Properties of Solids and Liquids — Practice Questions with Answers

Stress is the internal restoring force per unit area developed within a body when deformed: $\text{Stress} = F/A$.

Strain is the fractional change in dimension (length, volume, or shape) resulting from applied stress — a dimensionless ratio.

Hooke's law states that, within the elastic limit, stress is directly proportional to strain: $\text{Stress} = E \times \text{Strain}$, where $E$ is the modulus of elasticity (a constant for the material).

Young's modulus (Y): Ratio of longitudinal stress to longitudinal strain — applies to stretching/compressing a wire or rod along its length.

Bulk modulus (K): Ratio of volumetric stress (pressure) to volumetric strain — applies to uniform compression/expansion of volume, relevant to both solids and fluids.

Shear modulus (G, or modulus of rigidity): Ratio of shearing stress to shearing strain — applies to a change in shape without change in volume, relevant only to solids.

Surface tension is the property of a liquid surface that makes it behave like a stretched elastic membrane, tending to minimise its surface area — defined as force per unit length acting along the surface.

A liquid drop takes a spherical shape because, for a given volume, a sphere has the minimum possible surface area of any shape. Since surface tension acts to minimise surface area (and hence surface energy), the drop naturally adopts the spherical form, unless distorted by external forces like gravity.

Pascal's law states that pressure applied at any point in an enclosed, incompressible fluid is transmitted equally and undiminished in all directions throughout the fluid.

Application: A hydraulic lift/jack uses this principle — a small force applied to a small-area piston creates a pressure that is transmitted equally to a larger-area piston, producing a much larger output force ($F_2 = F_1 \times A_2/A_1$), allowing heavy loads to be lifted with modest input force.

Bernoulli's theorem states that for an ideal fluid in steady, streamline flow, the sum of pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline.

$P + \dfrac{1}{2}\rho v^2 + \rho gh = \text{constant}$

This is essentially a statement of energy conservation applied to fluid flow, explaining phenomena like lift on an aircraft wing and the functioning of a Venturi meter.

Viscosity is the internal frictional resistance a fluid offers to relative motion between its layers.

Liquids: Viscosity arises mainly from intermolecular cohesive forces; increased temperature increases molecular kinetic energy, weakening these cohesive forces' effectiveness, so liquid viscosity decreases with temperature.

Gases: Viscosity arises mainly from momentum transfer between molecules moving between layers; increased temperature increases molecular speed and hence momentum transfer, so gas viscosity increases with temperature.

Capillarity is the rise or fall of a liquid within a narrow tube due to the combined effects of surface tension and adhesive/cohesive forces.

Equating the upward force due to surface tension ($2\pi rT\cos\theta$) to the weight of the raised liquid column ($\pi r^2h\rho g$):

$2\pi rT\cos\theta = \pi r^2h\rho g$

$h = \dfrac{2T\cos\theta}{r\rho g}$, where $T$ is surface tension, $\theta$ is the angle of contact, $r$ is the tube radius, and $\rho$ is the liquid density.

Stokes' law gives the viscous drag force on a small sphere moving through a viscous fluid: $F = 6\pi\eta r v$, where $\eta$ is the coefficient of viscosity, $r$ is the sphere's radius, and $v$ is its velocity.

As the sphere falls, viscous drag (and buoyant force) increase with speed, while weight remains constant. Terminal velocity is reached when these upward forces exactly balance the downward weight, giving zero net force — the sphere then continues falling at this constant maximum speed.

For small drops, surface tension (which scales with surface area/radius, roughly $r^2$ or $r$) dominates over the effect of gravity (which scales with volume/weight, roughly $r^3$) — so surface tension effects dominate, and the drop remains spherical to minimise surface area.

As drop size increases, gravitational weight (scaling as $r^3$) grows much faster than surface tension effects (scaling as $r^2$ or less), so for larger drops, gravity dominates and flattens the drop against the supporting surface, reducing its height and departing from a perfect sphere.

A hydraulic brake system uses an incompressible brake fluid enclosed within a system of connected cylinders. When the driver presses the brake pedal, force is applied to a small-area master cylinder piston, generating pressure within the fluid.

By Pascal's law, this pressure is transmitted equally and undiminished throughout the fluid to larger-area cylinders at each wheel — since force = pressure × area, this transmitted pressure produces a much larger force at each wheel's larger piston, pressing the brake pads firmly against the wheels to decelerate the vehicle, using only a modest force applied at the pedal.

(a) Stress = $F/A = \dfrac{100}{2\times10^{-6}} = 5\times10^7\text{ N/m}^2$

(b) Strain = $\Delta L/L = \dfrac{0.5\times10^{-3}}{2} = 2.5\times10^{-4}$

(c) Young's modulus $Y = \dfrac{\text{Stress}}{\text{Strain}} = \dfrac{5\times10^7}{2.5\times10^{-4}} = \mathbf{2\times10^{11}\text{ N/m}^2}$

$h = \dfrac{2T\cos\theta}{r\rho g}$

With $T=0.073\text{ N/m}$, $\cos0°=1$, $r=0.5\times10^{-3}\text{ m}$, $\rho=1000\text{ kg/m}^3$, $g=10\text{ m/s}^2$:

$h = \dfrac{2\times0.073\times1}{0.5\times10^{-3}\times1000\times10} = \dfrac{0.146}{5}$

$h = \mathbf{0.0292\text{ m} = 2.92\text{ cm}}$

By Pascal's law, pressure is the same throughout: $\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}$

$F_2 = F_1 \times \dfrac{A_2}{A_1} = 50 \times \dfrac{500}{5} = 50\times100 = \mathbf{5000\text{ N}}$

Mechanical advantage = $F_2/F_1 = 5000/50 = \mathbf{100}$ — a 100-fold amplification of force.

(a) Using the equation of continuity: $A_1v_1 = A_2v_2$

$4\times2 = 1\times v_2 \Rightarrow v_2 = 8\text{ m/s}$

(b) Using Bernoulli's equation for horizontal flow (same height): $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$

$P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2) = \frac{1}{2}(1000)(64-4) = \frac{1}{2}(1000)(60) = \mathbf{30000\text{ Pa} = 30\text{ kPa}}$

Force applied: $F = mg = 10\times10 = 100\text{ N}$

From $Y = \dfrac{FL}{A\Delta L}$, rearranging: $\Delta L = \dfrac{FL}{AY}$

$\Delta L = \dfrac{100\times3}{1\times10^{-6}\times2\times10^{11}} = \dfrac{300}{2\times10^5}$

$\Delta L = 1.5\times10^{-3}\text{ m} = \mathbf{1.5\text{ mm}}$

At terminal velocity: Weight = Viscous drag + Buoyant force

$\dfrac{4}{3}\pi r^3\rho_{ball}g = 6\pi\eta rv + \dfrac{4}{3}\pi r^3\rho_{fluid}g$

Rearranging: $v = \dfrac{2r^2(\rho_{ball}-\rho_{fluid})g}{9\eta}$

With $r=1\times10^{-3}\text{ m}$, $\rho_{ball}=7800$, $\rho_{fluid}=1260$, $g=10$, $\eta=1.5$:

$v = \dfrac{2(1\times10^{-6})(6540)(10)}{9(1.5)} = \dfrac{0.1308}{13.5} \approx \mathbf{0.0097\text{ m/s} \approx 9.7\text{ mm/s}}$

For a soap bubble (two liquid surfaces, inner and outer), excess pressure inside is: $\Delta P = \dfrac{4T}{r}$

(compared to $\dfrac{2T}{r}$ for a single-surface liquid drop)

$\Delta P = \dfrac{4\times0.03}{2\times10^{-3}} = \dfrac{0.12}{0.002} = \mathbf{60\text{ Pa}}$

Pressure due to liquid column: $P = h\rho g = 5\times1000\times10 = 50000\text{ Pa} = 5\times10^4\text{ Pa}$

Total pressure = atmospheric pressure + liquid column pressure:

$P_{total} = 1.01\times10^5 + 0.5\times10^5 = \mathbf{1.51\times10^5\text{ Pa}}$

Volumetric strain = $\dfrac{\Delta V}{V} = 0.02\% = 0.0002 = 2\times10^{-4}$

Bulk modulus: $K = \dfrac{\Delta P}{\Delta V/V} = \dfrac{2\times10^6}{2\times10^{-4}}$

$K = \mathbf{1\times10^{10}\text{ Pa}}$

Volume conservation: $2\times\dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi R^3 \Rightarrow R^3 = 2r^3 \Rightarrow R = 2^{1/3}r$

Initial surface energy (two drops): $E_i = 2\times(4\pi r^2 T) = 8\pi r^2T$

Final surface energy (one drop): $E_f = 4\pi R^2T = 4\pi(2^{1/3}r)^2T = 4\pi \times 2^{2/3}r^2T$

Ratio: $\dfrac{E_i}{E_f} = \dfrac{8\pi r^2T}{4\pi\times2^{2/3}r^2T} = \dfrac{2}{2^{2/3}} = 2^{1/3} \approx \mathbf{1.26}$

Since $E_i > E_f$ (ratio > 1), surface energy decreases upon coalescence — this released energy typically appears as heat, since a single larger drop has less total surface area than two smaller drops of the same combined volume.