PhysicsICSE Class 9 · Chapter 5

Pressure in Fluids ICSE Class 9: Thrust, Liquid Pressure and Pascal's Law

Updated 15 October 2026 · Fact-checked

Pressure is thrust (normal force) per unit area, measured in pascals. In a liquid at rest, pressure increases with depth as p = hρg, is the same at one level and acts in all directions. Pascal's law says pressure applied to an enclosed liquid is transmitted equally throughout it. ICSE Class 9 applies these to dams and divers.

What Pressure in Fluids is about

A fluid is anything that can flow: liquids and gases. Fluids exert pressure on the walls of their container and on anything immersed in them.

The CISCE 2028 Class IX syllabus covers: thrust and pressure and their units; the pressure exerted by a liquid column, $p = h ρ g$ , and how pressure changes with depth; simple daily-life examples and factors in a stationary fluid, such as the broad base of a dam and a diver's suit; other consequences of $p = h ρ g$ ; and the transmission of pressure in liquids: Pascal's law (concept and simple numericals), with examples. The working of hydraulic machines is not included.

In the school exam expect definitions and units, 'give reasons' questions (why dams are thicker at the bottom), 2 to 3-mark numericals on $p = \frac{F}{A}$ and $p = h ρ g$ , and 3-mark Pascal's-law numericals on two pistons.

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Physics. Class IX is examined by your school: one 80-mark paper of 2 hours plus 20 marks of internal assessment. Pressure in Fluids is in the 4. Fluids (i) part of the 2028 syllabus.Source: CISCE ICSE 2028 Physics syllabus, Class IX

Key concepts

Thrust and pressure

Thrust is the force acting normally (perpendicularly) on a surface. It is a force, so its SI unit is the newton (N).

Pressure is the thrust per unit area: $p = \frac{F}{A}$ . SI unit: pascal (Pa), where $1 Pa = 1 N m^{- 2}$ . Pressure is a scalar.

The same thrust gives more pressure on a smaller area. That is why knives are sharpened, nails are pointed, and why bags have broad straps, tractors have wide tyres and camels have broad feet (to reduce pressure).

Worked example

A brick measuring 20 cm × 10 cm × 5 cm weighs 20 N. Find the greatest and least pressure it can exert on a table.

Try it first, then show the solutionSolution

Smallest face: $10 \times 5 = 50$ cm² $= 0.005$ m²: $p = \frac{20}{0.005} = 4000$ Pa.
Largest face: $20 \times 10 = 200$ cm² $= 0.02$ m²: $p = \frac{20}{0.02} = 1000$ Pa.

Answer: 4000 Pa and 1000 Pa

Pressure exerted by a liquid column: p = hρg

Consider a column of liquid of height $h$ , density $ρ$ and base area $A$ .

Volume $= A h$ ; mass $= A h ρ$ ; weight (thrust on the base) $= A h ρ g$ .
Pressure $= \frac{thrust}{area} = \frac{A h ρ g}{A} = h ρ g$ .

So the pressure due to a liquid at depth $h$ is $p = h ρ g$ . It depends on depth and density (and $g$ ), but not on the area of the base or the shape of the container.

With $h$ in m, $ρ$ in kg m⁻³ and $g$ in m s⁻², $p$ is in Pa.

Worked example

Find the pressure due to water at a depth of 10 m ( $ρ = 1000$ kg m⁻³, $g = 9.8$ m s⁻²).

Try it first, then show the solutionSolution

$p = h ρ g = 10 \times 1000 \times 9.8$ .
$p = 98000$ Pa $= 9.8 \times 1 0^{4}$ Pa.

Answer: 9.8 × 10⁴ Pa

Laws of liquid pressure

In a liquid at rest:
1. Pressure increases with depth.
2. Pressure is the same at all points at the same horizontal level.
3. At any point, pressure acts equally in all directions.
4. Pressure at a given depth depends on the density of the liquid.
5. A liquid finds its own level: in connected vessels of any shape, it stands at the same height.

A classic demonstration: water from a tall can with holes at different heights spurts farthest from the lowest hole, showing greater pressure at greater depth.

Worked example

A U-tube contains water. Oil is poured into one arm until the oil column is 10 cm high, and then the water in the other arm stands 8 cm above the oil–water boundary. Find the density of the oil.

Try it first, then show the solutionSolution

Pressures are equal at the level of the boundary: $h_{o} ρ_{o} g = h_{w} ρ_{w} g$ .
$0.10 \times ρ_{o} = 0.08 \times 1000$ .
$ρ_{o} = 800$ kg m⁻³.

Answer: 800 kg m⁻³

Consequences: dams and divers

Broad base of a dam: water pressure increases with depth, so the wall must be thicker and stronger at the bottom to withstand the greater pressure.
Diver's suit: deep under the sea the water pressure is very high; a special strong suit protects the diver's body.
Water tanks are placed high (on rooftops or towers) so that the water reaches taps below with good pressure.
Blood pressure is greater in the feet than in the brain when standing, because of the greater depth of the blood column.
Submarines have thick steel walls to withstand pressure at depth.

Worked example

Find the pressure due to sea water on a diver 50 m below the surface ( $ρ = 1030$ kg m⁻³, $g = 9.8$ m s⁻²), and explain why a special suit is needed.

Try it first, then show the solutionSolution

$p = 50 \times 1030 \times 9.8 = 504700$ Pa $\approx 5.05 \times 1 0^{5}$ Pa.
This is a very large pressure on every square metre of the body; the suit withstands it so the diver's body is not harmed.

Answer: ≈ 5.05 × 10⁵ Pa

Pascal's law: transmission of pressure

Pascal's law: pressure applied at any point of a liquid enclosed in a vessel is transmitted equally and undiminished in all directions throughout the liquid.

If a force $F_{1}$ acts on a small piston of area $A_{1}$ , the extra pressure $\frac{F _{1}}{A _{1}}$ reaches every part of the liquid. On a larger piston of area $A_{2}$ it produces a force

$F_{2} = \frac{F _{1}}{A _{1}} \times A_{2}$ .

So a small force on a small area can give a large force on a large area: the force is multiplied in the ratio $\frac{A _{2}}{A _{1}}$ . (Liquids are nearly incompressible, which is why they transmit pressure so well.)

Worked example

The small piston of a liquid-filled device has area 10 cm² and the large piston 500 cm². A force of 50 N is applied on the small piston. Find the pressure transmitted and the force on the large piston.

Try it first, then show the solutionSolution

$p = \frac{50}{10 \times 1 0 ^{- 4}} = 5 \times 1 0^{4}$ Pa.
$F_{2} = 5 \times 1 0^{4} \times 500 \times 1 0^{- 4} = 2500$ N.

Answer: 5 × 10⁴ Pa; 2500 N

Physics in real life: Pressure in Fluids

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.

Pressure and area

Suppose a 600 N person stands on snow, first in ordinary shoes (total area 0.04 m²) and then on skis (total area 0.4 m²).

Compare the pressures, and explain why skis stop you sinking in.

Try it first, then show the solutionSolution

Shoes: $\frac{600}{0.04} = 15000$ Pa.
Skis: $\frac{600}{0.4} = 1500$ Pa: ten times less, so the snow is not pushed in as much.

Answer: 15,000 Pa vs 1,500 Pa.

Memory hook: Spread the weight, lower the pressure.

Overhead tank

Imagine the water level in a rooftop tank is 12 m above a ground-floor tap.

Find the water pressure at the tap due to this height ( $g = 9.8$ m s⁻²).

Try it first, then show the solutionSolution

$p = h ρ g = 12 \times 1000 \times 9.8$ .
$= 117600$ Pa $\approx 1.18 \times 1 0^{5}$ Pa.

Answer: About 1.18 × 10⁵ Pa.

Memory hook: The higher the tank, the stronger the tap.

Pascal's law

Suppose a mechanic pushes with 200 N on a small piston of area 4 cm² in a sealed oil system connected to a large piston of area 400 cm².

What load can the large piston just support?

Try it first, then show the solutionSolution

Pressure $= \frac{200}{4 \times 1 0 ^{- 4}} = 5 \times 1 0^{5}$ Pa.
$F_{2} = 5 \times 1 0^{5} \times 400 \times 1 0^{- 4} = 20000$ N.

Answer: 20,000 N (100 times the effort).

Memory hook: Area ratio = force ratio.

Topics in this chapter

Pressure in Fluids formula sheet

Pressure: $p = \frac{F}{A}$; F = thrust (normal force).
Unit: $1 Pa = 1 N m^{- 2}$; SI unit of pressure.
Liquid column: $p = h ρ g$; Independent of shape and base area.
Thrust on a base: $F = p A = h ρ g A$; Weight of the liquid column above.
Two liquids balancing: $h_{1} ρ_{1} = h_{2} ρ_{2}$; Same level, same pressure.
Pascal's law: $\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}}$; Pressure transmitted undiminished.

Pressure in Fluids: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Can with holes at different depths

Show that liquid pressure increases with depth; 2 marks.

Draw

A tall vessel full of water.
Three small holes at different heights on one side.
Curved jets of water from each hole, the lowest jet reaching farthest.
Arrows showing the direction of each jet.

Label

Water
Holes A (top), B, C (bottom)
Jet from C reaches farthest

Watch out: Write the conclusion beside the figure: pressure is greatest at the deepest hole.

Pascal's law with two pistons

Illustrate the transmission of pressure; 2 to 3 marks.

Draw

Two connected cylinders filled with liquid: one narrow, one wide.
A piston in each cylinder.
An arrow pushing down on the narrow piston (F₁) and an arrow pushing up on the wide piston (F₂).
Mark the areas A₁ and A₂.

Label

Small piston, area A₁
Large piston, area A₂
F₁
F₂
Liquid

Watch out: State that the same pressure F₁/A₁ acts on the large piston, so F₂ = (F₁/A₁) × A₂.

How the school exam asks Pressure in Fluids

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Thrust and pressure	Define and give units; explain sharp knives, wide straps, camel feet; find pressure from force and area.	p = F/A; smaller area, larger pressure.
p = hρg numericals	Pressure at a depth; thrust on the base of a tank; compare pressures in different liquids.	SI units: h in m, ρ in kg m⁻³.
Give reasons	Dams have broad bases; divers wear special suits; water tanks are placed high.	Pressure increases with depth (p = hρg).
Laws of liquid pressure	State the laws; explain the can-with-holes experiment; liquids finding their own level.	Depth, same level, all directions, density.
Two liquids in a U-tube	Find a density from heights of balancing columns.	Equal pressure at the same level: h₁ρ₁ = h₂ρ₂.
Pascal's law	State the law; find the force on a large piston or the pressure transmitted.	F₁/A₁ = F₂/A₂; convert cm² to m².

Tricks to remember Pressure in Fluids

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

Point hurts, flat floats
Same force: a point gives huge pressure; a flat, wide surface gives little.
Use it when: Thrust and pressure examples.
h-rho-g: how really goes deep
Pressure = h ρ g: Height (depth), Rho (density), g.
Use it when: Recalling p = hρg.
Deeper, thicker
Dam walls thicken as the water gets deeper.
Use it when: Dam questions.
Squeeze one, squeeze all
Push on an enclosed liquid anywhere and the push is felt everywhere equally.
Use it when: Pascal's law.

Pressure in Fluids: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Unit of pressure · 1 mark · MCQExam level

One pascal is equal to:
(a) 1 N m
(b) 1 N m⁻²
(c) 1 N m²
(d) 1 kg m⁻²

Try it first, then show the solutionSolution

$p = \frac{F}{A}$ : newton per square metre.

Answer: (b)

Example 2: Thrust and pressure · 2 marks · Short answerExam level

A thrust of 500 N acts on an area of 0.25 m². Find the pressure. What would the pressure be if the area were halved?

Try it first, then show the solutionSolution

$p = \frac{500}{0.25} = 2000$ Pa.
Half the area: $4000$ Pa (pressure doubles).

Answer: 2000 Pa; 4000 Pa

Example 3: Thrust on a tank floor · 3 marks · Short answerExam level

A tank 3 m long and 2 m wide contains water 2 m deep. Find the pressure on the bottom due to the water and the thrust on the bottom ( $g = 9.8$ m s⁻²).

Try it first, then show the solutionSolution

$p = 2 \times 1000 \times 9.8 = 19600$ Pa.
Area $= 6$ m².
Thrust $= 19600 \times 6 = 117600$ N (equal to the weight of the water).

Answer: 19,600 Pa; 117,600 N

Example 4: Comparing two liquids · 3 marks · Short answerHarder exam level

Find the depth of kerosene (density 800 kg m⁻³) that gives the same pressure as 40 cm of water. Which liquid needs the greater depth, and why?

Try it first, then show the solutionSolution

$h_{k} \times 800 = 0.40 \times 1000$ .
$h_{k} = 0.50$ m $= 50$ cm.
Kerosene, because it is less dense; $p \propto ρ$ at a given depth.

Answer: 50 cm; kerosene, being less dense

Example 5: Give reasons · 3 marks · Short answerHarder exam level

Give reasons: (a) the walls of a dam are made thicker at the bottom (b) a diver needs a special suit at great depth (c) water spurts farthest from the lowest hole of a tall can.

Try it first, then show the solutionSolution

(a) Water pressure increases with depth ( $p = h ρ g$ ), so the bottom of the dam must withstand the greatest pressure.
(b) Pressure of sea water is very large at great depth; the suit withstands it to protect the body.
(c) The lowest hole is deepest, where pressure is greatest, so water leaves it with the greatest speed.

Answer: All three follow from p increasing with depth.

Example 6: Pascal's law with diameters · 3 marks · Short answerHarder exam level

The pistons of a liquid-filled system have diameters 2 cm and 20 cm. A force of 40 N acts on the smaller piston. Find the force exerted on the larger piston.

Try it first, then show the solutionSolution

Area ratio $= (\frac{20}{2})^{2} = 100$ (area $\propto$ diameter²).
$F_{2} = 40 \times 100 = 4000$ N.

Answer: 4000 N

Example 7: Same depth, different shapes · 4 marks · Long answerHOTS / competency

Three vessels of different shapes (a wide bowl, a narrow tube and a cone) are filled with water to the same height of 0.5 m. Compare the pressures at their bases, and the thrusts if their base areas are 0.04 m², 0.01 m² and 0.02 m².

Try it first, then show the solutionSolution

$p = h ρ g = 0.5 \times 1000 \times 9.8 = 4900$ Pa in all three: pressure does not depend on shape.
Thrust $= p A$ : $4900 \times 0.04 = 196$ N; $4900 \times 0.01 = 49$ N; $4900 \times 0.02 = 98$ N.
So the thrust on the base depends on its area, not on the amount of water in the vessel.

Answer: Equal pressures (4900 Pa); thrusts 196 N, 49 N, 98 N

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 9-mark question split into 3 + 3 + 3 marks.

Building a water supply

A village stores water in a tank on a tower. The water surface in the tank is 15 m above the ground. Pipes carry water to taps at ground level and to a first-floor tap 4 m above the ground. Nearby, a small dam holds water 20 m deep. ( $ρ = 1000$ kg m⁻³, $g = 9.8$ m s⁻².)

(i) · 3 marks
Find the water pressure at the ground-floor tap and at the first-floor tap (due to the water only).
Answer
Answer: 147,000 Pa and 107,800 Pa
Ground: $15 \times 1000 \times 9.8 = 147000$ Pa. First floor: depth $11$ m: $11 \times 1000 \times 9.8 = 107800$ Pa.
(ii) · 3 marks
Find the pressure at the bottom of the dam, and explain the shape of the dam wall.
Answer
Answer: 196,000 Pa; broader at the base
$20 \times 1000 \times 9.8 = 196000$ Pa. Pressure increases with depth, so the wall is made broader and stronger at the bottom.
(iii) · 3 marks
A tap at ground level has an opening of area 2 cm². Find the force needed to hold a thumb over it against the water, and explain why a tap on a higher floor is easier to block.
Answer
Answer: 29.4 N
$F = p A = 147000 \times 2 \times 1 0^{- 4} = 29.4$ N. On a higher floor the depth below the tank's surface is less, so the pressure and hence the force are smaller.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Define thrust and pressure, and state the SI unit of each. (2 marks)

Step	Marks
Thrust: force acting normally on a surface; unit newton (N)	1
Pressure: thrust per unit area; unit pascal (Pa) = N m⁻²	1

Derive the expression $p = h ρ g$ for the pressure at a depth $h$ in a liquid of density $ρ$ . (3 marks)

Step	Marks
Column of height $h$ and base area $A$ : volume $A h$ , mass $A h ρ$	1
Weight (thrust on base) $= A h ρ g$	1
Pressure $= \frac{A h ρ g}{A} = h ρ g$	1

State Pascal's law. In a liquid system, the small piston has area 5 cm² and the large piston 250 cm². What force on the small piston will support a load of 5000 N on the large piston? (4 marks)

Step	Marks
Pressure applied to an enclosed liquid is transmitted equally and undiminished in all directions	1
$\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}}$	1
$F_{1} = 5000 \times \frac{5}{250}$	1
$F_{1} = 100$ N	1

Where students lose marks in Pressure in Fluids

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

Mistake: Using cm and g/cm³ in p = hρg with g in m s⁻²
Mixed units give answers off by factors of 10 or 1000.
Wrong working
$p = 40 \times 1 \times 9.8$
Right working
$p = 0.40 \times 1000 \times 9.8 = 3920$ Pa
Do this: Convert to m and kg m⁻³ first.
Usually costs: The answer.
Mistake: Not converting cm² to m²
1 cm² = 10⁻⁴ m², not 10⁻² m².
Do this: Write the conversion line explicitly.
Usually costs: The answer.
Mistake: Saying pressure depends on the shape or amount of liquid
At a given depth, pressure depends only on $h$ , $ρ$ and $g$ .
Do this: Quote $p = h ρ g$ when explaining.
Usually costs: 1 mark.
Mistake: Confusing thrust and pressure
Thrust is a force (N); pressure is force per area (Pa).
Do this: Thrust = pressure × area.
Usually costs: 1 mark.
Mistake: Describing a hydraulic machine's working
The working of hydraulic machines is not in the Class IX syllabus; only Pascal's law and simple numericals are.
Do this: Answer with the law and the force ratio only.
Usually costs: Time.

Practice questions on Pressure in Fluids

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 · MCQExam level
Pressure at a point inside a liquid does not depend on:
(a) depth
(b) density of the liquid
(c) shape of the container
(d) g
Hint
p = hρg.
Answer
(c)
Q2 · 1 mark · Short answerExam level
Why does a school bag have broad straps?
Hint
Area.
Answer
Broad straps spread the weight over a larger area, reducing the pressure on the shoulders.
Q3 · 2 marks · Short answerExam level
Find the pressure due to a 76 cm column of mercury (density 13600 kg m⁻³, $g = 9.8$ m s⁻²).
Hint
h = 0.76 m.
Answer
$0.76 \times 13600 \times 9.8 \approx 1.013 \times 1 0^{5}$ Pa.
Q4 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): Two vessels with water to the same height but different base areas have the same pressure at their bases.
Reason (R): Liquid pressure depends on depth and density, not on base area.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
p = hρg.
Answer
(c)
Q5 · 2 marks · Short answerHarder exam level
At what depth in sea water (density 1030 kg m⁻³) is the pressure due to the water $2.06 \times 1 0^{5}$ Pa? ( $g = 10$ m s⁻²)
Hint
h = p/(ρg).
Answer
$h = \frac{2.06 \times 1 0 ^{5}}{1030 \times 10} = 20$ m.
Q6 · 3 marks · Short answerHarder exam level
A U-tube contains mercury. Water is poured into one arm until the water column is 27.2 cm high. How far above the mercury surface in the water arm does the mercury in the other arm stand? (Mercury 13600 kg m⁻³.)
Hint
h_w ρ_w = h_m ρ_m.
Answer
$0.272 \times 1000 = h \times 13600$ , $h = 0.02$ m $= 2$ cm.
Q7 · 3 marks · Short answerHarder exam level
An elephant of weight 40,000 N stands on four feet, each of area 0.1 m². A woman of weight 500 N stands on one heel of area 1 cm². Who exerts the greater pressure on the floor?
Hint
Compare F/A.
Answer
Elephant: $\frac{40000}{0.4} = 1 0^{5}$ Pa. Woman: $\frac{500}{1 0 ^{- 4}} = 5 \times 1 0^{6}$ Pa. The woman, 50 times more.
Q8 · 3 marks · Short answerHarder exam level
In a Pascal's-law system, a force of 25 N on a piston of area 0.005 m² supports a car of weight 15,000 N. Find the area of the larger piston.
Hint
A₂ = F₂A₁/F₁.
Answer
$A_{2} = \frac{15000 \times 0.005}{25} = 3$ m².
Q9 · 2 marks · Short answerExam level
A swimmer dives from 2 m to 6 m below the surface of a lake. By how much does the water pressure on her increase? ( $g = 9.8$ m s⁻²)
Hint
Δp = Δh ρ g.
Answer
$4 \times 1000 \times 9.8 = 39200$ Pa.
Q10 · 4 marks · Long answerHarder exam level
A cylindrical jar of base area 50 cm² holds water to a height of 20 cm. Find the pressure and the thrust on the base, and show that the thrust equals the weight of the water ( $g = 10$ m s⁻²).
Hint
Volume 1000 cm³.
Answer
$p = 0.2 \times 1000 \times 10 = 2000$ Pa; thrust $= 2000 \times 50 \times 1 0^{- 4} = 10$ N. Water: 1000 cm³ = 1 kg, weight $= 10$ N ✓.
Q11 · 3 marks · Short answerHOTS / competency
Why is blood pressure measured with the cuff on the upper arm at the level of the heart, rather than at the ankle, when a person is standing?
Hint
p = hρg in a column of blood.
Answer
When standing, the ankle is about 1.2 m below the heart, so the extra pressure of the blood column $h ρ g$ makes the reading there much higher. At heart level this extra depth is zero, so the reading reflects the heart's pumping pressure.
Q12 · 4 marks · Long answerHOTS / competency
Two cylinders of a sealed liquid system have pistons of area 2 cm² and 100 cm². If the small piston is pushed down 10 cm, how far does the large piston rise? If a force of 30 N acts on the small piston, find the force on the large piston and compare the two ratios you have found.
Hint
Volume pushed in = volume pushed out.
Answer
Volume moved $= 2 \times 10 = 20$ cm³; rise $= \frac{20}{100} = 0.2$ cm. By Pascal's law, $F_{2} = 30 \times \frac{100}{2} = 1500$ N. The force is multiplied 50 times, but the large piston moves only $\frac{1}{50}$ of the distance the small one moves: the gain in force is paid for by a loss in distance.

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.Define pressure.Show answer
Thrust per unit area.
2.SI unit of pressure?Show answer
Pascal (Pa) = N m⁻².
3.Formula for pressure due to a liquid column?Show answer
p = hρg.
4.Does liquid pressure depend on the shape of the vessel?Show answer
No.
5.Why is a dam broader at the base?Show answer
Pressure increases with depth.
6.State Pascal's law.Show answer
Pressure applied to an enclosed liquid is transmitted equally and undiminished in all directions.
7.Why are knives sharpened?Show answer
A smaller area gives a greater pressure for the same force.
8.1 cm² in m²?Show answer
10⁻⁴ m².

Best order to study it

Learn thrust, pressure, units and everyday examples.
Derive p = hρg and practise depth calculations.
Learn the laws of liquid pressure and the U-tube balance.
Learn reasons for dams, divers, tanks.
Learn Pascal's law and practise piston numericals.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you define thrust and pressure with units?
Can you derive p = hρg?
Do you convert cm to m and cm² to m²?
Can you explain why pressure does not depend on the vessel's shape?
Can you give reasons for broad dams and diving suits?
Can you solve U-tube density questions?
Can you state Pascal's law and find the force on a large piston?

Pressure in Fluids: frequently asked questions

Is atmospheric pressure in the Class IX 2028 syllabus?

No. The 2028 Class IX syllabus leaves out atmospheric pressure (its variation with altitude, weather forecasting and the altimeter). This chapter covers thrust, pressure in liquids, p = hρg and Pascal's law.

Is the working of a hydraulic press or brake in the syllabus?

No. The working of hydraulic machines is excluded. You need Pascal's law and simple numericals on transmitted pressure.

Why does pressure act in all directions in a liquid?

Liquid particles push on each other and on surfaces in every direction, so at a point the pressure is the same whichever way a surface faces.

Why are liquids good at transmitting pressure?

They are nearly incompressible, so a push at one place is passed on without being used up in squashing the liquid.

Is pressure a vector?

No. Pressure is a scalar; the thrust it produces on a surface is a force and has a direction (normal to the surface).

Based on: CISCE ICSE 2028 Science (52) Physics syllabus (Classes IX and X); CISCE Regulations for ICSE Examination Year 2028