Mechanical Properties of Fluids
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Direct answer
Fluids cannot be pulled, only pushed — which is why pressure, a scalar acting equally in all directions, replaces force throughout fluid mechanics. The chapter runs on four pillars: Pascal's law for enclosed fluids, Archimedes for buoyancy, Bernoulli's P + ½ρv^2 + ρgh = constant for streamline flow of an ideal fluid, and the viscous-surface pair of Stokes' law and surface tension. Terminal velocity, capillary rise and the excess pressure inside drops and bubbles are where NEET-UG places its numericals.
What you must remember
- Pressure at depth h: P = P_atm + ρgh, with 1 atm = 1.013 × 10^5 Pa — equivalently 76 cm of mercury or about 10.3 m of water.
- Pascal's law: pressure applied to an enclosed fluid is transmitted undiminished everywhere — the hydraulic lift multiplies force by the area ratio A2/A1.
- Archimedes: upthrust equals the weight of displaced fluid; a floating body displaces exactly its own weight of fluid.
- Bernoulli (streamline, incompressible, non-viscous): P + ½ρv^2 + ρgh = constant — faster flow means lower pressure, which explains aerofoil lift, the spinning ball's swing and the atomiser.
- Torricelli's theorem: efflux speed under a head h is sqrt(2gh).
- Stokes' law F = 6πηrv gives terminal velocity v_t = 2r^2(ρ - σ)g/(9η); viscosity of liquids falls with temperature (gases rise); water at 20 °C has η ≈ 1 × 10^-3 Pa s.
- Reynolds number R_e = ρvD/η: pipe flow is streamline below about 2000 and turbulent above about 3000.
- Surface tension S (N/m) is also energy per unit area; capillary rise h = 2S cosθ/(rρg); excess pressure is 2S/r in a liquid drop but 4S/r in a soap bubble, which has two surfaces.
A typical exam case: terminal velocity
Consider a mist droplet of radius 5 × 10^-5 m falling in air (η ≈ 1.8 × 10^-5 Pa s). At terminal velocity, weight balances the Stokes drag plus buoyancy, giving v_t = 2r^2ρg/9η = 2 × (5 × 10^-5)^2 × 1000 × 9.8/(9 × 1.8 × 10^-5) ≈ 0.3 m/s — a gentle drift. Scale the radius up tenfold and the terminal velocity grows a hundredfold, past 30 m/s, which is why large raindrops must break apart in flight; rain arrives as droplets because v_t grows as r^2, and any drop big enough to fall dangerously shatters first. This single scaling argument also answers the exam's standard question: what happens to terminal velocity when the radius doubles? Quadruples — provided viscosity and densities are unchanged.
The same ledger governs the capillary: h = 2S cosθ/(rρg) says a narrower tube gives a higher column, water-glass has an acute contact angle (near 0°, cos θ ≈ 1) so water climbs, while mercury-glass is obtuse (about 135°) and the column depresses. Detergent enters by lowering S, which is precisely how it helps water wet into fabric gaps that plain water cannot penetrate.
Where students slip
Bernoulli's equation is licensed only for streamline flow of a non-viscous, incompressible fluid along a tube of flow — applying it across a suddenly opened tap or through a turbulent section is out of contract. The soap-bubble factor of two is the most-looted error: a bubble has an inner and an outer surface, so its excess pressure is 4S/r, double a drop's 2S/r, and the exam restates this nearly every year in some form. In capillary questions, students quote the rise formula but forget that the meniscus angle decides the sign — cos θ negative means depression, not rise. Pascal-law problems trip on which piston the effort and load sit: the force multiplies by the area ratio, but the work does not multiply, because the small piston travels proportionally farther. Finally, viscosity questions punish "thicker flows slower in summer": liquids flow easier when hot (η falls), while gases become more viscous.
Frequently asked questions
Why is the pressure inside a soap bubble twice that inside a liquid drop of the same radius?
A soap bubble has two surfaces, inner and outer, each contributing 2S/r; total excess pressure is therefore 4S/r against the drop's 2S/r.
How does terminal velocity depend on droplet radius?
v_t = 2r^2(ρ - σ)g/9η grows as the square of the radius: double the radius and the terminal velocity quadruples, assuming η and the densities stay fixed.
Why does water rise in a glass capillary but mercury depress?
Water-glass contact angle is acute (nearly 0°), making cos θ positive and the rise real; mercury-glass angle is obtuse (about 135°), so cos θ is negative and the meniscus is pushed down.
State Bernoulli's principle and its conditions.
Along a streamline of a steady, incompressible, non-viscous flow, P + ½ρv^2 + ρgh stays constant; the trade between pressure and speed explains lift and venturi behaviour.
What does the Reynolds number tell you?
R_e = ρvD/η compares inertial to viscous effects: below roughly 2000 pipe flow is laminar, above roughly 3000 it is turbulent, with a transition band between.