Buoyancy, Archimedes' Principle & Floatation

Every cargo ship's hull wears a Plimsoll line — a legally enforced set of marks showing the maximum safe waterline in summer seawater, winter North Atlantic, tropical fresh water and more, because equation (3) says the same ship floats deeper in warmer, fresher, less dense water. Submarines fly through the density contest deliberately: flood the ballast tanks to nudge average density above seawater's and dive; blow compressed air in to rise; hold it exactly equal and hover. Astronauts train for spacewalks in NASA's Neutral Buoyancy Laboratory, a 23-million-litre pool where suits are weighted to hover — buoyancy cancelling weight is the cheapest imitation of orbit we have. Fish beat engineers to all of it: the swim bladder is a biological ballast tank, inflated or deflated to hold depth without swimming. And meteorologists launch three thousand radiosonde balloons daily — hydrogen or helium filled, $\rho_{\text{air}}Vg$ doing the lifting all the way to the stratosphere.

Push a beach ball under water and hold it there. Your arms feel the fluid fighting back — tens of newtons of it. Where does that force actually come from? Not from some special "floating tendency": from plain pressure. The ball's lower surface sits deeper than its upper surface, and pressure grows with depth ($P = P_0 + \rho gh$). So the water presses up on the bottom harder than it presses down on the top. Horizontal pushes cancel by symmetry; the vertical mismatch survives. Buoyancy is the depth law's vertical imbalance, integrated over the body's skin.

Archimedes' genius was to skip the integration with a replacement argument. Freeze the fluid everywhere except a body-shaped region, and let that region be filled with fluid for a moment. That parcel of fluid floats in equilibrium — it doesn't sink or rise — so the surrounding fluid must be supplying an upward force exactly equal to the parcel's weight, acting through the parcel's centre of gravity. Now swap the parcel for the actual body. The surrounding fluid hasn't changed, so it exerts the same force: up, equal to the weight of the fluid that would fill that space — the displaced fluid. That is the whole principle, derived in a paragraph, valid for any shape, any fluid, any depth. Three consequences organize every problem you will meet. If the body is denser than the fluid, its weight beats the maximum possible upthrust (even fully submerged, $\rho_{\text{body}}Vg > \rho_f Vg$): it sinks — though it weighs less on the way down. If densities match, it hovers, indifferent to depth: neutral buoyancy, the submarine's and the scuba diver's operating point. If the body is less dense, full submersion would overpay: the body rises and pokes out of the surface, shedding submerged volume until the upthrust from what remains under water exactly matches its weight. Floating is equilibrium by partial withdrawal.

The buoyant force on a body displacing volume $V_{\text{disp}}$ of a fluid of density $\rho_f$ is

where: $\rho_f$ — the fluid's density (the most-tested substitution slip in the chapter: never the body's); $V_{\text{disp}}$ — the volume of fluid pushed aside, which is the body's full volume only if fully submerged; $g$ — as your board prescribes. Read the structure: the body's material, mass, and depth are absent. A 1-litre lump of lead and a 1-litre lump of wood, both held fully under water, feel the same 9.8 N upthrust — they respond differently only because their weights differ.

Problems then split into two bookkeeping modes — identify which before touching numbers.

Fully submerged (sunk, held, or hovering): $V_{\text{disp}} = V$, the body's whole volume. The net downward force is $W - F_B = (\rho_{\text{body}} - \rho_f)Vg$, and the apparent weight — what a spring balance or your muscles feel — is

where: the bracketed factor is the fraction of weight surviving immersion. The "loss of weight" is exactly $F_B$ — Archimedes' own crown test: weigh in air, weigh in water, and the difference hands you the volume, hence the density, of any irregular object. For aluminium ($\rho = 2700$) in water the factor is $1 - 1000/2700 = 0.63$: it "loses" 37% of its weight.

Floating: the body chooses how much to submerge. Equilibrium demands $F_B = W$:

where: $V_{\text{sub}}$ — the submerged volume. The fraction submerged is the density ratio, with $g$ cancelled (so the same iceberg floats identically on the Moon). Ice at $917\ \mathrm{kg/m^3}$ in seawater at $1025$: $917/1025 = 0.895$ — the famous nine-tenths of the iceberg below the waterline. Equivalently, a floating body displaces its own weight of fluid — the form that instantly answers "a 200,000-tonne ship displaces how much water?" (200,000 tonnes of it.)

1 · Board level — apparent weight of a submerged stone (CBSE/ICSE pattern, 3 marks)

A stone of mass $2.5\ \mathrm{kg}$ and density $2500\ \mathrm{kg/m^3}$ hangs from a spring balance, fully submerged in water ($\rho_f = 1000\ \mathrm{kg/m^3}$, $g = 9.8\ \mathrm{m/s^2}$). Find (a) the buoyant force and (b) the balance reading.

Step 1. Volume from mass and density: $V = m/\rho_b = 2.5/2500 = 1.0\times10^{-3}\ \mathrm{m^3}$ — one litre.

Step 2. (a) Upthrust, equation (1) with the fluid's density: $F_B = 1000 \times 1.0\times10^{-3} \times 9.8 = \boxed{9.8\ \mathrm{N}}$.

Step 3. (b) True weight $W = 2.5 \times 9.8 = 24.5\ \mathrm{N}$; apparent weight $W_{\text{app}} = 24.5 - 9.8 = \boxed{14.7\ \mathrm{N}}$ — the balance reads $1.5\ \mathrm{kg}$.

Step 4. Cross-check with equation (2): $W(1 - \rho_f/\rho_b) = 24.5(1 - 0.4) = 14.7$ N ✓. The stone "lost" 40% of its weight — exactly $\rho_f/\rho_b$.

2 · Competitive level — the iceberg and its cousins (NEET/JEE-Main pattern, ~90 seconds)

Ice has density $917\ \mathrm{kg/m^3}$. (a) What fraction of an iceberg lies below the surface in seawater ($1025\ \mathrm{kg/m^3}$)? (b) A block floats in water with one-third of its volume above the surface — find its density.

Step 1. (a) Equation (3): $\dfrac{V_{\text{sub}}}{V} = \dfrac{917}{1025} = \boxed{0.895}$ — about 90% hidden. (In fresh water, $917/1000 = 0.917$: even more submerged. Denser fluid = higher ride.)

Step 2. (b) One-third above means two-thirds submerged: $\dfrac{\rho_b}{\rho_f} = \dfrac{2}{3}$, so $\rho_b = \dfrac{2}{3}\times1000 = \boxed{667\ \mathrm{kg/m^3}}$. Timing note: fraction-floating questions are pure ratio — no $g$, no volumes, 30 seconds each. The reliable trap is "above" vs "below": underline which fraction the question states, then convert. An option matching the wrong side (333 kg/m³ here) is always planted.

3 · Competitive level — two weighings, two liquids (JEE-Main pattern, ~3 minutes)

A metal piece weighs $7.86\ \mathrm{N}$ in air, $6.86\ \mathrm{N}$ fully submerged in water, and $7.06\ \mathrm{N}$ fully submerged in an unknown oil. Take $\rho_{\text{water}} = 1000\ \mathrm{kg/m^3}$, $g = 10\ \mathrm{m/s^2}$. Find (a) the metal's volume and density, (b) the oil's density.

Step 1. Water's upthrust = weight loss in water: $F_B^{w} = 7.86 - 6.86 = 1.00\ \mathrm{N}$.

Step 2. (a) Volume from $F_B^w = \rho_w V g$: $V = \dfrac{1.00}{1000\times10} = \boxed{1.0\times10^{-4}\ \mathrm{m^3}}$ (100 cm³). Mass $= 7.86/10 = 0.786$ kg, so $\rho_b = 0.786/10^{-4} = \boxed{7860\ \mathrm{kg/m^3}}$ — iron.

Step 3. (b) Oil's upthrust: $7.86 - 7.06 = 0.80\ \mathrm{N}$ on the same volume, so $\rho_{\text{oil}} = \dfrac{0.80}{10^{-4}\times10} = \boxed{800\ \mathrm{kg/m^3}}$.

Step 4. Shortcut worth memorizing: $\dfrac{\rho_{\text{oil}}}{\rho_w} = \dfrac{\text{loss in oil}}{\text{loss in water}} = \dfrac{0.80}{1.00}$ — relative density by two weighings, no volume needed. Timing note: this is Archimedes' crown method in modern dress and a JEE regular. The weight-loss-ratio shortcut (step 4) turns a 3-minute solve into 60 seconds; it works because the same $V$ and $g$ cancel.

Wrong density in $F_B$. The formula takes the fluid's density and the displaced volume. Writing $\rho_{\text{body}}$ in the upthrust (or the body's full $V$ for a floating object) is the chapter's top mark-loser. Say it as a sentence — "weight of displaced fluid" — before substituting.

Floating vs submerged bookkeeping. Floating: force balance $F_B = W$, displaced weight = body's weight. Submerged: $V_{\text{disp}} = V$, apparent weight $= W - \rho_f Vg$. Misfiling the problem costs every subsequent line.

"Above" vs "below" fractions. $V_{\text{sub}}/V = \rho_b/\rho_f$ gives the fraction below. Questions quote the fraction visible; subtract from 1 first. Both numbers appear in the options.

Buoyancy doesn't grow with depth. In an incompressible fluid the upthrust on a fully submerged body is the same at 1 m and 100 m ($d$ cancelled in the derivation). Distractors claiming "sinks faster because buoyancy weakens" are false at HS level (real-world compressibility subtleties are beyond board scope).

Buoyancy needs a pressure gradient, i.e. effective gravity. In a freely falling lift (or orbiting station), $g_{\text{eff}} = 0$: no depth law, no upthrust — a cork released underwater in free fall stays put. In an accelerating lift, replace $g \to g + a$ in both weight and upthrust; the fraction submerged of a floater is unchanged (both scale together — a favourite assertion–reason item).

Air is a fluid too. A helium balloon rises because $\rho_{\text{air}}Vg > W$; a body's "weight in air" is already minutely reduced by air's upthrust (precision chemistry corrects for it). If a question compares vacuum and air weighings, that correction is the question.

The upthrust's line of action. Through the centre of buoyancy — the centroid of the displaced fluid — not the body's centre of gravity. For a uniform fully submerged body the two coincide; for a floating ship they don't, and their interplay decides stability (archetype territory).

Which $g$? CBSE 9.8; JEE/NEET usually 10 as stated; A-Level 9.81. Fraction-submerged answers are $g$-free — if your fraction contains $g$, you've made an algebra slip.

Texts. NCERT Physics Class 11, Ch. 9 · HC Verma Vol. 1, Ch. 13 · Archimedes, On Floating Bodies (Prop. 5–7, in translation — genuinely readable).

Provenance. All results standard. A verifier should recompute: the derivation's $F_B = \rho_f ghA$; example 1's 9.8 N and 14.7 N; example 2's 0.895 and 667 kg/m³; example 3's $10^{-4}$ m³, 7860 kg/m³ (iron ✓), 800 kg/m³.

Prerequisites. Fluids at Rest: Pressure & Pascal's Law. Leads to. Floating-Body Problems (archetype) · Continuity & Bernoulli · The Bobbing Float (SHM archetype).