Hydrostatics — pressure = ρgh, derivation: Worked examples

2.2.5 · D3 · Physics › Fluid Mechanics › Hydrostatics — pressure = ρgh, derivation

Scenario matrix

Problems karne se pehle, chaliye har distinct tarah ki situation list karte hain jo formula $P = P_{0} + ρ g h$ produce kar sakta hai. Har row ek "case class" hai. Neeche ke worked examples us cell ke saath tagged hain jo woh cover karte hain, aur milake woh har row ko hit karte hain.

#	Case class	Isme kya khaas hai	Covered by
A	Plain gauge pressure	Sirf $ρ g h$ , ek liquid, "extra" pressure poochho	Ex 1
B	Absolute pressure	$P_{0}$ add karna yaad rakhna hai	Ex 2
C	Backwards solve for $h$	Pressure diya hai, depth dhundho	Ex 3
D	Backwards solve for $ρ$	Height + pressure diya, density dhundho	Ex 4
E	Do stacked liquids	Har layer ka $ρ g h$ alag se add karo	Ex 5
F	Slanted / bent tube	Sirf vertical height count hoti hai, path length nahi	Ex 6
G	Zero / degenerate input	$h = 0$ (surface pe), ya vacuum top ( $P_{0} = 0$ )	Ex 7
H	Limiting / real-world	Bahut deep ocean; $ρ g h$ ko $P_{0}$ se compare karo	Ex 8
I	Exam twist	Same-depth trick / hydrostatic paradox with numbers	Ex 9

Recall Shuru karne se pehle quick self-test

Ek tube bent hai jisme paani 5 m slanted path pe travel karta hai lekin sirf 3 m vertically utha hai. $ρ g h$ mein kaun sa number jaayega? ::: Vertical rise, 3 m. Path length irrelevant hai.

Example 1 — Plain gauge pressure (Cell A)

Worked example 3 m pe extra squeeze

Ek diver $h = 3 m$ ki depth pe fresh water mein jaata hai, $ρ = 1000 kg/m^{3}$ , $g = 9.8 m/s^{2}$ . Gauge pressure (sirf paani ki wajah se pressure) dhundho.

Forecast: Roughly ek atmosphere ka teesra hissa? Poora atmosphere? Aage padhne se pehle power of ten guess karo.

Step 1 — Formula ka sahi piece chuno. Gauge pressure matlab "atmosphere se upar," jo exactly $ρ g h$ hai — koi $P_{0}$ nahi. Ye step kyun? Gauge word humein batata hai $P_{0}$ drop karo; sirf water column matter karta hai.

Step 2 — SI numbers plug in karo. $P_{gauge} = ρ g h = 1000 \times 9.8 \times 3 = 2.94 \times 1 0^{4} Pa$ Ye step kyun? Har quantity already SI mein hai, toh product directly pascals mein aata hai.

Verify: Units: $\frac{kg}{m ^{3}} \cdot \frac{m}{s ^{2}} \cdot m = \frac{kg}{m \cdot s ^{2}} = \frac{N}{m ^{2}} = Pa$ ✓. Aur $2.94 \times 1 0^{4}$ roughly $0.29$ atmosphere hai — sirf 3 m ke liye sensible hai (yaad karo $\sim 10 m$ = 1 atm).

Example 2 — Absolute pressure (Cell B)

Example 3 — Depth ke liye backwards solve karo (Cell C)

Example 4 — Density ke liye backwards solve karo (Cell D)

Worked example Tube mein mystery liquid

$h = 0.758 m$ height ka liquid column atmospheric pressure $P_{0} = 1.01 \times 1 0^{5} Pa$ balance karta hai (ek barometer jisme upar vacuum hai). Uski density kya hai? Kaun sa liquid hai?

Forecast: Paani se denser hai ya lighter? (Hint: column chhota hai.)

Step 1 — Balance likho. Upar vacuum hai matlab top pressure $0$ hai, toh column ka $ρ g h$ poora atmosphere supply karna hoga: $ρ g h = P_{0}$ Ye step kyun? Ye barometer condition hai — dekho Atmospheric Pressure & Barometer.

Step 2 — $ρ$ ke liye solve karo. $ρ = \frac{P _{0}}{g h} = \frac{1.01 \times 1 0 ^{5}}{9.8 \times 0.758}$ Ye step kyun? Ab $ρ$ akela unknown hai; ise divide karke nikalo.

Step 3 — Compute karo. $ρ = \frac{1.01 \times 1 0 ^{5}}{7.43} \approx 1.36 \times 1 0^{4} kg/m^{3}$

Verify: $1.36 \times 1 0^{4} = 13600 kg/m^{3}$ = mercury ✓ (dekho Density and Specific Gravity). Ek chhota heavy column — hamare "paani se denser" expectation ke saath consistent.

Example 5 — Do stacked liquids (Cell E)

Figure — Hydrostatics — pressure = ρgh, derivation

Worked example Paani pe float karta oil

Ek tank mein $h_{1} = 2 m$ oil ( $ρ_{1} = 800 kg/m^{3}$ ) hai jo $h_{2} = 3 m$ paani ( $ρ_{2} = 1000 kg/m^{3}$ ) ke upar float kar raha hai. Bilkul bottom pe gauge pressure dhundho.

Forecast: 5 m pure paani se zyada hoga ya kam? Pehle guess karo.

Step 1 — Har layer ko alag treat karo. Pressure har layer mein neeche utarte waqt add hoti jaati hai. Figure dekho: blue oil green paani pe baitha hai. $P_{bottom} = oil se guzarke ρ_{1} g h_{1} + paani se guzarke ρ_{2} g h_{2}$ Ye step kyun? Har layer apna $ρ g h$ contribute karta hai. Oil–water boundary pe pressure neeche water layer ke liye "surface pressure" ban jaata hai, toh contributions simply add ho jaate hain.

Step 2 — Har piece compute karo. $ρ_{1} g h_{1} = 800 \times 9.8 \times 2 = 1.568 \times 1 0^{4} Pa$ $ρ_{2} g h_{2} = 1000 \times 9.8 \times 3 = 2.94 \times 1 0^{4} Pa$ Ye step kyun? Har layer mein independently SI numbers plug in karo.

Step 3 — Add karo. $P_{bottom} = 1.568 \times 1 0^{4} + 2.94 \times 1 0^{4} = 4.508 \times 1 0^{4} Pa$

Verify: Agar saare 5 m paani hote, toh hume milta $1000 \times 9.8 \times 5 = 4.9 \times 1 0^{4} Pa$ . Hamara answer $4.508 \times 1 0^{4}$ thoda kam hai — correct, kyunki top ke 2 m lighter oil hai ✓.

Example 6 — Slanted tube (Cell F)

Worked example Tilted straw trap

Paani ek seedhi tube mein bhara hai jo horizontal se $3 0^{\circ}$ neeche slanted chalti hai. Paani tube ke saath path length $L = 4 m$ travel karta hai. Neeche wale end pe gauge pressure dhundho.

Forecast: Kya hum $4 m$ use karenge, ya kuch chhota?

Step 1 — Sahi vertical depth identify karo. $ρ g h$ mein sirf vertical drop jaati hai. Figure se, vertical height hai $h = L sin 3 0^{\circ} = 4 \times 0.5 = 2 m$ Ye step kyun? $ρ g h$ mein $h$ surface ke neeche vertical depth define ki gayi thi. Angle $θ$ pe slanted path jiska length $L$ hai, woh $L sin θ$ drop karta hai — figure mein red vertical arrow, blue slanted wala nahi.

Step 2 — Formula apply karo. $P_{gauge} = ρ g h = 1000 \times 9.8 \times 2 = 1.96 \times 1 0^{4} Pa$ Ye step kyun? Sahi vertical $h$ ke saath, ye ordinary one-liquid formula hai.

Verify: Agar hum (galti se) $L = 4 m$ use karte, toh milta $3.92 \times 1 0^{4} Pa$ — exactly double, jo classic mistake hai. Physical answer mein $h = 2 m$ use hona chahiye, jo deta hai $1.96 \times 1 0^{4} Pa$ ✓.

Example 7 — Degenerate inputs (Cell G)

Worked example Do "nothing" cases

(a) Free surface par gauge pressure kya hai, $h = 0$ ? (b) Mercury barometer mein column ke upar ki jagah vacuum hai. Wahan $P_{0}$ (top pressure) kya hai?

Forecast: Dono ek specific tidy number hona chahiye — kya?

Step 1 — Case (a): $h = 0$ set karo. $P_{gauge} = ρ g \times 0 = 0 Pa$ Ye step kyun? Surface pe koi fluid tumhare upar nahi baitha, toh zero extra squeeze hai. Ye poore topic ka sanity anchor hai.

Step 2 — Case (b): vacuum mein koi pressure nahi hoti. $P_{0} = 0 Pa$ Ye step kyun? "Vacuum" ka literally matlab hai koi gas nahi, koi molecules push nahi kar rahe, isliye zero pressure. Isliye Example 4 mein poore atmosphere ko sirf $ρ g h$ se balance karna pada.

Verify: (a) Surface pe absolute pressure tab $P_{0} + 0 = P_{0}$ = atmosphere ✓, jo everyday life se match karta hai (pool ki surface pe air pressure). (b) $P_{0} = 0$ set karne se $P = P_{0} + ρ g h$ ban jaata hai $P = ρ g h$ , exactly woh barometer equation jo humne use ki ✓.

Example 8 — Limiting / real-world (Cell H)

Worked example Deep ocean crush

Mariana Trench ke bottom pe, $h \approx 11000 m$ , seawater $ρ \approx 1030 kg/m^{3}$ . Gauge pressure dhundho, aur ek atmosphere se compare karo.

Forecast: Tens of atmospheres? Hundreds? Thousands?

Step 1 — $ρ g h$ compute karo. $P_{gauge} = 1030 \times 9.8 \times 11000 = 1.110 \times 1 0^{8} Pa$ Ye step kyun? Wahi one-liquid formula — depth bas enormous hai. Seawater fresh water se thoda denser hai, isliye $1030$ .

Step 2 — Atmospheres mein express karo. $\frac{P _{gauge}}{P _{0}} = \frac{1.110 \times 1 0 ^{8}}{1.01 \times 1 0 ^{5}} \approx 1099$ Ye step kyun? $P_{0}$ se divide karne par pata chalta hai kitne atmospheres ka squeeze hai — intuitive scale.

Verify: Lagbhag 1100 atmospheres. 10 m ≈ 1 atm benchmark use karke: $11000/10 = 1100$ atm ✓. Benchmark aur exact formula agree karte hain, aur $P_{0}$ ( $1 0^{5}$ ) $1 0^{8}$ ke aage bilkul negligible hai — isliye deep-sea problems atmosphere ko ignore karte hain.

Example 9 — Exam twist: hydrostatic paradox with numbers (Cell I)

Worked example Do containers, ek trick

Container X ek wide barrel hai jisme $500 L$ paani hai, depth $h = 1.2 m$ tak bhara hai. Container Y ek thin pipe hai jisme sirf $0.5 L$ paani hai, woh bhi depth $h = 1.2 m$ tak bhara hai. Dono ke bottom pe gauge pressure compare karo.

Forecast: 500 L barrel zyada press karega... hai na?

Step 1 — X compute karo. $P_{X} = ρ g h = 1000 \times 9.8 \times 1.2 = 1.176 \times 1 0^{4} Pa$ Ye step kyun? Formula mein sirf depth aati hai, toh $h = 1.2 m$ use karo.

Step 2 — Y compute karo. $P_{Y} = ρ g h = 1000 \times 9.8 \times 1.2 = 1.176 \times 1 0^{4} Pa$ Ye step kyun? Same depth, same $ρ$ , same $g$ → same pressure. Volume kabhi aaya hi nahi.

Step 3 — Resolution state karo. $P_{X} = P_{Y}$ . Extra $499.5 L$ barrel mein wide base aur walls ne utha rakha hai, na ki ek bottom point ke upar stack kiya hua hai (figure dekho: sirf ek point ke upar narrow red column us pe press karta hai). Ye step kyun? Ye hydrostatic paradox hai — exam ka favourite "gotcha."

Verify: $P_{X} / P_{Y} = 1$ exactly ✓. $1000 \times$ volume difference ke bawajood, pressure identical hai kyunki dono share karte hain $h = 1.2 m$ . Equal-depth reasoning kahan kaam aata hai ye dekhne ke liye Manometers aur Buoyancy & Archimedes' Principle dekho.

Connections

Parent: Hydrostatics derivation — jahan $P = P_{0} + ρ g h$ prove hua tha.
Pascal's Law — kyun surface pressure $P_{0}$ har neeche ke point tak pahunchti hai.
Atmospheric Pressure & Barometer — Examples 4 & 7 action mein.
Manometers — pressure ko height difference ki tarah padhta hai (multi-liquid, jaise Example 5).
Buoyancy & Archimedes' Principle — depth ke saath pressure ka difference use karta hai.
Bernoulli's Equation — fluid static hone par $ρ g h$ mein reduce ho jaata hai.
Density and Specific Gravity — woh $ρ$ supply karta hai jo yahan har jagah use hoti hai.