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WBB • Class XI • Physics • Ch 7
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Properties of Bulk Matter

Properties of Bulk Matter investigates the macroscopic physical and thermal behavior of solids, liquids, and gases as continuous media. Rather than treating substances as isolated particles, bulk matter physics examines how interatomic and intermolecular forces manifest at observable engineering scales. The chapter commences with the mechanical response of elastic solids under mechanical loading, establishing Hooke's Law, stress-strain constitutive relations, elastic moduli, and strain energy density. Moving to the fluid state, we formulate hydrostatics through Pascal's transmission principle, hydrostatic pressure variation, and Archimedes' buoyant equilibrium. We then transition into hydrodynamics, analyzing Newtonian viscosity, Stokes' viscous drag on spheres, terminal velocity, and the governing conservation laws of fluid dynamics: the Equation of Continuity and Bernoulli's Principle. We explore interfacial phenomena including cohesive and adhesive forces, surface tension, excess Laplace pressure in bubbles and drops, contact angle wettability, and capillary ascent. Finally, we establish the thermodynamic and heat transfer foundations of bulk matter, analyzing linear and cubical thermal expansion, calorimetry, Fourier conduction, Stefan-Boltzmann radiation, and Newton's empirical law of cooling. This comprehensive curriculum equips West Bengal Board (WBCHSE) Class 11 students with a rigorous analytical framework indispensable for higher secondary board examinations and competitive engineering and medical entrance tests like WBJEE, JEE Main, and NEET.

Why This Chapter Matters

Every structural and technological system in human civilization operates under the principles of bulk matter physics. Civil and mechanical engineers rely on Young's modulus, shear modulus, and factor of safety to design earthquake-resistant skyscrapers, suspension bridges, and aircraft fuselages that deform elastically without catastrophic fracture. Hydraulic presses, vehicle braking systems, and heavy excavators multiply forces using Pascal's principle. Aerodynamic wing lift, carburetors, and medical ventilators function directly on Bernoulli's theorem and laminar fluid dynamics. Capillary action enables trees to draw water from subterranean roots over a hundred meters up to canopy leaves. Thermal expansion tolerances dictate the expansion gaps in railway tracks and highway bridges, while heat transfer calculations govern power plant steam boilers, microchip heat sinks, and atmospheric weather patterns.

Chapter Roadmap & Progression

1 1. Elastic Behavior of Solids, Stre...
2 2. Fluid Statics, Pascal's Principl...
3 3. Viscosity, Stokes' Drag Law & Te...
4 4. Hydrodynamics: Continuity Equati...
5 5. Surface Tension, Surface Energy...
6 6. Thermal Expansion, Calorimetry &...

Complete Concept Guide (100% Curriculum Coverage)

1. Elastic Behavior of Solids, Stress-Strain Relations & Moduli

Concept of Elasticity and Plasticity

Elasticity is the intrinsic property of a material body by virtue of which it regains its original shape and size upon removal of deforming external forces. When deforming forces displace atoms from their equilibrium lattice positions, internal restorative interatomic forces develop. Plasticity is the property whereby a body fails to regain its initial dimensions, retaining permanent deformation.

Stress and Strain Formulations

Stress ($\\sigma$): The internal restoring force developed per unit cross-sectional area of the deformed body:

$$\sigma = \frac{F_{\text{restoring}}}{A} \approx \frac{F_{\text{external}}}{A}$$

SI Unit: $\text{N/m}^2$ or Pascal ($\text{Pa}$). Dimensional formula: $[\text{M}\text{L}^{-1}\text{T}^{-2}]$. Types of stress include:

  • Longitudinal Stress: Normal force per unit area producing change in length (Tensile stress causes elongation; Compressive stress causes shortening).
  • Shearing / Tangential Stress: Deforming force acting tangentially to the surface, altering shape at constant volume.
  • Volumetric / Hydraulic Stress: Uniform perpendicular fluid pressure acting inward over the entire surface, changing volume without changing shape.

Strain ($\epsilon$): The fractional change in dimension produced in a body by deforming forces:

$$\text{Strain} = \frac{\text{Change in dimension}}{\text{Original dimension}}$$

Strain is a dimensionless and unitless ratio. Longitudinal strain $\epsilon_l = \frac{\Delta L}{L}$; Shearing strain $\theta = \frac{\Delta x}{h} = \tan\theta$; Volume strain $\epsilon_v = \frac{\Delta V}{V}$.

Hooke's Law & The Complete Stress-Strain Curve

Hooke's Law: Within the elastic limit of a material, stress is directly proportional to strain:

$$\sigma \propto \epsilon \implies \frac{\sigma}{\epsilon} = E \quad (\text{Modulus of Elasticity})$$
Key Stages of the Stress-Strain Curve for a Ductile Metal (e.g., Mild Steel):
  1. Proportional Limit (OA): Stress is strictly proportional to strain. Hooke's law holds perfectly. The slope gives Young's modulus $Y$.
  2. Elastic Limit / Yield Point (AB): Material deforms elastically. Up to point B, the material completely recovers its original dimensions upon unloading. Stress at B is the yield strength ($S_y$).
  3. Plastic Deformation (BC): Beyond B, strain increases rapidly with little stress increase. Unloading from any point in this region leaves a permanent set (residual strain $0.1\%\text{--}0.2\%$).
  4. Ultimate Tensile Strength (D): The peak stress the material can sustain before necking starts. Beyond D, localized thinning occurs.
  5. Fracture / Breaking Point (E): The wire breaks. If D and E are far apart, the material is ductile (copper, aluminum); if D and E are very close, the material is brittle (cast iron, glass).
The Three Elastic Moduli & Poisson's Ratio
ModulusMathematical FormulaPhysical Meaning
Young's Modulus ($Y$)$$Y = \frac{\text{Longitudinal Stress}}{\text{Longitudinal Strain}} = \frac{F/A}{\Delta L/L} = \frac{F L}{A \Delta L}$$Resistance to stretching or compression (solids only).
Bulk Modulus ($B$)$$B = -\frac{\Delta P}{\Delta V / V} = -V\frac{\Delta P}{\Delta V}$$Resistance to volume compression under hydraulic pressure. Negative sign indicates volume decreases as pressure increases. Compressibility $K = 1/B$.
Shear Modulus / Rigidity ($\eta$)$$\eta = \frac{\text{Shearing Stress}}{\text{Shearing Strain}} = \frac{F/A}{\theta}$$Resistance to shape distortion at constant volume (solids only; fluids have $\eta = 0$).

Poisson's Ratio ($\sigma$): The ratio of lateral strain to longitudinal strain:

$$\sigma = \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} = -\frac{\Delta d / d}{\Delta L / L}$$

Theoretical limits: $-1 \le \sigma \le 0.5$. Practical limits for most engineering materials: $0 \le \sigma \le 0.5$ (e.g., steel $\approx 0.28$, rubber $\approx 0.49$).

Elastic Potential Energy & Strain Energy Density

When a wire of length $L$ and area $A$ is stretched by length $\Delta L$, work is done against internal elastic forces. For an elongation $x$, restoring force is $F = \frac{Y A x}{L}$:

$$W = \int_0^{\Delta L} F\,dx = \int_0^{\Delta L} \frac{YA x}{L}\,dx = \frac{1}{2}\frac{YA(\Delta L)^2}{L} = \frac{1}{2} F \Delta L$$

The Elastic Strain Energy Density ($u$, energy per unit volume):

$$u = \frac{W}{A L} = \frac{1}{2}\left(\frac{F}{A}\right)\left(\frac{\Delta L}{L}\right) = \frac{1}{2} \times \text{Stress} \times \text{Strain} = \frac{\sigma^2}{2Y} = \frac{1}{2}Y \epsilon^2$$

2. Fluid Statics, Pascal's Principle & Archimedes' Law

Fluid Pressure & Hydrostatic Variation

A fluid is a substance that cannot sustain a static shear stress and deforms continuously under tangential forces. Normal force exerted by a fluid per unit surface area is hydrostatic pressure:

$$P = \frac{dF_\perp}{dA}$$

SI Unit: Pascal ($1\,\text{Pa} = 1\,\text{N/m}^2$). Standard atmospheric pressure: $1\,\text{atm} = 1.01325 \times 10^5\,\text{Pa} = 760\,\text{mm of Hg} = 1.013\,\text{bar}$.

Pressure Variation with Depth: Consider a cylindrical fluid element of cross-sectional area $A$ and height $h$ in a static fluid of uniform density $\rho$. Vertical force equilibrium requires:

$$P_2 A - P_1 A - m g = 0 \implies P_2 A = P_1 A + \rho (A h) g \implies P_2 = P_1 + \rho g h$$

If point 1 is at the open liquid surface exposed to atmospheric pressure $P_0$, the absolute pressure at depth $h$ is:

$$P = P_0 + \rho g h$$

Gauge Pressure: The excess pressure over atmospheric pressure: $P_{\text{gauge}} = P - P_0 = \rho g h$.

Pascal's Law & Hydraulic Force Multiplication

Pascal's Principle: Any change in pressure applied to an enclosed, incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of its containing vessel.

Hydraulic Lift Mechanics: Consider two interconnected vertical cylinders fitted with watertight frictionless pistons of cross-sectional areas $A_1$ (small) and $A_2$ (large). Applying a downward force $F_1$ on piston 1 generates pressure $\Delta P = \frac{F_1}{A_1}$. By Pascal's principle, this identical pressure is exerted upward on piston 2:$$\Delta P = \frac{F_2}{A_2} \implies F_2 = F_1 \left(\frac{A_2}{A_1}\right)$$
Since $A_2 \gg A_1$, the output force is magnified by the area ratio $\frac{A_2}{A_1}$. By volume conservation, $A_1 d_1 = A_2 d_2$, meaning work done is conserved: $W_1 = F_1 d_1 = F_2 d_2 = W_2$ (no violation of energy conservation).
Archimedes' Principle & The Laws of Flotation

Archimedes' Principle: When a body is wholly or partially immersed in a fluid at rest, it experiences an upward buoyant force (upthrust) equal in magnitude to the weight of the fluid displaced by the body:

$$F_B = V_{\text{submerged}} \cdot \rho_{\text{fluid}} \cdot g$$

The buoyant force acts vertically upward through the Center of Buoyancy, which is the center of gravity of the displaced fluid.

Conditions for Flotation: For a body of mass $M$, volume $V$, and average density $\rho_b$ placed in a fluid of density $\rho_f$:

  • Sinks: $\rho_b > \rho_f \implies W > F_{B,\max}$. The body accelerates downward to the bottom. Apparent weight: $W_{\text{app}} = W - F_B = V(\rho_b - \rho_f)g$.
  • Neutral Equilibrium: $\rho_b = \rho_f \implies W = F_{B,\max}$. The body remains suspended at any depth where it is placed.
  • Floats: $\rho_b < \rho_f$. The body floats partially submerged such that weight equals buoyant force: $V \rho_b g = V_{\text{sub}} \rho_f g \implies \frac{V_{\text{sub}}}{V} = \frac{\rho_b}{\rho_f}$.

3. Viscosity, Stokes' Drag Law & Terminal Velocity

Viscosity and Newton's Law of Viscous Force

Viscosity is the internal friction of a fluid that resists relative motion between adjacent parallel layers. In streamline laminar flow over a solid flat boundary, fluid layers move with velocities increasing with distance $x$ from the boundary.

Newton's Law of Viscosity: The tangential viscous force $F$ between two adjacent liquid layers of contact area $A$ is directly proportional to the area $A$ and the transverse velocity gradient $\frac{dv}{dx}$:

$$F = -\eta A \frac{dv}{dx}$$

where $\eta$ is the coefficient of dynamic viscosity. The negative sign denotes that viscous drag opposes relative motion.

  • SI Unit: $\text{N}\cdot\text{s/m}^2 = \text{Pa}\cdot\text{s}$ or Decapoise / Poiseuille ($\text{Pl}$).
  • CGS Unit: $\text{dyn}\cdot\text{s/cm}^2 = \text{Poise} \implies 1\,\text{Pa}\cdot\text{s} = 10\,\text{Poise}$.
  • Dimensional Formula: $[\eta] = \frac{[F]}{[A][dv/dx]} = \frac{[\text{MLT}^{-2}]}{[\text{L}^2][\text{LT}^{-1}/\text{L}]} = [\text{M}\text{L}^{-1}\text{T}^{-1}]$.
  • Temperature Dependence: For liquids, $\eta$ decreases exponentially with temperature (molecules gain thermal energy to overcome cohesive bonds). For gases, $\eta$ increases with temperature ($\eta \propto \sqrt{T}$, due to increased molecular momentum transport).
Stokes' Law for Viscous Drag on a Sphere

Sir George Gabriel Stokes showed that when a small smooth spherical body of radius $r$ moves with velocity $v$ through an infinite, homogeneous, stationary viscous medium of viscosity $\eta$, the retarding viscous force is:

$$F_v = 6\pi \eta r v$$

Conditions for validity: The sphere must be rigid and smooth, the fluid medium must be boundless compared to the sphere, and the motion must be non-turbulent with no slipping at the solid-fluid boundary.

Rigorous Derivation of Terminal Velocity ($v_t$)

Consider a small sphere of radius $r$ and density $\rho$ falling vertically under gravity through a viscous fluid of density $\sigma$ and viscosity $\eta$. As the sphere accelerates, downward speed $v$ increases, increasing the upward viscous drag $F_v$.

The three forces acting on the sphere are:

  1. Downward true gravitational weight: $W = m g = \left(\frac{4}{3}\pi r^3 \rho\right)g$
  2. Upward hydrostatic buoyant upthrust: $F_B = \left(\frac{4}{3}\pi r^3 \sigma\right)g$
  3. Upward retarding Stokes viscous drag: $F_v = 6\pi \eta r v$

Net downward force: $F_{\text{net}} = W - F_B - F_v = m\frac{dv}{dt}$.

When the retarding forces grow to balance net downward gravity, net acceleration becomes zero ($\\frac{dv}{dt} = 0$), and the sphere continues falling at a constant maximum speed called the terminal velocity ($v_t$):

$$W = F_B + F_v \implies \frac{4}{3}\pi r^3 \rho g = \frac{4}{3}\pi r^3 \sigma g + 6\pi \eta r v_t$$$$6\pi \eta r v_t = \frac{4}{3}\pi r^3 (\rho - \sigma) g$$$$v_t = \frac{2}{9}\frac{r^2(\rho - \sigma)g}{\eta}$$
Key Insights:
  • Terminal velocity is directly proportional to the square of the radius ($v_t \propto r^2$). Larger raindrops fall much faster than fine mist droplets.
  • If $\rho > \sigma$, $v_t > 0$ (the body falls downward). If $\rho < \sigma$ (e.g., air bubbles in water), $v_t < 0$, meaning the bubble ascends with terminal speed.
Reynolds Number & Turbulence Transition

The nature of flow through a pipe of diameter $D$ is characterized by the dimensionless Reynolds Number ($R_e$), which represents the ratio of inertial forces to viscous forces:

$$R_e = \frac{\text{Inertial Force}}{\text{Viscous Force}} = \frac{\rho v D}{\eta}$$
  • $R_e < 1000$: Flow is purely laminar (streamline, orderly).
  • $1000 < R_e < 2000$: Flow is unsteady and transitional.
  • $R_e > 2000$: Flow becomes turbulent (chaotic, eddying, mixing).

4. Hydrodynamics: Continuity Equation & Bernoulli's Principle

Streamline Flow & The Equation of Continuity

In streamline flow, every particle passing a given point follows the exact same path and possesses the exact same velocity as preceding particles. A bundle of streamlines forms a tube of flow across which fluid cannot cross.

Equation of Continuity: For steady, streamline flow of an incompressible non-viscous fluid through a tube of non-uniform cross-section:

$$A_1 v_1 = A_2 v_2 = \text{constant} \implies A v = \text{constant}$$

This expresses the Conservation of Mass: the mass flow rate $\frac{dm}{dt} = \rho A v$ remains invariant along the stream. Where the pipe narrows, flow velocity increases proportionally ($v \propto 1/A$).

Bernoulli's Theorem: Energy Conservation in Fluid Flow

Statement: For the steady, streamline flow of an ideal fluid (incompressible, non-viscous, and irrotational), the total mechanical energy per unit volume—comprising pressure energy, kinetic energy, and gravitational potential energy—remains strictly constant along any streamline:

$$P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}$$

Dividing throughout by $\rho g$ yields the Head Formulation (each term with dimensions of length $[\text{L}]$):

$$\frac{P}{\rho g} + \frac{v^2}{2g} + h = \text{constant}$$
  • $\frac{P}{\rho g} = \text{Pressure Head}$
  • $\frac{v^2}{2g} = \text{Velocity Head}$
  • $h = \text{Elevation / Potential Head}$
Proof of Bernoulli's Theorem via Work-Energy Theorem:
Consider a pipe section from position 1 ($A_1, P_1, v_1, h_1$) to position 2 ($A_2, P_2, v_2, h_2$). In small time $\Delta t$, mass $\Delta m = \rho A_1 v_1 \Delta t = \rho A_2 v_2 \Delta t$ enters at 1 and exits at 2.
Work done by pressure forces: $W = P_1 A_1 (v_1 \Delta t) - P_2 A_2 (v_2 \Delta t) = (P_1 - P_2)\frac{\Delta m}{\rho}$.
Change in kinetic energy: $\Delta K = \frac{1}{2}\Delta m (v_2^2 - v_1^2)$.
Change in potential energy: $\Delta U = \Delta m g (h_2 - h_1)$.
By the Work-Energy Theorem, $W = \Delta K + \Delta U$:
$$(P_1 - P_2)\frac{\Delta m}{\rho} = \frac{1}{2}\Delta m (v_2^2 - v_1^2) + \Delta m g (h_2 - h_1)$$
Dividing by $\Delta m / \rho$ and rearranging:
$$P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2 \implies P + \frac{1}{2}\rho v^2 + \rho g h = \text{const}$$
Major Applications of Bernoulli's Theorem
  1. Venturimeter: A device to measure liquid flow rate in a horizontal pipe ($h_1 = h_2$). By continuity $A_1 v_1 = A_2 v_2$. Bernoulli yields $P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2) = \rho_m g h$. The volume rate of flow is:$$Q = A_1 v_1 = A_1 A_2 \sqrt{\frac{2gh}{\left(\frac{\rho_m}{\rho}\right)(A_1^2 - A_2^2)}}$$
  2. Torricelli's Law of Efflux: For an open tank with a small orifice at depth $h$ below the free surface, the speed of efflux is identical to the speed acquired in free fall from height $h$:$$v = \sqrt{2gh}$$

    Horizontal range of efflux on ground at depth $H$: $R = v t = \sqrt{2gh}\sqrt{\frac{2(H-h)}{g}} = 2\sqrt{h(H-h)}$, with maximum range $R_{\max} = H$ achieved when the hole is drilled at mid-depth $h = H/2$.

  3. Aerodynamic Dynamic Lift: Airfoil geometry forces air to travel faster over the curved top surface ($v_{\text{top}} > v_{\text{bottom}}$), lowering pressure on top ($P_{\text{top}} < P_{\text{bottom}}$) and generating net upward dynamic lift.
  4. Magnus Effect: A spinning cricket or tennis ball drags air layers, creating velocity differential and asymmetric pressure that deflects the ball's trajectory in flight.

5. Surface Tension, Surface Energy & Capillarity

Molecular Theory of Surface Tension

Liquid molecules exert mutual attractive intermolecular forces within a sphere of molecular activity (radius $\approx 10^{-9}\,\text{m}$).

  • Interior Molecule: Completely surrounded by neighbors, experiencing zero net resultant intermolecular force.
  • Surface Film Molecule: Lacks upward neighbors, experiencing a net inward downward cohesive pull toward the bulk liquid.

Consequently, the liquid surface contracts to minimize its surface area, behaving like a stretched elastic membrane. Surface Tension ($T$ or $S$) is defined as the force acting per unit length of an imaginary line drawn on the liquid surface perpendicular to the line:

$$T = \frac{F}{L}$$

SI Unit: $\text{N/m}$. Dimensions: $[\text{M}\text{T}^{-2}]$.

Surface Energy & Work Done in Expanding Interfaces

Bringing molecules from the interior to the surface against the net inward pull requires work, stored as potential energy in the surface film. Surface energy per unit area is numerically equal to surface tension:

$$W = T \Delta A$$

Units: $\text{J/m}^2 = \text{N/m}$.

Excess Pressure (Young-Laplace Equation)

Because of surface tension, the pressure on the concave side of a curved liquid surface is always strictly greater than on the convex side.

Curved Interface SystemNumber of Free SurfacesExcess Pressure ($\Delta P$)
Liquid Droplet in Air1 (outer liquid-air boundary)$$\Delta P = P_{\text{in}} - P_{\text{out}} = \frac{2T}{R}$$
Air Bubble inside Liquid1 (inner air-liquid boundary)$$\Delta P = \frac{2T}{R}$$
Soap Bubble in Air2 (inner and outer liquid-air films)$$\Delta P = \frac{4T}{R}$$
Angle of Contact & Capillary Ascent (Jurin's Law)

Angle of Contact ($\theta$): The angle between the tangent to the liquid surface at the point of contact and the solid surface inside the liquid.

  • $\theta < 90^\circ$ (Acute): Adhesive force exceeds cohesive force. Liquid wets the solid, forming a concave meniscus (e.g., pure water on clean glass $\theta \approx 0^\circ$).
  • $\theta > 90^\circ$ (Obtuse): Cohesive force dominates. Liquid does not wet solid, forming a convex meniscus (e.g., mercury on glass $\theta \approx 135^\circ\text{--}140^\circ$).

Capillary Ascent Formula: When a clean narrow capillary tube of radius $r$ is dipped vertically into a wetting liquid of density $\rho$, the upward vertical component of surface tension supports the weight of the elevated liquid column:

$$2\pi r T \cos\theta = m g = (\pi r^2 h \rho)g \implies h = \frac{2T\cos\theta}{r \rho g}$$

Jurin's Law: $h \cdot r = \frac{2T\cos\theta}{\rho g} = \text{constant}$. The narrower the capillary bore, the higher the liquid rises.

Capillary Tube of Insufficient Length ($l < h$): If a capillary tube is cut to a length $l$ shorter than the equilibrium ascent height $h$, liquid will never fountain or overflow! Instead, the liquid reaches the top rim and flattens its meniscus to a larger radius of curvature $R'$ such that $h R = l R'$, preserving hydrostatic equilibrium.

6. Thermal Expansion, Calorimetry & Modes of Heat Transfer

Thermal Expansion of Solids & The 1:2:3 Ratio

Thermal expansion occurs because increasing temperature enhances atomic vibrational amplitudes about an asymmetric potential energy well.

  • Linear Expansion: $\Delta L = L_0 \alpha \Delta T \implies L = L_0(1 + \alpha \Delta T)$
  • Superficial (Area) Expansion: $\Delta A = A_0 \beta \Delta T \implies A = A_0(1 + \beta \Delta T)$
  • Cubical (Volume) Expansion: $\Delta V = V_0 \gamma \Delta T \implies V = V_0(1 + \gamma \Delta T)$

For an isotropic solid body, differentiating dimensions reveals:

$$\beta = 2\alpha, \quad \gamma = 3\alpha \implies \alpha : \beta : \gamma = 1 : 2 : 3$$

Anomalous Expansion of Water: Between $0^\circ\text{C}$ and $4^\circ\text{C}$, water contracts on heating (negative volume expansion coefficient $\gamma < 0$). Water reaches its maximum density at $4^\circ\text{C}$ ($1000\,\text{kg/m}^3$). This anomaly causes frozen ice ($0^\circ\text{C}$) to float on top of lakes, insulating $4^\circ\text{C}$ water below and preserving aquatic ecosystems through harsh winters.

Calorimetry & Latent Heat

Principle of Calorimetry: In an isolated, thermally insulated system, heat lost by warmer bodies equals heat gained by colder bodies until thermal equilibrium is established:

$$Q_{\text{lost}} = Q_{\text{gained}}$$

Sensible heat: $Q = m c \Delta T$. Latent heat of phase transformation: $Q = m L$. For water: Latent heat of fusion $L_f = 80\,\text{cal/g} = 3.36 \times 10^5\,\text{J/kg}$; Latent heat of vaporization $L_v = 540\,\text{cal/g} = 2.26 \times 10^6\,\text{J/kg}$.

Modes of Heat Transfer: Conduction, Convection & Radiation

1. Thermal Conduction (Fourier's Law): Heat flows down a temperature gradient via lattice vibrations (phonons) and free electron collisions without macroscopic material transport:

$$\frac{dQ}{dt} = -K A \frac{dT}{dx} = \frac{K A (T_1 - T_2)}{L}$$

where $K$ is thermal conductivity (SI unit: $\text{W}/(\text{m}\cdot\text{K})$). Thermal resistance: $R_{\text{th}} = \frac{L}{KA}$ (analogous to Ohm's Law $\Delta T = I_{\text{heat}} R_{\text{th}}$).

2. Thermal Convection: Heat transfer accompanied by actual bulk movement of fluid matter driven by buoyant density gradients (natural convection) or external blowers/pumps (forced convection).

3. Thermal Radiation: Emission of electromagnetic energy requiring no material medium.

  • Stefan-Boltzmann Law: Total radiant energy emitted per second per unit area by a black body at absolute temperature $T$ is $E = \sigma T^4$, where $\sigma = 5.67 \times 10^{-8}\,\text{W}/(\text{m}^2\cdot\text{K}^4)$. For an imperfect emitter of emissivity $\epsilon$ ($0 < \epsilon < 1$):$$\frac{dQ}{dt} = \epsilon \sigma A (T^4 - T_0^4)$$
  • Wien's Displacement Law: The wavelength $\lambda_{\max}$ corresponding to peak spectral emittance is inversely proportional to absolute temperature:$$\lambda_{\max} T = b \approx 2.898 \times 10^{-3}\,\text{m}\cdot\text{K}$$
  • Newton's Law of Cooling: For small temperature difference between body and surroundings ($\Delta T = T - T_0 \ll T_0$):$$\frac{dT}{dt} = -k(T - T_0) \implies \ln\left(\frac{T_1 - T_0}{T_2 - T_0}\right) = k t$$

Key Formulas, Laws & Physical Constants

Young's Modulus & Elastic Energy Density
Y = (F·L) / (A·ΔL), u = 1/2 · Stress · Strain = σ² / (2Y)
Fluid Pressure & Pascal's Hydraulic Lift
$$P = P₀ + ρgh, F₂ / A₂ = F₁ / A₁, F_B = V_sub · ρ_f · g$$
Stokes' Viscous Drag & Terminal Velocity
$$F_v = 6πηrv, v_t = 2/9 · [r²(ρ - σ)g] / η$$
Bernoulli's Theorem & Torricelli's Efflux
$$P + 1/2·ρv² + ρgh = const, A₁v₁ = A₂v₂, v_efflux = √(2gh)$$
Surface Tension, Excess Pressure & Capillarity
$$W = T·ΔA, ΔP_drop = 2T/R, ΔP_bubble = 4T/R, h = (2T·cosθ)/(r·ρ·g)$$
Thermal Expansion, Conduction & Radiation
$$α = β/2 = γ/3, dQ/dt = KA(T₁ - T₂)/L, E = εσT⁴, λ_max·T = b$$

Conceptual Solved Examples & Case Studies

Example 1
A steel wire of length 2 m and cross-sectional area 2 mm² is stretched by 1 mm. Calculate the work done in stretching the wire. (Young's modulus of steel Y = 2 × 10¹¹ N/m²).
Step-by-Step Solution:
Step 1: Convert given values to SI units:
Length $L = 2\,\text{m}$
Area $A = 2\,\text{mm}^2 = 2 \times 10^{-6}\,\text{m}^2$
Elongation $\Delta L = 1\,\text{mm} = 10^{-3}\,\text{m}$
Young's modulus $Y = 2 \times 10^{11}\,\text{N/m}^2$

Step 2: Formula for elastic strain energy (work done):
$$W = \frac{1}{2} \frac{Y A (\Delta L)^2}{L}$$
Step 3: Numerical computation:
$$W = \frac{1}{2} \times \frac{(2 \times 10^{11}) \times (2 \times 10^{-6}) \times (10^{-3})^2}{2}$$
$$W = \frac{1}{2} \times \frac{4 \times 10^5 \times 10^{-6}}{2} = \frac{0.4}{4} = 0.1\,\text{J}$$
Final Answer: The work done in stretching the wire is $0.1\,\text{J}$ (or $0.10\,\text{Joules}$).
Example 2
Calculate the work done in blowing a soap bubble from a radius of 2 cm to 4 cm. The surface tension of the soap solution is 0.03 N/m. (Take π = 3.1416).
Step-by-Step Solution:
Step 1: Identify the number of free surfaces:
A soap bubble in air has two free interfaces (an inner air-liquid interface and an outer liquid-air interface).
Total initial surface area: $A_1 = 2 \times (4\pi r_1^2) = 8\pi r_1^2$.
Total final surface area: $A_2 = 2 \times (4\pi r_2^2) = 8\pi r_2^2$.

Step 2: Increase in surface area:
$$\Delta A = A_2 - A_1 = 8\pi (r_2^2 - r_1^2)$$
Given: $r_1 = 2\,\text{cm} = 0.02\,\text{m}$, $r_2 = 4\,\text{cm} = 0.04\,\text{m}$.
$$\Delta A = 8\pi [(0.04)^2 - (0.02)^2] = 8\pi [0.0016 - 0.0004] = 8\pi \times 0.0012 = 0.0096\pi\,\text{m}^2$$
Step 3: Work done:
$$W = T \cdot \Delta A = 0.03 \times (0.0096\pi) = 0.000288\pi\,\text{J} = 2.88\pi \times 10^{-4}\,\text{J}$$
$$W = 2.88 \times 3.1416 \times 10^{-4} \approx 9.05 \times 10^{-4}\,\text{J}$$
Final Answer: The work done is approximately $9.05 \times 10^{-4}\,\text{J}$.
Example 3
A spherical raindrop of radius 0.3 mm falls through air under gravity. If the density of water is 1000 kg/m³, density of air is negligible, and viscosity of air is 1.8 × 10⁻⁵ Pa·s, calculate the terminal velocity of the raindrop. (g = 9.8 m/s²).
Step-by-Step Solution:
Step 1: Write down the terminal velocity formula:
$$v_t = \frac{2}{9}\frac{r^2 (\rho - \sigma) g}{\eta}$$
Step 2: Given data in SI units:
Radius $r = 0.3\,\text{mm} = 3 \times 10^{-4}\,\text{m}$
Density of water $\rho = 1000\,\text{kg/m}^3$
Density of air $\sigma \approx 0\,\text{kg/m}^3$ (negligible)
Viscosity $\eta = 1.8 \times 10^{-5}\,\text{Pa}\cdot\text{s}$
Acceleration $g = 9.8\,\text{m/s}^2$

Step 3: Substitute and compute:
$$v_t = \frac{2}{9} \times \frac{(3 \times 10^{-4})^2 \times (1000 - 0) \times 9.8}{1.8 \times 10^{-5}}$$
$$v_t = \frac{2}{9} \times \frac{(9 \times 10^{-8}) \times 9800}{1.8 \times 10^{-5}} = \frac{2 \times 10^{-8} \times 9800}{1.8 \times 10^{-5}}$$
$$v_t = \frac{1.96 \times 10^{-4}}{1.8 \times 10^{-5}} = \frac{1.96}{0.18} \approx 10.89\,\text{m/s}$$
Final Answer: The terminal velocity of the falling raindrop is approximately $10.89\,\text{m/s}$.
Example 4
Water flows through a horizontal Venturimeter. The diameter of the pipe at the inlet is 10 cm and at the throat is 5 cm. A vertical U-tube manometer contains mercury, and the difference in mercury levels between the inlet and throat is 4 cm. Calculate the rate of flow of water through the pipe. (Density of water = 1000 kg/m³, density of mercury = 13600 kg/m³, g = 9.8 m/s²).
Step-by-Step Solution:
Step 1: Calculate cross-sectional areas:
Inlet radius $r_1 = 5\,\text{cm} = 0.05\,\text{m} \implies A_1 = \pi r_1^2 = \pi (0.05)^2 = 2.5 \times 10^{-3}\pi\,\text{m}^2$.
Throat radius $r_2 = 2.5\,\text{cm} = 0.025\,\text{m} \implies A_2 = \pi r_2^2 = \pi (0.025)^2 = 6.25 \times 10^{-4}\pi\,\text{m}^2$.
Area ratio: $\frac{A_1}{A_2} = \left(\frac{r_1}{r_2}\right)^2 = (2)^2 = 4 \implies A_1 = 4A_2$.

Step 2: Pressure difference from manometer:
$$P_1 - P_2 = h (\rho_m - \rho_w) g = (0.04\,\text{m}) \times (13600 - 1000) \times 9.8$$
$$P_1 - P_2 = 0.04 \times 12600 \times 9.8 = 504 \times 9.8 = 4939.2\,\text{Pa}$$
Step 3: Apply Bernoulli's equation for horizontal pipe:
$$P_1 - P_2 = \frac{1}{2}\rho_w (v_2^2 - v_1^2) = \frac{1}{2}\rho_w \left[(4v_1)^2 - v_1^2\right] = \frac{1}{2}\rho_w (15 v_1^2)$$
$$4939.2 = \frac{1}{2}(1000)(15 v_1^2) = 7500 v_1^2 \implies v_1^2 = \frac{4939.2}{7500} \approx 0.65856$$
$$v_1 = \sqrt{0.65856} \approx 0.8115\,\text{m/s}$$
Step 4: Rate of flow (Volume discharge Q):
$$Q = A_1 v_1 = (2.5 \times 10^{-3}\pi) \times 0.8115 = 2.5 \times 3.1416 \times 10^{-3} \times 0.8115$$
$$Q \approx 6.374 \times 10^{-3}\,\text{m}^3/\text{s} = 6.374\,\text{liters/second}$$
Final Answer: The volume flow rate of water is $6.37 \times 10^{-3}\,\text{m}^3/\text{s}$ (approx. $6.37\,\text{L/s}$).
Example 5
(a) Derive the ascent formula for the height h to which a liquid rises in a capillary tube of radius r. (b) Water rises to a height of 6 cm in a capillary tube. If the radius of the tube is halved, to what height will water rise in the new tube?
Step-by-Step Solution:
Part (a): Derivation of Ascent Formula:
Consider a capillary tube of internal radius $r$ dipped vertically into a wetting liquid of density $\rho$ and surface tension $T$.
The meniscus is concave with angle of contact $\theta$.
The surface tension acts along the circumference of contact ($L = 2\pi r$) at an angle $\theta$ to the vertical tube wall.
The upward vertical component of surface tension force is:
$$F_{\text{up}} = T \cos\theta \times 2\pi r = 2\pi r T \cos\theta$$
The horizontal components cancel out symmetrically around the perimeter.
This upward force supports the weight of the elevated cylindrical liquid column of height $h$ (ignoring the tiny meniscus cap volume):
$$W = m g = (\pi r^2 h \rho) g$$
In hydrostatic equilibrium:
$$F_{\text{up}} = W \implies 2\pi r T \cos\theta = \pi r^2 h \rho g$$
Dividing both sides by $\pi r$ and solving for $h$:
$$h = \frac{2T\cos\theta}{r \rho g}$$
This is the classical Capillary Ascent Formula (Jurin's Law). Hence derived.

Part (b): Calculation:
From Jurin's Law, for the same liquid and temperature:
$$h \cdot r = \text{constant} \implies h_1 r_1 = h_2 r_2 \implies h_2 = h_1 \left(\frac{r_1}{r_2}\right)$$
Given $h_1 = 6\,\text{cm}$ and $r_2 = \frac{r_1}{2} \implies \frac{r_1}{r_2} = 2$:
$$h_2 = 6\,\text{cm} \times 2 = 12\,\text{cm}$$
Final Answer: Water will rise to a height of $12\,\text{cm}$ in the narrower tube.
Example 6
(a) State and prove Bernoulli's Theorem for the steady streamline flow of an ideal incompressible fluid. (b) A cylindrical water tank of height H is completely filled with water. A small hole is drilled in the side wall at depth h below the surface. (i) Derive the speed of efflux. (ii) Find the horizontal range R of the water jet on the ground, and (iii) Determine the depth h at which the hole should be drilled to achieve maximum horizontal range.
Step-by-Step Solution:
Part (a): Statement and Proof of Bernoulli's Theorem:
Statement: For the steady, streamline flow of an ideal (incompressible and non-viscous) fluid, the total energy per unit volume (sum of pressure energy, kinetic energy, and potential energy) is constant along any streamline:
$$P + \frac{1}{2}\rho v^2 + \rho g y = \text{constant}$$
Proof: Consider flow through a pipe between cross-section $A_1$ (height $y_1$, pressure $P_1$, velocity $v_1$) and $A_2$ (height $y_2$, pressure $P_2$, velocity $v_2$).
In small time $\Delta t$, mass $\Delta m = \rho A_1 v_1 \Delta t = \rho A_2 v_2 \Delta t$ flows through.
Work done on fluid at inlet: $W_1 = P_1 A_1 (v_1 \Delta t) = P_1 \frac{\Delta m}{\rho}$.
Work done by fluid at outlet: $W_2 = -P_2 A_2 (v_2 \Delta t) = -P_2 \frac{\Delta m}{\rho}$.
Net external work done: $W_{\text{net}} = (P_1 - P_2)\frac{\Delta m}{\rho}$.
Change in kinetic energy: $\Delta K = \frac{1}{2}\Delta m (v_2^2 - v_1^2)$.
Change in potential energy: $\Delta U = \Delta m g (y_2 - y_1)$.
By the Work-Energy Theorem, $W_{\text{net}} = \Delta K + \Delta U$:
$$(P_1 - P_2)\frac{\Delta m}{\rho} = \frac{1}{2}\Delta m(v_2^2 - v_1^2) + \Delta m g(y_2 - y_1)$$
Dividing by $\Delta m$ and multiplying by $\rho$:
$$P_1 - P_2 = \frac{1}{2}\rho v_2^2 - \frac{1}{2}\rho v_1^2 + \rho g y_2 - \rho g y_1$$
$$P_1 + \frac{1}{2}\rho v_1^2 + \rho g y_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g y_2 \implies P + \frac{1}{2}\rho v^2 + \rho g y = \text{constant}$$
Part (b): Efflux Velocity and Maximum Range:
(i) Speed of Efflux: Applying Bernoulli between surface ($P_0, v_1 \approx 0, y_1 = H$) and orifice ($P_0, v_2 = v, y_2 = H - h$):
$$P_0 + 0 + \rho g H = P_0 + \frac{1}{2}\rho v^2 + \rho g (H - h) \implies \frac{1}{2}\rho v^2 = \rho g h \implies v = \sqrt{2gh}$$
(ii) Horizontal Range: The jet emerges horizontally from height $(H - h)$ above the ground.
Time of fall to ground: $H - h = \frac{1}{2}g t^2 \implies t = \sqrt{\frac{2(H - h)}{g}}$.
Horizontal range $R = v \cdot t = \sqrt{2gh} \times \sqrt{\frac{2(H - h)}{g}} = 2\sqrt{h(H - h)}$.
(iii) Maximum Range Condition: To maximize $R$, we maximize $f(h) = h(H - h) = Hh - h^2$.
Differentiating with respect to $h$: $\frac{df}{dh} = H - 2h = 0 \implies h = \frac{H}{2}$.
Second derivative $\frac{d^2f}{dh^2} = -2 < 0$ (maximum verified).
Maximum range: $R_{\max} = 2\sqrt{\frac{H}{2}\left(H - \frac{H}{2}\right)} = 2\sqrt{\frac{H^2}{4}} = H$.
Final Answer: (i) $v = \sqrt{2gh}$; (ii) $R = 2\sqrt{h(H-h)}$; (iii) Hole must be drilled at mid-depth $h = H/2$, yielding $R_{\max} = H$.

Common Misconceptions & Examiner Traps

Common Misconception

Confusing the Proportional Limit with the Elastic Limit on the stress-strain curve.

Scientific Reality & Correction

Hooke's Law (Stress ∝ Strain) holds strictly ONLY up to the Proportional Limit (Point A). Between the Proportional Limit and the Elastic Limit (Point B), the graph is slightly curved (non-linear), yet the material will still fully recover its original shape when unloaded.

Common Misconception

Applying ΔP = 2T/R to soap bubbles instead of ΔP = 4T/R.

Scientific Reality & Correction

A liquid droplet in air has ONLY ONE free liquid-gas interface, so ΔP = 2T/R. A soap bubble consists of a hollow thin liquid film surrounded by air inside and outside, having TWO free liquid-gas interfaces, producing double the excess pressure: ΔP = 2(2T/R) = 4T/R.

Common Misconception

Believing water will fountain or overflow if a capillary tube is shorter than the ascent height.

Scientific Reality & Correction

Water will NEVER overflow out of an insufficient capillary tube (which would violate perpetual motion / energy conservation). When the liquid reaches the upper edge, its meniscus flattens, increasing its radius of curvature R' such that h·r = l·R' remains constant in hydrostatic equilibrium.

Common Misconception

Using Celsius temperatures instead of absolute Kelvin temperatures in Stefan's Law and Wien's Law.

Scientific Reality & Correction

Radiation emission E = εσT⁴ and Wien's law λ_max·T = b require absolute thermodynamic temperature in Kelvin (K = °C + 273.15). Doubling temperature from 100°C to 200°C is NOT a 2-fold increase; it is 373 K to 473 K (a 1.27-fold increase).

Common Misconception

Assuming terminal velocity depends linearly on the radius of the falling sphere.

Scientific Reality & Correction

Stokes' viscous drag force is proportional to r (F_v = 6πηrv), but gravitational weight is proportional to volume (r³). Balancing forces yields terminal velocity proportional to r squared: v_t ∝ r², NOT r.

Properties of Bulk Matter & Fluid Dynamics Infographic

7 Properties of Bulk Matter & Fluid Dynamics WBCHSE Class 11 Physics — Chapter 7: Mechanical & Thermal Properties Elasticity & Hooke's Law Stress-Strain Profile & Elastic Moduli Young's Modulus: Y = (F·L) / (A·ΔL) Elastic Energy Density: u = 1/2 · σ · ε Hooke: Stress = Y · Strain (Within Limit) Fluid Statics & Viscosity Pascal's Law, Viscous Drag & Terminal Speed Hydraulic Lift: F₂/A₂ = F₁/A₁ Stokes & Terminal Velocity: v_t ∝ r²(ρ - σ)g/η Viscous Force: F = 6πηrv (Stokes' Law) Bernoulli's Theorem & Surface Tension Streamline Dynamics & Capillary Action Bernoulli: P + 1/2·ρv² + ρgh = const Capillary Ascent: h = (2T·cosθ) / (r·ρ·g) Excess Bubble Pressure: ΔP = 4T/R Thermal Expansion & Heat Transfer Calorimetry, Conduction & Radiation T₁ (Hot) T₂ (Cold) Expansion Ratios: α : β : γ = 1 : 2 : 3 Stefan-Boltzmann Law: E = ε·σ·T⁴ Conduction: dQ/dt = KA(T₁ - T₂)/L

Chapter Summary & 10 Key Takeaways

Takeaway 1
Hooke's Law states that within the proportional limit, stress is directly proportional to strain (σ = E·ε).
Takeaway 2
Young's modulus applies to elongation of solids, Bulk modulus applies to volume compression of all states of matter, and Shear modulus applies to shape distortion of solids.
Takeaway 3
Elastic strain energy density stored in a deformed solid is u = 1/2 · stress · strain = σ² / (2Y).
Takeaway 4
Pascal's Principle states that pressure applied to an enclosed fluid is transmitted undiminished in all directions, enabling hydraulic force multiplication (F₂/A₂ = F₁/A₁).
Takeaway 5
Archimedes' Principle establishes that buoyant upthrust equals the weight of displaced fluid: F_B = V_sub · ρ_f · g.
Takeaway 6
Newton's Law of Viscosity (F = -ηA dv/dx) defines viscous drag, where dynamic viscosity η has SI unit Pa·s (Poiseuille) and CGS unit Poise (1 Pa·s = 10 Poise).
Takeaway 7
Stokes' Law (F = 6πηrv) leads to terminal velocity v_t = 2/9 · [r²(ρ - σ)g] / η, which scales with the square of the radius (r²).
Takeaway 8
Bernoulli's Theorem expresses mechanical energy conservation along a streamline: P + 1/2 ρv² + ρgh = constant.
Takeaway 9
Surface tension T = F/L = W/ΔA produces excess pressure ΔP = 2T/R for liquid drops and ΔP = 4T/R for soap bubbles with two free surfaces.
Takeaway 10
Capillary rise is given by Jurin's Law h = (2T cosθ)/(rρg); in an insufficient tube, the liquid never overflows but flattens its meniscus.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Why is steel considered more elastic than rubber in physics, contrary to everyday intuition?
Reveal Answer & Explanation
Answer:
Compare the elongation produced in steel and rubber under the same stretching load.
2
Two soap bubbles of radii r1 and r2 (with r1 > r2) coalesce under isothermal conditions to form a single bubble of radius R. Find an expression for R in terms of r1 and r2.
Reveal Answer & Explanation
Answer:
Apply Boyle's Law P₁V₁ + P₂V₂ = PV with excess pressure P = 4T/r in vacuum.
3
Explain why hot soup tastes more delicious than cold soup using the concept of surface tension.
Reveal Answer & Explanation
Answer:
Consider the relationship between liquid temperature, surface tension, and spreading across the tongue.
4
A metallic sphere cools from 80°C to 60°C in 10 minutes in an environment at 30°C. How much further time will it take to cool from 60°C to 40°C?
Reveal Answer & Explanation
Answer:
Apply (T₁ - T₂)/t = K[(T₁ + T₂)/2 - T₀] for both cooling stages.
5
What happens to the level of water in a beaker containing a floating ice cube when the ice cube completely melts?
Reveal Answer & Explanation
Answer:
Compare the volume of displaced water with the volume of water formed by melting.
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