Properties of Solids and Liquids — Easy Practice (Set 1)

Quick Recap — Elasticity & Basics

• Stress $= \dfrac{F}{A}$ (unit Pa); strain $= \dfrac{\Delta L}{L}$ (dimensionless).
• Hooke's law: within the elastic limit, stress $\propto$ strain.
• Young's modulus $Y = \dfrac{\text{stress}}{\text{strain}} = \dfrac{FL}{A\,\Delta L}$; the bulk modulus governs volume change and the rigidity modulus governs shape (shear).
• Elastic energy density $= \tfrac12 \times \text{stress} \times \text{strain}$.
• Fluid pressure $P = \rho g h$; thermal expansion $\Delta L = L\alpha\Delta T$, with $\beta = 2\alpha$ and $\gamma = 3\alpha$.

Beyond-NCERT JEE Formulae

Consolidated, exam-ready results for the properties of solids and liquids — each tagged with when to reach for it and the trap examiners like to set.

1. Elasticity — moduli, energy, thermal stress, self-weight

The three moduli, matched to the kind of deformation: $$Y=\dfrac{FL}{A\,\Delta L},\qquad B=-\dfrac{\Delta P}{\Delta V/V},\qquad \eta=\dfrac{F/A}{x/L}$$

• Use $Y$ for stretch or compression along a length, $B$ for a uniform squeeze from all sides, and the rigidity $\eta$ for a tangential shear.
• Elastic PE per unit volume (area under the stress-strain line): $u=\tfrac12\times\text{stress}\times\text{strain}=\dfrac{\sigma^2}{2Y}=\tfrac12 Y\varepsilon^2$, with total stored energy $U=\tfrac12 F\,\Delta L$.
• Poisson's ratio $\sigma=\dfrac{-\text{lateral strain}}{\text{longitudinal strain}}$, lying between $0$ and $0.5$ for common solids; a stretched wire's volume change is $\dfrac{\Delta V}{V}=(1-2\sigma)\dfrac{\Delta L}{L}$.
• Thermal stress in a rod clamped between rigid walls is $\sigma_{th}=Y\alpha\,\Delta T$ and the wall force is $F=YA\alpha\,\Delta T$ — neither depends on the rod's length.
• Self-weight elongation of a wire hanging from one end (mass $m$) is $\Delta L=\dfrac{mgL}{2AY}=\dfrac{\rho g L^2}{2Y}$; a hung load acts with only half the wire's own weight added.

[JEE Tip] Modulus is not stiffness. $Y$ is a fixed property of the material, but the wire's spring constant $k=\dfrac{YA}{L}$ depends on its shape — cutting a wire in half doubles $k$ and leaves $Y$ untouched.

2. Surface tension — excess pressure and capillarity

For a curved surface of surface tension $T$: $$\text{drop: }\Delta P=\dfrac{2T}{r},\qquad\text{soap bubble: }\Delta P=\dfrac{4T}{r},\qquad\text{capillary: }h=\dfrac{2T\cos\theta}{\rho g r}$$

• Reach for these with drops, bubbles, films and thin tubes; the excess pressure goes as $\dfrac{1}{r}$, so smaller drops and bubbles hold the higher internal pressure.
• Jurin's law gives $h\propto\dfrac{1}{r}$, so a narrower bore lifts liquid higher; for a non-wetting liquid ($\theta>90^\circ$, mercury on glass) $\cos\theta<0$ and the level is depressed.
• Surface energy $=T\,\Delta A$: breaking one drop of radius $R$ into $n$ equal droplets needs work $W=4\pi T R^2\left(n^{1/3}-1\right)$.
• Two soap bubbles ($r_1<r_2$) that join share a curved film of radius $R=\dfrac{r_1 r_2}{r_2-r_1}$, bulging into the larger bubble.

[JEE Tip] The classic slip is the factor of 2: a soap bubble in air has two liquid surfaces, so $\dfrac{4T}{r}$; a liquid drop or an air bubble inside a liquid has one, so $\dfrac{2T}{r}$.

3. Ideal-fluid flow — continuity, Bernoulli, Torricelli, venturi

For steady, incompressible, non-viscous streamline flow: $$A_1 v_1=A_2 v_2,\qquad P+\tfrac12\rho v^2+\rho g h=\text{constant}$$

• Continuity is mass conservation; Bernoulli is energy per unit volume conserved along one streamline.
• Torricelli: a hole a depth $h$ below the surface jets out at $v=\sqrt{2gh}$; sitting a height $y$ above the ground the jet has horizontal range $x=2\sqrt{h\,y}$.
• Venturi meter (horizontal constriction): joining Bernoulli with continuity gives the flow rate $Q=A_1 A_2\sqrt{\dfrac{2\,\Delta P}{\rho\left(A_1^2-A_2^2\right)}}$.
• Faster flow means lower pressure — the physics of aerofoil lift, the atomiser and a spinning ball's swing.

[JEE Tip] Bernoulli holds only along a single streamline for ideal flow. On a horizontal pipe the $\rho g h$ terms cancel, but never drop the $\tfrac12\rho v_1^2$ of fluid that was already moving before the constriction.

4. Viscosity — Stokes, terminal velocity, Reynolds, Poiseuille

For real (viscous) fluids and the slow motion of a small sphere: $$F=6\pi\eta r v,\qquad v_t=\dfrac{2r^2(\rho-\sigma)g}{9\eta},\qquad Re=\dfrac{\rho v D}{\eta}$$

• Stokes' drag $F=6\pi\eta r v$ acts on a small sphere of radius $r$; the SI unit of $\eta$ is Pa s (1 Pa s $=10$ poise).
• Terminal velocity follows from weight $=$ buoyancy $+$ drag, so $v_t\propto r^2$ and it uses the density difference $(\rho-\sigma)$; if $\rho<\sigma$ the body rises, as an air bubble does in water.
• Reynolds number sets the regime: $Re\lesssim1000$ is laminar and $Re\gtrsim2000$ turbulent in a pipe of diameter $D$.
• Poiseuille's law for laminar pipe flow is $Q=\dfrac{\pi\,\Delta P\,r^4}{8\eta L}$, a fluid resistance $\dfrac{8\eta L}{\pi r^4}$ that adds in series like resistors.

[JEE Tip] Terminal velocity uses the difference $(\rho-\sigma)$, not the sphere's density alone. When $n$ equal droplets coalesce, volume conservation gives $R=n^{1/3}r$, so the merged drop's terminal speed jumps to $n^{2/3}$ times the original.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Elastic energy in a stretched wire. A steel wire of length 3 m and cross-section 2 mm$^2$ ($Y=2\times10^{11}$ Pa) is stretched by 1.5 mm within its elastic limit. Find the stretching force, the elastic PE stored, and the energy per unit volume.

• Treat the wire as a spring: $k=\dfrac{YA}{L}=\dfrac{\left(2\times10^{11}\right)\left(2\times10^{-6}\right)}{3}=1.33\times10^{5}$ N/m, so $F=k\,\Delta L=\left(1.33\times10^{5}\right)\left(1.5\times10^{-3}\right)=200$ N.
• Energy $U=\tfrac12 F\,\Delta L=\tfrac12(200)\left(1.5\times10^{-3}\right)=0.15$ J; density $u=\dfrac{U}{AL}=\dfrac{0.15}{\left(2\times10^{-6}\right)(3)}=2.5\times10^{4}$ J/m$^3$.
• Cross-check: stress $=\dfrac{200}{2\times10^{-6}}=1\times10^{8}$ Pa and strain $=\dfrac{1.5\times10^{-3}}{3}=5\times10^{-4}$, so $u=\tfrac12\left(10^{8}\right)\left(5\times10^{-4}\right)=2.5\times10^{4}$ J/m$^3$.

Answer: $F=200$ N, $U=0.15$ J and $u=2.5\times10^{4}$ J/m$^3$. The factor $\tfrac12$ is the trap — dropping it doubles the energy, because the stress builds up linearly from zero.

Example 2 — Thermal stress in a clamped rod. A steel rod of cross-section 5 mm$^2$ is held between two rigid walls and heated through 100 K. Taking $Y=2\times10^{11}$ Pa and $\alpha=1.2\times10^{-5}$ K$^{-1}$, find the compressive stress and the force on each wall.

• The walls forbid expansion, so the whole thermal strain $\alpha\,\Delta T$ turns into elastic stress: $\sigma=Y\alpha\,\Delta T=\left(2\times10^{11}\right)\left(1.2\times10^{-5}\right)(100)=2.4\times10^{8}$ Pa.
• Force: $F=\sigma A=\left(2.4\times10^{8}\right)\left(5\times10^{-6}\right)=1.2\times10^{3}$ N.

Answer: $\sigma=2.4\times10^{8}$ Pa and $F=1200$ N. Both are independent of the rod's length — only the temperature rise and the material fix the stress.

Example 3 — Capillary rise of water. A clean glass capillary of internal radius 0.2 mm stands vertically in water ($T=0.072$ N/m, $\rho=1000$ kg/m$^3$, contact angle zero). With $g=10$ m/s$^2$, how high does the water climb?

• Jurin's law: $h=\dfrac{2T\cos\theta}{\rho g r}$, with $\cos\theta=1$.
• Denominator $\rho g r=(1000)(10)\left(2\times10^{-4}\right)=2$, so $h=\dfrac{2(0.072)}{2}=0.072$ m.

Answer: $h=0.072$ m, i.e. 7.2 cm. Halving the bore doubles the rise, since $h\propto\dfrac{1}{r}$.

Example 4 — Soap bubble versus drop. A soap bubble of radius 3 mm is blown from a solution of surface tension 0.03 N/m. Find the excess pressure inside it, and compare with a droplet of the same solution and the same radius.

• Soap bubble (two surfaces): $\Delta P=\dfrac{4T}{r}=\dfrac{4(0.03)}{3\times10^{-3}}=40$ Pa.
• Droplet (one surface): $\Delta P=\dfrac{2T}{r}=\dfrac{2(0.03)}{3\times10^{-3}}=20$ Pa.

Answer: 40 Pa inside the bubble and 20 Pa inside the drop — exactly the factor-of-2 difference that the bubble's second surface produces.

Example 5 — Terminal velocity in glycerine. A metal ball of radius 1 mm and density 8000 kg/m$^3$ is dropped into a deep jar of glycerine of density 1250 kg/m$^3$ and viscosity 1.0 Pa s. With $g=10$ m/s$^2$, find its terminal velocity.

• At terminal velocity weight $=$ buoyancy $+$ Stokes drag, giving $v_t=\dfrac{2r^2(\rho-\sigma)g}{9\eta}$.
• Density difference $\rho-\sigma=8000-1250=6750$ kg/m$^3$; numerator $=2\left(10^{-3}\right)^2(6750)(10)=0.135$, so $v_t=\dfrac{0.135}{9(1.0)}=0.015$ m/s.

Answer: $v_t=0.015$ m/s, i.e. 1.5 cm/s. Using the ball's density alone in place of the difference $(\rho-\sigma)$ overstates the speed.

Example 6 — Torricelli efflux and discharge. A large tank holds water to a depth of 5 m. A small hole of area 2 cm$^2$ is opened low in the side wall, 5 m below the free surface. With $g=10$ m/s$^2$, find the jet speed and the volume leaving per second.

• Torricelli (Bernoulli with the broad surface almost at rest): $v=\sqrt{2gh}=\sqrt{2(10)(5)}=\sqrt{100}=10$ m/s.
• Discharge: $Q=a v=\left(2\times10^{-4}\right)(10)=2\times10^{-3}$ m$^3$/s.

Answer: $v=10$ m/s and $Q=2\times10^{-3}$ m$^3$/s, i.e. 2 litre per second. The efflux speed is that of a body freely fallen through the head $h$, and does not depend on the hole's size.

Example 7 — Bernoulli in a venturi. Water ($\rho=1000$ kg/m$^3$) flows steadily through a horizontal pipe that narrows from 10 cm$^2$ to 5 cm$^2$; in the wide part the speed is 2 m/s. Find the speed in the throat and the pressure drop between the two sections.

• Continuity: $v_2=\dfrac{A_1}{A_2}v_1=\dfrac{10}{5}(2)=4$ m/s.
• Bernoulli (horizontal): $P_1-P_2=\tfrac12\rho\left(v_2^2-v_1^2\right)=\tfrac12(1000)(16-4)=6000$ Pa.

Answer: $v_2=4$ m/s and the pressure falls by 6000 Pa (6 kPa) in the throat. Keep the $v_1^2$ term — the water was already moving before the constriction.

Example 8 — Reynolds number and flow regime. Water ($\rho=1000$ kg/m$^3$, $\eta=1.0\times10^{-3}$ Pa s) flows at 0.2 m/s through a pipe of diameter 2 cm. Find the Reynolds number and classify the flow.

• $Re=\dfrac{\rho v D}{\eta}=\dfrac{(1000)(0.2)(0.02)}{1.0\times10^{-3}}=\dfrac{4}{10^{-3}}=4000$.

Answer: $Re=4000$; being well above the pipe-flow threshold of about 2000, the flow is turbulent. The number is dimensionless and weighs inertial against viscous forces.