Visual walkthrough — Manometers, barometers

2.2.8 · D2 · Physics › Fluid Mechanics › Manometers, barometers

Yeh page parent note Manometers & Barometers ka picture-by-picture version hai. Ise dheere padho; har figure apne aap mein ek argument hai.

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Step 1 — Pressure kya hota hai? (pushing ka ek picture)

KYA. Figure dekho: wahi fist-sized push ek badi plate par spread ho toh barely kuch hoga, lekin pin-tip par concentrate ho toh puncture ho jayega. Same force $F$, alag area $A$ — alag pressure.

KYUN ise force per area define karte hain? Kyunki ek fluid kabhi "total force" ke roop mein push nahi karta; woh jis bhi surface mein dip karo usi par push karta hai. Ek thin tube ko wide dish se compare karne ke liye hum per unit area ki baat karni padegi, aur — baad ke liye crucial — yahi reason hai ki tube ki width cancel ho jayegi.

Recall Khud ko check karo

Agar aap area double kar do lekin force same rakho, toh pressure ka kya hoga? ::: Woh half ho jayega — pressure force divided by area hota hai.

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Step 2 — Zyada gehraai mein jaane par zyada pressure kyun hota hai?

KYA. Figure mein ek still tank mein do depths mark ki gayi hain. Deeper red dot ke upar upper teal dot ki tulna mein paani ka zyada lamba column hai.

KYUN yeh badhnee chahiye. Fluid move nahi kar raha. Toh kisi bhi patch ke upar ka fluid hold up ho raha hai fluid ke upar se. Ek hi cheez weight ko hold up kar sakti hai woh hai neeche se push — pressure. Upar zyada weight ⇒ neeche zyada push chahiye ⇒ gehraai ke saath zyada pressure. (Dekho Hydrostatic Pressure.)

PICTURE. Arrows notice karo: gravity (plum, down) column par, aur support pressure (orange, up) neeche se. Inhe balance karna padega kyunki kuch bhi accelerate nahi kar raha.

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Step 3 — Amount pakka karo: ek cylinder par force balance

Ab hum "zyada" ko exact formula mein badlenge.

KYA. Ek imaginary cylinder of fluid cut karo: cross-section area $A$, uski top depth $h_1$ par, uski bottom depth $h_2$ par (deeper, isliye $h_2>h_1$). Teen vertical forces lagte hain, figure mein draw kiye gaye hain:

• Top face neeche push hoti hai upar ke fluid se: force $=P_{1}A$.
• Bottom face upar push hoti hai neeche ke fluid se: force $=P_{2}A$.
• Weight cylinder ka khud neeche kheencha jaata hai: $W=\rho A(h_2-h_1)g$.

KYUN cylinder aur kyun balance? Kyunki fluid static hai (equilibrium), is chunk par up-pushes aur down-pulls ka sum zero hona chahiye: $\underset{\text{up}}{\underbrace{P_{2}A}}=\underset{\text{down}}{\underbrace{P_{1}A}}+\underset{\text{weight down}}{\underbrace{\rho A(h_2-h_1)g}}$

Ab har term ko $A$ se divide karo. KYUN kar sakte hain: pressure per area hai, isliye area actually point tha hi nahi — yeh cleanly cancel ho jaata hai: $\boxed{P_2=P_1+\rho g(h_2-h_1)}$

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Step 4 — Same fluid, same level, same pressure

Yeh choti si fact har U-tube ka hinge hai.

KYA. Figure mein, ek connected fluid ki bent tube mein do points $L$ aur $R$ same horizontal dashed line par draw kiye gaye hain.

KYUN woh equal hain. Law $\Delta P=\rho gh$ mein, $L$ aur $R$ ke beech height difference $h$ zero hai — woh same level par baithe hain. Zero height ⇒ zero pressure difference ⇒ $P_L=P_R$.

PICTURE karo alternative: agar $P_L$ bada hota, toh extra push fluid ko sideways $L$ se $R$ ki taraf shove karta — lekin fluid still hai. Isliye pressures zaroor match karni chahiye. Yeh Pascal's Law kaam kar raha hai.

Recall Horizontal bend kyun matter nahi karta?

Kyunki $\Delta P=\rho gh$ sirf vertical $h$ use karta hai; sideways travel karne se koi height nahi badti, isliye pressure change nahi hota. ::: Sahi — sirf upar/neeche movement pressure change karta hai.

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Step 5 — Manometer banao aur reference line choose karo

KYA. Gas left arm mein liquid par push karta hai, use neeche drive karta hai aur right arm ko upar lift karta hai. Do free surfaces alag-alag heights par end hoti hain; unka vertical gap $h$ hai.

KYUN reference line choose karte hain. Step 4 se, fluid mein horizontal line dono arms par equal pressure deti hai. Hum dashed line lower free surface par choose karte hain (yahan left/gas arm). Us se neeche sab same connected liquid hai, isliye hum dono sides ko wahan equate kar sakte hain.

PICTURE. Figure mein: dashed reference line left surface ke through; right arm mein, height $h$ ka liquid column us same line ke upar khada hai.

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Step 6 — Reference line tak har arm se neeche walk karo

KYA — left (gas) arm. Gas se shuru karo, jo directly left liquid surface par $P_{\text{gas}}$ press karta hai — aur woh surface hi hamari reference line hai. Toh left se line par pressure: $P_{\text{left}}=P_{\text{gas}}$

KYA — right (open) arm. Open top se shuru karo: atmosphere right surface par $P_{\text{atm}}$ press karta hai. Reference line tak pahunchne ke liye hum height $h$ se neeche jaate hain, us column ka weight add karte hain (Step 3, "neeche jaane par add karo"): $P_{\text{right}}=P_{\text{atm}}+\rho gh$

Term by term: $P_{\text{atm}}$ upar air ka push hai; $\rho gh$ line ke upar baithe $h$-tall liquid slug ka extra push hai ($\rho$ uski density, $g$ gravity, $h$ uski vertical height).

KYUN equal set karte hain. Dono same fluid mein same reference line par pressure describe karte hain, isliye Step 4 se: $P_{\text{left}}=P_{\text{right}}⟹P_{\text{gas}}=P_{\text{atm}}+\rho gh$

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Step 7 — Doosra case: gas atmospheric se neeche (sign flip)

Koi bhi case kabhi dikhaaye bina mat chodo. Kya hoga agar atmosphere stronger wala ho?

KYA. Ab gas arm higher surface hai. Reference line lower (open-arm) surface par rakho.

• Right/open arm line par: $P_{\text{right}}=P_{\text{atm}}$ (open surface hi line hai).
• Left/gas arm: gas surface se line tak pahunchne ke liye $h$ neeche jao: $P_{\text{left}}=P_{\text{gas}}+\rho gh$.

Equate karo: $P_{\text{gas}}+\rho gh=P_{\text{atm}}⟹\boxed{P_{\text{gas}}=P_{\text{atm}}-\rho gh}$

KYUN minus. Gas itna weak hai ki apni side neeche nahi rakh sakta, isliye atmosphere open side ko neeche push karta hai aur gas side upar uthti hai. Gas side ka uthna ⇒ $\rho gh$ subtract karo.

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Step 8 — Gauge vs absolute (number ka matlab kya hai)

KYA. Figure mein $0$ se upar ek bar stack hua hai: pehle $P_{\text{atm}}$, phir manometer ka extra $\rho gh$ uske upar.

KYUN matter karta hai. Manometer sirf difference $\rho gh$ gas aur atmosphere ke beech deta hai — us difference ko gauge pressure kehte hain: $P_{\text{gauge}}=P_{\text{abs}}-P_{\text{atm}}=\rho gh$ True absolute pressure paane ke liye atmosphere wapas add karna padta hai: $P_{\text{abs}}=P_{\text{atm}}+\rho gh.$ (Barometer is se alag hai kyunki uska top vacuum hai, isliye woh directly absolute read karta hai — parent note dekho.)

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Ek-picture summary

Yeh single figure saare aath steps compress karti hai: fluid column ka weight ⇒ pressure gehraai ke saath badhta hai ($\Delta P=\rho gh$) ⇒ same level = same pressure ⇒ dono arms ko reference line par equate karo ⇒ height gap ko $P_{\text{gas}}=P_{\text{atm}}\pm \rho gh$ ke roop mein padho.

Recall Feynman retelling — ek dost ko batao

Fluid bhaari hota hai, isliye jitna gehra jao utna zyada stuff tumhare upar stack hota hai aur utna zyada hard press karta hai — woh "harder press" hi pressure hai, aur yeh exactly fluid ke upar ke weight se badhta hai per unit area, jo $h$ height ke liye $\rho gh$ nikalta hai. Ab is liquid ki tube ko U mein bend karo. Kyunki liquid connected aur still hai, koi bhi level line dono sides par same pressure rakhti hai — agar nahi rakhti, toh liquid tab tak slosh karta jab tak nahi karta. Gas ko ek side par rakho aur open air ko doosri side par. Jo bhi push stronger hai woh apni side ka liquid neeche shove karta hai aur doosri side upar lift karta hai, ek height gap $h$ chhodke. Har arm se shared line tak neeche walk karo, har baar neeche jaate waqt $\rho gh$ add karte jao, aur dono sides equal set karo. Nikalta hai $P_{\text{gas}}=P_{\text{atm}}\pm \rho gh$: plus jab open side upar ride kare (strong gas), minus jab gas side upar ride kare (weak gas), aur dead level jab equal hoon. Manometer kabhi bhi poora pressure nahi batata — sirf kitna air ko beat ya trail karta hai, gauge pressure — isliye agar absolute value chahiye toh atmosphere wapas add karo.

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Connections

• Hydrostatic Pressure — Step 3 ka $\rho gh$ yahan prove kiya gaya hai.
• Pascal's Law — Step 4 ka "same level, same pressure".
• Atmospheric Pressure — $P_{\text{atm}}$ jo open arm feel karta hai.
• Density and Specific Gravity — kyun $\rho$ (mercury vs water) height set karta hai.
• Buoyancy and Archimedes Principle — yeh bhi depth-dependent pressure par build hai.
• Bernoulli's Equation — pressure ko moving fluids tak extend karta hai.