Prerequisites
Before studying fluid pressure, make sure you understand:
- Newton’s Laws — Force and equilibrium
- Stress and Strain — Pressure concept
What is a Fluid?
A fluid is a substance that can flow and takes the shape of its container.
Types:
- Liquids: Fixed volume, variable shape
- Gases: Variable volume and shape
Key property: Fluids cannot sustain shearing stress (shear modulus = 0). Any shear stress causes continuous flow.
When you step on water, you apply a perpendicular force (normal stress) which water can support momentarily. But any tangential force (shearing stress) causes the water to flow away from under your feet. Solids can resist both, but fluids can’t resist shear — that’s why they flow!
Fun fact: The Basilisk lizard can run on water by slapping the surface so fast that it doesn’t sink before pushing off!
Pressure in Fluids
Pressure is the force per unit area acting perpendicular to a surface.
$$\boxed{P = \frac{F}{A}}$$SI Unit: Pascal (Pa) = N/m²
Other common units:
- 1 atm (atmosphere) = 1.013 × 10⁵ Pa
- 1 bar = 10⁵ Pa
- 1 torr = 133.3 Pa
- 1 mm Hg = 133.3 Pa
Dimensions: [ML⁻¹T⁻²] (same as stress)
Although defined as F/A (force is vector), pressure in fluids is a scalar quantity.
Why? In a fluid at rest, pressure at a point is same in all directions. The force acts perpendicular to any surface placed at that point, regardless of orientation.
Atmospheric Pressure
Atmospheric pressure is the pressure exerted by the weight of air above us.
At sea level:
- P₀ = 1 atm = 1.013 × 10⁵ Pa ≈ 10⁵ Pa (for calculations)
- Equivalent to 76 cm of Hg column
- Equivalent to 10.3 m of water column
Air presses on your body with force ≈ 100,000 N (weight of a truck!)
Why don’t we feel crushed?
Because pressure inside our body (blood, air in lungs) equals atmospheric pressure outside. Net force = 0!
When you go to high altitudes, external pressure drops but internal doesn’t adjust immediately — that’s why your ears “pop” and you feel pressure!
Pascal’s Law
Pascal’s Law states: Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of the container.
$$\boxed{\Delta P = \text{constant throughout the fluid}}$$Applications of Pascal’s Law
1. Hydraulic Lift (Hydraulic Press)
Two cylinders of different areas connected by fluid.
Principle:
- Force applied on smaller piston (area A₁)
- Transmitted to larger piston (area A₂)
- Mechanical advantage achieved
Since pressure is same: P₁ = P₂
Interactive Demo: Visualize Fluid Pressure
See how pressure changes with depth and how hydraulic systems work.
At a car service station, a mechanic easily lifts a 1500 kg car using a hydraulic lift!
How? Small piston area = 10 cm², large piston area = 1000 cm²
Mechanical advantage = 1000/10 = 100
Force needed = (1500 × 10)/100 = 150 N (just 15 kg weight!)
The mechanic pushes down 100 cm to lift the car 1 cm (work is conserved: F₁d₁ = F₂d₂).
Important: Work input = Work output (no free energy!)
$$F_1 \cdot d_1 = F_2 \cdot d_2$$where d₁, d₂ are distances moved.
2. Hydraulic Brakes
Pressing brake pedal creates pressure in brake fluid, transmitted equally to all wheel cylinders, applying brakes to all wheels simultaneously.
Advantage: Same braking force to all wheels, regardless of their position.
Hydrostatic Pressure (Pressure Due to Liquid Column)
Pressure at depth h below the surface of a liquid:
$$\boxed{P = P_0 + \rho gh}$$where:
- P₀ = atmospheric pressure at surface
- ρ = density of liquid
- g = acceleration due to gravity
- h = depth below surface
Gauge pressure (pressure above atmospheric):
$$\boxed{P_{gauge} = P - P_0 = \rho gh}$$Absolute pressure = P₀ + P_gauge
- Depends only on depth h, not on shape of container
- Same at all points at same horizontal level in connected fluid
- Increases linearly with depth
- Independent of container shape (Hydrostatic Paradox)
Find pressure at 100 m depth in ocean. (ρ_seawater = 1030 kg/m³, P₀ = 10⁵ Pa, g = 10 m/s²)
Solution:
$$P = P_0 + \rho gh$$ $$= 10^5 + 1030 \times 10 \times 100$$ $$= 10^5 + 1.03 \times 10^6$$ $$= 1.13 \times 10^6 \text{ Pa} = \boxed{11.3 \text{ atm}}$$At 100 m depth, pressure is about 11 times atmospheric pressure!
Hydrostatic Paradox
Paradox: Pressure at a point depends only on vertical depth, not on the shape or amount of liquid above it.
Three containers of different shapes but same height h have same pressure at the bottom, even though they contain different amounts of water!
Explanation: Pressure is due to vertical component of weight. Slanted walls provide support, reducing effective weight contribution.
Barometer (Measuring Atmospheric Pressure)
Torricelli’s barometer: A tube filled with mercury, inverted in a mercury dish.
Principle:
$$P_0 = \rho_{Hg} \cdot g \cdot h$$At sea level: h ≈ 76 cm of Hg
$$P_0 = 13600 \times 10 \times 0.76 = 1.03 \times 10^5 \text{ Pa} ≈ 1 \text{ atm}$$Why mercury?
- High density (13,600 kg/m³) → short column (76 cm vs 10.3 m for water!)
- Doesn’t wet glass
- Low vapor pressure
“76 cm Hg = 1 atm” is a must-remember value for JEE!
Alternatively: “760 mm Hg = 1 atm”
Manometer (Measuring Gauge Pressure)
Open-tube manometer: U-tube with one end open to atmosphere, other connected to gas container.
Pressure difference:
$$\boxed{P_{gas} - P_0 = \rho g h}$$where h = height difference between liquid levels.
- If h is positive (gas side lower): P_gas > P₀
- If h is negative (gas side higher): P_gas < P₀
A manometer connected to a gas cylinder shows mercury level 20 cm higher on the open side. Find gas pressure. (P₀ = 76 cm Hg)
Solution:
Gas side is lower → P_gas > P₀
$$P_{gas} = P_0 + \rho_{Hg} g h$$In cm of Hg: P_gas = 76 + 20 = 96 cm Hg
Or: P_gas = 1.26 atm
Archimedes’ Principle and Buoyancy
Archimedes’ Principle: When a body is partially or fully immersed in a fluid, it experiences an upward force (buoyant force) equal to the weight of fluid displaced.
$$\boxed{F_B = \rho_{fluid} \cdot V_{displaced} \cdot g}$$where:
- ρ_fluid = density of fluid
- V_displaced = volume of fluid displaced
- g = acceleration due to gravity
Direction: Always upward (opposite to gravity)
Origin: Pressure difference between top and bottom surfaces of immersed body.
A steel ship weighs thousands of tons. Steel density (7800 kg/m³) » water density (1000 kg/m³).
How does it float?
The ship’s hollow structure displaces huge volume of water. As long as:
Weight of ship < Weight of water displaced
the ship floats!
Buoyant force = Weight of water displaced > Weight of ship → Ship floats!
Three Cases of Floating/Sinking
For a body of volume V and density ρ_body in fluid of density ρ_fluid:
Weight of body: W = ρ_body × V × g
Maximum buoyant force: F_B,max = ρ_fluid × V × g (when fully submerged)
Case 1: Floating (ρ_body < ρ_fluid)
Body floats partially submerged.
At equilibrium: Buoyant force = Weight
$$\rho_{fluid} \cdot V_{submerged} \cdot g = \rho_{body} \cdot V_{total} \cdot g$$ $$\boxed{\frac{V_{submerged}}{V_{total}} = \frac{\rho_{body}}{\rho_{fluid}}}$$Fraction submerged = density ratio
Density of ice = 920 kg/m³, seawater = 1030 kg/m³
Fraction submerged = 920/1030 = 0.893 = 89.3%
So about 90% of an iceberg is underwater — “tip of the iceberg”!
Case 2: Just Floating (ρ_body = ρ_fluid)
Body is fully submerged but doesn’t sink.
Neutral buoyancy — like fish or submarines adjusting their density.
Case 3: Sinking (ρ_body > ρ_fluid)
Body sinks to bottom.
Net downward force:
$$F_{net} = W - F_B = (\rho_{body} - \rho_{fluid}) \cdot V \cdot g$$This is the apparent weight in fluid.
$$\boxed{W_{apparent} = W_{actual} - F_B}$$ $$\boxed{W_{apparent} = V g (\rho_{body} - \rho_{fluid})}$$A 2 kg metal block (density 8000 kg/m³) is immersed in water. Find apparent weight.
Solution:
Volume = m/ρ = 2/8000 = 2.5 × 10⁻⁴ m³
F_B = ρ_water × V × g = 1000 × 2.5 × 10⁻⁴ × 10 = 2.5 N
W_actual = 2 × 10 = 20 N
W_apparent = 20 - 2.5 = 17.5 N
The block feels lighter in water!
Common Mistakes to Avoid
Wrong: Wider container has more pressure at bottom.
Correct: Pressure at a depth h depends only on h, not on container width or shape (Hydrostatic Paradox).
Wrong: Buoyant force always equals weight of the immersed body.
Correct: Buoyant force = Weight of fluid displaced, NOT weight of body.
Only when floating: F_B = W (equilibrium condition)
Wrong: Pascal’s law works for all fluids.
Correct: Pascal’s law applies only to fluids at rest (hydrostatics). For moving fluids, use Bernoulli’s equation.
Practice Problems
Level 1: JEE Main Basics
Find pressure at a depth of 10 m in water. (ρ_water = 1000 kg/m³, g = 10 m/s², P₀ = 10⁵ Pa)
Solution:
$$P = P_0 + \rho gh$$ $$= 10^5 + 1000 \times 10 \times 10$$ $$= 10^5 + 10^5 = 2 \times 10^5 \text{ Pa} = \boxed{2 \text{ atm}}$$A hydraulic lift has pistons of area 10 cm² and 100 cm². What force must be applied to smaller piston to lift 500 kg?
Solution:
Mechanical advantage = A₂/A₁ = 100/10 = 10
F₂ = 500 × 10 = 5000 N
$$F_1 = \frac{F_2}{10} = \frac{5000}{10} = \boxed{500 \text{ N}}$$(Equal to weight of 50 kg!)
Level 2: JEE Main/Advanced
A wooden block of mass 2 kg and density 800 kg/m³ floats in water. Find: (a) Fraction of volume submerged (b) Mass to be added so it just sinks
Solution:
(a) Fraction submerged:
$$\frac{V_{sub}}{V_{total}} = \frac{\rho_{wood}}{\rho_{water}} = \frac{800}{1000} = \boxed{0.8 = 80\%}$$(b) Mass to add:
Volume of block: V = m/ρ = 2/800 = 2.5 × 10⁻³ m³
For just sinking, total mass must equal mass of water displaced when fully submerged:
m_total = ρ_water × V = 1000 × 2.5 × 10⁻³ = 2.5 kg
Mass to add = 2.5 - 2 = 0.5 kg
A U-tube contains water and oil (density 900 kg/m³). The oil column is 20 cm high. Find height of water column for equilibrium.
Solution:
At the same horizontal level (interface), pressure must be equal:
$$P_{water} = P_{oil}$$ $$\rho_{water} \cdot g \cdot h_{water} = \rho_{oil} \cdot g \cdot h_{oil}$$ $$1000 \times h_{water} = 900 \times 20$$ $$h_{water} = \frac{18000}{1000} = \boxed{18 \text{ cm}}$$Denser liquid (water) has shorter column!
Level 3: JEE Advanced
A cubical block of edge 10 cm and density 800 kg/m³ floats in water with a 4 cm high layer of oil (density 900 kg/m³) on top. Find how much of the cube is in oil and how much in water.
Solution:
Let x = height in oil, y = height in water
Total height: x + y = 10 cm (assuming it doesn’t float above oil)
Weight = Buoyant force:
$$\rho_{cube} \cdot V_{cube} \cdot g = \rho_{oil} \cdot V_{oil} \cdot g + \rho_{water} \cdot V_{water} \cdot g$$Area of cube face = 100 cm² = 10⁻² m²
$$800 \times (10 \times 10^{-2}) = 900 \times (x \times 10^{-2}) + 1000 \times (y \times 10^{-2})$$ $$800 \times 10 = 900x + 1000y$$ $$8000 = 900x + 1000y$$… (1)
Also: x + y = 10 … (2)
From (2): x = 10 - y
Substitute in (1):
$$8000 = 900(10 - y) + 1000y$$ $$8000 = 9000 - 900y + 1000y$$ $$8000 = 9000 + 100y$$ $$y = -10 \text{ cm}$$This is impossible!
Error: Oil layer is only 4 cm high. Let’s check if cube extends above oil:
Let h = height above oil surface
Then: h + 4 + y = 10
Weight = Buoyant forces:
$$800 \times 10 = 900 \times 4 + 1000 \times y$$ $$8000 = 3600 + 1000y$$ $$y = 4.4 \text{ cm}$$h = 10 - 4 - 4.4 = 1.6 cm
Answer:
- 1.6 cm above oil surface
- 4 cm in oil (entire oil layer)
- 4.4 cm in water
A cylindrical vessel of height 30 cm is filled with water up to 20 cm. A wooden cylinder of height 10 cm, cross-section equal to vessel, and density 600 kg/m³ is placed in water. Find: (a) Height of cylinder above water (b) Rise in water level
Solution:
(a) Floating depth:
Fraction submerged = ρ_wood/ρ_water = 600/1000 = 0.6
Height submerged = 0.6 × 10 = 6 cm
Height above water = 10 - 6 = 4 cm
(b) Rise in water level:
Volume of cylinder submerged = Volume of water displaced
Let A = cross-sectional area, Δh = rise in level
Volume displaced = A × Δh
Volume submerged = A × 6 cm
But wait! Water level rises, so actual submerged depth changes.
Let final water level be at height h from bottom.
Initial water: 20 cm
After adding cylinder:
- Cylinder floats with 6 cm submerged
- Water level = h
- Submerged portion is below h
Let’s say cylinder bottom is at height (h - 6) from vessel bottom.
Volume conservation: Initial water volume = A × 20
After: Water volume = A × h - A × 6 (cylinder displaces water)
Since water volume is conserved: A × 20 = A × h - A × 6
This doesn’t make sense. Let me reconsider.
Actually: When cylinder is placed, water level rises by Δh.
New water level = 20 + Δh
For cylinder to float with 6 cm submerged: Bottom of cylinder is at (20 + Δh - 6) = (14 + Δh) from bottom
Water displaced = Volume submerged = A × 6
This displaced water raises level by Δh: A × Δh = A × 6
Δh = 6 cm
Answer: Water level rises by 6 cm (from 20 to 26 cm)
But then: Total height in vessel = 26 cm < 30 cm ✓ (fits!)
Related Topics
Within Properties of Matter
- Elastic Moduli — Bulk modulus and compressibility
- Viscosity — Resistance in flowing fluids
- Bernoulli’s Theorem — Pressure in moving fluids
- Surface Tension — Surface effects in liquids
Connected Chapters
- Newton’s Laws — Force equilibrium
- Gravitation — Variation of g with depth