Buoyancy — Archimedes' principle, derivation from pressure difference: Worked examples

2.2.7 · D3 · Physics › Fluid Mechanics › Buoyancy — Archimedes' principle, derivation from pressure d

Yeh D3 companion hai Buoyancy topic note ka. Parent note ne idea build kiya aur $F_{B} = ρ_{f} V_{disp} g$ derive kiya. Yahan hum har tarah ki situation dhundhte hain jo buoyancy throw kar sakti hai aur ek-ek solve karte hain — taaki koi bhi exam problem kabhi surprise na kare.

Scenario matrix

Har buoyancy question inhi cells mein se kisi ek mein aata hai. Neeche har worked example us cell ke saath tagged hai jise woh cover karta hai.

Cell	Case class	Kya cheez ise alag banati hai	Example
A	Fully submerged, $ρ_{o} > ρ_{f}$ (sinks)	$F_{B} < W$ ; net force / acceleration nikalo	Ex 1
B	Floating, partly submerged	$F_{B} = W$ ; fraction below surface nikalo	Ex 2
C	Apparent weight (fluid mein weighing)	Scale read karta hai $W - F_{B}$ ; $ρ_{o}$ nikalo	Ex 3
D	Do stacked fluids (layered)	Object ek boundary pe straddling karta hai; do displaced volumes	Ex 4
E	Buoyancy in a gas (balloon)	$ρ_{f}$ bahut chhota; lift = payload jo carry kar sakta hai	Ex 5
F	Degenerate: $ρ_{o} = ρ_{f}$ (neutral)	Object kahin bhi hover karta hai; $f = 1$ exactly	Ex 6
G	Limiting case: $ρ_{o} \to 0$ ya object poora air	$f \to 0$ ; ship-hull / "steel kyun tairta hai"	Ex 7
H	Exam twist: floating object load pakde / ice melting	Conservation of displaced volume	Ex 8

Upar ki table ek lambi list hai, lekin saath ke saath teeno cells actually ek single yes/no question se branch karti hain. Neeche ka decision tree woh branch drawn out hai — har problem se pehle isse trace karo, aur yeh batata hai kaun sa example imitate karna hai.

Recall Har cell ke peeche woh ek decision tree

Compare karo $F_{B}^{full} = ρ_{f} V g$ (buoyancy agar POORA object andar hota) ko $W = ρ_{o} V g$ se.

Agar $F_{B}^{full} \geq W$ toh floats → float branch use karo: Cell B (Ex 2), Cell F (Ex 6), Cell G (Ex 7), Cell H (Ex 8); submerged fraction $= ρ_{o} / ρ_{f}$ .
Agar $F_{B}^{full} < W$ toh sinks → sink branch use karo: Cell A (Ex 1), Cell C (Ex 3); poora volume displaced.
Layered fluids (Cell D, Ex 4) aur gases (Cell E, Ex 5) same test run karte hain lekin sahi $ρ_{f}$ har layer ke liye.

Yeh har table cell ko exactly ek worked example se map karta hai.

Ex 1 — Sinking cube: net force aur acceleration (Cell A)

Step 1 — Cube ka volume. $V = (0.20)^{3} = 8.0 \times 1 0^{- 3} m^{3}$ Yeh step kyun? Fully submerged matlab displaced volume = full volume; dono forces ke liye $V$ chahiye.

Step 2 — Buoyant force (FLUID density use karta hai). $F_{B} = ρ_{f} V g = 1000 \times 8.0 \times 1 0^{- 3} \times 9.8 = 78.4 N$ Yeh step kyun? Paani pusher hai, toh $ρ_{f}$ — brass nahi — yahan jaata hai.

Step 3 — Weight (OBJECT density use karta hai). $W = ρ_{o} V g = 8500 \times 8.0 \times 1 0^{- 3} \times 9.8 = 666.4 N$ Yeh step kyun? Weight woh hai jo gravity actual brass par pull karti hai, toh uski apni density.

Step 4 — Cube ki mass. $m = ρ_{o} V = 8500 \times 8.0 \times 1 0^{- 3} = 68 kg$ Yeh step kyun? Newton's law ko mass chahiye, weight nahi, $a = F_{net} / m$ ke denominator mein; mass hai density times volume (upar $m$ ki definition dekho).

Step 5 — Net force aur acceleration. Newton's second law se, net down = $W - F_{B}$ : $F_{net} = 666.4 - 78.4 = 588 N (down), a = \frac{F _{net}}{m} = \frac{588}{68} = 8.65 m/s^{2}$ Yeh step kyun? Object accelerate karta hai kyunki forces balance nahi hain ( $F_{B} < W$ ); net force ko Step 4 ki mass se divide karne par acceleration milta hai.

Verify: Buoyancy hai $78.4/666.4 = 11.8%$ of weight — chhota lekin zero nahi, toh cube vacuum mein se dheere doobta hai ( $8.65 < 9.8$ ). Units: $N / kg = m/s^{2}$ . ✓

Ex 2 — Floating block: fraction submerged (Cell B)

Figure s01 — ek floating wooden block jo water line se cut hua hai. Horizontal teal line $y = 4$ par paani ki surface hai. Us line ke neeche block ko halke orange mein draw kiya gaya hai — yeh dooba hua hissa hai jo paani displace karta hai. Line ke upar block ko solid orange mein draw kiya gaya hai — woh hissa jo hawa mein jhank raha hai. LEFT par, ek plum double-headed arrow upar-wali height ko span karta hai jis par likha hai "40% above (4.8 cm)"; RIGHT par, ek teal double-headed arrow neeche-wali height ko span karta hai jis par likha hai "60% below (displaces water)" — woh right-hand teal length exactly woh $f$ hai jo hamara formula compute karta hai. Block ke centre se guzarte do single arrows balanced forces dikhate hain: teal $F_{B}$ upar ki taraf, orange $W$ neeche ki taraf, equal length mein kyunki block float karta hai.

Figure — Buoyancy — Archimedes' principle, derivation from pressure difference

Step 1 — Floating equilibrium condition. Sirf submerged part paani displace karta hai: $ρ_{f} (f V) g = ρ_{o} V g$ Yeh step kyun? Floating matlab $F_{B} = W$ ; buoyancy sirf doobe hue volume $f V$ ko count karti hai (Figure s01 mein teal-arrow region).

Step 2 — $V g$ cancel karo, $f$ solve karo. $f = \frac{ρ _{o}}{ρ _{f}} = \frac{600}{1000} = 0.60$ Yeh step kyun? $V$ aur $g$ dono sides par hain — dekho Density and relative density: yeh purely ek density ratio hai. Yeh $f = 0.60$ Figure s01 mein teal-arrow ki length hai.

Step 3 — Upar wali height. Upar ka fraction $= 1 - 0.60 = 0.40$ hai, toh $0.40 \times 12 cm = 4.8 cm above water .$ Yeh step kyun? Uniform cross-section ke prism ke liye, volume fraction = height fraction — yeh plum arrow ka $4.8 cm$ hai.

Verify: $60%$ underwater intuition se match karta hai (lakdi paani se thodi halki hai, toh kaafi neeche ride karti hai). Agar $ρ_{o}$ $999$ hota, toh $f \to 0.999$ (muskil se tairi). ✓

Ex 3 — Apparent weight, density nikalo (Cell C)

Step 1 — Weight drop se buoyant force. $F_{B} = W_{air} - W_{apparent} = 147 - 107 = 40 N$ Yeh step kyun? Scale read karta hai $W - F_{B}$ ; woh $40 N$ missing exactly paani ka upthrust hai.

Step 2 — $F_{B} = ρ_{f} V g$ se volume. $V = \frac{F _{B}}{ρ _{f} g} = \frac{40}{1000 \times 9.8} = 4.08 \times 1 0^{- 3} m^{3}$ Yeh step kyun? Buoyancy formula invert karo — statue ki apni density unknown hai, lekin paani ki nahi.

Step 3 — Mass aur density. $m = \frac{W _{air}}{g} = \frac{147}{9.8} = 15.0 kg, ρ_{o} = \frac{m}{V} = \frac{15.0}{4.08 \times 1 0 ^{- 3}} = 3676 kg/m^{3}$ Yeh step kyun? Density hai mass per volume — dono ab known hain.

Verify: $ρ_{o} = 3676 > 1000$ , toh yeh doobni chahiye — consistent hai isse andar pakad ke rakhne ki zaroorat se. Relative density $\approx 3.7$ (ek stone/ceramic, solid metal nahi). ✓

Ex 4 — Object do fluids ke beech straddling (Cell D)

Figure s02 — do-fluid jar aur pressure-slice hint. Jar mein do coloured bands hain: ek halka orange OIL band upar ( $ρ_{1} = 800$ ) aur ek gahra teal WATER band neeche ( $ρ_{2} = 1000$ ), horizontal "oil–water boundary" line se separated, upar "oil surface" line ke saath. RIGHT par, do chhote arrows dikhate hain kyun do layers se buoyancy ADD hoti hai: ek chhota orange "oil push" arrow aur ek lamba teal "water push (bigger)" arrow, plum mein tagged "pushes ADD" — yeh Step 1 ki physics hai. Plum cube ( $ρ = 1500$ ) jar ke FLOOR par resting draw kiya gaya hai, plum annotation ke saath "sinks: rho_o > rho_water, no boundary float" — yeh woh answer hai jo hum prove karne wale hain, pehle se dikha diya taaki figure aur narrative agree karein: assumed boundary-float hota nahi hai.

Step 1 — DONO fluids se total buoyancy set up karo (boundary float assume karke). $A = (0.10)^{2} = 0.01 m^{2}$ maano. Suppose height $x$ paani mein hai aur $(0.10 - x)$ oil mein: $F_{B} = ρ_{1} A (0.10 - x) g + ρ_{2} A xg$ Yeh step kyun? Har fluid apna $ρ_{f} V_{disp} g$ contribute karta hai; total unka sum hai kyunki hydrostatic pressure layer by layer build up hoti hai — oil slice apna push add karta hai, paani slice ek bada push add karta hai. Yeh Figure s02 mein "pushes ADD" wali do-arrow hint hai.

Step 2 — Floating equilibrium: $F_{B} = W$ . $[ρ_{1} (0.10 - x) + ρ_{2} x] A g = ρ_{o} A (0.10) g$ Yeh step kyun? Agar yeh boundary par float karta hai, toh net force zero hai.

Step 3 — $A g$ cancel karo aur $x$ solve karo. $800 (0.10 - x) + 1000 x = 1500 \times 0.10 \Rightarrow 80 + 200 x = 150 \Rightarrow x = \frac{70}{200} = 0.35 m$ Yeh step kyun? Algebra — lekin $x = 0.35 m > 0.10 m$ ek $0.10 m$ cube ke liye impossible hai!

Step 4 — Impossible answer padho. $x >$ cube height matlab even fully paani mein rahne par yeh support nahi ho sakta: $ρ_{o} = 1500 > ρ_{2} = 1000$ . Cube jar ke floor par doob jaata hai — exactly jaisa Figure s02 pehle se dikha raha hai. Assumed boundary-float kabhi hoti nahi. Yeh step kyun? Hamesha sanity-check karo ki $0 \leq x \leq height$ . Ek degenerate answer tumhe physics bata raha hai: yeh cell Cell A (sinking) mein reduce ho jaati hai.

Verify: Max buoyancy $9.8 N <$ weight $14.7 N$ ⇒ sinks. Answer: koi equilibrium float nahi; cube jar ke floor par rest karta hai (figure se consistent). ✓

Ex 5 — Gas mein buoyancy: ek helium balloon (Cell E)

Step 1 — Displaced AIR se buoyant force. $F_{B} = ρ_{air} V g = 1.29 \times 5.0 \times 9.8 = 63.21 N$ Yeh step kyun? Balloon air ko side mein dhakelta hai; same $F_{B} = ρ_{f} V g$ apply hota hai, $ρ_{f} = ρ_{air}$ ke saath.

Step 2 — Andar ki gas aur skin ka weight. $W_{He} = ρ_{He} V g = 0.18 \times 5.0 \times 9.8 = 8.82 N, W_{skin} = 0.20 \times 9.8 = 1.96 N$ Yeh step kyun? Har woh cheez jo balloon carry karta hai neeche pull karti hai: helium ka bhi real (chhota) weight hota hai.

Step 3 — Payload ke liye net lift. $F_{lift} = F_{B} - W_{He} - W_{skin} = 63.21 - 8.82 - 1.96 = 52.43 N$ $m_{payload} = \frac{52.43}{9.8} = 5.35 kg$ Yeh step kyun? Balloon+gas ko float karwane ke baad jo bacha woh ek load hoist kar sakta hai.

Verify: Positive lift ⇒ balloon utha, expected jaisa kyunki $ρ_{He} ≪ ρ_{air}$ . Agar helium ka weight bhool jaate toh lift $8.82 N$ zyada bolte. Units: N/(m/s²)=kg. ✓

Ex 6 — Neutral buoyancy, degenerate case (Cell F)

Step 1 — Fully submerged hone par buoyant force. $F_{B} = ρ_{f} V g = 1025 \times 2.0 \times 9.8 = 20090 N$ Yeh step kyun? Pehle poochho ki fluid kitni zyada push karta hai jab poori capsule andar ho — yeh sabse bada upthrust available hai, aur yeh fluid density $ρ_{f}$ use karta hai.

Step 2 — Weight. $W = ρ_{o} V g = 1025 \times 2.0 \times 9.8 = 20090 N$ Yeh step kyun? $F_{B}$ se compare karne ke liye downward pull chahiye; weight object ki apni density $ρ_{o}$ use karta hai.

Step 3 — Net force = zero, everywhere. $F_{net} = F_{B} - W = 0$ . Submerged fraction hai $f = \frac{ρ _{o}}{ρ _{f}} = \frac{1025}{1025} = 1.$ Yeh step kyun? $f = 1$ matlab object exactly, poori tarah paani ke andar hai zero net force ke saath — yeh hover karta hai kisi bhi depth par (neutral buoyancy, jaise submarines aur machhliyan apni depth tune karti hain).

Verify: $F_{B} = W$ exactly ⇒ $a = 0$ ⇒ jahan bhi release karo wahan ruk jaata hai. Yeh float ( $ρ_{o} < ρ_{f}$ ) aur sink ( $ρ_{o} > ρ_{f}$ ) ke beech ki boundary hai. ✓

Ex 7 — Limiting case: steel ka jahaaz kyun taairta hai (Cell G)

Step 1 — Poori boat ka weight. $W = m g = 2000 \times 9.8 = 19600 N$ Yeh step kyun? Neeche ki pull ke liye sirf total mass matter karta hai, material se nahi; yahan mass $m = 2000 kg$ directly given hai.

Step 2 — Maximum available buoyancy (poori hull andar). $F_{B}^{m a x} = ρ_{f} V_{hull} g = 1000 \times 3.0 \times 9.8 = 29400 N$ Yeh step kyun? Yeh sabse bada upthrust hai jo hull shape generate kar sakta hai. $W$ se compare karo.

Step 3 — Compare: $F_{B}^{m a x} = 29400 > W = 19600$ ⇒ floats. Ise sirf itna displace karna hai taaki weight match ho: $V_{disp} = \frac{W}{ρ _{f} g} = \frac{19600}{1000 \times 9.8} = 2.0 m^{3}, f = \frac{2.0}{3.0} = 0.667.$ Yeh step kyun? Equilibrium mein boat utni gehra doobti hai jitne mein displaced-water weight uske apne weight ke barabar ho — yahan hull ka $2/3$ .

Verify: Average density $667 < 1000$ ⇒ floats. $f = ρ_{o} / ρ_{f} = 667/1000 = 0.667$ Step 3 se match karta hai. ✓

Ex 8 — Exam twist: glass mein ice melting (Cell H)

Step 1 — Floating ice se displaced paani. Floating ⇒ buoyancy = ice weight: $ρ_{w} V_{disp} g = m_{ice} g \Rightarrow V_{disp} = \frac{m _{ice}}{ρ _{w}} .$ Yeh step kyun? Ice ka dooba hua hissa exactly utne paani ka weight displace karta hai jitna poori ice ka weight hai.

Step 2 — Ice pighalti hai toh bana paani. Ice ban jaati hai $m_{ice}$ liquid water, jisme jagah leti hai $V_{melt} = \frac{m _{ice}}{ρ _{w}} .$ Yeh step kyun? Mass conserved hai; pighi hui paani ki density $ρ_{w}$ hai.

Step 3 — Compare karo. $V_{disp} = V_{melt}$ exactly ⇒ pighli hui paani exactly woh jagah fill karti hai jo ice ke submerged part ne occupy ki thi. Level same rehta hai — koi overflow nahi. Yeh step kyun? Do volumes algebraically identical hain; "displaced" hole aur "produced" paani match karte hain.

Step 4 — Iron wala twist. Iron ( $ρ_{iron} = 7800 > ρ_{w}$ ) apne aap float nahi karta. Jab ice mein frozen hota hai toh aas-paas ki ice usse upar rakhti hai, toh poora lump float karta hai aur paani displace karta hai iron ke weight ke barabar: $V_{float} = m_{iron} / ρ_{w}$ . Pighalte hi, iron loose ho ke floor par gir jaata hai aur ab sirf apna volume displace karta hai: $V_{sunk} = m_{iron} / ρ_{iron}$ , jo chhota hai (kyunki $ρ_{iron} > ρ_{w}$ ). Kam displacement ⇒ paani ka level fall hota hai. Yeh step kyun? Ek dense sinker weight se displace karta hai jab floating (upar rakha hua), lekin sirf apne volume se jab doob jaata hai — floating hamesha zyada displace karta hai. Toh iron wali ice cube ka pighalna level girata hai, unlike plain ice jo ise constant rakhti hai.

Verify: Pure ice ke liye, $V_{disp} - V_{melt} = 0$ ⇒ unchanged. Iron ke liye: floating displacement $\propto 1/ ρ_{w} = 1/1000$ vs sunk displacement $\propto 1/ ρ_{iron} = 1/7800$ ; kyunki $1/1000 > 1/7800$ , floating ne zyada displace kiya, toh pighalte hi level gir jaata hai. ✓

Active Recall

Impossible x matlab koi float nahi — object doob jaata hai; max buoyancy vs weight recompute karo

Sahi, yeh Cell A mein reduce ho jaata hai.

Balloon gas ka real weight hota hai compared to tiny air buoyancy; paani mein object ki apni mass mein sab kuch already included hai

Sahi — gas mein fluid aur contents ki comparable densities hoti hain.

Floating iron weight se displace karta hai (zyada); sunk iron apne chhote volume se displace karta hai

Level girta hai.

Connections

Parent topic (Hinglish)
Pressure in fluids — hydrostatic pressure
Density and relative density
Apparent weight and weighing methods
Floating bodies and stability — metacentre
Newton's laws — equilibrium of forces
Pascal's principle