JEE Mains Physics · Mechanical Properties of Fluids

Pressure, Pascal's Law and Buoyancy

Pressure in a still liquid grows with depth as P = P₀ + ρgh, a pressure added anywhere in an enclosed liquid reaches every point of it, and a floating body sinks until it displaces its own weight of liquid.

All Mechanical Properties of Fluids notes JEE Mains Physics strategy

Why this matters

Fifteen PYQs, eleven of them multiple choice, and three from 2026. Six use pressure at a depth: an absolute pressure, a force on a base or a door, the work gravity does when two vessels share their water, and whether pressure exists inside a still fluid at all; four are Pascal's law and the hydraulic lift; five are floating bodies, from the submerged height of a block to a cube resting across two liquids.

Concept 1 of 3: Pressure at a depth, absolute and gauge

The liquid above a point presses down on it with its weight: a column of height h and density ρ adds

\rho g h

. The air above the surface adds

P_0

on top of that. A gauge reads only the part due to the liquid. Pressure acts in every direction at every point inside the liquid, not only on the walls, and at one level in one connected still liquid it is the same everywhere.

Definition

Absolute pressure: $P = P_0 + \rho g h$ . Gauge pressure: $P - P_0 = \rho g h$ .
Same level in one connected still liquid: same pressure, whatever the shape of the vessel.
Doubling the depth doubles $\rho g h$ , not $P$ : the pressure goes from $P_0 + \rho g h$ to $P_0 + 2\rho g h$ .
Force on a flat base: $F = P \times A$ . Include $P_0$ when the stem gives it.
A window in a partition with a liquid on each side: $P_0$ acts on both sides and cancels, so the net force is $(\rho_1 - \rho_2) g h A$ at the window's depth h.
Two vessels joined at the bottom settle to a common level. A column of height h and base area A has potential energy $\rho A g h^{2}/2$ ; gravity's work is the loss in this energy.

Pressure at depth h

P = P_0 + \rho g h

Worked example

A diver feels an absolute pressure of

2.5 \times 10^{5}

Pa. Atmospheric pressure is

1.0 \times 10^{5}

Pa. The diver goes three times as deep. Find the new absolute pressure and its percentage increase.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 5 · Q4Moderate

Example 1 · Mechanical Properties of Fluids · Pressure, Pascal's Law and Buoyancy

The pressure acting on a submarine is

3 \times10^{5}

Pa

at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be : (Assume that atmospheric pressure is

1 \times10^{5}\text{ }Pa

density of water is

10^{3}\text{ }kg{\text{ }m}^{- 3},\text{ }g = 10{\text{ }ms}^{- 2}

)

Doubling the depth does not double the pressure

Only ρgh doubles. The atmosphere's share stays the same, so the absolute pressure grows by less than 100%. Doubling the whole of P counts the atmosphere twice.

Keep P₀ in a force on a base when the stem gives it

For the force a liquid exerts on the bottom of a tube open to the air, JEE has keyed (P₀ + ρgh) × A when it gives P₀. The smaller option, ρgh × A, leaves the atmosphere out.

Across a partition, the atmosphere cancels

Air presses on both free surfaces, so only the difference of the liquid pressures pushes on a door in the partition: (ρ₁ − ρ₂)ghA.

Drill 5 more on pressure at a depth, absolute and gauge

Concept 2 of 3: Pascal's law and the hydraulic lift

A push on an enclosed liquid raises the pressure everywhere in it by the same amount. A small force on a small piston therefore makes the same pressure under a large piston, and there it gives a large force. Energy is not created: the small piston has to move farther, by the ratio of the areas.

Definition

Pascal's law: a change in pressure applied to an enclosed incompressible fluid reaches every point of the fluid and the walls undiminished.
Hydraulic lift: $\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}$ , so $F_2 = F_1\dfrac{A_2}{A_1}$ .
The same volume moves on both sides: $A_1 d_1 = A_2 d_2$ . So $F_1 d_1 = F_2 d_2$ : work in equals work out.
The pressure under the small piston equals the load divided by the LARGE area, $Mg/A_2$ .
Squeezing a toothpaste tube, hydraulic brakes and hydraulic presses all use Pascal's law.

Hydraulic lift

\frac{F_1}{A_1} = \frac{F_2}{A_2}, \qquad F_1 d_1 = F_2 d_2

Worked example

A hydraulic press has pistons of area

10\ \text{cm}^{2}

and

400\ \text{cm}^{2}

. A 50 N push moves the small piston down 8 cm. Find the force on the large piston, how far it rises and the work done.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q24Moderate

Example 2 · Mechanical Properties of Fluids · Pressure, Pascal's Law and Buoyancy

In a hydraulic lift, the surface area of the input piston is

6{\text{ }cm}^{2}

and that of the output piston is

1500{\text{ }cm}^{2}

. If 100 N force is applied to the input piston to raise the output piston by 20 cm , then the work done is

\_\_\_\_\_

kJ .

The small piston carries the same pressure

Pressure is the same throughout the enclosed fluid. The small piston has a smaller FORCE on it, but the pressure on it is still the load divided by the large area.

A lift multiplies force, not work

The force gain A₂/A₁ is paid for in distance: the small piston moves A₂/A₁ times farther. Work out equals work in.

Drill 3 more on pascal's law and the hydraulic lift

Concept 3 of 3: Floating bodies and Archimedes' principle

A floating body sinks until the liquid it pushes aside weighs as much as it does. So the fraction under the surface is its density divided by the liquid's. Put a load on it and it sinks further, by just enough extra liquid to carry the load.

Definition

Archimedes' principle: upthrust = weight of liquid displaced = $\rho_l V_{\text{sub}} g$ .
Floating: $\rho_b V = \rho_l V_{\text{sub}}$ , so $\dfrac{V_{\text{sub}}}{V} = \dfrac{\rho_b}{\rho_l}$ . For a block of height H the submerged height is $h = H\dfrac{\rho_b}{\rho_l}$ .
A load of mass m displaces an extra $m/\rho_l$ of liquid. A block of base area A sinks a further $\Delta h = \dfrac{m}{\rho_l A}$ .
Hollow sphere (outer diameter D, cavity d, material of relative density σ) that just floats: $\sigma(D^{3} - d^{3}) = D^{3}$ , so $\dfrac{D}{d} = \left(\dfrac{\sigma}{\sigma - 1}\right)^{1/3}$ .
A body floating across two liquids: $V_1\rho_1 + V_2\rho_2 = (V_1 + V_2)\rho_b$ .

Submerged fraction

\frac{V_{\text{sub}}}{V} = \frac{\rho_b}{\rho_l}

Worked example

A wooden cylinder of height 12 cm and density

750\ \text{kg/m}^{3}

floats upright in water (

1000\ \text{kg/m}^{3}

). How much of its height is under water? How much would be under in oil of density

900\ \text{kg/m}^{3}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 2 · Q12Moderate

Example 3 · Mechanical Properties of Fluids · Pressure, Pascal's Law and Buoyancy

A cubical block of density

\rho_{b}= 600\text{ }kg/m^{3}

floats in a liquid of density

\rho_{e}= 900\text{ }kg/m^{3}

. If the height of block is

H = 8.0\text{ }cm

then height of the submerged part is

\_\_\_\_

cm.

Body over liquid, not liquid over body

The submerged height is H × ρ_body/ρ_liquid, which is less than H. Inverting the ratio gives a height bigger than the block itself.

A load adds displaced liquid, not displaced wood

The extra depth is m/(ρ_liquid × A). The density of the floating block plays no part in it.

Drill 4 more on floating bodies and archimedes' principle

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Pressure at a depth, absolute and gauge
Pressure at depth h
$P = P_0 + \rho g h$
Pascal's law and the hydraulic lift
Hydraulic lift
$\frac{F_1}{A_1} = \frac{F_2}{A_2}, \qquad F_1 d_1 = F_2 d_2$
Floating bodies and Archimedes' principle
Submerged fraction
$\frac{V_{\text{sub}}}{V} = \frac{\rho_b}{\rho_l}$

Watch out for (7)

Doubling the depth does not double the pressure
→ Pressure at a depth, absolute and gauge
Keep P₀ in a force on a base when the stem gives it
→ Pressure at a depth, absolute and gauge
Across a partition, the atmosphere cancels
→ Pressure at a depth, absolute and gauge
The small piston carries the same pressure
→ Pascal's law and the hydraulic lift
A lift multiplies force, not work
→ Pascal's law and the hydraulic lift
Body over liquid, not liquid over body
→ Floating bodies and Archimedes' principle
A load adds displaced liquid, not displaced wood
→ Floating bodies and Archimedes' principle

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