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Physics 3% exam weight

Fluid Mechanics

Part of the NEET UG study roadmap. Physics topic phy-009 of Physics.

By Last updated 3% exam weight

Fluid Mechanics

🟢 Lite — Quick Review (1h–1d)

Pressure at depth h in a fluid of density ρ: P = ρgh, where g = 9.8 m/s². Absolute pressure = atmospheric pressure + gauge pressure (P₀ + ρgh). Pascal’s law: pressure applied to an enclosed fluid transmits equally in all directions — used in hydraulic lifts.

Buoyancy and Archimedes’ principle: upthrust equals weight of displaced fluid. Floatation condition: a body floats when its density ≤ fluid density; fraction immersed = ρ_body/ρ_fluid.

Fluid flow: Continuity equation A₁v₁ = A₂v₂ (incompressible fluid). Bernoulli’s theorem (along a streamline, steady, non-viscous flow): P + ½ρv² + ρgh = constant. Reynolds number Re = ρvd/η classifies flow: Re < 2000 → laminar, Re > 3000 → turbulent.

Viscosity: F = ηA(dv/dx) (Newton’s law). Stokes’ law: v = (2/9)(r²g(ρ−σ)/η) for sphere falling through fluid.

Surface tension γ = F/L; excess pressure in soap bubble = 4γ/r (air bubble = 2γ/r). Capillary rise h = (2γ cosθ)/(rρg).

NEET pointers: Questions frequently trap students on absolute vs gauge pressure, Bernoulli application only along a streamline, and temperature decreasing liquid viscosity but increasing gas viscosity. Units matter — convert carefully.


🟡 Standard — Regular Study (2d–2mo)

Pressure and Hydrostatics

Pressure is force per unit area perpendicular to the surface: P = F/A. For a static liquid column, pressure at depth h below the free surface is P = P₀ + ρgh, where ρ is liquid density in kg/m³ and h in metres. Note that pressure is scalar — it acts equally in all directions at a point.

Pascal’s law states that external pressure applied to an enclosed fluid transmits undiminished to every portion of the fluid and the walls. Hydraulic brakes and hydraulic presses exploit this: force multiplication ratio = A₂/A₁.

Buoyancy and Archimedes’ Principle

The buoyant force (upthrust) on a body immersed in fluid equals the weight of the displaced fluid: Fb = ρ_fluid × V_displaced × g. Apparent weight = actual weight − upthrust. A body floats when ρ_body ≤ ρ_fluid; the immersed volume fraction equals the density ratio.

Fluid Dynamics

Streamline flow (laminar): fluid particles follow smooth paths; velocity at any point constant in time. Turbulent flow: irregular, chaotic motion.

Continuity equation for incompressible flow: A₁v₁ = A₂v₂, where A is cross-sectional area and v is flow speed. Volumetric flow rate Q = Av (m³/s).

Bernoulli’s equation applies to steady, incompressible, non-viscous flow along a streamline: P + ½ρv² + ρgh = constant. Total mechanical energy per unit volume remains constant. Venturimeters use this to measure discharge.

Viscosity and Surface Tension

Dynamic viscosity η measures fluid resistance to flow. Newton’s law: F = ηA(dv/dx). Kinematic viscosity ν = η/ρ.

Stokes’ law governs viscous drag on a small sphere: terminal velocity v = (2/9)(r²g(ρ − σ)/η), where r is sphere radius, ρ sphere density, σ fluid density.

Surface tension γ = F/L = energy per unit area. Excess pressure inside a soap bubble (two surfaces) = 4γ/r; inside an air bubble (one surface) = 2γ/r. Capillary rise h = (2γ cosθ)/(rρg) — inversely proportional to tube radius.

Exam Pattern

NEET typically sets 1–2 MCQs from this chapter (≈3% weightage). Numerical problems often combine pressure–depth with Archimedes’ principle or Bernoulli with continuity. Assertion–reason type questions frequently test Bernoulli’s streamline requirement or viscosity temperature dependence.


🔴 Extended — Deep Study (3mo+)

Depth in Pressure Derivation

Pressure at depth h derives from hydrostatic equilibrium: consider a cylindrical element of liquid with base area A and height dh. Downward weight = ρgA dh balances upward pressure difference dP × A, giving dP = ρg dh. Integrating from surface (P = P₀) to depth h yields P = P₀ + ρgh. This derivation works only for incompressible fluids; for compressible fluids (gases), density varies with pressure and the integration is more complex.

Buoyancy and Metacentre

When a floating body tilts, its centre of buoyancy shifts. Stability requires the metacentre (point where vertical line through new centre of buoyancy meets the original vertical axis) to lie above the centre of gravity. The metacentric height determines the period of oscillation — a concept used in ship design.

Bernoulli’s Assumptions and Limitations

Bernoulli’s equation assumes steady flow (parameters time-independent at any point), incompressible fluid, non-viscous flow (no energy dissipation), and application along a streamline only. Real fluids lose energy due to viscosity; the equation overestimates speed at lower pressures in such cases. A correction term (head loss) is added in practical hydraulic engineering.

Viscosity Temperature Dependence

Liquid viscosity decreases with temperature increase (molecules move apart, intermolecular forces weaken — e.g., honey flows easier when warm). Gas viscosity increases with temperature (greater molecular momentum transfer at higher speeds). This is a common NEET trap.

Surface Tension Anomalies

Surface tension γ = F/L = dW/dA (work done per unit area increase). It explains why insects walk on water, capillary action in plants, and the shape of liquid droplets. The contact angle θ determines whether capillary rise (θ < 90°, wetting fluid like water in glass) or depression (θ > 90°, non-wetting like mercury in glass) occurs.

Common Mistakes to Avoid

  1. Confusing gauge pressure with absolute pressure — many students forget to add atmospheric pressure (1.01 × 10⁵ Pa).
  2. Applying Bernoulli between arbitrary points — it is valid only along the same streamline.
  3. Swapping η (dynamic) and ν (kinematic) viscosity — dimensionally distinct (η: Pa·s; ν: m²/s).
  4. Neglecting that ρ in P = ρgh must be in kg/m³, not g/cm³.

Practice Prompts

  1. A pipe of cross-sectional area 10 cm² narrows to 4 cm². Fluid flows at 2 m/s in the wide section. Find speed in narrow section and volumetric flow rate.
  2. A sphere of radius 2 mm and density 8000 kg/m³ falls through oil (η = 0.5 Pa·s, ρ = 900 kg/m³). Calculate terminal velocity using Stokes’ law.

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