1. Chapter at a Glance
This chapter follows the JEE unit "Properties of Solids and Liquids". It covers elasticity (how solids deform), fluids (pressure, Pascal's law, buoyancy, Bernoulli's principle), viscosity and surface tension, and thermal properties (expansion, calorimetry, and heat transfer by conduction, convection and radiation). JEE Main usually asks two questions from this wide unit.
- Young's modulus and extension of wires.
- Bernoulli's equation, speed of efflux, venturi meter.
- Terminal velocity (Stokes' law) and capillary rise.
- Calorimetry with change of state; Stefan's and Wien's laws; Newton's law of cooling.
2. Elasticity
- Stress = restoring force per area (N m⁻²); strain = fractional change in size (no unit).
- Hooke's law: within the elastic limit, stress ∝ strain; the constant is the modulus of elasticity.
- Young's modulus Y = (F/A)/(ΔL/L); bulk modulus B = −ΔP/(ΔV/V) (compressibility = 1/B); shear modulus η = shear stress/shear angle.
- Poisson's ratio σ = lateral strain/longitudinal strain (between 0 and 0.5 in practice).
- Elastic energy stored per unit volume = ½ × stress × strain; total energy in a stretched wire = ½FΔL.
3. Fluid Statics and Dynamics
- Pressure at depth h: P = P₀ + ρgh. Pascal's law: pressure applied to an enclosed fluid is transmitted equally — the hydraulic lift gives F₂/F₁ = A₂/A₁.
- Archimedes' principle: buoyant force = weight of fluid displaced. A floating body displaces its own weight: fraction submerged = ρbody/ρfluid.
- Equation of continuity: A₁v₁ = A₂v₂ (incompressible flow).
- Bernoulli's equation: P + ½ρv² + ρgh = constant along a streamline. Applications: speed of efflux v = √(2gh) (Torricelli), venturi meter, lift on aircraft wings, atomisers.
- Streamline flow becomes turbulent when the Reynolds number Re = ρvD/η exceeds about 2000.
4. Viscosity and Surface Tension
- Viscous force F = ηA(dv/dx); η in Pa s (poise in CGS). Viscosity of liquids falls with temperature; that of gases rises.
- Stokes' law: F = 6πηrv. Terminal velocity vt = 2r²(ρ − σ)g/(9η) ∝ r².
- Surface tension T = force per unit length = surface energy per unit area (N m⁻¹).
- Excess pressure: liquid drop or air bubble in liquid 2T/r; soap bubble (two surfaces) 4T/r.
- Capillary rise h = 2T cosθ/(rρg); for mercury θ > 90°, so it falls. Work to form a soap bubble of radius r = 8πr²T.
5. Thermal Properties
- Expansion: ΔL = LαΔT, ΔA = AβΔT, ΔV = VγΔT with β = 2α, γ = 3α. Water contracts from 0 °C to 4 °C (anomalous expansion).
- Heat: Q = mcΔT (no change of state); Q = mL (change of state; Lice = 80 cal g⁻¹, Lsteam = 540 cal g⁻¹). Calorimetry: heat lost = heat gained.
- Conduction: dQ/dt = KAΔT/L; thermal resistance R = L/KA; rods in series add R, in parallel add 1/R.
- Radiation: Stefan–Boltzmann law P = σeAT⁴; Wien's law λmT = b ≈ 2.9 × 10⁻³ m K; a perfect black body has e = 1 and is the best absorber and emitter.
- Newton's law of cooling (small temperature differences): rate of cooling ∝ (T − Ts). Average form: (T₁ − T₂)/t = k[(T₁ + T₂)/2 − Ts].
6. Formula Master Table
| Quantity | Formula |
|---|---|
| Young's modulus | Y = FL/(AΔL) |
| Hydraulic lift | F₂/F₁ = A₂/A₁ |
| Bernoulli | P + ½ρv² + ρgh = const |
| Efflux | v = √(2gh) |
| Terminal velocity | vt = 2r²(ρ − σ)g/9η |
| Capillary rise | h = 2T cosθ/(rρg) |
| Excess pressure | drop 2T/r; soap bubble 4T/r |
| Conduction | Q/t = KAΔT/L |
| Radiation | P = σeAT⁴; λmT = b |
7. Mind Map
- Y, B, η
- Stress–strain curve
- Energy stored
- Pascal, Archimedes
- Continuity
- Bernoulli
- Stokes, terminal v
- Excess pressure
- Capillarity
- Expansion
- Calorimetry
- Conduction, radiation, cooling
8. Tips and Tricks
- For wires of the same material and load, ΔL ∝ L/r².
- In ice–water mixing, first check whether the hot water has enough heat to melt all the ice.
- Terminal velocity ∝ r²; if eight equal drops merge, the radius doubles and vt becomes 4 times.
- Emitted power ∝ T⁴ — always convert to kelvin first.
- Using 2T/r for a soap bubble — it has two surfaces, so 4T/r.
- Using °C in Stefan's law.
- Forgetting that a larger modulus means a stiffer (less stretchable) material — steel is more elastic than rubber.
9. Important One-liners
- Steel is more elastic than rubber because its Young's modulus is larger.
- Surface tension decreases with rise in temperature.
- Bernoulli's principle is conservation of energy for flowing fluids.
- A good absorber of radiation is also a good emitter (Kirchhoff's law).
- Water has maximum density at 4 °C.
- Viscosity of a gas increases with temperature.
10. Solved Numericals: Average to Advanced
Five average-level and five advanced-level problems. Click to reveal each answer and its full working. Take g = 10 m s⁻² and ρwater = 1000 kg m⁻³.
Average level
A 2 m wire of cross-section 1 mm² stretches by 1 mm under a 10 kg load. Find Young's modulus.
Show answer & solution
Concept: Y = FL/(AΔL).
- F = 10 × 10 = 100 N; A = 1 × 10⁻⁶ m²; ΔL = 1 × 10⁻³ m.
- Y = (100 × 2)/(10⁻⁶ × 10⁻³) = 2 × 10¹¹ N m⁻².
Answer: 2 × 10¹¹ N m⁻²
In a hydraulic lift, the pistons have areas in the ratio 1 : 100. What load can be lifted by a 50 N force on the small piston?
Show answer & solution
Concept: Pascal's law: F₂/F₁ = A₂/A₁.
- F₂ = 50 × 100 = 5000 N.
Answer: 5000 N
Water flows out of a small hole of area 1 cm² located 5 m below the water surface in a large open tank. Find the speed of efflux and the volume flow rate.
Show answer & solution
Concept: Torricelli: v = √(2gh); flow rate = Av.
- v = √(2 × 10 × 5) = 10 m s⁻¹.
- Q = 10⁻⁴ × 10 = 10⁻³ m³ s⁻¹ = 1 litre per second.
Answer: 10 m s⁻¹; 1 L s⁻¹
Find the excess pressure inside (a) a soap bubble and (b) a water drop, each of radius 1 cm, if surface tension is 0.03 N m⁻¹ for both.
Show answer & solution
Concept: soap bubble has two surfaces (4T/r); a drop has one (2T/r).
- (a) 4 × 0.03/0.01 = 12 Pa.
- (b) 2 × 0.03/0.01 = 6 Pa.
Answer: 12 Pa; 6 Pa
A 10 m steel rail (α = 1.2 × 10⁻⁵ K⁻¹) is heated through 40 K. By how much does it lengthen?
Show answer & solution
Concept: ΔL = LαΔT.
- ΔL = 10 × 1.2 × 10⁻⁵ × 40 = 4.8 × 10⁻³ m.
- = 4.8 mm — this is why gaps are left between rails.
Answer: 4.8 mm
Advanced level
Water (T = 0.07 N m⁻¹, contact angle 0°) rises in a glass capillary of radius 0.5 mm. Find the height of rise. What happens if the tube is only 2 cm long above the water?
Show answer & solution
Concept: h = 2T cosθ/(rρg).
- h = 2 × 0.07/(5 × 10⁻⁴ × 1000 × 10) = 0.14/5 = 0.028 m = 2.8 cm.
- If the tube is only 2 cm long, the water rises to the top but does not overflow.
- Instead the meniscus becomes flatter (larger radius of curvature R) so that hR = constant.
Answer: 2.8 cm; no overflow — the meniscus flattens
A steel ball of radius 1 mm (density 8000 kg m⁻³) falls through a liquid of density 1000 kg m⁻³ and viscosity 1.0 Pa s. Find its terminal velocity.
Show answer & solution
Concept: at terminal velocity, weight = buoyancy + Stokes drag ⇒ vt = 2r²(ρ − σ)g/(9η).
- r² = 10⁻⁶ m²; ρ − σ = 7000 kg m⁻³.
- vt = 2 × 10⁻⁶ × 7000 × 10/(9 × 1.0) = 0.14/9.
- ≈ 0.0156 m s⁻¹ = 1.56 cm s⁻¹.
Answer: about 1.56 cm s⁻¹
100 g of ice at 0 °C is added to 200 g of water at 60 °C. Find the final temperature (Lice = 80 cal g⁻¹, cwater = 1 cal g⁻¹ °C⁻¹; ignore the container).
Show answer & solution
Concept: check whether all the ice can melt, then heat lost = heat gained.
- Heat to melt all ice = 100 × 80 = 8000 cal.
- Heat available if water cools to 0 °C = 200 × 60 = 12 000 cal > 8000, so all ice melts.
- Let final temperature be T: 200(60 − T) = 8000 + 100T.
- 12 000 − 200T = 8000 + 100T ⇒ 300T = 4000 ⇒ T ≈ 13.3 °C.
Answer: about 13.3 °C
A body cools from 80 °C to 60 °C in 5 minutes in a room at 20 °C. Using the average form of Newton's law of cooling, find the time taken to cool from 60 °C to 40 °C.
Show answer & solution
Concept: (T₁ − T₂)/t = k[(T₁ + T₂)/2 − Ts].
- First interval: 20/5 = k(70 − 20) ⇒ 4 = 50k ⇒ k = 0.08 min⁻¹.
- Second interval: 20/t = 0.08 × (50 − 20) = 2.4.
- t = 20/2.4 ≈ 8.33 min.
Answer: about 8.3 minutes
Water flows through a horizontal pipe whose cross-section narrows from 10 cm² to 5 cm². The speed in the wide part is 2 m s⁻¹. Find the speed in the narrow part and the pressure difference between the two parts.
Show answer & solution
Concept: continuity A₁v₁ = A₂v₂, then Bernoulli at the same height.
- v₂ = (10/5) × 2 = 4 m s⁻¹.
- P₁ − P₂ = ½ρ(v₂² − v₁²).
- = ½ × 1000 × (16 − 4) = 6000 Pa.
- The pressure is lower where the speed is higher (the principle of the venturi meter).
Answer: 4 m s⁻¹; 6000 Pa
11. PYQ-pattern Questions with Solutions
These questions are original and modelled on concepts frequently tested in JEE Main.
The excess pressure inside a soap bubble of radius r and surface tension T is:
- T/r
- 2T/r
- 4T/r
- 8T/r
Show solution
A soap film has two surfaces, so excess pressure = 2 × 2T/r = 4T/r.
Answer: (C)
A star appears bluish-white while another appears red. This shows the bluish star has:
- Lower temperature
- Higher temperature
- Larger size
- Same temperature
Show solution
Wien's law: λm ∝ 1/T. Peak emission at shorter (blue) wavelength means higher temperature.
Answer: (B)
A wire is replaced by another of the same material with double the length and half the radius. Under the same load, the extension becomes:
- Double
- Four times
- Eight times
- Half
Show solution
ΔL = FL/(πr²Y) ∝ L/r² = 2/(1/4) = 8 times.
Answer: (C)
Eight identical raindrops, each falling with terminal velocity v, merge into one drop. Its terminal velocity is:
- 2v
- 4v
- 8v
- v
Show solution
Volume is 8 times, so the radius doubles. vt ∝ r², giving 4v.
Answer: (B)
Two rods of equal length and area, with conductivities K and 2K, are joined end to end. The effective conductivity of the combination is:
- 3K/2
- 4K/3
- 2K/3
- 3K
Show solution
In series, resistances add: 2L/(KeffA) = L/(KA) + L/(2KA) ⇒ Keff = 2K₁K₂/(K₁ + K₂) = 4K/3.
Answer: (B)
If the absolute temperature of a black body is doubled, by what factor does its emitted power increase?
Show solution
P ∝ T⁴ = 2⁴ = 16.
Answer: 16
12. Quick Revision Summary
- Y = FL/AΔL; energy stored = ½FΔL.
- P = P₀ + ρgh; Pascal; buoyancy = weight of displaced fluid.
- A₁v₁ = A₂v₂; P + ½ρv² + ρgh = const; efflux √(2gh).
- vt ∝ r²; capillary h = 2T cosθ/rρg; bubble 4T/r.
- Q = mcΔT, Q = mL; conduction KAΔT/L; radiation ∝ T⁴; λmT = b.