PhysicsICSE Class 9 · Pressure in Fluids

Thrust, Pressure and Liquid Pressure p = hρg: Laws, Layered Liquids and Dams (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

Thrust is the force acting normally on a surface (N); pressure is thrust per unit area, p = F/A (pascal, Pa = N m⁻²). A liquid column of height h and density ρ exerts a pressure p = hρg on its base, which depends on depth and density but not on the shape or area of the vessel.

Notice that only the vertical height h sets the pressure at the base; the shape and width of the vessel make no difference.

Illustration drawn for QuestionBuddy.

Understand Thrust, Pressure and the Pressure Due to a Liquid Column (p = hρg)

Thrust and pressure. Thrust is the force acting perpendicular to a surface (newton). Pressure is the thrust on unit area:

$p = \frac{F}{A}, 1 Pa = 1 N m^{- 2} .$

The same thrust acting on a smaller area gives a larger pressure (a pin, a knife edge); spreading it over a larger area reduces the pressure (broad shoulder straps, a tractor’s wide tyres, snowshoes).

Pressure due to a liquid column. For a column of height $h$ , density $ρ$ and base area $A$ : volume $= A h$ , mass $= A h ρ$ , weight $= A h ρ g$ . Pressure on the base $= \frac{A h ρ g}{A}$ :

$p = h ρ g .$

$h$ is the vertical depth below the free surface, not the length along a slanting tube.
$ρ$ in kg m⁻³ (water 1000), $g$ in m s⁻² (10 or 9.8), $p$ in Pa.
$p$ does not depend on the base area or the shape of the vessel. Vessels of different shapes filled to the same height have the same pressure at the base, but different weights of liquid. (The thrust on the base is $p A$ , which can differ from the weight of the liquid.)

Laws of liquid pressure. At rest: pressure increases with depth; it is the same at all points at one level; it acts equally in all directions; it is greater for a denser liquid at the same depth.

Layered liquids. If liquids that do not mix lie one over another, the pressure at the bottom is the sum of the pressures of the layers: $p = h_{1} ρ_{1} g + h_{2} ρ_{2} g$ .

Balancing columns. Where two liquids meet at the same level in a U-tube, the pressures are equal there: $h_{1} ρ_{1} = h_{2} ρ_{2}$ (g cancels). The liquid of lower density stands higher.

Consequences.

The wall of a dam is thicker at the bottom because the pressure of the water increases with depth.
Deep-sea divers and submarines need strong suits or hulls.
A water tank is placed high so that taps below get water at greater pressure.
A liquid finds its own level in connected vessels.

Thrust on a flat wall. Because pressure changes steadily from zero at the top to the greatest value at the bottom, the thrust on a vertical wall can be found by taking the average pressure over its depth ( $\frac{1}{2} ρ g H$ for a wall from the surface down to depth $H$ ) times the wetted area.

Thrust, Pressure and the Pressure Due to a Liquid Column (p = hρg): formulas and key facts

Pressure: $p = \frac{F}{A}$; Pa = N m⁻².
Liquid column: $p = h ρ g$; h vertical depth.
Thrust on a base: $F = p A = h ρ g A$; Not always equal to the weight of the liquid.
Layers and balancing: $p = \sum h ρ g; h_{1} ρ_{1} = h_{2} ρ_{2}$; Non-mixing liquids.

How to solve Thrust, Pressure and the Pressure Due to a Liquid Column (p = hρg) questions

Identify depth, density and area, then choose p or F.

1Convert depths to metres and densities to kg m⁻³.
2Use $p = h ρ g$ for pressure in a liquid; $p = F / A$ for a solid’s thrust.
3Add the contributions of each layer for layered liquids.
4Multiply by area to get thrust, remembering this may differ from the liquid’s weight.
5For a U-tube equate $h ρ$ at the interface level.

Thrust, Pressure and the Pressure Due to a Liquid Column (p = hρg) in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A rooftop tank

Suppose a household tank on a roof has its water surface 12 m above a bathroom tap and a second household has its tank only 3 m above its tap.

Compare the water pressures at the two taps (ρ = 1000 kg m⁻³, g = 10 m s⁻²).

Try it first, then show the solutionSolution

$p_{1} = 12 \times 1000 \times 10 = 1.2 \times 1 0^{5}$ Pa; $p_{2} = 3 \times 1000 \times 10 = 3 \times 1 0^{4}$ Pa.
Ratio $= 4 : 1$ .

Answer: The first tap has four times the pressure.

Memory hook: Pressure grows with vertical height of the water above.

Tricks to remember Thrust, Pressure and the Pressure Due to a Liquid Column (p = hρg)

Short memory aids. Say each one aloud once, then use it on a question the same day.

hρg: height, rho, g
"Heavy Rock Gives" pressure: h ρ g.
Use it when: Remembering the formula.
Vertical depth only
Slanting tubes do not matter; measure straight down from the surface.
Use it when: Diagram questions.
Lighter liquid stands taller
In a U-tube hρ is equal on both sides, so a lower ρ means a larger h.
Use it when: Balancing columns.

Thrust, Pressure and the Pressure Due to a Liquid Column (p = hρg): solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Doubling the depth · 1 mark · MCQBoard level

In a liquid of fixed density, if the depth of a point is doubled, the pressure due to the liquid:
(a) is halved
(b) is unchanged
(c) is doubled
(d) becomes four times

Try it first, then show the solutionSolution

$p = h ρ g \propto h$ , so doubling $h$ doubles $p$ .

Answer: (c) is doubled

Example 2: One block, two faces · 2 marks · Short answerBoard level

A block weighing 200 N rests on a table with a face of area 0.04 m² down. Find the pressure it exerts. What is the pressure if it is turned on a face of area 0.01 m²?

Try it first, then show the solutionSolution

$p_{1} = \frac{200}{0.04} = 5000$ Pa. (1)
$p_{2} = \frac{200}{0.01} = 20000$ Pa. (1)

Answer: 5000 Pa; 20 000 Pa

Example 3: Thrust and weight · 3 marks · Short answerHarder board level

A straight-sided cylindrical vessel of base area 0.05 m² holds water ( $ρ = 1000$ kg m⁻³) to a depth of 0.8 m ( $g = 10$ m s⁻²). Find the pressure on the base, the thrust on the base and the weight of the water.

Try it first, then show the solutionSolution

$p = 0.8 \times 1000 \times 10 = 8000$ Pa. (1)
Thrust $= p A = 8000 \times 0.05 = 400$ N. (1)
Weight of water $= 0.05 \times 0.8 \times 1000 \times 10 = 400$ N, equal to the thrust for a straight-sided vessel. (1)

Answer: 8000 Pa; 400 N; 400 N

Example 4: Oil on water · 3 marks · Short answerHarder board level

A tall vertical tube holds water 0.40 m deep with oil of density 800 kg m⁻³ floating 0.25 m deep above it. Find the pressure due to the liquids at the bottom ( $ρ_{w a t er} = 1000$ kg m⁻³, $g = 10$ m s⁻²).

Try it first, then show the solutionSolution

Oil: $0.25 \times 800 \times 10 = 2000$ Pa. (1)
Water: $0.40 \times 1000 \times 10 = 4000$ Pa. (1)
Total $= 6000$ Pa. (1)

Answer: 6000 Pa

Example 5: Dam wall · 4 marks · Case-basedHOTS / competency

A reservoir behind a dam is 30 m deep and the wall is 100 m long ( $ρ = 1000$ kg m⁻³, $g = 10$ m s⁻²).
(a) Find the water pressure at 10 m and 30 m depth.
(b) State why the wall is thicker at the bottom.
(c) Find the average pressure over the wall and the total thrust on it.
(d) Does the thrust change if the reservoir stretches 5 km behind the wall instead of 1 km?

Try it first, then show the solutionSolution

(a) $p_{10} = 10 \times 1000 \times 10 = 1.0 \times 1 0^{5}$ Pa; $p_{30} = 3.0 \times 1 0^{5}$ Pa. (1)
(b) Pressure increases with depth, so the lower part must withstand more. (1)
(c) Average $= \frac{0 + 3 \times 1 0 ^{5}}{2} = 1.5 \times 1 0^{5}$ Pa; area $= 30 \times 100 = 3000$ m²; thrust $= 1.5 \times 1 0^{5} \times 3000 = 4.5 \times 1 0^{8}$ N. (1)
(d) No: pressure depends on depth, not on the horizontal extent of the water. (1)

Answer: (a) 1.0 × 10⁵ and 3.0 × 10⁵ Pa (c) 1.5 × 10⁵ Pa; 4.5 × 10⁸ N (d) no

Where students lose marks in Thrust, Pressure and the Pressure Due to a Liquid Column (p = hρg)

The slips examiners see most often, and what to write instead.

Mistake: Slant length used as depth
Depth is vertical.
Wrong working
h = length of the tilted tube.
Right working
h = vertical height from the free surface.
Do this: Drop a vertical line from the surface.
Usually costs: The answer.
Mistake: Pressure in cm of water and ρ in kg m⁻³ mixed
h must be in m for Pa.
Wrong working
p = 40 × 1000 × 10
Right working
p = 0.40 × 1000 × 10 = 4000 Pa
Do this: Convert first.
Usually costs: The answer.
Mistake: Thrust on base equated to weight of liquid for every vessel
True only for straight-sided vessels.
Wrong working
Thrust = weight in a conical flask.
Right working
Thrust = pA; weight may be more or less.
Do this: Compute pA separately.
Usually costs: The reasoning mark.

Exam tips

Write p = hρg and state that h is vertical depth.
Show unit conversion for h (cm to m).
Say "pressure does not depend on shape or area of the vessel" when comparing vessels.
Quote Pa = N m⁻² and use it when asked for units.

Practice questions on Thrust, Pressure and the Pressure Due to a Liquid Column (p = hρg)

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The pressure at a point inside a liquid at rest acts:
(a) only downward
(b) only upward
(c) equally in all directions
(d) only sideways
Hint
Law of liquid pressure.
Answer
(c)
Q2 · 1 mark · Very short answerBoard level
State the SI unit of thrust.
Hint
It is a force.
Answer
Newton (N).
Q3 · 2 marks · Short answerHarder board level
At what depth does a liquid of density 1200 kg m⁻³ produce a pressure of 6000 Pa ( $g = 10$ m s⁻²)?
Hint
h = p/(ρg).
Answer
$h = \frac{6000}{1200 \times 10} = 0.5$ m.
Q4 · 3 marks · Short answerHarder board level
The two arms of a U-tube contain mercury at the bottom. Liquid A (density 800 kg m⁻³) stands 30 cm high in one arm and liquid B (density 1200 kg m⁻³) in the other arm, and the mercury surfaces in the two arms are at the same level. Find the height of liquid B.
Hint
Equal pressure on the mercury at the same level.
Answer
$h_{B} \times 1200 = 0.30 \times 800$ , so $h_{B} = 0.20$ m $= 20$ cm.
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): A sharp knife cuts better than a blunt one.
Reason (R): For the same thrust, a smaller area gives a larger pressure.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Hint
p = F/A.
Answer
(a)
Q6 · 3 marks · Short answerHOTS / competency
The hatch of a submarine has an area of 0.5 m² and lies 120 m below the sea surface (sea water $ρ = 1030$ kg m⁻³, $g = 10$ m s⁻²). Find the extra pressure due to the water and the thrust on the hatch. Express the thrust as the weight of a mass in kg.
Hint
F = pA, then F = mg.
Answer
$p = 120 \times 1030 \times 10 = 1.236 \times 1 0^{6}$ Pa; $F = 1.236 \times 1 0^{6} \times 0.5 = 6.18 \times 1 0^{5}$ N; equal to the weight of $\frac{6.18 \times 1 0 ^{5}}{10} = 61800$ kg.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Unit of pressure?Show answer
Pascal (N m⁻²).
2.Formula for liquid pressure?Show answer
p = hρg.
3.Does it depend on the vessel’s shape?Show answer
No.
4.Why is a dam thicker at the bottom?Show answer
Pressure increases with depth.
5.What is thrust?Show answer
Force acting normally on a surface.

Thrust, Pressure and the Pressure Due to a Liquid Column (p = hρg): frequently asked questions

How does the pressure at a point in a liquid depend on depth and density?

p = hρg, where h is the vertical depth, ρ the density of the liquid and g the acceleration due to gravity.

Why does pressure increase with depth?

The pressure at a point is due to the weight of the liquid column above it, which grows with depth.

What is the difference between thrust and pressure?

Thrust is a force (newton); pressure is thrust per unit area (pascal).