Fluid Mechanics
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Fluid Mechanics — Key Facts for JEE Main • Core formula: Bernoulli’s equation — P + ½ρv² + ρgh = constant (along a streamline, for steady, incompressible, non-viscous flow) • Core concept: In a flowing fluid, an increase in flow speed always corresponds to a decrease in pressure, and vice versa — the trade-off between kinetic and potential energy. • Most common application: Venturi meter — a pipe that narrows, causing faster flow and lower pressure at the throat, used to measure flow rate or speed. • Key numerical value: Density of water = 1,000 kg/m³; standard atmospheric pressure = 1.01 × 10⁵ Pa; g = 9.8 m/s². • Most tested concept: Applying equation of continuity A₁v₁ = A₂v₂ combined with Bernoulli’s equation to relate pressure difference to velocity change (e.g., aeroplane wing lift, siphon, tank discharge). • Common mistake students make: Using gauge pressure in Bernoulli’s equation when absolute pressure is required — or forgetting to set reference heights correctly and mixing up points at different horizontal levels. ⚡ Exam tip: For any JEE Main problem with fluid flow through a pipe of varying cross-section, always write the continuity equation first (A₁v₁ = A₂v₂) to find one unknown, then plug into Bernoulli to find the other — never try to go straight to Bernoulli without the velocity relationship.
🟡 Standard — Regular Study (2d–2mo)
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Fluid Mechanics — JEE Main / Advanced Study Guide
Fluid statics deals with fluids at rest. Pressure at a depth h in a fluid of density ρ is P = P₀ + ρgh (P₀ = atmospheric pressure at the surface). Pascal’s law states that pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid — the principle behind hydraulic presses and brakes.
In fluid dynamics, we consider fluids in motion. Two fundamental equations govern ideal fluid flow:
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Equation of continuity: For an incompressible fluid, mass conservation gives A₁v₁ = A₂v₂, where A is cross-sectional area and v is flow speed. The product Av represents the volume flow rate (m³/s).
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Bernoulli’s equation: Along a streamline, for steady, incompressible, non-viscous flow: P + ½ρv² + ρgh = constant. Each term has units of pressure (Pa). At the same horizontal level (h₁ = h₂), the equation simplifies to P + ½ρv² = constant.
A key application is the Torricelli formula for a tank draining through a small hole at the bottom: v = √(2gh), where h is the height of the liquid surface above the hole. This is derived by applying Bernoulli between the liquid surface and the orifice.
Solved Example 1: Water flows through a horizontal pipe. At section A (diameter 10 cm), pressure is 200 kPa and speed is 3 m/s. At section B (diameter 5 cm), find the pressure.
- A₁ = π(0.05)² = 7.85 × 10⁻³ m²; v₁ = 3 m/s → Q = A₁v₁ = 2.36 × 10⁻² m³/s
- At B: A₂ = π(0.025)² = 1.96 × 10⁻³ m² → v₂ = Q/A₂ = 2.36 × 10⁻²/1.96 × 10⁻³ ≈ 12 m/s
- Bernoulli (h same): P₁ + ½ρv₁² = P₂ + ½ρv₂²
- P₂ = 200,000 + ½(1000)(3² − 12²) = 200,000 + 500(9 − 144) = 200,000 − 67,500 = 132,500 Pa = 132.5 kPa
Solved Example 2: A vertical tank 5 m tall is filled with water. A hole is punched 1 m above the base. Find the speed of water emerging from the hole.
- Height difference from surface to hole: h = 5 − 1 = 4 m
- v = √(2gh) = √(2 × 9.8 × 4) = √(78.4) ≈ 8.85 m/s
🔴 Extended — Deep Study (3mo+)
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Fluid Mechanics — Comprehensive JEE Notes
Derivation of Bernoulli from work-energy: Consider a fluid element of volume dV moving along a streamline. The net mechanical work done on the element equals its change in kinetic + potential energy. Forces doing work are pressure forces at the two ends (P₁dV − P₂dV) and gravity (weight = ρgdV × Δh). Setting net work equal to ΔK + ΔU gives, after dividing by dV and taking the limit as sections approach each other: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂. This confirms Bernoulli’s equation and clarifies its assumptions: steady flow, incompressible fluid, non-viscous (no shear stress), flow along a streamline.
Viscosity and the Navier–Stokes equation: Real fluids experience viscous dissipation. The viscosity η introduces a shear stress τ = η(dv/dy). The full momentum equation for a Newtonian fluid is the Navier–Stokes equation: ρ(Dv/Dt) = −∇P + η∇²v + ρg. For steady laminar flow through a circular pipe (Poiseuille flow), solving gives a parabolic velocity profile v(r) = (ΔP/4ηL)(R² − r²) and volumetric flow rate Q = (πR⁴ΔP)/(8ηL) — the famous Poiseuille law. Note the quartic dependence on radius: doubling the pipe radius increases flow rate by 16×, which is why arteries (despite blockages) dramatically reduce flow.
Surface tension and capillarity: At liquid–air interfaces, surface tension γ acts like a stretched membrane, tending to minimise surface area. The pressure difference across a curved surface is given by the Laplace–Young equation: ΔP = γ(1/R₁ + 1/R₂). For a soap bubble (two surfaces), ΔP = 4γ/R. Capillary rise follows h = 2γ cosθ/(ρgr) — important in plant physiology and thin porous media.
Advanced solved example: A capillary tube of radius r = 0.5 mm is dipped vertically into water (γ = 0.072 N/m, θ ≈ 0°). How high does water rise?
- h = 2γ/(ρgr) = 2(0.072)/(1000 × 9.8 × 5 × 10⁻⁴) = 0.144/(4.9) ≈ 0.0294 m = 2.94 cm
Rotating fluid dynamics: For a fluid in a rotating container (e.g., a rotating bucket), the free surface takes a paraboloid shape. The pressure at radius r and depth z is P = P₀ + ½ρω²r² + ρgz. The equipotential surfaces are paraboloids of revolution — key to understanding cyclonic weather systems and centrifugal separators.
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