Mechanical Properties of Fluids - Vardaan Master Notes

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References

Compiled from: NCERT Physics Part-2 (Mechanical Properties of Fluids) • H.C. Verma — Concepts of Physics Vol-1 (Ch. 13: Fluid Mechanics, Ch. 14: Some Mechanical Properties of Matter) • D.C. Pandey — Understanding Physics: Properties of Matter • Pradeep's Fundamental Physics XI • S.L. Arora • Cengage (B.M. Sharma) • Errorless Physics. Target exams: Boards • JEE Main • JEE Advanced • NEET • NDA.

1. Fluid Pressure & its Variation with Depth

A fluid (liquid or gas) cannot resist shear — it flows. At rest it exerts force only perpendicular to any surface. Density ρ = m/V (kg/m³); relative density (specific gravity) = ρ_substance/ρ_water (dimensionless); ρ_water = 1000 kg/m³; ρ_mercury = 13.6 × 10³ kg/m³.

Pressure $$P = \frac{F_{\perp}}{A} \qquad \text{At depth h: } \boxed{P = P_0 + \rho gh}$$

Scalar; SI unit pascal (Pa) = N/m²; dimension $[ML^{-1}T^{-2}]$. 1 atm = 1.013 × 10⁵ Pa = 760 mm of Hg = 760 torr; 1 bar = 10⁵ Pa. Gauge pressure = P − P_atm = ρgh; absolute pressure = P₀ + ρgh.

Key Facts

Pressure at a point in a static fluid is the same in all directions (that's why it is a scalar).
Pressure is the same at all points at the same horizontal level in a connected fluid — the basis of every U-tube problem (hydrostatic paradox: shape/amount of vessel doesn't matter, only depth).
Barometer (Torricelli): P_atm = ρ_Hg g h, with h = 76 cm; the space above mercury is a near-vacuum (Torricellian vacuum). Water barometer would need ≈ 10.3 m!
Manometer: measures gauge pressure of a gas via the height difference of liquid columns: P_gas = P₀ ± ρgh.
U-tube with two liquids: equate pressures at the common interface level: ρ₁h₁ = ρ₂h₂.

Image Placeholder — Fig. 1

Pressure vs depth, barometer, and the two-liquid U-tube (Landscape • Background #ffffff)

AI IMAGE PROMPT: Create a three-panel physics textbook diagram in a row, LANDSCAPE orientation (16:9), pure white background (#ffffff). PANEL 1 titled "P = P₀ + ρgh": a tall open beaker drawn as a thin black outline filled with light blue (#e3f2fd) liquid up to a surface line labelled "P₀ (atmosphere)"; a dashed vertical dimension arrow from the surface down to a marked point labelled "depth h"; at that point a small dot with four short red (#d32f2f) arrows pointing in all four directions, caption "pressure same in all directions"; boxed formula "P = P₀ + ρgh (depends only on depth, not on shape)". PANEL 2 titled "Mercury barometer": a dish of mercury (grey liquid) with an inverted closed glass tube standing in it; mercury column risen inside the tube with the empty top region labelled "Torricellian vacuum"; a dimension arrow along the column labelled "h = 76 cm"; equation below "P_atm = ρ_Hg g h = 1.013 × 10⁵ Pa". PANEL 3 titled "U-tube, two liquids": a U-shaped thin black tube; the left arm contains a taller column of light blue liquid labelled "ρ₁, height h₁" above the interface, the right arm a shorter column of orange (#fff1e8 fill with #d9480f outline) liquid labelled "ρ₂, height h₂"; a dashed horizontal line through the interface level labelled "same level ⟹ same pressure"; boxed relation "ρ₁h₁ = ρ₂h₂". Style: flat minimalist vector textbook figure, thin black outlines, light-blue and orange liquids exactly as specified, dashed reference lines, clean sans-serif labels, no gradients, no shadows, pure white background.

Three instruments of hydrostatics — the source of every Board's "define/derive/describe" question in this part.

✍ IN-TEXT PRACTICE 1.1 — Boards / NEET

Q. Find the absolute and gauge pressure at the bottom of a 10 m deep water tank. (P₀ = 1.01 × 10⁵ Pa, ρ = 1000 kg/m³, g = 10 m/s²)

Gauge: $\rho gh = 1000\times10\times10 = \mathbf{10^5 \text{ Pa}}$ (≈ 1 atm — every 10 m of water adds one atmosphere!)

Absolute: $P = 1.01\times10^5 + 10^5 = \mathbf{2.01\times10^5 \text{ Pa}}$

✍ IN-TEXT PRACTICE 1.2 — JEE Main (U-tube)

Q. Water is poured into one arm of a U-tube containing mercury. What height of water column balances a 2 mm rise of mercury level difference? (Rel. density of Hg = 13.6)

Balance at interface: $\rho_wh_w = \rho_{Hg}h_{Hg} \Rightarrow h_w = 13.6\times2 = \mathbf{27.2 \text{ mm}}$

2. Pascal's Law & Hydraulic Machines

Pascal's Law

Statement: Pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container.

Hydraulic lift (force multiplier): pistons of areas a (small) and A (large):

$$\frac{f}{a} = \frac{F}{A} \;\Rightarrow\; F = f\cdot\frac{A}{a}$$

Force is multiplied by A/a, but the small piston moves farther in the same ratio ($a\,d_1 = A\,d_2$, volume conserved) — work is NOT multiplied (energy conservation). Same principle: hydraulic brakes (equal braking effort transmitted to all wheels), hydraulic press/jack.

Image Placeholder — Fig. 2

Hydraulic lift — force multiplication with volume balance (Landscape • Background #ffffff)

AI IMAGE PROMPT: Create a professional physics textbook diagram, LANDSCAPE orientation (16:9), pure white background (#ffffff). Draw a U-shaped connected vessel (thin black outline) filled with light blue (#e3f2fd) liquid; the LEFT arm is NARROW with a small piston on top labelled "area a = 10 cm²", pushed down by a red (#d32f2f) downward arrow labelled "f = 100 N"; a small dashed double-headed arrow beside it labelled "moves down d₁"; the RIGHT arm is WIDE with a large piston carrying a simple car silhouette on a platform, pushed up by a green (#43a047) upward arrow labelled "F = f·A/a = 5000 N", area labelled "A = 500 cm²", with a shorter dashed double-headed arrow "moves up d₂ (a·d₁ = A·d₂)". Inside the liquid draw several small grey arrows along the tube pointing from the small piston toward the large piston labelled once "pressure transmitted undiminished (Pascal's law)". A dashed horizontal line connects the two liquid levels labelled "same pressure at same level: f/a = F/A". BOTTOM caption: "Force multiplied ×50, distance divided ÷50 — work is conserved". Style: flat minimalist vector textbook figure, thin black outlines, light-blue liquid, red input and green output arrows sized very differently, dashed guides, clean sans-serif labels, no gradients, no shadows, pure white background.

Numbers (10 cm², 500 cm², 100 N → 5000 N) form a ready worked example.

✍ IN-TEXT PRACTICE 2.1 — Boards / NDA

Q. In a hydraulic lift the piston areas are 5 cm² and 500 cm². What force on the small piston lifts a 1500 kg car? How far must it move to raise the car by 1 cm? (g = 10 m/s²)

Force: $f = F\cdot\dfrac{a}{A} = 15000\times\dfrac{5}{500} = \mathbf{150 \text{ N}}$

Distance: $d_1 = d_2\cdot\dfrac{A}{a} = 1\times100 = \mathbf{100 \text{ cm} = 1 \text{ m}}$ — the price of force multiplication.

3. Buoyancy, Archimedes' Principle & Floatation

Archimedes' Principle

A body wholly or partly immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced: $F_B = \rho_{fluid}\, V_{displaced}\,g$, acting through the centre of buoyancy (COM of displaced fluid). Cause: pressure is greater on the lower surface than the upper.

Floatation Results $$\text{Apparent weight} = W - F_B = V\rho_{body}g - V\rho_{fl}g$$ $$\text{Floating body: } W = F_B \;\Rightarrow\; \frac{V_{submerged}}{V_{total}} = \frac{\rho_{body}}{\rho_{fluid}}$$

Loss of weight in water = weight of water displaced; relative density of a body $= \dfrac{W_{air}}{W_{air} - W_{water}}$ (the classical lab method). Ice (ρ = 0.917) floats with ~92% of its volume under water (~8% visible) — the "tip of the iceberg".

⭐ CONCEPT TRAPS — ICE MELTING & SINK/FLOAT

Ice floating in water melts → level UNCHANGED (it becomes exactly the water it displaced).
Ice containing a stone melts → level FALLS (the sunken stone displaces only its own small volume afterwards).
Ice containing a trapped air bubble melts → level unchanged.
A boat carrying stones: stones thrown into the water → lake level falls (same reasoning).
Body floats if ρ_body ≤ ρ_fluid; a ship of steel floats because its average density (hollow hull) is less than water's.
Buoyant force is the same at every depth (for an incompressible fluid) — it depends on V and ρ_fl, not on how deep the body is.

Image Placeholder — Fig. 3

Archimedes' principle: sinking, suspended and floating bodies (Landscape • Background #ffffff)

AI IMAGE PROMPT: Create a three-panel physics textbook diagram in a row, LANDSCAPE orientation (16:9), pure white background (#ffffff). Each panel is a beaker (thin black outline) of light blue (#e3f2fd) water with a grey block, plus an FBD beside it with a red (#d32f2f) downward arrow "W = ρ_body V g" and a green (#43a047) upward arrow "F_B = ρ_fl V_disp g". PANEL 1 titled "Sinks (ρ_body > ρ_fl)": block resting at the bottom, red W arrow LONGER than green F_B; caption "apparent weight = W − F_B > 0". PANEL 2 titled "Just suspended (ρ_body = ρ_fl)": block fully submerged floating mid-water, arrows EQUAL; caption "W = F_B, apparent weight = 0". PANEL 3 titled "Floats (ρ_body < ρ_fl)": block floating at the surface with a fraction below the waterline; a dashed waterline through the block with dimension labels "submerged fraction = ρ_body/ρ_fl"; equal-length arrows W and F_B; small iceberg note: "ice in water: 92% submerged". Style: flat minimalist vector textbook figure, thin black outlines, light-blue water, grey blocks, red/green FBD arrows with lengths encoding the comparison, dashed waterlines, clean sans-serif labels, no gradients, no shadows, pure white background.

Arrow lengths = the entire sink/suspend/float logic.

✍ IN-TEXT PRACTICE 3.1 — NEET (Floating fraction)

Q. A block floats in water with 4/5 of its volume submerged. Find its density. In a liquid it floats with 2/3 submerged — find that liquid's density.

Water: $\rho_{body} = \dfrac45\times1000 = \mathbf{800 \text{ kg/m}^3}$

Liquid: $\dfrac{\rho_{body}}{\rho_{liq}} = \dfrac23 \Rightarrow \rho_{liq} = \dfrac{3}{2}\times800 = \mathbf{1200 \text{ kg/m}^3}$

✍ IN-TEXT PRACTICE 3.2 — Boards (Apparent weight)

Q. A metal piece weighs 210 g in air and 180 g in water. Find its relative density and the density of a liquid in which it weighs 186 g.

RD: $\dfrac{210}{210-180} = \dfrac{210}{30} = \mathbf{7}$

Liquid: loss in liquid/loss in water = $\dfrac{24}{30} = 0.8$ → $\rho_{liq} = \mathbf{800 \text{ kg/m}^3}$

4. Fluid Flow — Continuity Equation & Bernoulli's Theorem

4.1 Streamline vs Turbulent Flow

Streamline (laminar) flow: every particle passing a point follows the same path and velocity as its predecessor; streamlines never cross (two crossing lines would mean two velocities at one point — impossible). Speed below the critical velocity.
Turbulent flow: above critical velocity — eddies and whorls, energy dissipated.
Ideal fluid (for Bernoulli): incompressible, non-viscous, streamline, irrotational.

4.2 Equation of Continuity (Mass Conservation)

Continuity $$A_1v_1 = A_2v_2 \qquad (Av = \text{volume flow rate, constant})$$

Narrower tube → faster flow. Why a river speeds up in a gorge; why the falling water stream from a tap narrows as it accelerates (Av constant while v grows).

4.3 Bernoulli's Theorem (Full Derivation)

Statement: For streamline flow of an ideal fluid, the total energy per unit volume — pressure + kinetic + potential — is constant along a streamline.

Derivation (work–energy on a fluid element flowing from end 1 to end 2): net work by pressure forces = $P_1A_1v_1\Delta t - P_2A_2v_2\Delta t = (P_1 - P_2)\Delta V$. This equals the gain in KE + PE of the transferred mass $\Delta m = \rho\Delta V$:

$$(P_1-P_2)\Delta V = \tfrac12\rho\Delta V(v_2^2 - v_1^2) + \rho\Delta V g(h_2 - h_1)$$

Bernoulli's Equation $$\boxed{\,P + \tfrac12\rho v^{2} + \rho gh = \text{constant}\,}$$

Each term is energy per unit volume (= pressure). For a horizontal pipe: $P + \tfrac12\rho v^2$ = constant — faster flow ⟹ lower pressure, the key to every application.

4.4 Applications — the Full Exam List

Application	Explanation in one line
Venturimeter	Flow speed from the pressure drop at a constriction: $v_1 = A_2\sqrt{\dfrac{2(P_1-P_2)}{\rho(A_1^2-A_2^2)}}$ — measures flow rate in pipes
Atomizer / sprayer / Bunsen burner	Fast air over the tube lowers pressure → liquid/gas is pushed up and dragged along
Aerofoil (aeroplane lift)	Air moves faster over the curved top → lower pressure above → net upward lift
Magnus effect (spinning ball swings)	Spin makes air faster on one side → pressure difference curves the flight
Blowing over a paper / two boats colliding	Fast stream between/above → low pressure → paper rises, boats drift together
Roofs blown off in storms	Fast wind above the roof lowers pressure; higher pressure inside pushes the roof UP

4.5 Torricelli's Theorem — Speed of Efflux

Hole at depth h below the free surface of an open tank: apply Bernoulli between surface (v ≈ 0) and hole (both at P₀):

$$\boxed{v = \sqrt{2gh}}\ \text{— same as free fall from height h.}$$

Range of the jet (hole at height y from the ground, depth h from surface, tank on the ground of total head H = h + y): $R = 2\sqrt{hy}$; R is maximum = H when the hole is at the middle (h = y = H/2). Two holes at depths h and H − h give the same range — a JEE favourite pair.

Image Placeholder — Fig. 4

Continuity + Bernoulli tube + Torricelli efflux in one master figure (Landscape • Background #ffffff)

AI IMAGE PROMPT: Create a two-panel physics textbook diagram side by side, LANDSCAPE orientation (16:9), pure white background (#ffffff). LEFT PANEL titled "Continuity + Bernoulli": draw a pipe flowing left to right that starts WIDE at a lower level, rises, and becomes NARROW at a higher level (smooth taper), outlined in black and filled light blue (#e3f2fd); at the wide inlet mark "A₁, v₁, P₁, height h₁" with a short green (#43a047) flow arrow, at the narrow outlet mark "A₂, v₂, P₂, height h₂" with a LONGER green flow arrow labelled "narrower ⟹ faster (A₁v₁ = A₂v₂)"; draw thin grey streamlines inside the pipe that squeeze together in the narrow part; two small vertical pressure-gauge tubes on the pipe with liquid columns — TALLER column on the wide section, SHORTER on the narrow one, labelled "faster flow ⟹ lower pressure"; boxed equation below: "P + ½ρv² + ρgh = constant". RIGHT PANEL titled "Torricelli's efflux": a tall open tank of light blue water; free surface labelled "P₀, v ≈ 0"; a small hole in the side wall at depth marked "h below surface" and height "y above ground"; from the hole a blue parabolic jet curving to the ground with an initial horizontal green arrow labelled "v = √(2gh)"; ground distance dimension arrow labelled "R = 2√(hy), max when hole at mid-height"; a faint dashed second jet from the mirror depth labelled "hole at H − h: same range". Style: flat minimalist vector textbook figure, thin black outlines, light-blue fluid, grey streamlines, green velocity arrows with proportional lengths, dashed comparison jet, clean sans-serif labels, no gradients, no shadows, pure white background.

Squeezed streamlines + gauge columns encode both laws; the twin-jet shows the equal-range pair.

✍ IN-TEXT PRACTICE 4.1 — NEET (Continuity + Bernoulli numbers)

Q. Water flows at 2 m/s through a horizontal pipe of cross-section 8 cm², which narrows to 2 cm². Find the speed in the narrow part and the pressure drop. (ρ = 1000 kg/m³)

Continuity: $v_2 = \dfrac{A_1v_1}{A_2} = \dfrac{8\times2}{2} = \mathbf{8 \text{ m/s}}$

Bernoulli (horizontal): $\Delta P = \tfrac12\rho(v_2^2 - v_1^2) = 500(64-4) = \mathbf{3\times10^4 \text{ Pa}}$

✍ IN-TEXT PRACTICE 4.2 — JEE Main (Torricelli)

Q. A tank filled to height 5 m stands on the ground. A small hole is punched 1.25 m above the base. Find the efflux speed and the horizontal range of the jet. (g = 10 m/s²)

Depth of hole: $h = 5 - 1.25 = 3.75$ m → $v = \sqrt{2gh} = \sqrt{75} \approx \mathbf{8.66 \text{ m/s}}$

Range: $R = 2\sqrt{hy} = 2\sqrt{3.75\times1.25} = 2\sqrt{4.6875} \approx \mathbf{4.33 \text{ m}}$

5. Viscosity — Internal Friction of Fluids

Definition

Viscosity is the property by which a fluid opposes relative motion between its layers (internal friction). A layer drags its slower neighbour forward and is dragged backward by it. Newton's law of viscous flow:

Newton's Law of Viscosity $$F = -\eta A\frac{dv}{dx}$$

η = coefficient of viscosity; dv/dx = velocity gradient. SI unit of η: Pa·s (= poiseuille); CGS: poise; 1 Pa·s = 10 poise. Dimension $[ML^{-1}T^{-1}]$. Temperature effect: liquids — η decreases on heating (hot honey flows); gases — η increases on heating (a favourite MCQ reversal).

5.1 Stokes' Law & Terminal Velocity (Full Derivation)

Stokes' law: viscous drag on a sphere of radius r moving slowly at speed v through a fluid of viscosity η:

$$F = 6\pi\eta rv$$

Terminal velocity: a sphere falling in a fluid accelerates until weight = buoyancy + drag; thereafter it falls at constant $v_t$:

$$\tfrac43\pi r^3\rho g = \tfrac43\pi r^3\sigma g + 6\pi\eta rv_t$$ $$\boxed{\,v_t = \frac{2r^{2}(\rho - \sigma)g}{9\eta}\,}$$

ρ = sphere density, σ = fluid density. Note $v_t \propto r^2$ — big raindrops fall faster; tiny cloud droplets practically float. If σ > ρ (air bubble in water), v_t is negative — the bubble rises at terminal speed. Applications: parachute descent, millikan's oil-drop experiment, rain reaching ground at safe speed.

5.2 Poiseuille's Formula & Reynolds Number

Poiseuille (laminar flow through a tube): volume flow rate $Q = \dfrac{\pi P r^{4}}{8\eta L}$ — the fierce r⁴ dependence (halve the radius → 1/16 the flow; why slightly narrowed arteries matter so much).
Reynolds number: $R_e = \dfrac{\rho vD}{\eta}$ (dimensionless). $R_e \lesssim 1000$: laminar; $\gtrsim 2000$: turbulent; between: unstable. Critical velocity $v_c = \dfrac{R_e\eta}{\rho D}$.

Image Placeholder — Fig. 5

Terminal velocity: force balance and the v–t growth curve (Landscape • Background #ffffff)

AI IMAGE PROMPT: Create a two-panel physics textbook diagram side by side, LANDSCAPE orientation (16:9), pure white background (#ffffff). LEFT PANEL titled "Force balance at terminal velocity": a tall cylinder (thin black outline) filled with light blue (#e3f2fd) viscous liquid; a small grey sphere midway down with a short black downward arrow beside it labelled "falls at constant v_t"; FBD on the sphere: one red (#d32f2f) downward arrow "W = 4/3 πr³ρg" balanced by TWO upward arrows stacked — a blue (#1e88e5) arrow "F_B = 4/3 πr³σg (buoyancy)" and a green (#43a047) arrow "F_v = 6πηr v_t (Stokes drag)"; boxed result "W = F_B + F_v ⟹ v_t = 2r²(ρ − σ)g / 9η"; small note "v_t ∝ r² — bigger drops fall faster". RIGHT PANEL titled "Approach to terminal velocity": a first-quadrant graph, x-axis "time t", y-axis "speed v"; a blue curve rising steeply from the origin and smoothly flattening to a horizontal dashed asymptote labelled "v_t (terminal velocity)"; annotate the early part "a ≈ g initially" and the flat part "a = 0, net force zero". Style: flat minimalist vector textbook figure, thin black outlines, light-blue liquid, red/blue/green FBD arrows with the two upward arrows together matching the downward arrow's length, dashed asymptote, clean sans-serif labels, no gradients, no shadows, pure white background.

The saturation curve is itself an exam question: "sketch v vs t for a sphere dropped in a viscous liquid."

✍ IN-TEXT PRACTICE 5.1 — NEET (Terminal velocity scaling)

Q. A raindrop of radius r falls with terminal velocity 5 cm/s. Find the terminal velocity of a drop of radius 2r. If eight such small drops coalesce into one, what is the big drop's terminal velocity?

v_t ∝ r²: radius 2r → $v = 4\times5 = \mathbf{20 \text{ cm/s}}$

Coalescence: $8\cdot\tfrac43\pi r^3 = \tfrac43\pi R^3 \Rightarrow R = 2r$ → again $\mathbf{20 \text{ cm/s}}$ (i.e., $v \propto n^{2/3}v_0$ with n = 8).

6. Surface Tension — Complete Theory

6.1 Definition & Surface Energy

Surface Tension $$S = \frac{F}{L} \ (\text{force per unit length of an imaginary line on the surface})$$ $$S = \frac{\text{Surface Energy}}{\text{Area}} \ \Rightarrow\ W = S\,\Delta A$$

SI unit N/m = J/m²; dimension $[MT^{-2}]$. Cause: unbalanced inward cohesive pull on surface molecules — the surface behaves like a stretched elastic membrane and minimises its area (drops and bubbles are spherical: minimum area for a given volume). A soap film/bubble has TWO surfaces: work to blow a bubble of radius R: $W = S\cdot2\cdot4\pi R^2 = 8\pi R^2S$. Splitting one big drop into n small ones costs energy: $W = 4\pi S(nr^2 - R^2)$ with $R = n^{1/3}r$; coalescing releases the same energy (as heat).

Temperature: S decreases with rising temperature (zero at critical temperature). Hot soup tastes better — lower S spreads it over the tongue.
Impurities: detergents/soap lower S of water (better wetting/cleaning); highly soluble salts raise it slightly. Camphor dances on water: it dissolves unevenly, lowering S locally — stronger S elsewhere tugs it around.
Examples: needle floats on water; insects walk on ponds; hair of a paintbrush cling together when wet.

6.2 Angle of Contact

Pair	Angle of contact θ	Behaviour
Water–clean glass	≈ 0° (acute)	Wets the surface; meniscus concave; liquid RISES in capillary
Mercury–glass	≈ 135–140° (obtuse)	Does not wet; meniscus convex; liquid FALLS in capillary
Water–waxy/lotus leaf	Obtuse	Water beads up (does not wet)

θ depends only on the pair of materials (and temperature/impurities) — not on the tube's radius or inclination. Detergents work by making θ small (better wetting); waterproofing agents make θ large.

6.3 Excess Pressure Inside Drops & Bubbles

Excess Pressure (concave side is at higher pressure) $$\text{Liquid drop / air bubble in liquid (1 surface): } \Delta P = \frac{2S}{R}$$ $$\text{Soap bubble in air (2 surfaces): } \Delta P = \frac{4S}{R}$$

Smaller bubble → larger excess pressure. Connect a small and a big soap bubble by a tube: air flows from small to big — the small one shrinks further (classic counter-intuitive MCQ). Two bubbles r₁ < r₂ in contact: the common interface bulges toward the bigger one with radius $r = \dfrac{r_1r_2}{r_2 - r_1}$ (JEE).

6.4 Capillarity (Full Derivation of the Ascent Formula)

Derivation (pressure method): in a tube of radius r, the concave meniscus has radius $R = r/\cos\theta$. Just below the meniscus the pressure is lower than atmospheric by $2S/R$. Liquid rises height h until $\rho gh$ restores the balance:

$$\rho gh = \frac{2S\cos\theta}{r}$$

Capillary Ascent Formula $$\boxed{\,h = \frac{2S\cos\theta}{r\rho g}\,}$$

h ∝ 1/r (thinner tube, higher rise — Jurin's law). Mercury (θ obtuse, cos θ < 0) is depressed. Tube shorter than h: the liquid does NOT overflow — the meniscus flattens (R increases) so that the reduced ΔP matches the available height (asked in JEE & Boards). Applications: rise of oil in a wick, water in soil and plant xylem, blotting paper, towel absorbing water.

Image Placeholder — Fig. 6

Excess pressure family + capillary rise and fall (Landscape • Background #ffffff)

AI IMAGE PROMPT: Create a two-panel physics textbook diagram side by side, LANDSCAPE orientation (16:9), pure white background (#ffffff). LEFT PANEL titled "Excess pressure ΔP = 2S/R or 4S/R": three circles in a row — (1) a filled light-blue circle labelled "liquid drop: 1 surface, ΔP = 2S/R"; (2) a white circle with a light-blue ring outline inside a light-blue rectangle background labelled "air bubble in liquid: 1 surface, ΔP = 2S/R"; (3) a circle drawn with a DOUBLE outline (two concentric close circles) labelled "soap bubble in air: 2 surfaces, ΔP = 4S/R"; below them a small sketch of a SMALL and a LARGE soap bubble connected by a thin tube with a grey arrow inside the tube pointing from small to large, captioned "air flows small → large (smaller bubble has higher pressure)". RIGHT PANEL titled "Capillarity": a wide dish of light-blue water with TWO thin vertical glass tubes standing in it — the LEFT tube shows water RISEN above the dish level with a CONCAVE meniscus, a dimension arrow labelled "h = 2S cos θ / (rρg)" and a small angle mark at the wall labelled "θ ≈ 0° (wets)"; the RIGHT tube stands in a separate small dish of grey mercury with the level DEPRESSED below the dish level and a CONVEX meniscus, labelled "mercury: θ obtuse ⟹ depression"; caption below "thinner tube → higher rise (h ∝ 1/r)". Style: flat minimalist vector textbook figure, thin black outlines, light-blue water and grey mercury, double-outline soap bubble, correct concave/convex menisci, dimension arrows, clean sans-serif labels, no gradients, no shadows, pure white background.

The 1-surface vs 2-surface distinction and the two menisci are exactly what examiners check.

✍ IN-TEXT PRACTICE 6.1 — NEET (Excess pressure)

Q. Find the excess pressure inside (a) a water drop of radius 1 mm (S = 0.072 N/m), (b) a soap bubble of radius 1 cm (S_soap = 0.025 N/m).

(a) Drop: $\Delta P = \dfrac{2S}{R} = \dfrac{2\times0.072}{10^{-3}} = \mathbf{144 \text{ Pa}}$

(b) Soap bubble: $\Delta P = \dfrac{4S}{R} = \dfrac{4\times0.025}{10^{-2}} = \mathbf{10 \text{ Pa}}$

✍ IN-TEXT PRACTICE 6.2 — Boards / JEE (Capillary rise)

Q. Water rises 6 cm in a capillary tube. How high will it rise in a tube of half the radius? What happens in the original tube on the Moon (g/6)?

h ∝ 1/r: half radius → $h = \mathbf{12 \text{ cm}}$

h ∝ 1/g: on the Moon → $6\times6 = \mathbf{36 \text{ cm}}$ (if the tube is long enough; else the meniscus flattens at the top without overflowing).

7. Solved Examples — Every Question Type in the Chapter

One worked model of each distinct question type asked from this chapter, tagged by exam. Attempt each yourself first, then check.

✍ TYPE 1 — Boards · State, Prove & Derive

Q. State and prove Bernoulli's theorem (5 marks). / Derive the terminal velocity formula (3 marks). / Derive the capillary ascent formula (3 marks). / State Pascal's law and explain the hydraulic lift (3 marks).

Where to answer from: Section 4.3 (work–energy derivation with Fig. 4), Section 5.1 (three-force balance with Fig. 5), Section 6.4 (pressure method with Fig. 6), Section 2 with Fig. 2. Always state assumptions (ideal fluid, streamline flow) for full marks.

✍ TYPE 2 — NEET · Force on a Dam / Wall (Average Pressure)

Q. Find the net horizontal thrust on a dam wall of width w holding water to depth H.

Average gauge pressure: $\bar{P} = \dfrac{0 + \rho gH}{2}$ (linear variation).

Thrust: $F = \bar{P}\times(wH) = \mathbf{\tfrac12\rho gH^2w}$, acting at H/3 above the base (centroid of the pressure triangle) — why dams are thick at the bottom.

✍ TYPE 3 — JEE Main · Floating Between Two Liquids

Q. A cube of density 800 kg/m³ floats at the interface of water (below) and oil of density 600 kg/m³ (above), fully submerged. What fraction of its volume is in water?

Weight = total buoyancy: $800V = 1000V_w + 600(V - V_w)$

$200V = 400V_w \Rightarrow \dfrac{V_w}{V} = \mathbf{\dfrac12}$ — half in water, half in oil.

✍ TYPE 4 — NEET / NDA · Balloon / Density Mixtures

Q. Equal masses of two liquids of densities 2 g/cc and 3 g/cc are mixed. Find the density of the mixture. What if equal volumes are mixed instead?

Equal masses (harmonic-type): $\rho = \dfrac{2\rho_1\rho_2}{\rho_1+\rho_2} = \dfrac{2\times6}{5} = \mathbf{2.4 \text{ g/cc}}$

Equal volumes (arithmetic): $\rho = \dfrac{\rho_1+\rho_2}{2} = \mathbf{2.5 \text{ g/cc}}$ — mirror of the average-speed rules!

✍ TYPE 5 — JEE Main · Venturimeter / Aerofoil Numbers

Q. Air (ρ = 1.3 kg/m³) flows at 60 m/s over the top of a wing of area 25 m² and at 45 m/s below it. Find the lift.

Bernoulli: $\Delta P = \tfrac12\rho(v_t^2 - v_b^2) = 0.65(3600-2025) = 1024$ Pa

Lift: $F = \Delta P\times A = 1024\times25 \approx \mathbf{2.56\times10^4 \text{ N}}$

✍ TYPE 6 — JEE · Tank Emptying / Hole Pair

Q. A tank of head H = 4 m has two holes, at depths 1 m and 3 m below the surface. Compare their jet ranges on the ground.

R = 2√(hy): hole 1: $2\sqrt{1\times3} = 2\sqrt3$; hole 2: $2\sqrt{3\times1} = 2\sqrt3$ — equal ranges (depths h and H − h always pair up). Max range would need the hole at 2 m: R = H = 4 m.

✍ TYPE 7 — NEET · Splitting a Drop (Surface Energy)

Q. A water drop of radius 3 mm is split into 27 identical droplets. Find the work done. (S = 0.072 N/m)

Small radius: $r = R/n^{1/3} = 3/3 = 1$ mm.

Work = SΔA: $W = 4\pi S(27r^2 - R^2) = 4\pi(0.072)(27 - 9)\times10^{-6} = 4\pi(0.072)(18\times10^{-6}) \approx \mathbf{1.63\times10^{-5} \text{ J}}$

✍ TYPE 8 — JEE Main · Soap Bubble Radii Combination

Q. Two soap bubbles of radii 3 cm and 4 cm coalesce in vacuum isothermally into one bubble. Find its radius.

In vacuum, surface energy area is conserved via air content — the neat result: $R^2 = r_1^2 + r_2^2 = 9 + 16 = 25$

R = 5 cm. (With outside pressure, use $P_0(V_1+V_2) + \ldots$ — the vacuum case is the standard MCQ.)

✍ TYPE 9 — Boards · Poiseuille Scaling

Q. The radius of a tube is reduced to half while the pressure head and length stay the same. How does the volume flow rate change?

Q ∝ r⁴: new rate $= \left(\tfrac12\right)^4 = \mathbf{\dfrac{1}{16}}$ of the original — the r⁴ law.

✍ TYPE 10 — JEE Advanced Level · Capillary + Energy Subtlety

Q. Water rises to height h in a capillary of radius r. Show that the surface-tension force does work 2× the PE gained by the risen column. Where does the rest go?

Work by ST force: upward pull $2\pi rS\cos\theta$ acts through height h: $W = 2\pi rS\cos\theta\,h = \pi r^2\rho g h^2$ (using the ascent formula).

PE gained: mass $\pi r^2h\rho$ rises to CM height h/2: $U = \tfrac12\pi r^2\rho gh^2$ — exactly half of W.

The other half is dissipated as heat (viscous losses during the rise) — a beautiful energy audit asked in JEE Advanced.

8. Common Mistakes & Misconceptions — Final Checklist

Night Before Exam

Read this the night before the exam:

Pressure depends only on depth (and ρ, g) — never on the vessel's shape or the amount of liquid (hydrostatic paradox).
Gauge = ρgh; absolute = P₀ + ρgh. Manometers and most gauges read GAUGE pressure.
Hydraulic machines multiply force, never work — the small piston must travel proportionally farther.
Buoyant force = weight of DISPLACED fluid, not of the body; it is independent of depth.
Floating fraction = ρ_body/ρ_fluid; floating ice melting leaves the level unchanged (stone inside → level falls).
Continuity: narrower ⟹ faster. Bernoulli: faster ⟹ lower pressure. Chain them in that order.
Bernoulli needs an ideal fluid along a streamline — don't apply it across turbulent or viscous regions.
Torricelli speed √(2gh) uses depth below the FREE SURFACE, not height above the ground; holes at h and H − h have equal ranges.
Terminal velocity ∝ r²; coalescing n drops: v → n^(2/3)·v.
η of liquids falls on heating, of gases RISES — opposite behaviours.
Poiseuille's Q ∝ r⁴ — tiny radius changes cause huge flow changes.
Soap bubble has TWO surfaces: ΔP = 4S/R (drop: 2S/R); smaller bubble has HIGHER inside pressure — air flows small → big.
Capillary rise h ∝ 1/r and ∝ 1/g; a too-short tube does NOT overflow — the meniscus flattens.
Angle of contact depends on the material pair, not on tube radius or tilt.
Surface tension DECREASES with temperature; detergents lower it (that's the cleaning mechanism).

9. Rapid Revision — One-Page Formula Sheet

Concept	Formula	Condition / Note
Pressure at depth	$P = P_0 + \rho gh$	1 atm = 1.013×10⁵ Pa = 76 cm Hg
U-tube balance	$\rho_1h_1 = \rho_2h_2$	Equate P at the same level
Hydraulic lift	$F = f\,A/a$	$a\,d_1 = A\,d_2$ (work conserved)
Buoyant force	$F_B = \rho_{fl}V_{disp}g$	Apparent wt = W − F_B
Floating fraction	$\frac{V_{sub}}{V} = \frac{\rho_{body}}{\rho_{fl}}$	RD = W_air/(W_air − W_water)
Continuity	$A_1v_1 = A_2v_2$	Falling tap stream narrows
Bernoulli	$P + \tfrac12\rho v^2 + \rho gh = $ const	Ideal fluid, along a streamline
Venturimeter	$v_1 = A_2\sqrt{\frac{2\Delta P}{\rho(A_1^2-A_2^2)}}$	Flow-rate meter
Torricelli	$v = \sqrt{2gh}$	Range $R = 2\sqrt{hy}$, max = H at mid-depth
Force on a dam	$F = \tfrac12\rho gH^2w$	Acts at H/3 from the base
Viscous force	$F = -\eta A\,dv/dx$	η: Pa·s = 10 poise
Stokes' law	$F = 6\pi\eta rv$	Small sphere, laminar
Terminal velocity	$v_t = \frac{2r^2(\rho-\sigma)g}{9\eta}$	∝ r²; n drops coalesce → n^{2/3}v
Poiseuille	$Q = \frac{\pi Pr^4}{8\eta L}$	Q ∝ r⁴
Reynolds number	$R_e = \rho vD/\eta$	<1000 laminar, >2000 turbulent
Surface tension	$S = F/L = W/\Delta A$	Bubble blow work = 8πR²S
Excess pressure	Drop: $2S/R$; soap bubble: $4S/R$	Smaller bubble → higher P
Drop splitting	$W = 4\pi S(nr^2 - R^2)$	$R = n^{1/3}r$; coalescing releases heat
Capillary rise	$h = \frac{2S\cos\theta}{r\rho g}$	h ∝ 1/r, 1/g; short tube: meniscus flattens

Study Plan

How to use these notes: Day 1: Sections 1–3 (pressure, Pascal, Archimedes) + Figs 1–3; drill the ice-melting cases aloud. Day 2: Section 4 (continuity + Bernoulli + Torricelli) + Fig 4; write the Bernoulli derivation and all six applications from memory. Day 3: Section 5 (viscosity) + Fig 5; derive v_t unaided. Day 4: Section 6 (surface tension) + Fig 6, then all ten Types of Section 7 without looking, followed by the mistakes checklist. Finish every session by writing the formula sheet from memory.