JEE Main & Advanced

Properties of Matter

Properties of Matter for JEE Main & Advanced

1
Module 1

Elasticity and Fluid Mechanics

Stress, Strain, Elastic ModuliTopic 1

Stress = restoring force per unit area: $\sigma = F/A$ (N/m² or Pa). Strain = fractional change in dimension (dimensionless).

TypeStressStrainModulus
Longitudinal (tensile/compressive)$F/A$$\Delta L/L$Young's Modulus $Y = \dfrac{FL}{A\Delta L}$
Volumetric$\Delta P$$-\Delta V/V$Bulk Modulus $B = -\dfrac{V\Delta P}{\Delta V}$
Shear (tangential)$F/A$$\theta$ (rad)Shear Modulus $\eta = \dfrac{F}{A\theta}$

Hooke's Law: Within elastic limit, stress $\propto$ strain. Compressibility $K = 1/B$. Poisson's ratio $\sigma = -\dfrac{\text{lateral strain}}{\text{longitudinal strain}}$; $-1 < \sigma < 0.5$.

Elastic Potential Energy stored per unit volume: $u = \dfrac{1}{2} \times \text{stress} \times \text{strain} = \dfrac{1}{2}Y(\text{strain})^2$.

For a wire: $U = \dfrac{1}{2} \times \dfrac{YA(\Delta L)^2}{L}$. Force constant of wire: $k = YA/L$.

Stress-Strain Curve: Proportional limit → Elastic limit → Yield point → Ultimate strength → Fracture point. Region beyond elastic limit shows permanent (plastic) deformation.

Worked Examples
1

A steel wire of length $2$ m and cross-section $1$ mm² is stretched by $0.1$ mm under a load. Find the load. $Y_{\text{steel}} = 2 \times 10^{11}$ Pa.

Show solution

$$F = \frac{YA\Delta L}{L} = \frac{2 \times 10^{11} \times 10^{-6} \times 10^{-4}}{2} = \frac{2 \times 10^{1}}{2} = 10 \text{ N}$$

Final Answer: $F = 10$ N.

2

A wire of length $L$, area $A$ is stretched by force $F$. Find the elastic energy stored if Young's modulus is $Y$.

Show solution

Strain $= F/(AY)$, $\Delta L = FL/(AY)$. $$U = \frac{1}{2}F \cdot \Delta L = \frac{1}{2}F \cdot \frac{FL}{AY} = \frac{F^2 L}{2AY}$$

Final Answer: $U = F^2L/(2AY)$.

✎ Self-Check — 5 questions0 / 5
Q1.

The dimensional formula for Young's modulus is:

Q2.

The Poisson's ratio of a material is the ratio of:

Q3.

A wire elongates by $1$ mm under a load. Another wire of same material, double the length, half the area, under the same load elongates by:

Q4.

The work done in stretching a wire is stored as:

Q5.

Bulk modulus is reciprocal of:

Pressure, Buoyancy, Bernoulli's EquationTopic 2

Pressure in a fluid at depth $h$: $P = P_0 + \rho gh$, where $P_0$ = atmospheric pressure. Pascal's law: Pressure applied to enclosed fluid transmits equally in all directions (basis of hydraulic press).

Archimedes' Principle: Buoyant force = weight of displaced fluid. $$F_B = V_{\text{submerged}} \cdot \rho_{\text{fluid}} \cdot g$$

For floating body: $\rho_{\text{body}} V_{\text{total}} = \rho_{\text{fluid}} V_{\text{submerged}}$.

Equation of Continuity (incompressible flow): $A_1 v_1 = A_2 v_2$ (mass conservation).

Bernoulli's Equation (steady, incompressible, non-viscous, streamline flow): $$P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}$$

Sum of pressure energy, kinetic energy, and potential energy per unit volume is constant along a streamline.

Torricelli's Law: Speed of efflux through small hole at depth $h$ below free surface: $$v = \sqrt{2gh}$$ (same as free-fall from height $h$).

Venturimeter: Used to measure flow rate. From Bernoulli + continuity: $$v_1 = a_2\sqrt{\frac{2gh}{a_1^2 - a_2^2}}$$ where $h$ = manometer height difference; $a_1, a_2$ = pipe and throat areas.

Worked Examples
1

A tank of height $5$ m is filled with water. A small hole is made at depth $4$ m. Find the speed of efflux. ($g = 10$ m/s²)

Show solution

By Torricelli: $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 4} = \sqrt{80} = 4\sqrt{5} \approx 8.94$ m/s.

Final Answer: $v \approx 8.94$ m/s.

2

A block of wood floats with $3/5$ of its volume submerged in water. Find its density.

Show solution

For floating: $\rho_{\text{wood}} V = \rho_w V_{\text{submerged}}$. $$\rho_{\text{wood}} \cdot V = 1000 \cdot \frac{3V}{5} \implies \rho_{\text{wood}} = 600 \text{ kg/m}^3$$

Final Answer: $\rho_{\text{wood}} = 600$ kg/m³.

✎ Self-Check — 5 questions0 / 5
Q1.

The unit of pressure in SI is:

Q2.

According to Bernoulli's equation, increasing fluid velocity:

Q3.

A body floats with $1/4$ of its volume above water. Its density:

Q4.

Pascal's law is used in:

Q5.

The velocity of efflux of a liquid through a hole at the bottom of a tank of height $h$ is:

2
Module 2

Viscosity, Surface Tension and Thermal Properties

Viscosity, Stokes' Law, Surface TensionTopic 1

Viscosity: Internal friction in a fluid. Newton's law of viscous flow: $F = \eta A \dfrac{dv}{dx}$, where $\eta$ = coefficient of viscosity (Pa·s = $10$ poise). Velocity gradient $dv/dx$ exists between adjacent fluid layers.

Stokes' Law: Viscous drag on a sphere of radius $r$ moving with velocity $v$ in fluid of viscosity $\eta$: $$F = 6\pi\eta r v$$

Terminal Velocity of a sphere falling in a viscous fluid (when net force = 0): $$v_T = \frac{2r^2(\rho - \sigma)g}{9\eta}$$ where $\rho$ = sphere density, $\sigma$ = fluid density.

Reynolds Number $Re = \dfrac{\rho v d}{\eta}$. Flow is laminar if $Re < 2000$, turbulent if $Re > 3000$.

Surface Tension ($T$ or $S$): Force per unit length acting on liquid surface, tangentially. Units: N/m. Equivalently, surface energy per unit area (J/m²).

Excess Pressure:

Drop typeExcess pressure
Liquid drop$\Delta P = 2T/R$
Bubble inside liquid$\Delta P = 2T/R$
Soap bubble (2 surfaces)$\Delta P = 4T/R$

Capillary Rise (Jurin's law): $h = \dfrac{2T\cos\theta}{\rho gr}$, where $r$ = capillary radius, $\theta$ = contact angle. Mercury falls in glass capillary ($\theta > 90°$).

Worked Examples
1

A spherical drop of water of radius $1$ mm falls in air. Coefficient of viscosity of air = $1.8 \times 10^{-5}$ Pa·s. Find terminal velocity. ($\rho_w = 1000$, $\rho_{\text{air}} \approx 0$, $g = 10$)

Show solution

$$v_T = \frac{2r^2 \rho g}{9\eta} = \frac{2 \times (10^{-3})^2 \times 1000 \times 10}{9 \times 1.8 \times 10^{-5}}$$ $$= \frac{2 \times 10^{-2}}{1.62 \times 10^{-4}} \approx 123.5 \text{ m/s}$$

(Note: In reality, air resistance becomes turbulent; Stokes' law breaks down — this is an idealised value.)

Final Answer: $v_T \approx 123$ m/s (Stokes' regime).

2

Find the excess pressure inside a soap bubble of radius $1$ cm. Surface tension of soap solution = $0.03$ N/m.

Show solution

For soap bubble (two surfaces): $$\Delta P = \frac{4T}{R} = \frac{4 \times 0.03}{10^{-2}} = 12 \text{ Pa}$$

Final Answer: $\Delta P = 12$ Pa.

✎ Self-Check — 5 questions0 / 5
Q1.

SI unit of coefficient of viscosity:

Q2.

Terminal velocity of a sphere in viscous fluid is proportional to:

Q3.

Excess pressure inside a soap bubble of radius $R$:

Q4.

The angle of contact for water on glass is:

Q5.

Two soap bubbles of radii $r$ and $2r$ are joined. The radius of the curvature of common interface:

Thermal Expansion, Calorimetry, Heat TransferTopic 2

Thermal Expansion:

TypeFormula
Linear$\Delta L = L_0 \alpha \Delta T$
Areal$\Delta A = A_0 \beta \Delta T$, $\beta \approx 2\alpha$
Volume$\Delta V = V_0 \gamma \Delta T$, $\gamma \approx 3\alpha$

Bimetallic strip bends due to different $\alpha$ values; used in thermostats.

Anomalous expansion of water: Maximum density at $4°$C; expansion below $4°$C.

Calorimetry:

  • Heat absorbed/released: $Q = mc\Delta T$, where $c$ = specific heat capacity (J/kg·K).
  • Latent heat: $Q = mL$ during phase change (constant temperature).
  • Principle of calorimetry: heat lost = heat gained (in mixing).
  • Specific heat of water = $4186$ J/kg·K; latent heat of fusion of ice = $3.36 \times 10^5$ J/kg; vaporisation of water = $2.26 \times 10^6$ J/kg.

Heat Transfer Mechanisms:

ModeDescriptionEquation
ConductionThrough solids$\dfrac{dQ}{dt} = -KA\dfrac{dT}{dx}$
ConvectionThrough fluids by bulk motion$\dfrac{dQ}{dt} = hA\Delta T$
RadiationEM waves; Stefan-Boltzmann$P = e\sigma A T^4$

Thermal resistance: $R_{\text{th}} = L/(KA)$, similar to electrical resistance. For series: $R_{\text{eq}} = R_1 + R_2$; parallel: $1/R_{\text{eq}} = 1/R_1 + 1/R_2$.

Wien's Displacement Law: $\lambda_{\max} T = b$, where $b = 2.898 \times 10^{-3}$ m·K.

Newton's Law of Cooling: $\dfrac{dT}{dt} = -k(T - T_s)$, valid for small temperature differences.

Worked Examples
1

A steel rod of length $1$ m at $20°$C is heated to $120°$C. Find the change in length. $\alpha_{\text{steel}} = 11 \times 10^{-6}$/°C.

Show solution

$$\Delta L = L_0 \alpha \Delta T = 1 \times 11 \times 10^{-6} \times 100 = 11 \times 10^{-4} \text{ m} = 1.1 \text{ mm}$$

Final Answer: $\Delta L = 1.1$ mm.

2

$100$ g of ice at $0°$C is added to $200$ g of water at $50°$C. Find the final temperature. ($L_f = 80$ cal/g, $c_w = 1$ cal/g·°C)

Show solution

Heat to melt ice: $Q_1 = 100 \times 80 = 8000$ cal. Heat available from water cooling to $0°$C: $Q_2 = 200 \times 1 \times 50 = 10000$ cal.

Since $Q_2 > Q_1$, all ice melts. Remaining heat = $10000 - 8000 = 2000$ cal.

This heats the $300$ g of water (now at $0°$C) to $T$: $$2000 = 300 \times 1 \times T \implies T = 6.67°\text{C}$$

Final Answer: $T \approx 6.67°$C.

✎ Self-Check — 5 questions0 / 5
Q1.

The coefficient of areal expansion is related to linear:

Q2.

SI unit of thermal conductivity:

Q3.

Specific heat of water in cal/g·°C:

Q4.

According to Stefan's law, the rate of energy emission is proportional to:

Q5.

Newton's law of cooling is a special case of:

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