Properties of Matter
Properties of Matter for JEE Main & Advanced
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).
| Type | Stress | Strain | Modulus |
|---|---|---|---|
| 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.
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.
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: $U = F^2L/(2AY)$.
The dimensional formula for Young's modulus is:
The Poisson's ratio of a material is the ratio of:
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:
The work done in stretching a wire is stored as:
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.
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.
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.
The unit of pressure in SI is:
According to Bernoulli's equation, increasing fluid velocity:
A body floats with $1/4$ of its volume above water. Its density:
Pascal's law is used in:
The velocity of efflux of a liquid through a hole at the bottom of a tank of height $h$ is:
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 type | Excess 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 ($\theta > 90°$).
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).
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.
SI unit of coefficient of viscosity:
Terminal velocity of a sphere in viscous fluid is proportional to:
Excess pressure inside a soap bubble of radius $R$:
The angle of contact for water on glass is:
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:
| Type | Formula |
|---|---|
| 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:
| Mode | Description | Equation |
|---|---|---|
| Conduction | Through solids | $\dfrac{dQ}{dt} = -KA\dfrac{dT}{dx}$ |
| Convection | Through fluids by bulk motion | $\dfrac{dQ}{dt} = hA\Delta T$ |
| Radiation | EM 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.
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.
$100$ 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.
The coefficient of areal expansion is related to linear:
SI unit of thermal conductivity:
Specific heat of water in cal/g·°C:
According to Stefan's law, the rate of energy emission is proportional to:
Newton's law of cooling is a special case of:
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