Elasticity is the property of a material that allows it to return to its original shape and size once a deforming force is removed. A perfectly elastic body fully recovers; a plastic body shows no tendency to recover at all and retains the deformation permanently.
Stress is the internal restoring force per unit cross-sectional area, developed within a material in response to an external deforming force: $\text{Stress} = F/A$, with SI unit $\text{N/m}^2$ (pascal).
Strain is the fractional change in dimension (length, volume, or shape) produced by the stress — a dimensionless ratio, since it's a ratio of two similar quantities (e.g., change in length / original length).
A wire of cross-sectional area $2 \times 10^{-6}\ \text{m}^2$ has a force of 40 N applied along its length. Calculate the stress in the wire.
$\text{Stress} = \dfrac{F}{A} = \dfrac{40}{2\times10^{-6}} = 2\times10^7\ \text{N/m}^2$
Within the elastic limit, stress is directly proportional to strain: $\text{Stress} = E \times \text{Strain}$, where $E$ is a constant called the modulus of elasticity — its specific value and name depend on the type of deformation involved.
Applies specifically to longitudinal deformation (stretching or compressing along one direction, as in a wire or rod):
$$Y = \dfrac{\text{Longitudinal stress}}{\text{Longitudinal strain}} = \dfrac{F/A}{\Delta L/L} = \dfrac{FL}{A\,\Delta L}$$Up to the proportional limit, stress and strain follow a straight-line (Hooke's law) relationship. Beyond the elastic limit, the material no longer fully returns to its original shape after the force is removed (permanent deformation begins). Continuing further leads to the yield point, then the ultimate tensile strength (maximum stress the material can withstand), and finally the fracture point, where the material breaks.
A wire of length 2 m and cross-sectional area $1\times10^{-6}\ \text{m}^2$ stretches by 1 mm under a force of 100 N. Calculate its Young's modulus.
$Y = \dfrac{FL}{A\Delta L} = \dfrac{(100)(2)}{(1\times10^{-6})(1\times10^{-3})} = \dfrac{200}{1\times10^{-9}} = 2\times10^{11}\ \text{N/m}^2$
When a wire is stretched, it also becomes slightly thinner. Poisson's ratio ($\mu$) is the ratio of lateral (sideways) strain to longitudinal strain: $\mu = \dfrac{\text{Lateral strain}}{\text{Longitudinal strain}}$ — a dimensionless quantity, typically between 0 and 0.5 for most common materials.
A wire of original diameter 2 mm and length 1 m is stretched, causing its length to increase by 0.5 mm and its diameter to decrease by 0.0004 mm. Calculate Poisson's ratio for this wire.
Longitudinal strain $= \dfrac{0.5\times10^{-3}}{1} = 5\times10^{-4}$
Lateral strain $= \dfrac{0.0004\times10^{-3}}{2\times10^{-3}} = 2\times10^{-4}$
$\mu = \dfrac{2\times10^{-4}}{5\times10^{-4}} = 0.4$
Applies to volume deformation, when a uniform pressure acts on all surfaces of an object (compressing its overall volume without changing its shape):
$$B = \dfrac{\text{Volume stress}}{\text{Volume strain}} = \dfrac{-\Delta P}{\Delta V/V}$$The negative sign accounts for the fact that increasing pressure ($\Delta P$ positive) causes a decrease in volume ($\Delta V$ negative), keeping $B$ itself a positive quantity.
Applies to shape deformation, when a tangential force causes one face of an object to shift relative to the opposite face (shearing), without changing volume:
$$G = \dfrac{\text{Shear stress}}{\text{Shear strain}} = \dfrac{F/A}{\theta}$$where $\theta$ is the angle of shear (in radians).
Solids can have all three moduli (Young's, bulk, shear), since they can resist changes in length, volume, and shape. Liquids and gases have only a bulk modulus, since they cannot resist shear or maintain a fixed shape — they flow freely under any tangential force rather than developing genuine shear stress.
A solid sphere of volume $0.5\ \text{m}^3$ experiences a pressure increase of $2\times10^5\ \text{Pa}$, causing its volume to decrease by $0.001\ \text{m}^3$. Calculate its bulk modulus.
$B = \dfrac{\Delta P}{\Delta V/V} = \dfrac{2\times10^5}{0.001/0.5} = \dfrac{2\times10^5}{0.002} = 1\times10^8\ \text{Pa}$
Work done in stretching a wire within its elastic limit is stored as elastic potential energy: $U = \tfrac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume} = \tfrac{1}{2}F\Delta L$ — analogous to the energy stored in a stretched spring.
A wire stretches by 2 mm under a force of 500 N. Calculate the elastic potential energy stored in the wire.
$U = \tfrac{1}{2}F\Delta L = \tfrac{1}{2}(500)(2\times10^{-3}) = 0.5\ \text{J}$
Pressure applied to an enclosed, incompressible fluid is transmitted equally and undiminished to every point of the fluid and to the walls of its container, regardless of the shape of the container.
A small force applied to a narrow piston creates pressure that is transmitted equally to a much wider piston, producing a much larger force there (since $P = F/A$ is the same at both pistons, a larger area gives a proportionally larger force): $\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}$.
In a hydraulic lift, the small piston has an area of $0.001\ \text{m}^2$ and the large piston has an area of $0.1\ \text{m}^2$. If a force of 50 N is applied to the small piston, calculate the force generated on the large piston.
$\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2} \Rightarrow F_2 = F_1 \times \dfrac{A_2}{A_1} = 50 \times \dfrac{0.1}{0.001} = 50 \times 100 = 5000\ \text{N}$
Viscosity is the internal friction within a fluid that resists relative motion between its adjacent layers — a measure of a fluid's "thickness" or resistance to flow.
where $\eta$ (eta) is the coefficient of viscosity, $A$ is the area of contact between fluid layers, and $dv/dx$ is the velocity gradient (rate of change of velocity with distance perpendicular to flow).
For liquids, viscosity decreases with increasing temperature (molecules move faster, weakening the intermolecular forces responsible for internal friction). For gases, viscosity increases with increasing temperature (more frequent molecular collisions increase momentum transfer between layers) — an important and often-tested contrast between the two states of matter.
Explain why honey flows much more slowly than water, in terms of viscosity.
Honey has a much higher coefficient of viscosity ($\eta$) than water, due to stronger intermolecular forces (hydrogen bonding and larger, more complex sugar molecules) that more strongly resist relative motion between adjacent fluid layers. This greater internal friction means a much larger force is needed to produce the same velocity gradient in honey compared to water, causing honey to flow far more slowly under the same conditions (e.g., simply pouring under gravity).
For a small sphere moving through a viscous fluid at low speed, the viscous drag (retarding) force is given by:
$$F = 6\pi\eta rv$$where $r$ is the sphere's radius, $v$ is its velocity relative to the fluid, and $\eta$ is the fluid's coefficient of viscosity.
When a sphere falls through a viscous fluid, it accelerates initially, but as speed increases, the upward viscous drag force also increases, until it exactly balances the net downward force (weight minus buoyancy) — at this point, acceleration becomes zero and the sphere falls at a constant terminal velocity.
where $\rho$ = density of the sphere, $\sigma$ = density of the fluid.
A small steel ball of radius 1 mm and density 8000 kg/m³ falls through oil of density 900 kg/m³ and viscosity 0.5 Pa·s. Calculate its terminal velocity. (g = 10 m/s²)
$v_t = \dfrac{2r^2(\rho-\sigma)g}{9\eta} = \dfrac{2(10^{-3})^2(8000-900)(10)}{9(0.5)}$
$= \dfrac{2(10^{-6})(7100)(10)}{4.5} = \dfrac{0.142}{4.5} \approx 0.0316\ \text{m/s}$
For an ideal fluid (incompressible, non-viscous) in steady flow, the sum of pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline — this is a direct consequence of energy conservation applied to flowing fluids.
At the same height, where fluid speed is higher, pressure is lower, and vice versa — this inverse relationship explains phenomena like the lift generated by an aeroplane wing (faster airflow over the curved top surface creates lower pressure there compared to the flatter bottom surface) and why a shower curtain gets pulled inward when water flows past it.
Water flows through a horizontal pipe that narrows from a wider section (area 4 cm², speed 2 m/s) to a narrower section (area 2 cm²). Using the continuity equation, find the speed in the narrower section, then use Bernoulli's principle to state qualitatively how pressure changes there.
By the continuity equation ($A_1v_1 = A_2v_2$): $(4)(2) = (2)v_2 \Rightarrow v_2 = 4\ \text{m/s}$
Since speed increases in the narrower section (and height is unchanged, being a horizontal pipe), Bernoulli's principle requires pressure to decrease correspondingly in that narrower section, to keep the total sum constant.
In streamline (laminar) flow, fluid particles move along smooth, well-defined, non-crossing paths — Bernoulli's equation applies cleanly to this regime. In turbulent flow, fluid motion becomes chaotic and irregular, with paths crossing and mixing unpredictably — this typically occurs above a critical speed, and Bernoulli's simple equation no longer accurately describes the flow.
Explain qualitatively why blood flow in narrowed arteries (due to plaque buildup) can become turbulent, causing an audible sound (bruit) detectable by a stethoscope.
As an artery narrows, the fluid (blood) must speed up considerably to maintain the same volume flow rate through the smaller cross-sectional area (by the continuity equation). Beyond a certain critical speed, smooth streamline flow becomes unstable and transitions into turbulent flow, with chaotic eddies and pressure fluctuations — this turbulence generates audible vibrations/sound, which is exactly the abnormal "whooshing" sound a doctor may detect with a stethoscope near a significantly narrowed artery.
Surface tension is the property of a liquid's surface that makes it behave like a stretched elastic membrane, causing it to minimise its surface area — arising because surface molecules experience a net inward pull from neighbouring molecules below (since there are no liquid molecules above the surface to balance this attraction).
where $F$ is the force acting along the surface (tangential to it), and $l$ is the length of the line along which this force acts. SI unit: N/m.
A molecule deep within the liquid experiences equal attractive forces from all surrounding neighbours (net force zero). A molecule at the surface has neighbours only below and to the sides, but none above — resulting in a net downward/inward force, pulling surface molecules inward and creating tension that resists any increase in surface area, favouring the shape (like a sphere) with minimum surface area for a given volume.
A wire ring of circumference 0.2 m is pulled out of a soap solution, requiring an additional force of 0.014 N (beyond the ring's own weight) due to the soap film formed. Calculate the surface tension of the soap solution.
A soap film has two surfaces (top and bottom), so the total length of contact is $2 \times 0.2 = 0.4\ \text{m}$.
$T = \dfrac{F}{l} = \dfrac{0.014}{0.4} = 0.035\ \text{N/m}$
The pressure inside a curved liquid surface is always greater than outside, due to surface tension trying to minimise surface area — this excess pressure increases as the radius decreases, meaning smaller drops/bubbles have proportionally higher internal pressure.
Calculate the excess pressure inside a soap bubble of radius 2 mm, given the surface tension of the soap solution is 0.03 N/m.
$\Delta P = \dfrac{4T}{r} = \dfrac{4(0.03)}{2\times10^{-3}} = \dfrac{0.12}{0.002} = 60\ \text{Pa}$
Capillarity (capillary action) is the rise or fall of a liquid within a narrow tube (capillary), caused by the combined effects of surface tension and the relative strength of adhesive (liquid-to-tube) versus cohesive (liquid-to-liquid) forces.
If adhesive forces exceed cohesive forces (e.g., water in a clean glass tube), the liquid wets the tube and rises — this corresponds to an acute angle of contact. If cohesive forces exceed adhesive forces (e.g., mercury in a glass tube), the liquid does not wet the tube and instead falls/depresses below the outer level — corresponding to an obtuse angle of contact.
where $T$ = surface tension, $\theta$ = angle of contact, $r$ = radius of the capillary tube, $\rho$ = density of the liquid.
Water rises to a height of 4 cm in a capillary tube of radius 0.25 mm. If the surface tension of water is 0.07 N/m and the angle of contact is 0° (cos0° = 1), calculate the density of water implied by this data. (g = 10 m/s²)
Rearranging $h = \dfrac{2T\cos\theta}{r\rho g}$ for $\rho$: $\rho = \dfrac{2T\cos\theta}{hrg} = \dfrac{2(0.07)(1)}{(0.04)(0.25\times10^{-3})(10)}$
$= \dfrac{0.14}{1\times10^{-4}} = 1400\ \text{kg/m}^3$ (this value is illustrative, based on the specific numbers given, and not meant to represent water's actual real-world density exactly)