Properties of Solids & Liquids

Elasticity describes a material's ability to resist and recover from deformation.
Stress is the restoring force per unit area: $\sigma$ = F/A [$M^{1} \; L^{-1} \; T^{-2}$] (Pa = N/$m^{2}$), where F is the force (N) and A is the cross-sectional area ($m^{2}$). Stress can be tensile/compressive (normal to surface) or shear (tangential to surface).
Strain is the fractional deformation (dimensionless): longitudinal strain = $\Delta$-L/L, volumetric strain = $\Delta$-V/V, and shear strain = tan($\phi$) ≈ $\phi$ for small angles. Hooke's law states that stress is proportional to strain within the elastic limit, with the constant of proportionality being the elastic modulus.

Three moduli characterise different deformation types.
Young's modulus Y = (F/A)/($\Delta$-L/L) = FL/(A $\Delta$-L) [$M^{1} \; L^{-1} \; T^{-2}$] (Pa) governs linear stretching.
Bulk modulus B = -V(dP/dV) [$M^{1} \; L^{-1} \; T^{-2}$] (Pa) governs volume change, with compressibility = 1/B.
Shear modulus (modulus of rigidity) G = shear stress/shear strain [$M^{1} \; L^{-1} \; T^{-2}$] (Pa). The stress-strain curve progresses through: proportional limit, elastic limit, yield point, ultimate stress (maximum), and breaking point. The region before the elastic limit is elastic (reversible); beyond it is plastic (permanent).

Pascal's law states that pressure applied to a confined fluid transmits equally in all directions.
The hydraulic press uses this: $\frac{F_{1}}{A_{1}}$ = $\frac{F_{2}}{A_{2}}$.
Pressure at depth h in a fluid: P = $P_{0}$ + $\rho$ g h [$M^{1} \; L^{-1} \; T^{-2}$] (Pa).

For fluid dynamics, the equation of continuity $A_{1} \; v_{1}$ = $A_{2} \; v_{2}$ ensures mass conservation in incompressible flow.
Bernoulli's equation P + $\frac{1}{2}$ $\rho \; v^{2}$ + $\rho$ g h = constant [$M^{1} \; L^{-1} \; T^{-2}$] (Pa) relates pressure, velocity, and height. A key consequence: higher velocity means lower pressure, explaining airplane lift and the Venturi effect.

Viscosity is internal friction in fluids.
Newton's viscous force law: F = $\eta$ A (dv/dx), where $\eta$ is the coefficient of viscosity [$M^{1} \; L^{-1} \; T^{-1}$] (Pa s).
Stokes' law gives the drag on a sphere: F = 6 $\pi \; \eta$ r v, where r is radius (m) and v is speed (m/s).
Terminal velocity is reached when weight equals buoyancy plus drag: $v_{t}$ = $2r^{2}$($\rho$ - $\sigma$)g / (9 $\eta$) [$M^{0} \; L^{1} \; T^{-1}$] (m/s), where $\rho$ is sphere density (kg/$m^{3}$) and $\sigma$ is fluid density (kg/$m^{3}$). Critically, $v_{t}$ is proportional to $r^{2}$ — doubling the radius quadruples the terminal velocity.

Surface tension S = F/L = Energy/Area [$M^{1} \; L^{0} \; T^{-2}$] (N/m).
Capillary rise: h = 2S cos($\theta$) / ($\rho$ g r), where $\theta$ is the contact angle, r is tube radius (m).
Excess pressure inside a liquid drop: $\Delta$-P = 2S/R (one surface).
Inside a soap bubble: $\Delta$-P = 4S/R (two free surfaces). Students must never confuse these — bubbles have twice the excess pressure of drops.

Thermal expansion coefficients relate as $\beta$ (volume) = 3 $\alpha$ (linear).
Heat conduction follows Fourier's law: Q/t = KA($\Delta$-T)/L, where K is thermal conductivity [$M^{1} \; L^{1} \; T^{-3} \; K^{-1}$] (W $m^{-1} \; K^{-1}$).
Stefan-Boltzmann radiation law: P = $\sigma$ A $T^{4}$, where $\sigma$ = 5.67 x $10^{-8}$ W $m^{-2} \; K^{-4}$.

Solved Numerical 1: Steel wire, length L = 2 m, area A = 1 $mm^{2}$ = 1 x $10^{-6} \; m^{2}$, load F = 100 N, Y = 2 x $10^{11}$ Pa.
Extension: $\Delta$-L = FL/(AY) = (100)(2) / ((1 x $10^{-6}$)(2 x $10^{11}$)) = 200 / (2 x $10^{5}$) = 1 x $10^{-3}$ m = 1 mm.
Dimensional check: [N][m]/([$m^{2}$][Pa]) = [M L $T^{-2}$][L] / ([$L^{2}$][M $L^{-1} \; T^{-2}$]) = [M $L^{2} \; T^{-2}$]/[M L $T^{-2}$] = [L].

Solved Numerical 2: Capillary rise: h = 10 cm = 0.1 m, r = 0.2 mm = 2 x $10^{-4}$ m, $\theta$ = 0 degrees (cos $\theta$ = 1), $\rho$ = 1000 kg/$m^{3}$, g = 10 m/$s^{2}$.
S = h r $\rho$ g / (2 cos $\theta$) = (0.1)(2 x $10^{-4}$)(1000)(10) / (2 x 1) = 0.2 / 2 = 0.1 N/m = 100 mN/m.

Solved Numerical 3: Steel ball in glycerine: r = 2 mm = 2 x $10^{-3}$ m, $\rho$ = 7800 kg/$m^{3}$, $\sigma$ = 1260 kg/$m^{3}$, $\eta$ = 0.8 Pa s, g = 10 m/$s^{2}$.
Terminal velocity: $v_{t}$ = $2r^{2}$($\rho$ - $\sigma$)g / (9 $\eta$) = 2(2 x $10^{-3}$)²(7800 - 1260)(10) / (9 x 0.8) = 2(4 x $10^{-6}$)(6540)(10) / 7.2 = 2(4 x $10^{-6}$)(65400) / 7.2 = 0.5232 / 7.2 = 0.0727 m/s = 7.27 cm/s.
Dimensional formula: [$L^{2}$][M $L^{-3}$][L $T^{-2}$] / [M $L^{-1} \; T^{-1}$] = [M $L^{-1} \; T^{-2}$] x [$L^{2}$] / [M $L^{-1} \; T^{-1}$] = [L $T^{-1}$].

The key testable concept is the distinction between excess pressure in a liquid drop (2S/R, one surface) and a soap bubble (4S/R, two surfaces), and terminal velocity's proportionality to $r^{2}$.