Pressure in Fluids and Pascal's Law

• Define pressure and calculate it for fluids at rest
• Apply the pressure-depth formula P = P₀ + ρgh
• State Pascal's law and explain how a hydraulic press works
• Distinguish between gauge pressure and absolute pressure
• Analyse U-tube manometer problems
• Apply Archimedes' principle to determine buoyant force and floating conditions

Level 1 — Foundations

Pressure

$P = \frac{F}{A}$

SI unit: Pascal (Pa) = N/m². Other units: bar (10⁵ Pa), atm (1.013 × 10⁵ Pa), mm Hg (133.3 Pa).

Pressure is isotropic in a fluid — it acts equally in all directions at a given depth.

Pressure at Depth h in a Fluid

$P = P_0 + \rho g h$

where P₀ = pressure at the surface (atmospheric pressure = 10⁵ Pa), ρ = fluid density, g = 9.8 ≈ 10 m/s², h = depth below surface.

• Pressure increases with depth
• Pressure does not depend on the shape of the container (hydrostatic paradox)
• Pressure depends only on the vertical depth, not the horizontal distance

Pascal's Law

"A change in pressure applied to an enclosed fluid is transmitted undiminished to every point in the fluid and to the walls of the container."

Hydraulic Press: Small force F₁ on small piston (area A₁) creates pressure P = F₁/A₁. Same pressure acts on large piston (area A₂), giving large force F₂ = P × A₂.

$\frac{F_1}{A_1} = \frac{F_2}{A_2} \Rightarrow F_2 = F_1 \cdot \frac{A_2}{A_1}$

Energy is conserved: F₁d₁ = F₂d₂ (small piston moves more than large piston).

Atmospheric Pressure

P_atm = 1.013 × 10⁵ Pa ≈ 10⁵ Pa (for JEE calculations) Mercury barometer: P_atm = ρ_Hg × g × h → h = 76 cm Hg at STP

Level 2 — JEE Depth

Absolute vs Gauge Pressure

• Absolute pressure: P_abs = P_atm + P_gauge (total pressure)
• Gauge pressure: P_gauge = P_abs − P_atm (pressure above atmospheric)

Tyre pressure gauges read gauge pressure. Vacuum: P_abs < P_atm, P_gauge < 0.

U-Tube Manometer Analysis

For a U-tube with two fluids:

• Write pressure at the bottom of the U-tube from both sides; they must be equal.
• Left side: P_unknown + ρ₁g h₁
• Right side: P_atm + ρ₂g h₂
• Equate and solve.

For differential manometer (both arms connected to flowing fluid), pressure difference = ρ_manometer × g × Δh.

Archimedes' Principle

"A body wholly or partially immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced."

$F_b = \rho_{fluid} \cdot V_{submerged} \cdot g$

where V_submerged = volume of body that is below the fluid surface.

Conditions for Floating:

Fraction submerged: For a floating body, weight = buoyant force: $\rho_{body} \cdot V \cdot g = \rho_{fluid} \cdot V_{sub} \cdot g$ $\frac{V_{sub}}{V} = \frac{\rho_{body}}{\rho_{fluid}}$

Apparent Weight:

W_apparent = W_actual − F_b = mg − ρ_fluid · V · g

If body is fully submerged: W_app = V g (ρ_body − ρ_fluid)

Pressure on a Tilted Surface: The pressure at a point depends only on its vertical depth below the free surface, regardless of the angle of the container walls.

Example 1: Find gauge pressure at depth 10 m in water (ρ = 1000 kg/m³, g = 10 m/s²)

Given: h = 10 m, ρ = 1000 kg/m³, g = 10 m/s²

Gauge pressure = P − P₀ = ρgh (pressure due to water column only)

P_gauge = 1000 × 10 × 10
P_gauge = 100,000 Pa = 10⁵ Pa = 1 atm = 1 bar

Absolute pressure = P₀ + P_gauge = 10⁵ + 10⁵ = 2 × 10⁵ Pa = 2 atm

Physical insight: Every 10 m depth in water adds 1 atm of pressure.
A scuba diver at 30 m depth experiences 4 atm absolute pressure.

Example 2: Wooden block of density 600 kg/m³ and volume 0.5 m³ — find buoyant force and whether it floats

Given: ρ_wood = 600 kg/m³, V = 0.5 m³, ρ_water = 1000 kg/m³, g = 10 m/s²

Step 1: Weight of block
  W = ρ_wood × V × g = 600 × 0.5 × 10 = 3000 N

Step 2: Check if it floats
  Since ρ_wood (600) < ρ_water (1000), the block floats.

Step 3: Volume submerged (when floating, W = F_b)
  F_b = ρ_water × V_sub × g = W
  1000 × V_sub × 10 = 3000
  V_sub = 0.3 m³

Step 4: Fraction submerged
  V_sub/V = 0.3/0.5 = 0.6 = 60%
  (Or: ρ_wood/ρ_water = 600/1000 = 0.6 ✓)

Buoyant force = W = 3000 N (when floating, F_b equals weight)

If pushed fully underwater:
  F_b = 1000 × 0.5 × 10 = 5000 N > W (3000 N) → net upward force, pops back up.

• Q1 Find absolute and gauge pressure at the bottom of a 5 m deep swimming pool. (ρ = 1000 kg/m³, P₀ = 10⁵ Pa)
• Q2 A hydraulic jack has input piston area 2 cm² and output area 100 cm². What input force is needed to lift a 5000 N load?
• Q3 A steel sphere (ρ = 8000 kg/m³) of volume 50 cm³ is submerged in water. Find (a) buoyant force, (b) apparent weight.
• Q4 An ice cube of density 900 kg/m³ floats in water. What fraction is above the water surface?
• Stretch: Q5 In a U-tube open at both ends, one arm has water (ρ = 1000 kg/m³) to height 15 cm. The other arm has an unknown liquid to height 20 cm. Find the density of the unknown liquid.

NCERT Chapter 9 link: Chapter 9 (Mechanical Properties of Fluids) covers pressure, Pascal's law, Archimedes' principle in Sections 9.2–9.4. NCERT derivation of buoyancy and the hydraulic machine are directly exam-relevant. Study the worked examples in NCERT carefully — JEE Main often uses similar setups.

Exam connections: JEE Main: pressure at depth, Pascal's law numerical, buoyancy and apparent weight, fraction submerged. JEE Advanced: U-tube manometer with two immiscible liquids, block floating at interface of two liquids, pressure on inclined surfaces (net force calculations).

Study strategy: Master the single equation P = P₀ + ρgh — it underlies all fluid statics. For buoyancy problems, draw a free-body diagram showing weight downward and buoyant force upward; write the equilibrium or net-force equation. For floating-at-interface problems (JEE Advanced), write separate buoyancy contributions from each liquid.

• Use the platform-native live simulation: a fluid statics sandbox where you adjust depth, fluid density, and body density to observe pressure readouts and floating/sinking behaviour.
• Mirror / body / home activity: fill a bucket with water, carefully push an empty plastic bottle underwater, feel the upward force — estimate buoyant force from the volume of the bottle.
• Voice or text reflection with AI Mentor: explain to a family member why a ship made of steel floats while a steel bolt sinks.

• "Explain Pascal's Law using the example of a hydraulic braking system in a car — how does one pedal stop all four wheels?"
• "What is one common mistake students make when calculating pressure at depth, and how would you catch yourself?"
• Stretch: "How do submarines control their depth without using a propeller to push up or down? What fluid physics principle is involved?"

• Complete the core practice + one extension activity (photo, table, short reflection, or mini-project) for base XP + topic badge.
• 5-7 day streak or family discussion note = multiplier + visible artifact in parent/principal dashboard.
• Best real-world application stories (anonymised) featured on class or national leaderboard.

• Build a simple hydraulic arm using syringes connected by tubing — demonstrate Pascal's law and calculate the mechanical advantage from piston area ratios.
• Future Skill track: Green Tech / Sustainable Living — understanding fluid pressure is core to designing water supply systems, dams, and irrigation networks.
• Coding extension: Write a programme that takes depth and fluid density as inputs and outputs absolute pressure, gauge pressure, and buoyant force on a given object.

This material emphasises experiential "learning by doing", competency (apply/create/analyse), vocational exposure, critical thinking, and multidisciplinary connections. Designed to feed live worlds, AI Mentor (with memory), gamification, robotics, parent analytics, and future skills — not just exam prep.

Portfolio Evidence Idea: Your photo/table/reflection/project + one sentence on "How this helps me in real life or a possible future path."

Open the Practice tab for aligned questions (easy/medium/hard + case-based) with full AI scaffolding.

See curriculum for cross-links and the full future-skills/robotics chapters.