Hydrostatics — pressure = ρgh, derivation: Exercises

2.2.5 · D4 · Physics › Fluid Mechanics › Hydrostatics — pressure = ρgh, derivation

Shuru karne se pehle, ek baar har symbol yaad kar lo, taaki kuch bhi unexplained na lage:

Level 1 — Recognition (kya tum sahi formula spot karke plug in kar sakte ho?)

L1.1 — Ek depth par pressure

Paani ek tank mein bhara hai. Surface se $4 m$ neeche gauge pressure nikalo.

Recall Solution

KYA chahiye: paani se aane wala extra pressure, isliye $P_{gauge} = ρ g h$ use karo. YEH formula KYU: "gauge" matlab atmosphere se upar, aur yahi ek height $h$ ka fluid column contribute karta hai. $P_{gauge} = ρ g h = 1000 \times 9.8 \times 4 = 3.92 \times 1 0^{4} Pa .$ Answer: $3.92 \times 1 0^{4} Pa$ (lagbhag $39 kPa$ ).

L1.2 — Absolute vs gauge

Usi $4 m$ depth par absolute pressure kya hoga?

Recall Solution

KYA chahiye: total pressure = atmosphere + water column. $P_{0}$ KYU add karein: "absolute" matlab poora push, aur atmosphere surface par pehle se baitha hai kisi bhi paani ke count hone se pehle, isliye uska pressure poora neeche tak jaata hai aur paani ke $ρ g h$ mein add ho jaata hai. $P_{abs} = P_{0} + ρ g h = 1.01 \times 1 0^{5} + 3.92 \times 1 0^{4} = 1.402 \times 1 0^{5} Pa .$ Answer: $\approx 1.40 \times 1 0^{5} Pa$ .

L1.3 — Units padho

Ek pressure $50 kPa$ diya gaya hai. Ise $N/m^{2}$ mein express karo aur paani ki depth ke roop mein bhi.

Recall Solution

KYA: unit rewrite karo, phir matching depth nikalne ke liye $ρ g h$ invert karo. $ρ g h$ KYU invert karein: same pressure ko paani ki height mein "bola" ja sakta hai; $ρ g h = P$ ko $h$ ke liye solve karna pressure ko equivalent water column mein translate karta hai. $50 kPa = 50000 N/m^{2}$ (kyunki $1 Pa = 1 N/m^{2}$ aur $k = 1000$ ). Is gauge pressure ko dene wali paani ki depth: $ρ g h = 50000$ solve karo. $h = \frac{50 000}{1000 \times 9.8} = 5.102 \dots m .$ Kyunki data ( $50 kPa$ ) sirf do significant figures carry karta hai, hum $h \approx 5.1 m$ tak round karte hain. Answer: $5.0 \times 1 0^{4} N/m^{2}$ , jo lagbhag $5.1 m$ paani ke barabar hai (2 s.f.).

Level 2 — Application (values choose karo, units convert karo, ek extra step)

L2.1 — Mixed units

Ek diver $1.03 g/cm^{3}$ density wale seawater mein $12 m$ gehri hai. Gauge pressure?

Recall Solution

PEHLE KYA: density ko SI mein convert karo, kyunki $ρ g h$ ko $kg/m^{3}$ chahiye. $1.03 g/cm^{3} = 1030 kg/m^{3} .$ KYU convert karein: $g/cm^{3}$ ko metres ke saath mix karne par units nonsense ho jaate hain. $P_{gauge} = 1030 \times 9.8 \times 12 = 1.2113 \times 1 0^{5} Pa .$ Answer: $\approx 1.21 \times 1 0^{5} Pa$ (ek atmosphere se thoda zyada — sense banta hai, seawater har ~10 m par ~1 atm add karta hai).

L2.2 — Depth ke liye solve karo

Freshwater mein wo depth nikalo jahan gauge pressure do atmospheres ( $2 P_{0}$ ) ke barabar ho.

Recall Solution

KYA: $ρ g h = 2 P_{0}$ set karo aur $h$ ke liye solve karo. $h = \frac{2 P _{0}}{ρ g} = \frac{2 \times 1.01 \times 1 0 ^{5}}{1000 \times 9.8} = 20.6 m .$ Answer: $\approx 20.6 m$ . (Parent se mile "~10 m per atmosphere" benchmark se consistent.)

L2.3 — Do stacked liquids

Ek tube mein $0.20 m$ oil ( $ρ_{o} = 800 kg/m^{3}$ ) hai jo $0.30 m$ paani par float kar raha hai. Bottom par gauge pressure?

Figure — Hydrostatics — pressure = ρgh, derivation

Figure L2.3 — Ek vertical tube: neeche wala $0.30 m$ (deep teal) paani hai, upar wala $0.20 m$ (burnt orange) oil hai jo uspar float kar raha hai. Do double-headed arrows har layer ki height mark karte hain. Base par plum arrow likhta hai "P = rho_o g h_o + rho_w g h_w": bottom pressure dono layers ke contributions ka sum hai, paani ke upar oil ka.

Recall Solution

KYA: har layer apna $ρ g h$ add karti hai; pressures stack hote hain kyunki har column neeche wale par baitha hota hai. KYU add karein: oil–water boundary par pressure $ρ_{o} g h_{o}$ hai; phir paani uske upar $ρ_{w} g h_{w}$ add karta hai. $P = ρ_{o} g h_{o} + ρ_{w} g h_{w} .$ $= 800 \times 9.8 \times 0.20 + 1000 \times 9.8 \times 0.30$ $= 1568 + 2940 = 4508 Pa .$ Answer: $\approx 4.51 \times 1 0^{3} Pa$ .

Level 3 — Analysis (geometry, direction, ya kisi subtlety ke baare mein sochna)

L3.1 — Hydrostatic paradox, numerically

Do open containers mein paani same depth $h = 0.5 m$ tak bhara hai: container A ek wide barrel hai (base area $1.0 m^{2}$ ), container B ek thin tube hai (base area $0.001 m^{2}$ ). Base par pressure aur base par force compare karo.

Recall Solution

Pressure: sirf depth par depend karta hai, area par nahi. $P = ρ g h = 1000 \times 9.8 \times 0.5 = 4900 Pa dono ke liye.$ Force $= P \times A$ alag hai: $F_{A} = 4900 \times 1.0 = 4900 N, F_{B} = 4900 \times 0.001 = 4.9 N .$ KYU alag hain: pressure force per area hai; same pressure ek bade base par zyada total force deta hai. Paradox sirf pressure ke baare mein hai, jo genuinely equal hai. Answer: same pressure $4900 Pa$ ; forces $4900 N$ vs $4.9 N$ .

L3.2 — Slanted depth

Ek straight pipe surface se $3 0^{\circ}$ angle par horizontal se neeche pool mein jaati hai. Ek point surface se pipe ke saath $6 m$ along the pipe door hai. Woh kitna gauge pressure feel karta hai?

Figure L3.2 — Pale teal region paani hai jiska "free surface" upar hai. Thick ink line pipe hai jo $3 0^{\circ}$ below horizontal utarti hai; burnt-orange arrow uski slant length $L = 6 m$ label karta hai. Vertical plum arrow true depth $h = L sin 3 0^{\circ} = 3 m$ dikhata hai — slant se chhota. $ρ g h$ mein sirf yahi vertical drop count hota hai.

Recall Solution

KYA: $ρ g h$ mein $h$ vertical depth hai, slant ke saath distance nahi. KYU: sirf fluid ka vertical stack ek point par neeche weighs karta hai; ek tilted length overcounts karta hai. Vertical depth: pipe $3 0^{\circ}$ below horizontal par $6 m$ utarti hai, isliye $h = 6 sin 3 0^{\circ} = 6 \times 0.5 = 3.0 m .$ $P = ρ g h = 1000 \times 9.8 \times 3.0 = 2.94 \times 1 0^{4} Pa .$ Answer: $\approx 2.94 \times 1 0^{4} Pa$ (woh $5.88 \times 1 0^{4}$ nahi jo raw $6 m$ se milta).

L3.3 — Degenerate case: zero depth

Free surface par exactly, $h = 0$ par, kaun sa pressure (gauge aur absolute) act karta hai?

Recall Solution

$P_{gauge} = ρ g (0) = 0, P_{abs} = P_{0} + 0 = P_{0} = 1.01 \times 1 0^{5} Pa .$ YEH KYU matter karta hai: formula boundary par sensibly behave karta hai — fluid apni khud ki surface par kuch add nahi karta, aur sirf atmosphere bachta hai. Yeh woh "anchor" hai jisse poora linear law grow karta hai. Answer: gauge $= 0$ ; absolute $= P_{0}$ .

Level 4 — Synthesis (hydrostatics ko kisi aur idea ke saath combine karo)

L4.1 — Force on a dam wall (pressure depth ke saath vary karta hai)

Ek rectangular dam $H = 5 m$ gehre paani ko $w = 8 m$ wide wall ke against rok raha hai. Paani wall par jo total force lagata hai woh nikalo (gauge, yaani sirf paani).

Figure L4.1 — Wall thick ink line hai left par; paani (pale teal) rightward press karta hai. Burnt-orange arrows depth ke saath lambe hote jaate hain, dikhate hain pressure top par $0$ se bottom par $ρ g H$ tak — ek plum dashed triangle. Kyunki profile ek straight triangle hai, average push mid-value $\frac{1}{2} ρ g H$ hai, jise hum wetted area se multiply karte hain.

Recall Solution

KYA naya hai: wall par pressure constant nahi hai — woh top par $0$ se bottom par $ρ g H$ tak linearly badhta hai. Isliye hum ek $P$ se $P \times A$ nahi kar sakte. Average KYU kaam karta hai: kyunki pressure profile ek straight line (triangle) hai, depth par uska average mid-value $\frac{1}{2} ρ g H$ hai. $\overset{ˉ}{P} = \frac{1}{2} ρ g H .$ Total force = average pressure $\times$ wetted area $A = w H$ : $F = \overset{ˉ}{P} \cdot w H = \frac{1}{2} ρ g H \cdot w H = \frac{1}{2} ρ g w H^{2} .$ $F = \frac{1}{2} \times 1000 \times 9.8 \times 8 \times 5^{2} = 9.8 \times 1 0^{5} N .$ Answer: $F = 9.8 \times 1 0^{5} N$ (lagbhag $980 kN$ ).

L4.2 — Simple manometer

Ek U-tube dono arms mein air ke liye open hai aur paani contain karta hai. Ek arm ek gas se connected hai jiska pressure doosri arm mein paani ko $Δ h = 0.15 m$ height difference se raise karta hai. Gas gauge pressure nikalo.

Recall Solution

KYA: gas ek column ko neeche aur doosre ko upar push karta hai; height difference $Δ h$ pressure gap measure karta hai. $ρ g Δ h$ KYU: lower common level par dono arms ka pressure match karna chahiye; difference exactly height $Δ h$ ka ek water column hai. $P_{gas, gauge} = ρ g Δ h = 1000 \times 9.8 \times 0.15 = 1470 Pa .$ Answer: $\approx 1.47 \times 1 0^{3} Pa$ . (General rule ke liye Manometers dekho.)

Level 5 — Mastery (reasoning khud design karo; multi-step)

L5.1 — Unknown density ke liye two-liquid manometer

Ek U-tube mein mercury hai ( $ρ_{H g} = 13600 kg/m^{3}$ ). Ek unknown oil left arm mein daala jaata hai, $H_{o} = 0.20 m$ height ka column banaata hai. Right arm mein mercury left arm ke mercury se $0.012 m$ upar rise karta hai. Oil ki density nikalo.

Recall Solution

KYA kya balance karta hai: left arm mein mercury–oil interface ka level lo. Wahan pressure same horizontal level par right arm mein pressure ke barabar hai.

Left arm interface par: oil column neeche push karta hai, $P = ρ_{o} g H_{o}$ (interface ke upar gauge).
Us same level par right arm: $Δ h = 0.012 m$ height ka mercury column uske upar hai, $P = ρ_{H g} g Δ h$ .

KYU equal: rest mein connected fluid ka pressure equal heights par equal hota hai (same fluid path). $ρ_{o} g H_{o} = ρ_{H g} g Δ h .$ $g$ KYU remove ho sakta hai: yeh dono sides ko identically multiply karta hai, isliye dono sides ko $g$ se divide karne par equation unchanged rehti hai — gravity dono columns ko equally affect karti hai, isliye uski strength density ratio mein kabhi nahi aati. $ρ_{o} = ρ_{H g} \frac{Δ h}{H _{o}} = 13600 \times \frac{0.012}{0.20} = 816 kg/m^{3} .$ Answer: $ρ_{o} \approx 816 kg/m^{3}$ (ek typical light oil). Density and Specific Gravity dekho.

L5.2 — Trapped gas correction wala barometer

Ek "faulty" mercury barometer $760 mm$ ki jagah $735 mm$ read karta hai kyunki thoda air mercury ke upar trapped hai jo pressure $P_{gas}$ exert kar raha hai. Agar true atmospheric pressure $760 mm Hg$ correspond karta hai, toh $P_{gas}$ pascals mein nikalo.

Recall Solution

KYA: ek ideal barometer mein, atmosphere = mercury ka $ρ g h$ (upar vacuum). Yahan trapped gas upar pressure add karta hai, isliye mercury ko utna upar push nahi karna padta. Dish surface par balance: $P_{atm} = P_{gas} + ρ_{H g} g h_{read} .$ Heights ko metres mein convert karo: true $760 mm = 0.760 m$ , read $735 mm = 0.735 m$ . $P_{atm} = ρ_{H g} g (0.760) = 13600 \times 9.8 \times 0.760 = 1.0129 \times 1 0^{5} Pa .$ $P_{gas} = P_{atm} - ρ_{H g} g (0.735) = ρ_{H g} g (0.760 - 0.735) .$ $ρ_{H g} g$ KYU factor out hota hai: $P_{atm}$ aur read column dono same fluid ( $ρ_{H g}$ ) aur same gravity $g$ share karte hain, isliye unka difference sirf $ρ_{H g} g$ times height gap $(0.760 - 0.735)$ hai — common factor cleanly bahar aa jaata hai. $= 13600 \times 9.8 \times 0.025 = 3332 Pa .$ Answer: $P_{gas} \approx 3.33 \times 1 0^{3} Pa$ (lagbhag $25 mm Hg$ ke barabar). Atmospheric Pressure & Barometer dekho.

L5.3 — Pressure difference se buoyancy

Ek cube jiska side $a = 0.10 m$ hai paani mein fully submerged hai, uska top face depth $d = 0.30 m$ par hai. Sirf top aur bottom faces ke beech pressure difference use karke net upward (buoyant) force nikalo.

Recall Solution

KYA mechanism hai: bottom face top se $a$ zyada gehri hai, isliye woh zyada pressure feel karta hai. Woh difference, times area, buoyant force hai. Step 1 — har horizontal face par pressure (gauge; charon side faces par horizontal forces symmetry se cancel hote hain, isliye hum unhe ignore karte hain):

Top face depth $d$ : $P_{top} = ρ g d = 1000 \times 9.8 \times 0.30 = 2940 Pa$ .
Bottom face depth $d + a = 0.40 m$ : $P_{bot} = ρ g (d + a) = 1000 \times 9.8 \times 0.40 = 3920 Pa$ .

Step 2 — pressures ko face area $A = a^{2} = (0.10)^{2} = 0.01 m^{2}$ par forces mein badlo:

Bottom par upward push: $F_{bot} = P_{bot} A = 3920 \times 0.01 = 39.2 N$ .
Top par downward push: $F_{top} = P_{top} A = 2940 \times 0.01 = 29.4 N$ .

Step 3 — net upward force = up − down: $F = F_{bot} - F_{top} = 39.2 - 29.4 = 9.8 N .$ $d$ KYU drop out hua: algebraically, $F = (P_{bot} - P_{top}) a^{2} = ρ g [(d + a) - d] a^{2} = ρ g a \cdot a^{2} = ρ g a^{3} .$ Sirf depth ka difference ( $a$ ) survive karta hai; shared depth $d$ cancel ho jaata hai. Yeh Buoyancy & Archimedes' Principle hai jo seedha $P = ρ g h$ se nikal raha hai. Answer: $F = ρ g a^{3} = 1000 \times 9.8 \times (0.10)^{3} = 9.8 N$ , exactly displaced paani ke weight ke barabar.

Recall Master check — one-line self-quiz

Kya container ki shape ek given depth par pressure change karti hai? ::: Nahi — pressure sirf vertical depth $h$ par depend karta hai (hydrostatic paradox). Ek straight dam wall par force nikalne ke liye, area se kaun sa pressure multiply karte ho? ::: Average pressure, $\frac{1}{2} ρ g H$ , bottom value nahi. Ek manometer mein, height difference kya measure karta hai? ::: Pressure difference, $ρ g Δ h$ . Buoyant force $ρ g a^{3}$ kyun hai aur depth par dependent kyun nahi? ::: Sirf top-vs-bottom depth difference survive karta hai; common depth cancel ho jaata hai.

Connections

2.2.5 Hydrostatics — pressure = ρgh, derivation — parent derivation jise ye problems drill karte hain.
Manometers — L4.2 aur L5.1 manometer balances hain.
Atmospheric Pressure & Barometer — L5.2 barometer correction.
Buoyancy & Archimedes' Principle — L5.3 pressure se buoyancy derive karta hai.
Pascal's Law · Density and Specific Gravity · Bernoulli's Equation