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PhysicsModerate Weightage β˜…β˜…β˜…Class 11

Properties of Matter

Elasticity, viscosity, and surface tension β€” a conceptual chapter with direct formula MCQs. Expect 2–3 EAPCET questions.

2–3Questions in EAPCET
~2%Paper Weightage
8Core Formulas
3Mistake Traps

Concept Core

Elasticity, viscosity, and surface tension β€” the three pillars of properties of matter.

Elasticity β€” Stress, Strain, and Young's Modulus
Stress = Force / Area (N/mΒ²) Strain = Change in dimension / Original dimension (dimensionless) Young's Modulus Y = Tensile Stress / Tensile Strain = FL/(AΞ”L) Bulk Modulus K = Pressure / Volumetric Strain = βˆ’P/(Ξ”V/V) Modulus of Rigidity Ξ· = Shear Stress / Shear Strain

Hooke's Law: Stress ∝ Strain (within elastic limit). Beyond elastic limit β†’ permanent deformation.

Elastic Energy Stored in a Wire
Energy U = Β½ Γ— Stress Γ— Strain Γ— Volume U = Β½ Γ— F Γ— Ξ”L (since F = YΒ·AΒ·Ξ”L/L) Energy density = Β½ Γ— Stress Γ— Strain = StressΒ²/(2Y)
Viscosity β€” Stokes' Law
Viscous force (Stokes): F = 6πηrv Terminal velocity: vβ‚œ = 2rΒ²(Οβˆ’Οƒ)g / (9Ξ·) Poiseuille's Law: Q = Ο€r⁴ΔP/(8Ξ·L) (flow through pipe)

Ξ· = coefficient of viscosity (NΒ·s/mΒ²). Terminal velocity when weight = buoyancy + viscous drag. For liquids, Ξ· decreases with temperature. For gases, Ξ· increases with temperature.

Surface Tension
Surface tension T = Force / Length (N/m) Excess pressure inside bubble: Ξ”P = 4T/r (soap bubble, 2 surfaces) Excess pressure inside drop: Ξ”P = 2T/r (one surface) Capillary rise: h = 2T cosΞΈ/(ρgr)

Soap bubble has 2 surfaces (inner and outer), hence 4T/r. Liquid drop has 1 surface, hence 2T/r.

Bernoulli's Equation & Continuity
Continuity: A₁v₁ = Aβ‚‚vβ‚‚ (incompressible fluid) Bernoulli: P + ½ρvΒ² + ρgh = constant Torricelli: v = √(2gh) (efflux from a hole)
Thermal Expansion
Linear: Ξ”L = Ξ±LΞ”T β†’ L = Lβ‚€(1+Ξ±Ξ”T) Area: Ξ”A = 2Ξ±AΞ”T (Ξ² = 2Ξ±) Volume: Ξ”V = 3Ξ±VΞ”T (Ξ³ = 3Ξ±) Relation: Ξ± : Ξ² : Ξ³ = 1 : 2 : 3

Formula Vault

Properties of matter formulas for EAPCET.

Young's Modulus
Y = FL/(AΞ”L)
F = force; L = length; A = area
Bulk Modulus
K = βˆ’PV/Ξ”V
Resistance to compression
Elastic Energy
U = Β½ Γ— F Γ— Ξ”L = Β½ Γ— Stress Γ— Strain Γ— V
V = volume of wire
Stokes' Law
F = 6πηrv
Ξ· = viscosity; r = radius; v = speed
Terminal Velocity
vβ‚œ = 2rΒ²(Οβˆ’Οƒ)g/(9Ξ·)
ρ = sphere density; Οƒ = fluid density
Surface Tension
T = F/l (N/m)
F = force along length l
Excess P in Soap Bubble
Ξ”P = 4T/r
Two surfaces β†’ factor 4
Excess P in Drop
Ξ”P = 2T/r
One surface β†’ factor 2
Capillary Rise
h = 2T cosθ/(ρgr)
ΞΈ = contact angle; r = tube radius
Thermal Expansion
Ξ±:Ξ²:Ξ³ = 1:2:3
Linear, area, volume coefficients

Worked Examples

5 problems β€” Young's modulus, terminal velocity, surface tension, Bernoulli, and a trap.

EasyFind extension of wire: Y=2Γ—10ΒΉΒΉ Pa, L=1m, A=10⁻⁢mΒ², F=200Nβ–Ύ
A steel wire (Y = 2Γ—10ΒΉΒΉ N/mΒ², L = 1 m, A = 10⁻⁢ mΒ²) is loaded with F = 200 N. Find the extension.
1
Y = FL/(AΞ”L) β†’ Ξ”L = FL/(YA)
2
Ξ”L = 200 Γ— 1/(2Γ—10ΒΉΒΉ Γ— 10⁻⁢) = 200/(2Γ—10⁡) = 10⁻³ m = 1 mm
βœ“  Extension = 1 mm
EasyFind excess pressure inside a soap bubble of radius 5cm (T=0.03 N/m)β–Ύ
Find the excess pressure inside a soap bubble of radius 5 cm. Surface tension = 0.03 N/m.
1
Soap bubble (2 surfaces): Ξ”P = 4T/r = 4 Γ— 0.03/(0.05) = 0.12/0.05 = 2.4 Pa
βœ“  Excess pressure = 2.4 Pa
MediumFind terminal velocity of sphere: r=0.5mm, ρ=8000 kg/mΒ³, Οƒ=1000 kg/mΒ³, Ξ·=0.1 PaΒ·sβ–Ύ
Find terminal velocity of a steel sphere (r = 0.5 mm, ρ = 8000 kg/mΒ³) falling through oil (Οƒ = 1000 kg/mΒ³, Ξ· = 0.1 PaΒ·s). g = 10 m/sΒ².
1
vβ‚œ = 2rΒ²(Οβˆ’Οƒ)g/(9Ξ·)
2
= 2Γ—(0.5Γ—10⁻³)Β²Γ—(8000βˆ’1000)Γ—10/(9Γ—0.1)
3
= 2Γ—2.5Γ—10⁻⁷×70000/0.9 = 3.5Γ—10⁻²/0.9 β‰ˆ 0.039 m/s β‰ˆ 3.9 cm/s
βœ“  Terminal velocity β‰ˆ 3.9 cm/s
EAPCET LevelWater flows from radius 4cm pipe into radius 2cm pipe β€” find speed ratioβ–Ύ
Water flows from a pipe of radius 4 cm into one of radius 2 cm. Find the ratio of speeds.
1
Continuity equation: A₁v₁ = Aβ‚‚vβ‚‚
2
Ο€(4)Β²v₁ = Ο€(2)Β²vβ‚‚
3
16v₁ = 4vβ‚‚ β†’ vβ‚‚/v₁ = 4
4
The smaller pipe has 4Γ— higher speed.
βœ“  vβ‚‚ : v₁ = 4 : 1
Trap QuestionSoap bubble vs liquid drop β€” same formula for excess pressure?β–Ύ
A soap bubble and a liquid drop both have radius r = 2 cm, surface tension T = 0.04 N/m. Compare their excess pressures.
1
The trap: Same formula doesn't apply to both.
2
Soap bubble: 2 surfaces (inner and outer). Ξ”P = 4T/r = 4Γ—0.04/0.02 = 8 Pa
3
Liquid drop: 1 surface (the outer surface only). Ξ”P = 2T/r = 2Γ—0.04/0.02 = 4 Pa
4
Soap bubble has exactly double the excess pressure of a liquid drop of the same radius and surface tension.
βœ“  Soap bubble: 8 Pa; Drop: 4 Pa β€” bubble is double (two surfaces vs one)

Mistake DNA

3 properties of matter errors from EAPCET distractor analysis.

🫧
Using 2T/r for Soap Bubble (Should Be 4T/r)
A soap bubble has TWO surfaces (inner and outer), so the pressure formula is 4T/r, not 2T/r.
❌ Wrong
Soap bubble Ξ”P = 2T/r βœ— (one-surface formula; only valid for drop)
βœ“ Correct
Soap bubble: 4T/r βœ“ (two surfaces contribute) Liquid drop: 2T/r βœ“ (only outer surface)
Remember: soap film has inner and outer surfaces. Each contributes 2T/r. Total = 4T/r. A liquid drop has only one curved surface: Ξ”P = 2T/r.
🌑️
Ξ· of Liquids Increases with Temperature (Like Gases)
For liquids, viscosity DECREASES with temperature (molecules have more KE, layers slide more easily). For gases, Ξ· INCREASES with temperature.
❌ Wrong
Heating oil β†’ more viscous (Ξ· increases) βœ— (that's for gases!)
βœ“ Correct
Liquids: Ξ· decreases with T βœ“ (honey flows easier when warm) Gases: Ξ· increases with T βœ“ (more molecular collisions)
Physical intuition: in liquids, viscosity arises from molecular attraction. Higher T β†’ molecules overcome attraction β†’ Ξ· decreases. In gases, viscosity arises from molecular momentum transfer between layers, which increases with T.
πŸ“
Ξ±:Ξ²:Ξ³ = 1:2:3 Means Ξ² = 2 and Ξ³ = 3
The ratio is Ξ±:Ξ²:Ξ³ = 1:2:3, meaning Ξ² = 2Ξ± and Ξ³ = 3Ξ±. Not Ξ² = 2 and Ξ³ = 3 in absolute terms.
❌ Wrong
Ξ± = 12Γ—10⁻⁢/Β°C β†’ Ξ² = 2, Ξ³ = 3 βœ— (should be 2Ξ±, 3Ξ±)
βœ“ Correct
Ξ² = 2Ξ± = 24Γ—10⁻⁢/Β°C βœ“ Ξ³ = 3Ξ± = 36Γ—10⁻⁢/Β°C βœ“ The ratio is preserved not the absolute value
The thermal expansion coefficients maintain ratio 1:2:3 always. If linear expansion is Ξ±, area expansion is 2Ξ± and volume expansion is 3Ξ±.

Chapter Intelligence

Properties of matter integrates with fluids, thermal physics, and elasticity.

EAPCET Weightage (2019–2024)
Surface tension (bubbles, drops, capillary)
~7
Young's modulus and stress/strain
~6
Viscosity and terminal velocity
~5
Bernoulli's equation
~4
Thermal expansion Ξ±:Ξ²:Ξ³
~3
High-Yield PYQ Patterns
Excess pressure in soap bubble vs dropTerminal velocity calculationYoung's modulus extension of wireBernoulli: speed ratio in pipesCapillary rise heightΞ±:Ξ²:Ξ³ = 1:2:3 propertyElastic energy in stretched wire
Exam Strategy
  • Soap bubble (2T/r each surface): Ξ”P = 4T/r. Liquid drop (1 surface): Ξ”P = 2T/r. The factor-of-2 difference is tested directly every few years.
  • Terminal velocity: all factors in the formula vβ‚œ = 2rΒ²(Οβˆ’Οƒ)g/(9Ξ·). Key insight: vβ‚œ ∝ rΒ² β€” doubling radius β†’ 4Γ— terminal velocity.
  • Thermal expansion: Ξ³ = 3Ξ± always. This is a direct fact question. Volume expands 3Γ— more than linear for same Ξ”T.
  • Continuity equation: A₁v₁ = Aβ‚‚vβ‚‚. Narrower pipe β†’ faster flow. This is the basis of the Venturi meter and atomiser.
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