Physics Lab

Unit 7: Properties of Solids and Liquids

This is a content-heavy unit covering both mechanical properties of solids (elasticity) and mechanics of fluids (hydrostatics, hydrodynamics, viscosity, surface tension). NEET sets 2–3 MCQs here, and questions are mostly formula-recall with light arithmetic. Surface tension and capillary rise yield assertion-reason questions almost every year.

Concept Map

  • Elasticity
    • Stress, strain
    • Young's, bulk, shear moduli; Poisson's ratio
    • Stress-strain curve
    • Elastic PE
  • Hydrostatics
    • Pressure variation with depth
    • Pascal's law
    • Archimedes' principle, buoyancy
  • Hydrodynamics
    • Streamline vs turbulent flow
    • Equation of continuity
    • Bernoulli's equation and applications
  • Viscosity
    • Newton's law of viscosity
    • Stokes' law, terminal velocity
    • Reynolds number
  • Surface tension
    • Capillary rise
    • Excess pressure (drops, bubbles)

Topic 1: Elasticity

Sub-topic A: Stress and Strain

Stress =F/A= F/A (force per unit area), SI unit N/m² == Pa.

Three types:

  • Longitudinal (tensile/compressive): along length.
  • Volumetric: uniform pressure.
  • Shear (tangential): force parallel to surface.

Strain == change/original (dimensionless):

  • Longitudinal: ΔL/L\Delta L/L.
  • Volumetric: ΔV/V\Delta V/V.
  • Shear: tanϕϕ\tan\phi \approx \phi (angular deformation).

Sub-topic B: Hooke's Law and Elastic Moduli

Within the elastic limit: stress \propto strain. The proportionality constant is a modulus.

ModulusDefinition
Young's modulus YYlongitudinal stress / longitudinal strain =(F/A)/(ΔL/L)= (F/A)/(\Delta L/L)
Bulk modulus KKvolumetric stress / volumetric strain =P/(ΔV/V)= -P/(\Delta V/V)
Shear (rigidity) modulus η\etashear stress / shear strain =(F/A)/ϕ= (F/A)/\phi

Compressibility =1/K= 1/K. Liquids have KK only (no YY, no η\eta).

Sub-topic C: Poisson's Ratio

When stretched along length, lateral dimensions contract. Define

σ=lateral strainlongitudinal strain.\sigma = -\frac{\text{lateral strain}}{\text{longitudinal strain}}.

Typically 0σ0.50 \le \sigma \le 0.5. For most metals σ0.3\sigma \approx 0.3.

Relations among moduli (theoretical limits):

Y=2η(1+σ),Y=3K(12σ),9Y=1K+3η.Y = 2\eta(1 + \sigma), \quad Y = 3K(1 - 2\sigma), \quad \frac{9}{Y} = \frac{1}{K} + \frac{3}{\eta}.

Sub-topic D: Stress-Strain Curve

For a typical ductile material:

  • Proportional limit: stress ∝ strain holds.
  • Elastic limit / yield point: beyond here, permanent deformation.
  • Plastic region: strain hardening.
  • Ultimate stress: max stress before necking.
  • Fracture point.

Brittle materials (cast iron, glass) fracture near the elastic limit. Ductile (steel, copper) show extensive plastic region.

Sub-topic E: Elastic PE

Energy stored per unit volume:

u=12stress×strain=12Yϵ2=(stress)22Y.u = \tfrac{1}{2}\,\text{stress}\,\times\,\text{strain} = \tfrac{1}{2} Y \epsilon^2 = \tfrac{(\text{stress})^2}{2Y}.

For a wire of natural length LL, area AA, stretched by xx: U=12(YA/L)x2U = \tfrac{1}{2} (Y A/L) x^2, i.e. behaves like a spring with k=YA/Lk = Y A/L.

Topic 2: Hydrostatics

Sub-topic A: Pressure

P=F/A,[P]=[ML1T2].P = F/A, \quad [P] = [M L^{-1} T^{-2}].

Atmospheric pressure 1.013×105 Pa=760 mm Hg=1 atm\approx 1.013 \times 10^5\ \text{Pa} = 760\ \text{mm Hg} = 1\ \text{atm}.

Sub-topic B: Pressure in a Fluid Column

P=P0+ρgh.P = P_0 + \rho g h.

The pressure depends only on depth, not shape of vessel. This explains the hydrostatic paradox.

Sub-topic C: Pascal's Law

Pressure applied to a confined fluid is transmitted equally in all directions. Used in hydraulic press, lift, brakes.

Hydraulic press: Effort F1F_1 on small piston of area A1A_1 produces load F2F_2 on big piston of area A2A_2:

F2=F1(A2/A1).F_2 = F_1 (A_2/A_1).

Sub-topic D: Archimedes' Principle

A body wholly or partly submerged experiences an upward buoyant force equal to the weight of fluid displaced:

Fb=Vsubmergedρfluidg.F_b = V_\text{submerged} \rho_\text{fluid} g.

Floating body: total buoyancy = weight, so

VsubmergedVtotal=ρbodyρfluid.\frac{V_\text{submerged}}{V_\text{total}} = \frac{\rho_\text{body}}{\rho_\text{fluid}}.

An iceberg (ρice=0.917ρwater\rho_\text{ice} = 0.917 \rho_\text{water}) floats with ~92% submerged.

Topic 3: Hydrodynamics

Sub-topic A: Streamline and Turbulent Flow

  • Streamline: velocity at each point is constant in time (steady flow), particles follow smooth paths.
  • Turbulent: chaotic, swirling.
  • Transition controlled by Reynolds number:
Re=ρvd/η.\text{Re} = \rho v d/\eta.

Re < 2000 streamline; Re > 4000 turbulent; in between, unstable.

Sub-topic B: Equation of Continuity

For an incompressible fluid:

A1v1=A2v2=constant.A_1 v_1 = A_2 v_2 = \text{constant}.

Narrower pipe ⟹ faster flow.

Sub-topic C: Bernoulli's Equation

For steady, incompressible, non-viscous flow along a streamline:

P+12ρv2+ρgh=constant.P + \tfrac{1}{2} \rho v^2 + \rho g h = \text{constant}.

Three terms: pressure energy, kinetic energy per volume, potential energy per volume.

Sub-topic D: Applications of Bernoulli

  • Venturi meter: pressure difference ΔP=12ρ(v22v12)\Delta P = \tfrac{1}{2} \rho (v_2^2 - v_1^2).
  • Pitot tube: measures fluid speed.
  • Aerodynamic lift: faster flow over curved wing top ⟹ lower pressure on top.
  • Spinning ball curve (Magnus effect).
  • Torricelli's theorem: speed of efflux from a small hole at depth hh below the free surface:
v=2gh.v = \sqrt{2 g h}.

For a vessel of cross-section AA, hole of area aa, time to drain from height h0h_0 to h1h_1 is

t=Aa2/g(h0h1).t = \frac{A}{a}\sqrt{2/g}\,(\sqrt{h_0} - \sqrt{h_1}).

Topic 4: Viscosity

Sub-topic A: Newton's Law of Viscous Force

For a fluid with velocity gradient dv/dydv/dy across layers,

F=ηAdvdy,F = -\eta A \frac{dv}{dy},

where η\eta is the coefficient of viscosity (dynamic viscosity). SI unit Pa·s = poise/10. Dimensions [ML1T1][M L^{-1} T^{-1}].

Viscosity of liquids decreases with temperature; of gases increases.

Sub-topic B: Stokes' Law

For a small sphere of radius rr moving with speed vv through a fluid:

Fdrag=6πηrv.F_\text{drag} = 6\pi\eta r v.

Sub-topic C: Terminal Velocity

When the sphere falls under gravity through a viscous fluid, it reaches terminal speed when weight - buoyancy == drag:

vt=2r2(ρsρf)g9η.v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}.

Hence vtr2v_t \propto r^2 — bigger particles fall faster.

Topic 5: Surface Tension

Sub-topic A: Definition

Surface tension TT is the force per unit length on any line drawn in the surface, acting tangentially:

T=F/L.T = F/L.

Equivalently, TT is the surface energy per unit area (units N/m or J/m²). Dimensions [MT2][M T^{-2}].

Surface tension decreases with temperature and falls to zero at the critical temperature.

Sub-topic B: Angle of Contact

The angle between the tangent to the liquid surface and the solid wall (measured inside the liquid). Examples:

  • Water-glass: θ0°\theta \approx 0° (wets, rises in capillary).
  • Mercury-glass: θ138°\theta \approx 138° (does not wet, depresses).

Sub-topic C: Capillary Rise

A capillary tube of radius rr dipped vertically in a liquid:

h=2Tcosθrρg.h = \frac{2 T \cos\theta}{r \rho g}.

The smaller the tube, the higher the rise (Jurin's law). For water in glass (T0.073 N/mT \approx 0.073\ \text{N/m}, θ=0\theta = 0), in a 1 mm tube, h1.5h \approx 1.5 cm.

If the tube is too short for the predicted height, the liquid does not overflow — the meniscus radius adjusts.

Sub-topic D: Excess Pressure

The pressure inside a curved surface exceeds outside by:

SurfaceΔP\Delta P
Liquid drop (one surface)2T/R2T/R
Air bubble inside liquid2T/R2T/R
Soap bubble in air (two surfaces)4T/R4T/R

When two bubbles of radii r1r_1 and r2r_2 (r1<r2r_1 < r_2) merge through a tube, the smaller one collapses into the larger one (lower pressure inside larger).

Sub-topic E: Work Done to Form a Drop / Bubble

Work done by external agent to form a drop = increase in surface area ×T\times T:

  • Spherical drop of radius RR: W=4πR2TW = 4\pi R^2 T.
  • Soap bubble: W=8πR2TW = 8\pi R^2 T (two surfaces).
  • Splitting drop of radius RR into nn identical droplets of radius rr (volume conservation gives R3=nr3R^3 = n r^3): surface energy increases by ΔW=4πT(nr2R2)=4πTR2(n1/31)\Delta W = 4\pi T (n r^2 - R^2) = 4\pi T R^2 (n^{1/3} - 1).

NEET Pattern MCQ Tips

  • Young's modulus: thin wire problem, ΔL=FL/(AY)\Delta L = FL/(AY).
  • Bulk modulus: pressure-induced volume change.
  • Elastic PE: U=12stressstrainVU = \tfrac{1}{2} \cdot \text{stress} \cdot \text{strain} \cdot V.
  • Pascal/hydraulic press: force ratio = area ratio.
  • Archimedes: floating fraction = ρbody/ρliquid\rho_\text{body}/\rho_\text{liquid}.
  • Bernoulli / Torricelli: v=2ghv = \sqrt{2gh}.
  • Capillary: smaller tube ⟹ higher rise; bubbles inside a liquid.
  • Excess pressure: drop 2T/R2T/R, soap bubble 4T/R4T/R.
  • Stokes & terminal velocity: vtr2v_t \propto r^2.

Common Confusions and Traps

  • YY, KK, ηshear\eta_\text{shear} apply to solids; for fluids only KK exists.
  • The capillary rise formula assumes the meniscus is hemispherical. The smaller the tube radius, the higher the rise.
  • A soap bubble in air has TWO surfaces (inner and outer), so excess pressure is 4T/R4T/R. A liquid drop in air or air bubble in liquid has ONE surface, so 2T/R2T/R.
  • Two bubbles merged: smaller transfers air to larger (smaller has higher pressure).
  • Stokes' law applies to small spheres in laminar flow; large/fast objects experience inertial drag.
  • Reynolds number is dimensionless — students sometimes give it units.
  • Surface tension decreases with temperature (and with surfactants like soap), eventually vanishing at TcT_c.

Quick Revision Card

  • Y=Y = stress/strain (longitudinal); K=K = stress / fractional volume change; ηshear=\eta_\text{shear} = tangential stress / shear angle.
  • Elastic PE per unit volume: 12stress×strain\tfrac{1}{2}\,\text{stress}\,\times\,\text{strain}.
  • Hydrostatic: P=P0+ρghP = P_0 + \rho g h.
  • Buoyancy: F=VρfluidgF = V \rho_\text{fluid} g; floating fraction = ρbody/ρliquid\rho_\text{body}/\rho_\text{liquid}.
  • Continuity: Av=A v = const.
  • Bernoulli: P+12ρv2+ρgh=P + \tfrac{1}{2}\rho v^2 + \rho g h = const.
  • Torricelli: v=2ghv = \sqrt{2gh}.
  • Stokes: F=6πηrvF = 6\pi\eta r v; vt=2r2(ρsρf)g/(9η)v_t = 2 r^2 (\rho_s - \rho_f) g/(9\eta).
  • Capillary: h=2Tcosθ/(rρg)h = 2T\cos\theta/(r\rho g).
  • Drop: ΔP=2T/R\Delta P = 2T/R; bubble (air-liquid): ΔP=2T/R\Delta P = 2T/R; soap bubble (in air): ΔP=4T/R\Delta P = 4T/R.

Worked NEET Examples

Example 1: Strain Energy in Wire

A steel wire (Y = 2 × 10¹¹ N/m²), length 1 m, cross-section 1 mm². Stretched by 1 mm. Strain energy?

U=(1/2)Yε2V=(1/2)(2×1011)(103)2(106×1)=(1/2)(2×1011)(106)(106)=101U = (1/2)Y\varepsilon^2 V = (1/2)(2 \times 10^{11})(10^{-3})^2 (10^{-6} \times 1) = (1/2)(2 \times 10^{11})(10^{-6})(10^{-6}) = 10^{-1} J = 0.1 J.

Example 2: Pressure at Depth

Pressure at 100 m below water surface: P=P0+ρgh=105+103×10×100=1.1×106P = P_0 + \rho g h = 10^5 + 10^3 \times 10 \times 100 = 1.1 \times 10^6 Pa 11\approx 11 atm.

Example 3: Capillary Tube in Water

Water (T = 0.073 N/m, contact angle ≈ 0°) in glass capillary of 0.5 mm radius. Rise:

h=2Tcosθ/(rρg)=2×0.073/(5×104×103×10)=0.146/(5)=0.029 m=2.9 cm.h = 2T\cos\theta/(r\rho g) = 2 \times 0.073/(5 \times 10^{-4} \times 10^3 \times 10) = 0.146/(5) = 0.029\ \text{m} = 2.9\ \text{cm}.

Example 4: Terminal Velocity of Droplet

A water droplet of radius 0.1 mm falls through air (η = 1.8 × 10⁻⁵ Pa s).

vt=2r2(ρwρa)g/(9η)2(104)2(103)(10)/(9×1.8×105)v_t = 2 r^2 (\rho_w - \rho_a) g/(9\eta) \approx 2(10^{-4})^2(10^3)(10)/(9 \times 1.8 \times 10^{-5}) =2×108×104/(1.62×104)1.2 m/s.= 2 \times 10^{-8} \times 10^4 /(1.62 \times 10^{-4}) \approx 1.2\ \text{m/s}.

Example 5: Bernoulli on Tank Hole

A water tank with surface 5 m above a small hole at the side. Speed of water exiting:

v=2gh=2×10×5=10 m/s.v = \sqrt{2gh} = \sqrt{2 \times 10 \times 5} = 10\ \text{m/s}.

Derivations

Bernoulli's Equation

For steady, incompressible, non-viscous flow, work-energy theorem on a fluid element gives:

P+12ρv2+ρgh=const.P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{const}.

Interpretation: pressure energy + KE per volume + PE per volume is constant along a streamline.

Stokes' Law

For laminar flow past a sphere, dimensional analysis gives drag ηrv\propto \eta r v. Detailed solution (no derivation expected at NEET level) gives prefactor 6π6\pi:

Fdrag=6πηrv.F_\text{drag} = 6\pi\eta r v.

Terminal Velocity

A sphere of density ρs\rho_s falling through a fluid of density ρf\rho_f experiences three forces:

  1. Weight: (4/3)πr3ρsg(4/3)\pi r^3 \rho_s g (down).
  2. Buoyancy: (4/3)πr3ρfg(4/3)\pi r^3 \rho_f g (up).
  3. Drag: 6πηrv6\pi\eta r v (up).

At terminal velocity, net force = 0:

(4/3)πr3(ρsρf)g=6πηrvt.(4/3)\pi r^3 (\rho_s - \rho_f) g = 6\pi\eta r v_t.

Solving: vt=(2/9)r2(ρsρf)g/ηv_t = (2/9) r^2 (\rho_s - \rho_f) g/\eta.

Capillary Rise

Surface tension TT acts along the contact line 2πr2\pi r at angle θ\theta to the wall. Vertical component supports the column of weight:

T2πrcosθ=πr2hρg,T \cdot 2\pi r \cos\theta = \pi r^2 h \rho g,

giving Jurin's law:

h=2Tcosθ/(rρg).h = 2T\cos\theta/(r\rho g).

Excess Pressure in a Drop

Consider half of a drop of radius RR. Surface tension along the equator pulls inward with force T2πRT \cdot 2\pi R. Excess pressure inside pushes the half outward over area πR2\pi R^2. At equilibrium:

PexcessπR2=T2πR    Pexcess=2T/R.P_\text{excess} \cdot \pi R^2 = T \cdot 2\pi R \implies P_\text{excess} = 2T/R.

Soap bubble has two surfaces (inner and outer), so excess pressure is 4T/R4T/R.

Special Hydrodynamic Cases

Vena Contracta

Water exiting a sharp-edged orifice constricts to about 62% of the orifice area before re-expanding. The discharge coefficient is ~0.62 for sharp edges, ~0.97 for rounded.

Venturi Meter

A constriction in a horizontal pipe. From continuity and Bernoulli (h constant):

A1v1=A2v2,P1+12ρv12=P2+12ρv22.A_1 v_1 = A_2 v_2, \quad P_1 + \tfrac{1}{2}\rho v_1^2 = P_2 + \tfrac{1}{2}\rho v_2^2.

So ΔP=12ρ(v22v12)>0\Delta P = \tfrac{1}{2}\rho(v_2^2 - v_1^2) > 0 (pressure drop at constriction). Measure ΔP\Delta P to deduce flow rate.

Pitot Tube

A bent tube facing the flow. Stagnation pressure P0=P+12ρv2P_0 = P + \tfrac{1}{2}\rho v^2. By measuring P0PP_0 - P, the flow speed is

v=2(P0P)/ρ.v = \sqrt{2(P_0 - P)/\rho}.

Surface Tension Phenomena

Why Insects Walk on Water

Water surface acts like a stretched membrane with T0.073T \approx 0.073 N/m. A water strider's legs push down on the surface, deforming it; surface tension provides upward force without breaking the surface.

Soap Films vs Water Films

Surfactants in soap lower water's surface tension (from 0.073 to about 0.025 N/m), allowing stable thin films. The drop in TT also makes droplets larger (capillary rise is less).

Sphericality of Drops

Free drops are spherical because a sphere minimises surface area for a given volume — surface tension minimises the surface energy.

Viscosity in Daily Life

  • Engine oils have viscosity ratings (SAE 10W-40).
  • Honey: η10 Pa s\eta \sim 10\ \text{Pa s}.
  • Water at 20 °C: η=103 Pa s\eta = 10^{-3}\ \text{Pa s}.
  • Air at 20 °C: η1.8×105 Pa s\eta \sim 1.8 \times 10^{-5}\ \text{Pa s}.

Viscosity of liquids decreases with temperature (molecules slide past each other more easily). Viscosity of gases increases with temperature (more momentum exchange between layers).

Formula Sheet

QuantityFormula
Stressσ=F/A\sigma = F/A
Longitudinal strainϵ=ΔL/L\epsilon = \Delta L/L
Young's modulusY=(FL)/(AΔL)Y = (FL)/(A\Delta L)
Bulk modulusK=VΔP/ΔVK = -V\Delta P/\Delta V
Compressibility1/K1/K
Poisson's ratioσ=(lat strain)/(long strain)\sigma = -(\text{lat strain})/(\text{long strain})
Elastic PE / volu=12σϵu = \tfrac{1}{2}\,\sigma\epsilon
Pressure variationP=P0+ρghP = P_0 + \rho g h
Pascal (hydraulic)F2/F1=A2/A1F_2/F_1 = A_2/A_1
ArchimedesFb=VρfgF_b = V\rho_f g
ContinuityA1v1=A2v2A_1 v_1 = A_2 v_2
BernoulliP+12ρv2+ρgh=P + \tfrac{1}{2}\rho v^2 + \rho g h = const
Torricelli effluxv=2ghv = \sqrt{2gh}
Reynolds numberRe=ρvd/η\text{Re} = \rho v d/\eta
Stokes dragF=6πηrvF = 6\pi\eta r v
Terminal velocityvt=2r2(ρsρf)g/(9η)v_t = 2r^2(\rho_s - \rho_f)g/(9\eta)
Capillary riseh=2Tcosθ/(rρg)h = 2T\cos\theta/(r\rho g)
Excess pressure dropΔP=2T/R\Delta P = 2T/R
Excess pressure soap bubbleΔP=4T/R\Delta P = 4T/R
Surface energyW=TΔAW = T \cdot \Delta A

Sub-topics

5 pages

Practice quiz

Quiz
NEET Unit 7: Properties of Solids and Liquids — Quiz
15 questions · pick the best answer
Q1

A wire of length 2 m and area 0.1 mm² is stretched by 0.5 mm under a load of 25 N. Young's modulus is:

Q2

Excess pressure inside a soap bubble of radius R is:

Q3

A solid floats with 1/4 of its volume above water. Density of solid is:

Q4

Capillary rise of water in a tube of radius r is h. If the tube radius is halved, the new rise is:

Q5

Terminal velocity of a small sphere falling through a viscous fluid is proportional to:

Q6

Two soap bubbles of radii r₁ and r₂ (r₁ < r₂) are connected. What happens?

Q7

A liquid flows through a pipe of varying cross-section. At a point where area is A and speed v, what is the speed where area is A/4?

Q8

Velocity of efflux from a hole at depth 5 m below the free surface (g = 10 m/s²):

Q9

Assertion: Two raindrops of different radii falling through air reach different terminal velocities. Reason: Stokes' drag depends on radius.

Q10

Energy stored per unit volume in a stretched wire with stress σ and strain ε is:

Q11

The pressure at the bottom of a column of mercury (ρ = 13.6 × 10³ kg/m³) of height 76 cm is approximately:

Q12

A hydraulic lift has pistons of areas 5 cm² and 500 cm². To lift a load of 1000 N on the large piston, the force on the small piston is:

Q13

Surface tension of a liquid varies with temperature as:

Q14

Assertion: Liquids have only bulk modulus, not Young's modulus. Reason: Liquids resist compression but cannot sustain shear stress.

Q15

A water drop of radius R is broken into n equal droplets. The work done against surface tension is: