NEET UG Physics · Chapter 7

Mechanical Properties of Solids and Fluids

Stress, strain and Young’s modulus; pressure, Pascal’s law, Bernoulli’s principle, viscosity and surface tension.

NCERT Class 11NEET: 2 Qs per paper 11 min read

1. Chapter at a Glance

This chapter joins two NCERT chapters: Mechanical Properties of Solids (how solids stretch and spring back) and Mechanical Properties of Fluids (how liquids and gases push, flow and form drops and bubbles). You will study stress, strain and Young's modulus; fluid pressure, Pascal's law and buoyancy; streamline flow, the equation of continuity and Bernoulli's principle; viscosity and terminal velocity; and surface tension, capillarity and excess pressure. NEET usually asks two questions here.

What NEET asks most
  • Young's modulus and extension of a wire; stress–strain curve.
  • Pressure at depth, hydraulic lift and Archimedes' principle.
  • Continuity equation and Bernoulli's theorem (Torricelli, venturi).
  • Terminal velocity (Stokes' law).
  • Capillary rise and excess pressure in drops and bubbles.

2. Elasticity of Solids

  • Elasticity is the property of regaining original shape after the deforming force is removed. Plasticity is the lack of it (putty, wet clay).
  • Stress = restoring force per unit area (N m⁻² = Pa). Tensile, compressive, shearing and hydraulic (volume) stress.
  • Strain = change in dimension/original dimension (no unit).
  • Hooke's law: within the elastic limit, stress ∝ strain. The ratio is the modulus of elasticity.
  • Young's modulus Y = (F/A)/(ΔL/L) = FL/(AΔL). Bulk modulus B = −P/(ΔV/V); compressibility = 1/B. Shear modulus G = shear stress/shear strain.
  • Steel is more elastic than rubber because it has a larger Young's modulus (needs larger stress for the same strain).
  • Elastic energy stored in a stretched wire U = ½ × F × ΔL; energy per unit volume = ½ × stress × strain.
proportional limityield pointultimate strengthfracture slope = Y stressstrain
Stress–strain curve of a ductile metal. Up to the proportional limit the graph is straight and its slope is Young's modulus.

3. Pressure, Pascal's Law and Buoyancy

  • Pressure P = F/A (scalar). In a liquid at depth h: P = Pa + ρgh. Gauge pressure = P − Pa = ρgh.
  • Pressure is the same at all points at the same depth in a liquid at rest — independent of the shape of the container (hydrostatic paradox).
  • Pascal's law: pressure applied to an enclosed fluid is transmitted equally in all directions. Hydraulic lift/brakes: F₂/F₁ = A₂/A₁.
  • Archimedes' principle: a body immersed in a fluid is pushed up by a force equal to the weight of fluid displaced. A floating body displaces its own weight; fraction submerged = ρbody/ρliquid (about 90% for ice in water).

4. Fluid Flow, Viscosity and Surface Tension

  • Streamline flow: every particle follows the path of the one before it; streamlines never cross. Beyond a critical speed flow becomes turbulent.
  • Equation of continuity: A₁v₁ = A₂v₂ — water speeds up where a pipe narrows (conservation of mass).
  • Bernoulli's principle: P + ½ρv² + ρgh = constant along a streamline (conservation of energy). Faster flow ⇒ lower pressure.
  • Applications: speed of efflux v = √(2gh) (Torricelli), venturi meter, lift on aircraft wings, blowing-off of roofs in storms, atomisers.
  • Viscosity is fluid friction: F = ηA(dv/dx), η in Pa s (poiseuille). Liquids become less viscous when heated; gases become more viscous.
  • 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 or J m⁻²). It decreases with temperature and with detergents.
  • Excess pressure: drop 2T/r; bubble in a liquid 2T/r; soap bubble in air (two surfaces) 4T/r.
  • Capillary rise h = 2T cosθ/(rρg). Water rises (θ < 90°); mercury falls (θ > 90°). h ∝ 1/r.

5. Formula Master Table

QuantityFormula
Young's modulusY = FL/(AΔL)
Bulk modulusB = −PV/ΔV
Elastic energyU = ½FΔL; per volume ½ × stress × strain
Pressure at depthP = Pa + ρgh
Hydraulic liftF₂/F₁ = A₂/A₁
ContinuityA₁v₁ = A₂v₂
BernoulliP + ½ρv² + ρgh = constant
Terminal velocityvt = 2r²(ρ − σ)g/9η
Excess pressuredrop 2T/r; soap bubble 4T/r
Capillary riseh = 2T cosθ/(rρg)

6. Master Summary Table

ConceptKey pointNEET Note
Young's modulusDepends only on materialNot on length or area of the wire
Pressure in liquidρgh, same at same depthIndependent of vessel shape
BuoyancyUpthrust = weight displacedFloating: fraction inside = ρb/ρl
BernoulliHigher speed, lower pressureEnergy conservation for fluids
Viscosityvt ∝ r²Liquid η falls, gas η rises with temperature
Surface tensionh ∝ 1/r in capillariesSoap bubble has two surfaces

7. Mind Map

Solids & Fluids
Elasticity
  • Stress, strain
  • Y, B, G
  • Stress–strain curve
Fluid statics
  • P = Pa + ρgh
  • Pascal's law
  • Archimedes
Fluid flow
  • Continuity
  • Bernoulli
  • Viscosity, Stokes
Surface tension
  • Excess pressure
  • Capillarity
  • Angle of contact

8. Tips and Tricks

Shortcuts
  • For wires of the same material and load, ΔL ∝ L/r² (or L/d²).
  • When 8 identical drops merge into one, the radius doubles: terminal velocity becomes 4 times.
  • In a capillary, hr is constant: half the radius gives twice the rise; on the Moon, h becomes 6 times.
  • Pipe radius halves ⇒ area becomes ¼ ⇒ speed becomes 4 times.
Common NEET traps
  • Using 2T/r for a soap bubble (it has two surfaces — use 4T/r).
  • Thinking a thicker wire has a larger Young's modulus — Y is a material constant.
  • Believing water overflows from a capillary tube that is too short — it does not; the meniscus flattens.

9. Important One-liners

  • Steel is more elastic than rubber.
  • Stress has the same units and dimensions as pressure.
  • Pressure at a point in a liquid acts equally in all directions.
  • Bernoulli's principle expresses conservation of energy in fluid flow.
  • Viscosity of liquids decreases on heating.
  • Surface tension decreases as temperature rises.
  • Small drops are spherical because a sphere has the least surface area for a given volume.

10. Solved Numericals: Average to Advanced

Five average and five advanced numericals. Take g = 10 m s⁻², ρwater = 1000 kg m⁻³, Pa = 1 × 10⁵ Pa. Click to reveal each answer and working.

Average level

N1Average

A 2 m wire of cross-section 2 mm² stretches by 0.5 mm under a 100 N load. Find the stress, strain and Young's modulus.

Show answer & solution

Concept: stress = F/A, strain = ΔL/L, Y = stress/strain.

  1. Stress = 100/(2 × 10⁻⁶) = 5 × 10⁷ Pa.
  2. Strain = 0.5 × 10⁻³/2 = 2.5 × 10⁻⁴.
  3. Y = 5 × 10⁷/2.5 × 10⁻⁴ = 2 × 10¹¹ Pa.

Answer: 5 × 10⁷ Pa; 2.5 × 10⁻⁴; 2 × 10¹¹ Pa

N2Average

Find the total pressure 10 m below the surface of a lake.

Show answer & solution

Concept: P = Pa + ρgh.

  1. ρgh = 1000 × 10 × 10 = 1 × 10⁵ Pa.
  2. P = 1 × 10⁵ + 1 × 10⁵ = 2 × 10⁵ Pa (about 2 atmospheres).

Answer: 2 × 10⁵ Pa

N3Average

A hydraulic lift has pistons of diameter 2 cm and 20 cm. What force on the small piston lifts a 1000 kg car on the large one?

Show answer & solution

Concept: Pascal's law, F₁/F₂ = A₁/A₂ = (d₁/d₂)².

  1. Area ratio = (2/20)² = 1/100.
  2. Weight of car = 1000 × 10 = 10 000 N.
  3. F₁ = 10 000/100 = 100 N.

Answer: 100 N

N4Average

Water flows at 1 m s⁻¹ through a pipe of radius 2 cm, which narrows to radius 1 cm. Find the speed in the narrow part.

Show answer & solution

Concept: A₁v₁ = A₂v₂, A ∝ r².

  1. v₂ = v₁(r₁/r₂)² = 1 × (2/1)².
  2. v₂ = 4 m s⁻¹.

Answer: 4 m s⁻¹

N5Average

Find the excess pressure inside a soap bubble of radius 2 cm if the surface tension of the soap solution is 0.04 N m⁻¹.

Show answer & solution

Concept: soap bubble has two surfaces: ΔP = 4T/r.

  1. ΔP = 4 × 0.04/0.02.
  2. = 8 Pa.

Answer: 8 Pa

Advanced level

N6Advanced

Water flows through a horizontal pipe. At one point the speed is 2 m s⁻¹ and at another it is 6 m s⁻¹. Find the pressure difference between the two points.

Show answer & solution

Concept: Bernoulli at equal height: P₁ − P₂ = ½ρ(v₂² − v₁²).

  1. v₂² − v₁² = 36 − 4 = 32 m² s⁻².
  2. ΔP = ½ × 1000 × 32 = 16 000 Pa.
  3. The pressure is lower where the water moves faster.

Answer: 1.6 × 10⁴ Pa

N7Advanced

A raindrop of radius 0.1 mm falls through air (η = 1.8 × 10⁻⁵ Pa s). Neglecting the buoyancy of air, find its terminal velocity.

Show answer & solution

Concept: vt = 2r²ρg/(9η).

  1. r² = (10⁻⁴)² = 10⁻⁸ m².
  2. vt = 2 × 10⁻⁸ × 1000 × 10/(9 × 1.8 × 10⁻⁵).
  3. = 2 × 10⁻⁴/1.62 × 10⁻⁴ ≈ 1.23 m s⁻¹.

Answer: about 1.23 m s⁻¹

N8Advanced

Water rises 6 cm in a capillary tube on Earth. How high will it rise (a) in a tube of half the radius, (b) in the original tube on the Moon (gmoon = g/6)?

Show answer & solution

Concept: h = 2T cosθ/(rρg), so h ∝ 1/(rg).

  1. (a) Radius halved ⇒ h doubles: 12 cm.
  2. (b) g becomes g/6 ⇒ h becomes 6 times: 36 cm.

Answer: 12 cm; 36 cm

N9Advanced

How much work is done in blowing a soap bubble of radius 5 cm if the surface tension is 0.03 N m⁻¹?

Show answer & solution

Concept: W = T × increase in surface area; a bubble has two surfaces, so area = 2 × 4πr².

  1. Area = 8πr² = 8π × 0.0025 ≈ 0.0628 m².
  2. W = 0.03 × 0.0628 ≈ 1.88 × 10⁻³ J.

Answer: about 1.9 × 10⁻³ J

N10Advanced

A 2 m wire of cross-section 1 mm² stretches by 2 mm under a 200 N load. Find the elastic energy stored and the energy per unit volume.

Show answer & solution

Concept: U = ½FΔL; volume = AL.

  1. U = ½ × 200 × 2 × 10⁻³ = 0.2 J.
  2. Volume = 10⁻⁶ × 2 = 2 × 10⁻⁶ m³.
  3. Energy density = 0.2/2 × 10⁻⁶ = 1 × 10⁵ J m⁻³.

Answer: 0.2 J; 1 × 10⁵ J m⁻³

11. NEET-pattern Questions with Detailed Answers

These questions are original and modelled on concepts repeatedly asked in NEET UG.

Q1Easy

The Young's modulus of a perfectly rigid body is:

  1. Zero
  2. One
  3. Infinite
  4. Depends on its length
Show solution

A perfectly rigid body has zero strain for any stress, so Y = stress/strain → infinity.

Answer: (C)

Q2Easy

Bernoulli's theorem is based on the conservation of:

  1. Mass
  2. Momentum
  3. Energy
  4. Angular momentum
Show solution

Each term (P, ½ρv², ρgh) is energy per unit volume; their sum is constant. The continuity equation expresses mass conservation.

Answer: (C)

Q3Medium

Two wires of the same material have lengths in the ratio 1 : 2 and radii in the ratio 2 : 1. Under the same load, their extensions are in the ratio:

  1. 1 : 8
  2. 8 : 1
  3. 1 : 4
  4. 2 : 1
Show solution

ΔL ∝ L/r²: (1/4) : (2/1) = 1/4 : 2 = 1 : 8.

Answer: (A)

Q4Medium

Water flows out of a hole 5 m below the water surface in a wide tank. The speed of efflux is (g = 10 m s⁻²):

  1. 5 m s⁻¹
  2. 10 m s⁻¹
  3. 50 m s⁻¹
  4. √50 m s⁻¹
Show solution

Torricelli: v = √(2gh) = √100 = 10 m s⁻¹.

Answer: (B)

Q5Medium

Eight identical raindrops, each with terminal velocity 5 cm s⁻¹, combine into one drop. Its terminal velocity is:

  1. 10 cm s⁻¹
  2. 20 cm s⁻¹
  3. 40 cm s⁻¹
  4. 5 cm s⁻¹
Show solution

Volume × 8 ⇒ radius × 2 ⇒ vt ∝ r² gives × 4: 20 cm s⁻¹.

Answer: (B)

Q6Medium

A block of ice floats in water. The fraction of its volume above the water is about (ρice = 900 kg m⁻³):

  1. 0.9
  2. 0.5
  3. 0.1
  4. 0.2
Show solution

Fraction submerged = 900/1000 = 0.9, so 0.1 (10%) stays above water.

Answer: (C)

Q7Hard

The excess pressure inside soap bubble A is twice that inside soap bubble B. The ratio of their volumes VA : VB is:

  1. 1 : 2
  2. 1 : 4
  3. 1 : 8
  4. 8 : 1
Show solution

ΔP = 4T/r ⇒ r ∝ 1/ΔP, so rA : rB = 1 : 2 and volumes (∝ r³) = 1 : 8.

Answer: (C)

Q8Hard

When a liquid is heated, its viscosity and surface tension respectively:

  1. Increase, increase
  2. Decrease, decrease
  3. Increase, decrease
  4. Decrease, increase
Show solution

Heating weakens intermolecular attraction in a liquid, lowering both viscosity and surface tension. (Gas viscosity, in contrast, rises with temperature.)

Answer: (B)

12. Quick Revision Summary

Remember before the exam
  • Y = FL/AΔL (material constant); energy = ½FΔL.
  • P = Pa + ρgh; Pascal: F₂/F₁ = A₂/A₁; upthrust = weight displaced.
  • A₁v₁ = A₂v₂; P + ½ρv² + ρgh = const; efflux √(2gh).
  • Stokes: 6πηrv; vt ∝ r².
  • Drop 2T/r, soap bubble 4T/r; capillary h = 2T cosθ/rρg.
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