Elasticity and fluids for JEE Main

This chapter, "properties of solids and liquids" in the syllabus, is a collection of short topics rather than one long argument. That makes it scoring. Each topic has two or three formulas, a derivation you can do in a few lines and a handful of standard traps. Aspirants who skip it because it looks like "just formulas" give away easy marks.

The syllabus lists elastic behaviour and the stress–strain relationship, Hooke's law, Young's, bulk and rigidity moduli, fluid pressure and Pascal's law, viscosity, Stokes' law and terminal velocity, streamline and turbulent flow, Bernoulli's principle, and surface tension with drops, bubbles and capillary rise.

Stress, strain and the moduli

Stress is the restoring force per unit area; strain is the fractional change in shape or size and has no unit. Within the elastic limit, stress ∝ strain (Hooke's law), and the constant is a modulus of elasticity.

Compressibility is 1/B. Liquids and gases have no Young's modulus or rigidity, only a bulk modulus.

Rearranging Young's modulus gives the result behind most wire problems: ΔL = FL/(AY). For a given material and load, ΔL ∝ L/r².

On a stress–strain curve, the straight part ends at the proportional limit. Beyond the elastic limit the wire no longer returns to its length; it then yields and finally breaks. A ductile material (copper) has a long plastic region; a brittle one (glass) breaks soon after the elastic limit.

Elastic energy: U = ½ × F × ΔL, or per unit volume, ½ × stress × strain.

Fluids at rest

• Pressure at depth h: P = P₀ + ρgh. It depends only on the depth, not on the shape of the container.
• Gauge pressure is P − P₀.
• Pascal's law: pressure applied to an enclosed fluid is transmitted undiminished. In a hydraulic lift, F₁/A₁ = F₂/A₂.
• Archimedes' principle: buoyant force = weight of fluid displaced = ρ_fluid × V_submerged × g. A floating body has fraction submerged = ρ_body/ρ_fluid.

Fluids in motion

For steady, streamline flow of an incompressible, non-viscous fluid:

• Continuity: A₁v₁ = A₂v₂. The fluid speeds up where the pipe narrows.
• Bernoulli: P + ½ρv² + ρgh is constant along a streamline. It is energy conservation per unit volume.

Torricelli's theorem. For a hole at depth h below the free surface of a wide tank, Bernoulli between the surface and the hole gives v = √(2gh), the speed of free fall from height h.

If the tank's water column is H tall and stands on the ground, a jet from depth h lands at horizontal range 2√[h(H − h)]. This is largest, equal to H, when the hole is halfway down.

Viscosity

Viscous force between layers: F = ηA (dv/dx), where η is the coefficient of viscosity (unit Pa s).

Stokes' law: a sphere of radius r moving slowly at speed v feels a drag of 6πηrv.

Terminal velocity. A sphere of density ρ falls through a fluid of density σ. At terminal speed, weight = buoyancy + drag:
(4/3)πr³ρg = (4/3)πr³σg + 6πηrv, so v_t = 2r²(ρ − σ)g / (9η).
The terminal velocity is proportional to r². If ρ < σ, as for an air bubble in water, v_t is negative: the bubble rises.

Flow is streamline at low speed and turns turbulent above a critical velocity. The Reynolds number, ρvD/η, measures which regime you are in; low values mean streamline flow.

Surface tension

Surface tension T is the force per unit length along a line in the surface, and equally the surface energy per unit area (N/m = J/m²).

Capillary rise: h = 2T cos θ / (rρg). The angle of contact θ is less than 90° for water on clean glass, so water rises. For mercury on glass θ > 90°, so cos θ < 0 and mercury is depressed. Since h ∝ 1/r, narrower tubes give a higher rise.

The work done to blow a soap bubble of radius r is 8πr²T, because a soap film has two surfaces. When small drops merge into a big one, the surface area falls and energy is released.

Worked problems

Problem 1 (numerical answer): wire extension. A steel wire 2 m long with a cross-section of 1 mm² carries a 10 kg load. Y = 2 × 10¹¹ Pa and g = 10 m/s². Find the extension in mm and the energy stored.

ΔL = FL/(AY) = (100 × 2)/(10⁻⁶ × 2 × 10¹¹) = 200/(2 × 10⁵) = 10⁻³ m = 1 mm.
Energy = ½ × 100 × 10⁻³ = 0.05 J.

Problem 2: Bernoulli in a pipe. Water flows through a horizontal pipe that narrows from 10 cm² to 5 cm². The speed in the wide part is 2 m/s. Find the pressure difference.

Continuity: v₂ = 2 × 10/5 = 4 m/s.
Bernoulli: P₁ − P₂ = ½ρ(v₂² − v₁²) = ½ × 1000 × (16 − 4) = 6,000 Pa. The pressure is lower in the narrow section.

Problem 3: floating block. A wooden block floats in water with 3/5 of its volume submerged. What fraction is submerged in oil of density 800 kg/m³?

Density of wood = 0.6 × 1000 = 600 kg/m³. In oil, fraction = 600/800 = 3/4.

Problem 4: capillary rise. Find the rise of water in a clean glass tube of radius 0.1 mm (T = 0.07 N/m, θ = 0°, g = 10 m/s²).

h = 2 × 0.07/(10⁻⁴ × 1000 × 10) = 0.14/1 = 0.14 m = 14 cm.

Practice set

1. A raindrop of radius r falls with terminal velocity v. Eight such drops merge into one. Find the new terminal velocity.
2. Water rises to height h in a capillary of radius r. Find the rise in a tube of radius r/2.
3. Two wires of the same material carry the same load. The second is twice as long and has twice the radius. Compare the extensions.
4. In a hydraulic lift the small piston has an area of 10 cm² and the large one 500 cm². What force on the small piston lifts a 1,000 kg car (g = 10 m/s²)?
5. A tank stands on the ground with water 20 m deep. A small hole is made 5 m below the surface. Find the speed of efflux and the horizontal range of the jet (g = 10 m/s²).
6. Find the excess pressure inside a soap bubble of radius 2 cm (T = 0.03 N/m).
7. Water has B = 2.2 × 10⁹ Pa. By what percentage does its volume fall under an extra pressure of 10⁷ Pa?
8. Compare the extension of a wire hanging under its own weight with that of the same wire (treated as massless) carrying the same weight at its end.

Answers

1. Volume is 8 times, so the radius is 2r. v ∝ r², so 4v.
2. h ∝ 1/r, so 2h.
3. ΔL ∝ L/r² = 2/4, so the second extends by half as much.
4. F = 10,000 × 10/500 = 200 N.
5. v = √(2 × 10 × 5) = 10 m/s. Range = 2√(5 × 15) = 2√75 = 10√3 ≈ 17.3 m.
6. 4T/r = 0.12/0.02 = 6 Pa.
7. ΔV/V = 10⁷/(2.2 × 10⁹) ≈ 4.5 × 10⁻³, about 0.45%.
8. MgL/(2AY) against MgL/(AY): half.

What to do next

• Derive terminal velocity, Torricelli's result and the excess pressure in a bubble once each.
• Make a one-page table of every proportionality (h ∝ 1/r, v ∝ r², ΔL ∝ L/r²).
• Solve 20 mixed problems across elasticity, Bernoulli and surface tension.
• Revise work, energy and power (Bernoulli is its fluid form), then move on to thermal properties of matter.