Properties of Solids and Liquids – JEE Main Physics Formula Sheet | Elasticity, Surface Tension, Bernoulli, Viscosity, Thermal Expansion & All Formulas

Elasticity — Stress, Strain, Young's Modulus, Bulk Modulus, Rigidity, Poisson's Ratio

Why Young's Modulus, Wire as Spring, and Thermal Stress Are the Most-Tested Elasticity JEE Main Formulas

Stress (from Aakash PDF — Properties of Solids and Liquids):

Stress = Restoring force per unit area = F/A. Unit: N/m² = Pa. Three types:

(1) Normal (Tensile/Compressive) Stress: Force perpendicular to cross-section. = F/A.

(2) Shear (Tangential) Stress: Force parallel to surface. = F/A.

(3) Hydraulic Stress: Equal normal force from all sides = pressure P.

Strain (from Aakash PDF — Properties of Solids and Liquids):

(1) Longitudinal strain = Δl/l (change in length/original length; dimensionless).

(2) Volumetric strain = ΔV/V

(3) Shear strain = Δx/l = tanφ ≈ φ (for small angles; φ = shear angle)

Moduli of Elasticity (from Aakash PDF — Properties of Solids and Liquids JEE Main):

Young's Modulus (Y): Tensile stress / Longitudinal strain

Y = (F/A)/(Δl/l) = Fl/(AΔl)

Rearranged for extension: Δl = Fl/(AY). Unit: N/m² = Pa. [ML⁻¹T⁻²].

Bulk Modulus (B): Normal stress / Volumetric strain = –P/(ΔV/V)

B = –PV/ΔV (negative sign because volume decreases when pressure increases)

Compressibility = 1/B. For water: B ≈ 2×10⁹ Pa; for ideal gas: B = P (isothermal) or γP (adiabatic).

Modulus of Rigidity (η): Shear stress / Shear strain

η = (F/A)/φ = F/(Aφ)

Poisson's Ratio (σ) (from Aakash PDF — Properties of Solids and Liquids JEE Main):

When a rod is stretched longitudinally, it contracts laterally:

σ = –(ΔR/R)/(Δl/l) = –(Lateral strain)/(Longitudinal strain)

Theoretical range: –1 ≤ σ ≤ 0.5. Practical range: 0 ≤ σ ≤ 0.5. When σ = 0.5: density is constant (incompressible).

Relation: Y = 3B(1–2σ) = 2η(1+σ). Also: η = Y/2(1+σ); B = Y/3(1–2σ).

Wire as Spring (from Aakash PDF — Properties of Solids and Liquids JEE Main):

A wire of length l, cross-section A, Young's modulus Y behaves like a spring with:

k = YA/l

Extension under force F: Δl = Fl/AY = F/k.

Work done stretching wire: W = ½FΔl = ½kΔl² = F²l/2AY.

Elastic PE stored = ½FΔl.

Energy density (energy per unit volume) = ½ × stress × strain = (stress)²/2Y = Y(strain)²/2

Thermal Stress (from Aakash PDF — Properties of Solids and Liquids JEE Main):

Rod fixed between rigid supports. Temperature increased by ΔT. Rod cannot expand freely → thermal stress develops:

Thermal strain = αΔT (where α = linear thermal expansion coefficient)

Thermal stress = Y × αΔT = YαΔT

Force = YAαΔT (compressive force if temperature increases, since rod tries to expand)

If rod is free (not fixed): Δl = lαΔT; no stress develops.

Rod hanging under own weight (from Aakash PDF — Properties of Solids and Liquids):

Uniform rod of mass M, length l, area A, modulus Y hanging from ceiling:

Extension due to own weight: Δl = Mgl/2AY (factor 2 from average stress = Mg/2A)

Download the Free PDF for Properties of Solids and Liquids for all elasticity examples for JEE Main.

Fluid Statics — Pressure, Pascal's Law, Archimedes' Principle, U-Tube

Why Pressure-Depth Formula, Archimedes' Principle, and U-Tube Problems Are Essential Fluid Statics Results

Pressure in a Fluid (from Aakash PDF — Properties of Solids and Liquids JEE Main):

Pressure at depth h in a fluid of density ρ:

P = P₀ + ρgh

where P₀ = atmospheric pressure at surface. Pressure acts equally in all directions at any point in a fluid (Pascal's principle for static fluids).

Pascal's Law (from Aakash PDF — Properties of Solids and Liquids):

A pressure applied to an enclosed fluid is transmitted equally to all parts of the fluid and to the walls of the container. Application: hydraulic lift (F₁/A₁ = F₂/A₂ → F₂ = F₁A₂/A₁).

Archimedes' Principle (from Aakash PDF — Properties of Solids and Liquids JEE Main):

A body submerged (partially or fully) in a fluid experiences an upward buoyant force equal to the weight of fluid displaced:

F_buoyancy = ρ_liquid × V_submerged × g

Floating condition: weight of body = buoyant force

ρ_body × V_total × g = ρ_liquid × V_submerged × g

V_submerged/V_total = ρ_body/ρ_liquid

For floating body: fraction submerged = density ratio. If ρ_body < ρ_liquid: body floats with fraction ρ_body/ρ_liquid submerged. If ρ_body > ρ_liquid: body sinks completely.

U-Tube Problems (from Aakash PDF — Properties of Solids and Liquids JEE Main):

Two immiscible liquids in U-tube at equilibrium: pressures at the same horizontal level must be equal:

P₀ + ρ₁gh₁ = P₀ + ρ₂gh₂ → ρ₁h₁ = ρ₂h₂

When U-tube accelerates horizontally with acceleration a:

tan θ = a/g = h/L where θ = angle of liquid surface with horizontal, h = height difference, L = horizontal distance.

When U-tube rotates about one limb with angular velocity ω:

Height difference: h = ω²L²/2g; tan θ = ω²L/2g.

Download the Free PDF for Properties of Solids and Liquids for all fluid statics examples for JEE Main.

Surface Tension — Excess Pressure, Capillary Rise, Combining Bubbles

Why Excess Pressure Formulas (2T/r vs 4T/r) and Capillary Rise Are the Most-Tested Surface Tension JEE Main Questions

Surface Tension (from Aakash PDF — Properties of Solids and Liquids JEE Main):

Surface tension T = Force per unit length along the surface = F/l. Unit: N/m. Arises from net inward cohesive forces on surface molecules.

Surface energy = T × ΔA (T also = surface energy per unit area = J/m²).

Excess Pressure (from Aakash PDF — Properties of Solids and Liquids JEE Main):

(1) Liquid drop (one surface): ΔP = Pᵢ–Pₒ = 2T/r

(2) Soap bubble (two surfaces — inner and outer): ΔP = 4T/r

(3) Air bubble inside liquid (one surface): ΔP = 2T/r

Key distinction: soap bubble has TWO surfaces (inner and outer), so excess pressure is double that of a liquid drop. A liquid drop and an air bubble in liquid both have ONE surface each.

Capillary Rise (from Aakash PDF — Properties of Solids and Liquids JEE Main):

In a capillary tube of radius r, contact angle θ, surface tension T, liquid density ρ:

h = 2T cosθ/(ρgr)

In terms of radius of meniscus R (R = r/cosθ):

h = 2T/(ρgR)

Concave meniscus (wetting liquid like water-glass): θ < 90° → cosθ > 0 → h > 0 (rises).

Convex meniscus (non-wetting like mercury-glass): θ > 90° → cosθ < 0 → h < 0 (depressed).

Capillary rise formula also written as: T = ρghr/(2cosθ). Narrower tube → taller rise (h ∝ 1/r).

Combining Bubbles (from Aakash PDF — Properties of Solids and Liquids JEE Main):

If two soap bubbles coalesce in vacuum (P₀=0):

Surface energy conservation: T×4πr₁² + T×4πr₂² = T×4πr²

r² = r₁² + r₂² (radius of combined bubble)

If two soap bubbles come in contact (double bubble):

Radii r₁ and r₂ (r₁ > r₂). Radius of interface r:

1/r = 1/r₂ – 1/r₁ (interface bulges toward larger bubble since smaller bubble has higher pressure)

The interface is convex toward the larger bubble (P₂ > P₁ → smaller bubble has higher pressure).

Download the Free PDF for Properties of Solids and Liquids for all surface tension examples for JEE Main.

Bernoulli's Theorem, Torricelli, Venturimeter, and Flow Applications

Why Bernoulli's Theorem and Torricelli Efflux Are the Most Directly Tested Fluid Dynamics Formulas

Equation of Continuity (from Aakash PDF — Properties of Solids and Liquids JEE Main):

For an ideal (incompressible, non-viscous) fluid in steady flow:

a₁v₁ = a₂v₂ (conservation of mass/flow rate Q = av = constant)

where a = cross-sectional area, v = flow speed. Narrower tube → higher speed.

Bernoulli's Theorem (from Aakash PDF — Properties of Solids and Liquids JEE Main):

For ideal (non-viscous, incompressible) fluid in steady flow along a streamline:

P + ½ρv² + ρgh = constant

This is the energy per unit volume conservation: pressure energy + kinetic energy + potential energy = constant.

Energy heads: P/ρg (pressure head) + v²/2g (velocity head) + h (potential head) = constant.

Application: at two points on the same streamline:

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

Torricelli's Theorem (from Aakash PDF — Properties of Solids and Liquids JEE Main):

Hole at depth h below water surface in a tank, water surface area >> hole area:

Efflux velocity: vₑ = √(2gh) (same as free-fall velocity from height h)

Horizontal range of efflux jet: The jet hits the ground (at distance H–h below the hole) at horizontal distance:

x = vₑ × t = √(2gh) × √(2(H–h)/g) = 2√(h(H–h))

Maximum range: dx/dh = 0 → h = H/2 → x_max = H (at depth H/2, range = H = full tank height)

Equal ranges: holes at depths h and (H–h) give equal ranges (symmetric about midpoint).

Venturimeter (from Aakash PDF — Properties of Solids and Liquids JEE Main):

Pipe with wide section (area a₁, pressure P₁) and narrow throat (area a₂, pressure P₂). Both horizontal:

By continuity: a₁v₁ = a₂v₂. By Bernoulli: P₁+½ρv₁² = P₂+½ρv₂².

Flow rate Q = a₁a₂√(2(P₁–P₂)/ρ(a₁²–a₂²)) = a₁a₂√(2g(h₁–h₂)/(a₁²–a₂²))

where (h₁–h₂) = height difference in manometer limbs.

Magnus Effect and Aerofoil Lift (from Aakash PDF — Properties of Solids and Liquids):

Magnus effect: rotating cylinder in fluid develops lift due to Bernoulli effect (higher speed on one side = lower pressure). Aerofoil: upper surface curved (higher speed) → lower pressure; lower surface flat (lower speed) → higher pressure → net upward lift force. Download the Free PDF for Properties of Solids and Liquids for all Bernoulli examples for JEE Main.

Viscosity — Newton's Law, Stokes Law, Terminal Velocity, Poiseuille's Equation

Why Terminal Velocity and Poiseuille's Equation Are Direct Properties of Solids and Liquids JEE Main Formulas

Newton's Law of Viscosity (from Aakash PDF — Properties of Solids and Liquids JEE Main):

Viscous force between adjacent fluid layers:

F = ηA(dv/dy)

where η = coefficient of viscosity (Pa·s = kg/m·s = poise×10), A = area of layer, dv/dy = velocity gradient (rate of change of velocity with distance perpendicular to flow).

1 Pa·s = 10 poise; 1 poise = 1 dyne·s/cm² (CGS). Dimension of η: [ML⁻¹T⁻¹].

Viscosity decreases with temperature for liquids (molecular cohesion reduces); increases for gases (molecular momentum transfer increases).

Stokes' Law (from Aakash PDF — Properties of Solids and Liquids JEE Main):

For a small sphere of radius r moving with velocity v through a fluid of viscosity η:

F_viscous = 6πηrv (Stokes' drag force)

Terminal Velocity (from Aakash PDF — Properties of Solids and Liquids JEE Main):

When a sphere of radius r, density ρ falls through a fluid of density σ and viscosity η, terminal velocity vT is reached when net downward force = viscous drag:

mg – F_buoy = F_viscous → (4/3)πr³ρg – (4/3)πr³σg = 6πηrvT

vT = 2r²(ρ–σ)g / 9η

If ρ > σ: vT > 0 (body sinks). If ρ < σ: vT < 0 (body rises — bubbles in water).

vT ∝ r² (larger spheres reach higher terminal velocity); vT ∝ 1/η (less viscous → faster).

Poiseuille's Equation (from Aakash PDF — Properties of Solids and Liquids JEE Main):

Volume flow rate Q through a cylindrical pipe of radius r, length l, under pressure difference P, fluid viscosity η:

Q = dV/dt = πPr⁴/8ηl

Flow rate ∝ r⁴ (very sensitive to radius!). Double the radius → 16× the flow rate.

Hydraulic resistance: R_h = 8ηl/πr⁴ (analogous to electrical resistance R = ρl/A).

Series combination (same fluid): R_total = R₁+R₂ → l_effective/r⁴_effective = l₁/r₁⁴ + l₂/r₂⁴

Parallel combination: 1/R_total = 1/R₁ + 1/R₂

Reynolds Number (from Aakash PDF — Properties of Solids and Liquids JEE Main):

N_R = ρvD/η (dimensionless; D = diameter, v = average flow speed)

N_R < 2000: laminar (streamline) flow. N_R > 3000: turbulent flow. 2000 < N_R < 3000: unstable (transition).

Critical velocity (onset of turbulence): vₒ = 2000η/(ρD)

Download the Free PDF for Properties of Solids and Liquids for all viscosity examples for JEE Main.

Thermal Expansion, Heat Transfer, Stefan's Law, Newton's Cooling, Wien's Law

Why Thermal Resistance in Series/Parallel and Newton's Cooling Are the Most-Tested Thermal Properties Results

Thermal Expansion (from Aakash PDF — Properties of Solids and Liquids JEE Main):

Linear expansion: L = L₀(1 + αΔT) → ΔL = L₀αΔT

Area expansion: A = A₀(1 + βΔT) where β = 2α

Volume expansion: V = V₀(1 + γΔT) where γ = 3α

These relations hold for isotropic solids. For anisotropic: α, β, γ may differ in different directions.

Density on heating: ρ = ρ₀/(1+γΔT) ≈ ρ₀(1–γΔT)

Specific Heat and Latent Heat (from Aakash PDF — Properties of Solids and Liquids):

Q = mcΔT (where c = specific heat capacity J/kg/K)

Molar heat capacity C = Mc (M = molar mass); Q = nCΔT (n = moles).

Latent heat: Q = mL_f (fusion) or Q = mL_v (vaporisation) — no temperature change during phase transition.

For water: c = 1 cal/g/°C = 4200 J/kg/K; L_f = 80 cal/g = 336 J/g; L_v = 540 cal/g = 2268 J/g.

For ice: c_ice = 0.5 cal/g/°C = 2100 J/kg/K.

Fourier's Law of Thermal Conduction (from Aakash PDF — Properties of Solids and Liquids JEE Main):

Rate of heat flow through a slab of area A, thickness l, thermal conductivity k, temperature difference ΔT:

dQ/dt = kA(T₁–T₂)/l = kAΔT/l

Thermal resistance: R_th = l/(kA) (analogous to electrical R = ρl/A)

Series combination of rods (same area A, lengths l₁ and l₂, conductivities k₁ and k₂):

R = R₁+R₂ → l_total/k_eff = l₁/k₁ + l₂/k₂ → k_eff = (l₁+l₂)/(l₁/k₁+l₂/k₂)

For equal lengths: k_eff = 2k₁k₂/(k₁+k₂) (harmonic mean)

Junction temperature: T = (k₁T₁/l₁ + k₂T₂/l₂)/(k₁/l₁+k₂/l₂)

Parallel combination of rods (same length l, areas A₁ and A₂, conductivities k₁ and k₂):

k_eff = (k₁A₁+k₂A₂)/(A₁+A₂)

For equal areas: k_eff = (k₁+k₂)/2 (arithmetic mean)

Stefan-Boltzmann Law (from Aakash PDF — Properties of Solids and Liquids JEE Main):

P = σAεT⁴ where σ = 5.67×10⁻⁸ W/m²/K⁴ (Stefan's constant), ε = emissivity (0 to 1; ε=1 for black body).

Net heat loss rate for body at T in surroundings at T₀:

dQ/dt = σAε(T⁴–T₀⁴)

Rate of cooling (dT/dt): dT/dt = –(3σε/ρsr)(T⁴–T₀⁴) [for sphere of radius r, density ρ, specific heat s]

Newton's Law of Cooling (from Aakash PDF — Properties of Solids and Liquids JEE Main):

When temperature difference between body and surroundings is small (|T–T₀| small):

dT/dt = –b(T–T₀)

Solution: T = T₀ + (Tᵢ–T₀)e^(–bt)

Average form: (T₁–T₂)/t ∝ [(T₁+T₂)/2 – T₀] (useful for JEE Main comparison problems)

Log form: ln(T–T₀) = –bt + constant → straight line on ln(T–T₀) vs t graph.

Wien's Displacement Law (from Aakash PDF — Properties of Solids and Liquids JEE Main):

λₘT = b = 2.898×10⁻³ m·K

λₘ = wavelength of peak intensity. Higher temperature → shorter λₘ (body appears bluer at higher T). Area under spectral curve = total emissive power ∝ T⁴ (Stefan's law). Download the Free PDF for Properties of Solids and Liquids for all thermal examples for JEE Main.