A down minus \rho g A h down equals zero net force. The slab is in equilibrium because nothing around it is changing — the fluid is static.

Divide through by A:

P - P_0 - \rho g h = 0,

and rearrange:

\boxed{\; P = P_0 + \rho g h. \;}

This is the hydrostatic pressure formula. In words: the pressure at depth h below a free surface (where the pressure is P_0) is the surface pressure plus the weight of fluid per unit area in a column of height h.

Three things about this formula deserve attention.

(a) It does not depend on A. The cross-sectional area of the slab cancelled. The formula is the same whether you imagine a tall thin column or a short fat one; whether you are computing pressure in a drinking glass, in a swimming pool, or in a dam. Pressure at a given depth depends only on the depth and the fluid, not on the horizontal shape of the container.

(b) It is linear in h. Go twice as deep and the extra pressure doubles. At the top of a 10-metre-deep Andaman dive, the water adds about 1 \times 10^5 Pa (one extra atmosphere) on top of the atmospheric pressure already pressing down on the surface. Go to 20 m and it's two extra atmospheres. (This is why scuba instructors drill depth-pressure conversions into your head — a 1-atm-per-10 m rule of thumb is how they keep you alive.)

(c) It does depend on \rho. In mercury (\rho = 13{,}600 kg/m³), a depth of just 76 cm produces the same extra pressure as a 10-metre column of water — because mercury is 13.6 times denser. This is how Torricelli, in 1643, weighed the whole atmosphere with a glass tube full of mercury. (That story has its own article — Atmospheric Pressure and Barometers.)

The same-level rule

One immediate corollary: in a connected fluid at rest, all points at the same height are at the same pressure, regardless of the shape of the container. If two points are at the same depth h below the same free surface, both have pressure P = P_0 + \rho g h. The shape of the path between them is irrelevant.

U-tube illustrating that equal-height points have equal pressure A U-shaped tube is filled with water to the same level on both sides. Three points are marked: A at the left column, B in the connecting base, and C at the right column, all at the same depth h below the surface. All three points have the same pressure P = P₀ + ρ g h. $P_0$ $A$ $B$ $C$ $h$
All three marked points sit at the same depth $h$ below the common water surface. The pressure at A, B, and C is identical — $P_0 + \rho g h$ — even though the path between them bends.

This explains why rivers find a level, why a carpenter's water-level tool works, and why in a U-tube the two free surfaces line up when both columns contain the same fluid.

Gauge pressure vs absolute pressure

In a physics problem, when you write "P = 2 \times 10^5 Pa", which P do you mean? The one that includes the atmosphere always pressing on everything, or the one that is extra on top of it?

The distinction has a name.

The relation is simple:

P_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}}.

A car tyre inflated to "32 psi" is at 32 psi above atmospheric — that is a gauge pressure. In absolute terms, the air inside is at about 32 + 14.7 = 46.7 psi. A pressure cooker's safety valve is rated in gauge pressure — the "1 kg/cm²" marking on the side of an Indian cooker means about 1 atmosphere above atmospheric, i.e. 2 atm absolute.

When the problem says "the tyre is at 220 kPa", and no context is given, it almost always means gauge. When a physicist writes "the pressure inside the Marianas Trench is 1100 atm", they mean absolute.

When does the distinction matter?

Whenever the problem involves a closed system and you are computing things like forces on walls, buoyancy, or flow: use absolute pressure, because the wall or the float or the pump feels the full force including atmosphere.

Whenever the problem involves an open system and what you care about is pressure differences (an open-ended manometer, a Bernoulli calculation between two points both exposed to atmosphere): you can freely use gauge, because the atmosphere cancels from both sides.

JEE problems are almost always explicit about which they want. When in doubt, write the formula with absolute pressure, and convert at the end.

The units of pressure — a lookup table that repays memorising

The SI unit is the pascal (Pa) — one newton per square metre. But atmospheric-scale pressures come wrapped in four historical units that you will meet in JEE, competitive exams, and engineering handbooks for the rest of your life.

Unit Symbol SI equivalent Context
Pascal Pa 1 N/m² SI unit, small (used in cleanroom specs, pascal-second for viscosity)
Kilopascal kPa 10³ Pa Most tyre gauges, barometric weather reports (in some countries)
Bar bar 10⁵ Pa Engineering pressure vessels, diving gauges, Indian cooker ratings
Atmosphere (standard) atm 1.01325 × 10⁵ Pa Reference pressure at sea level (standardised by convention)
Millimetre of mercury mmHg 133.322 Pa Blood pressure, weather reports (India), medical gauges
Torr Torr 133.322 Pa Vacuum-chamber physics, scientific papers
Pound per square inch psi 6894.76 Pa Tyre gauges in India (inherited from British/American engineering)

Two conversions to have in memory:

1 \text{ atm} = 1.013 \times 10^5 \text{ Pa} = 1.013 \text{ bar} = 760 \text{ mmHg} = 760 \text{ torr} \approx 14.7 \text{ psi}.

Note that mmHg and torr are numerically equal — 1 mmHg = 1 torr. The torr is defined as 1/760 of a standard atmosphere, which is very nearly the pressure of a one-millimetre-tall mercury column at 0 °C. In practice you can treat the two symbols as interchangeable unless you are writing a research paper on high-precision manometry.

Why 760 mmHg?

Because at sea level, atmospheric pressure is just heavy enough to hold up a column of mercury about 76 cm tall in an evacuated glass tube. This is the Torricelli experiment, and it is the reason mmHg is a unit at all. The depth formula P = \rho g h says: a column of mercury (\rho = 13{,}600 kg/m³) of height h sitting in a tube exerts a pressure P = \rho g h at the bottom. Set this equal to atmospheric pressure:

h = \frac{P_{\text{atm}}}{\rho_{\text{Hg}} \, g} = \frac{1.013 \times 10^5}{13{,}600 \times 9.8} \approx 0.760 \text{ m} = 760 \text{ mm}.

The full story is in Atmospheric Pressure and Barometers — the point here is that "760 mmHg" and "1 atm" really mean the same physical pressure, routed through two different ways of measuring it.

Worked examples

Example 1: Pressure at the bottom of a Mumbai apartment water tank

The overhead water tank on top of a 12-storey Mumbai apartment building is 40 m above the ground-floor tap. The tank is open to the atmosphere (P_0 = 1.013 \times 10^5 Pa at the top), and water has density \rho = 1000 kg/m³. Take g = 9.8 m/s². Find the absolute and gauge pressure at the tap.

Mumbai apartment water tank connected to a ground-floor tap A water tank on top of a 40 m tall building, open to the atmosphere at the top. A pipe runs down to a tap at ground level. The height difference between the water surface and the tap is 40 m. The pressure at the tap equals atmospheric plus ρ g h. tank ($P_0$, open) tap $h = 40$ m
The water surface at the top of the tank is at atmospheric pressure $P_0$. The tap sits 40 m below the surface. The pressure at the tap is $P_0 + \rho g h$.

Step 1. Identify the reference pressure and the depth.

The free surface at the top of the tank sits at P_0 = 1.013 \times 10^5 Pa (atmospheric). The tap is h = 40 m below that surface.

Why: the only surface that "knows" about the atmosphere is the open top of the tank. The tap is connected to that surface by a continuous column of water; by the hydrostatic formula, its pressure is the surface pressure plus \rho g h.

Step 2. Apply P = P_0 + \rho g h.

P_{\text{abs}} = 1.013 \times 10^5 + (1000)(9.8)(40) = 1.013 \times 10^5 + 3.92 \times 10^5 = 5.03 \times 10^5 \text{ Pa}.

Why: the weight of a 40-metre tall column of water of 1 m² cross-section is \rho V g = 1000 \cdot 40 \cdot 9.8 = 3.92 \times 10^5 N. Spread over 1 m², that is the extra pressure the column contributes.

Step 3. Compute the gauge pressure.

P_{\text{gauge}} = P_{\text{abs}} - P_{\text{atm}} = 5.03 \times 10^5 - 1.013 \times 10^5 = 3.92 \times 10^5 \text{ Pa} \approx 3.9 \text{ bar}.

Why: subtracting the atmospheric contribution leaves only the excess due to the water column. A reading from a pressure gauge bolted to the tap would read about 3.9 bar, because the gauge compares against the atmospheric air that surrounds it.

Result: The absolute pressure at the tap is about 5.0 \times 10^5 Pa, and the gauge pressure is about 3.9 \times 10^5 Pa \approx 3.9 bar.

What this shows: Every 10 m of water height adds roughly 1 atmosphere (about 10^5 Pa) to the pressure. A 40 m tank builds up four atmospheres' worth of hydrostatic pressure on top of whatever is at the surface. This is why the ground-floor tap in a tall building hits with a much stronger jet than the top-floor tap — it is sitting at the bottom of a 40 m column of water.

Example 2: Pressure difference on the two sides of a Chennai dam

A small dam at a reservoir in Tamil Nadu has water behind it to a depth of 25 m. The water density is 1000 kg/m³ and g = 9.8 m/s². Take atmospheric pressure to be the same on both sides of the dam.

(a) Find the gauge pressure at the base of the dam. (b) Find the net horizontal force per unit width on the dam (i.e., per metre of dam length measured along the river).

Pressure distribution against a dam wall A dam wall holds back water of depth 25 m on the left and air on the right. The gauge pressure against the dam grows linearly from zero at the top of the water to ρ g H at the base. The resulting horizontal force per unit width is the area of a triangle, (1/2) ρ g H squared. pressure profile $P(y) = \rho g y$ (gauge) $H=25$ m dam air ($P_{\text{atm}}$)
The gauge pressure on the upstream face of the dam increases linearly with depth — zero at the top of the water, maximum at the base. The net horizontal force per unit width is the area of the pressure triangle.

Step 1. Gauge pressure at the base.

Using P_{\text{gauge}} = \rho g h with h = 25 m:

P_{\text{gauge}} = 1000 \times 9.8 \times 25 = 2.45 \times 10^5 \text{ Pa} \approx 2.5 \text{ atm}.

Why: the atmospheric contribution cancels from the two sides of the dam (air pushes on the downstream face with the same P_{\text{atm}}), so only the gauge part matters for the net horizontal force.

Step 2. Build the force integral.

At depth y below the water surface, the gauge pressure is P(y) = \rho g y. A thin horizontal strip of width 1 m (along the river) and height dy at depth y feels a horizontal force dF = P(y) \cdot 1 \cdot dy = \rho g y \, dy.

Why: force = pressure × area; the strip's area per unit width is just dy. The pressure varies with depth, so you have to add up the strip contributions by integration rather than multiplying by a single number.

Step 3. Integrate from y = 0 (surface) to y = H = 25 m.

\frac{F}{w} = \int_0^H \rho g y \, dy = \rho g \cdot \frac{H^2}{2}.

Why: the integrand \rho g y is a linear function of y, so the integral is \rho g \cdot y^2/2 evaluated between limits. Equivalently: the pressure profile is a triangle, and the area of a triangle with base \rho g H and height H is \tfrac{1}{2} \rho g H^2 — you can read the answer straight off the diagram.

Step 4. Plug in numbers.

\frac{F}{w} = \frac{1}{2} \times 1000 \times 9.8 \times 25^2 = \frac{1}{2} \times 1000 \times 9.8 \times 625 = 3.06 \times 10^6 \text{ N/m}.

Why: that is about 3 meganewtons pushing horizontally on every metre of dam length — the equivalent of 300 tonnes of weight. For a 100 m wide dam, the total horizontal force is 300 meganewtons. This is why dams are built massively thicker at the base than at the top — the base must resist this enormous hydrostatic push.

Result: (a) The gauge pressure at the base is 2.45 \times 10^5 Pa. (b) The net horizontal force per metre of dam length is about 3.06 \times 10^6 N/m.

What this shows: Because pressure grows linearly with depth, the force on a vertical wall grows with the square of the depth — a 50 m dam feels four times the force per unit width that a 25 m dam does, not twice. This quadratic growth is the reason tall dams have to be so disproportionately thick at the base.

Common confusions

If you are comfortable with the hydrostatic formula, gauge vs absolute, and the unit conversions, you already have everything you need for most JEE pressure problems. What follows is the calculus-based generalisation, a look at what happens when \rho or g varies, and a proof that the hydrostatic pressure at a point depends only on its height — not on the path you take to reach it.

Hydrostatic equilibrium as a differential equation

The derivation above used a finite slab of fluid. The same argument with an infinitesimal slab gives a differential equation that is more general.

Take a vertical slab of thickness dy at depth y below the surface, with cross-section A. The pressure on the top is P(y), and on the bottom is P(y + dy) = P + dP. The vertical force balance is

(P + dP) A - P A - \rho(y) \, g \, A \, dy = 0.

Why: pressure from below (acting up) minus pressure from above (acting down) minus the weight of the slab equals zero. Sign convention: y increases downward, so the bottom face has the larger pressure P + dP.

Cancelling A and dividing by dy:

\frac{dP}{dy} = \rho(y) \, g.

This is hydrostatic equilibrium as a differential equation. When \rho is constant (an incompressible fluid like water), you can integrate directly:

P(h) - P(0) = \int_0^h \rho g \, dy = \rho g h,

giving P = P_0 + \rho g h — our earlier result. When \rho varies with height (as with the atmosphere, where the density falls off exponentially), the integral is more interesting and produces the barometric formula that Atmospheric Pressure and Barometers derives.

Why the shape of the container does not matter — a path-independence argument

The rule "two points at the same height in a connected fluid have the same pressure" seems almost obvious in a U-tube, but what about a container with a shape that zig-zags? Here is the proof.

Pick two points A and B in the fluid, both at the same depth. Take any path through the fluid from A to B — it need not be straight. Parametrise the path as (x(s), y(s)). The pressure change along the path satisfies

\frac{dP}{ds} = \frac{\partial P}{\partial x} \frac{dx}{ds} + \frac{\partial P}{\partial y} \frac{dy}{ds}.

In a fluid at rest, \partial P / \partial x = 0 (no horizontal pressure gradient — otherwise fluid would flow), and \partial P / \partial y = \rho g. So

\frac{dP}{ds} = \rho g \, \frac{dy}{ds}.

Integrating from A to B:

P_B - P_A = \rho g \int_A^B \frac{dy}{ds} \, ds = \rho g (y_B - y_A).

Why: the integrand is a perfect derivative — integrating dy/ds along a path returns the net change in y, regardless of how the path twists. So the pressure difference between any two points in a static connected fluid depends only on their height difference, not on the route.

If y_A = y_B, then P_A = P_B. The same-level rule is a one-line corollary of hydrostatic equilibrium and the fact that horizontal gradients vanish in a static fluid.

A subtle case — does the formula assume constant g?

The derivation used g as a constant over the height of the fluid column. This is safe for any column less than a few kilometres tall, because g changes by only \sim 0.03% per kilometre of altitude near Earth's surface. For deep-ocean hydrostatics, constant g is fine at the 1% level. For atmospheric columns that stretch to the stratosphere, one should strictly use g(r) = GM/r^2 — but the correction is small until altitudes comparable to Earth's radius. For JEE problems, treat g as constant.

Pressure is a thermodynamic state variable

In thermodynamics, pressure appears alongside volume and temperature as a state variable of a fluid. The hydrostatic formula we derived is a mechanical equilibrium statement — it says that the fluid is not accelerating. A fluid can be in mechanical equilibrium without being in thermal equilibrium (a pool with a cold layer on top and warm water below is mechanically stable but thermally transient). Full equilibrium requires both.

A practical consequence: the density \rho in the hydrostatic formula can depend on temperature. Warm water is slightly less dense than cold water, so a stratified lake has a slightly non-linear pressure-depth profile. For JEE-level problems, this refinement is ignored — we treat water as incompressible and uniform. For oceanography, it matters.

Where this leads next