Fluids and Pressure
Introduction
Fluids (liquids and gases) exert pressure on objects immersed in them. Understanding fluid pressure is crucial for explaining phenomena from submarines to balloons. ICSE Class 9 covers pressure in liquids, atmospheric pressure, buoyancy, and Archimedes principle.
Thrust and Pressure
Thrust
Force applied perpendicular to a surface.
Pressure
Pressure = Thrust / Area = F/A
Unit: Pa (Pascal) = N/m²
Key Points:
- Pressure is inversely proportional to area (for the same force)
- Sharp knives have small area, producing high pressure
- Wide tyres have large area, reducing pressure
Pressure in Liquids
Formula
P = hρg
Where:
- P = pressure at depth h
- h = depth below the liquid surface
- ρ = density of the liquid
- g = acceleration due to gravity
Key Points:
- Pressure in a liquid increases with depth
- Pressure depends on the density of the liquid
- Pressure at the same depth is the same in all directions
- Pressure does NOT depend on the shape of the container
Pascals Law
Statement: Pressure applied to an enclosed fluid is transmitted equally in all directions throughout the fluid and to the walls of the container.
Applications
- Hydraulic lift: Used to lift heavy objects using a small force
- Hydraulic brakes: Transmit force through brake fluid
- Hydraulic press: Used for compressing materials
Hydraulic Lift Formula
F1/A1 = F2/A2
A small force F1 on a small area A1 produces a large force F2 on a large area A2.
Atmospheric Pressure
Atmospheric pressure is the pressure exerted by the atmosphere on all objects on Earth.
Value at sea level: 1 atm = 1.013 × 10⁵ Pa = 76 cm of Hg
Measurement: Using a barometer.
<ICSEExample title="Atmospheric Pressure Conversion"> Express 1 atmospheric pressure in terms of height of water column. (Density of mercury = 13600 kg/m³, density of water = 1000 kg/m³) <Solution> P = h₁ρ₁g = h₂ρ₂g 0.76 × 13600 × g = h₂ × 1000 × g h₂ = 0.76 × 13600/1000 = 10.34 m Thus, 1 atm = 10.34 m of water column. </Solution> </ICSEExample>Buoyancy and Archimedes Principle
Archimedes Principle
When a body is immersed fully or partially in a fluid, it experiences an upward force (buoyant force) equal to the weight of the fluid displaced by the body.
Buoyant force = Weight of fluid displaced = Volume of displaced fluid × density × g
Apparent Weight
Apparent weight = Actual weight - Buoyant force
<ICSEExample title="Apparent Weight"> A stone weighs 50 N in air and 30 N in water. Find the volume of the stone. (Density of water = 1000 kg/m³, g = 10 m/s²) <Solution> Loss of weight = 50 - 30 = 20 N Buoyant force = Weight of water displaced = 20 N V × ρ × g = 20 V × 1000 × 10 = 20 V = 2 × 10⁻³ m³ = 2000 cm³ </Solution> </ICSEExample>Floatation
Condition for Floatation:
- If density of object < density of liquid: object floats
- If density of object = density of liquid: object is suspended
- If density of object > density of liquid: object sinks
Law of Floatation: A floating body displaces its own weight of the fluid in which it floats.
Common Mistakes With Fixes
| Mistake | Correction |
|---|---|
| Pressure only depends on depth | True, but also on density of liquid |
| Atmospheric pressure is negligible | 1 atm = 101,325 Pa (very significant) |
| Buoyant force depends on weight of object | Buoyant force depends on weight of displaced fluid, not object |
| Density same as weight | Density is mass/volume; weight is mass × g |
ICSE Exam Focus
| Topic | Marks (approx.) | Frequency |
|---|---|---|
| Archimedes principle numericals | 4-5 marks | Very common |
| Pressure in liquids (P = hρg) | 3-4 marks | Common |
| Pascals law applications | 3 marks | Common |
| Atmospheric pressure | 2-3 marks | Occasionally asked |
Self-Test
Q1: Calculate the pressure exerted by a force of 100 N on an area of 2 m².
Q2: Find the pressure at a depth of 5 m in a liquid of density 1200 kg/m³.
Q3: A body weighs 80 N in air and 50 N in water. Find the volume of the body.
Q4: State Archimedes principle.
Q5: Why does an iron nail sink but an iron ship float?
