Manometers, barometers: Exercises — Hinglish deep dive

2.2.8 · D4 · Physics › Fluid Mechanics › Manometers, barometers

Shuru karne se pehle, woh symbols jo hum baar baar use karte hain:

Definition

$P$ , $Δ P$ , aur $ρ g h$ ke teen letters ka matlab

$P$ = pressure = fluid har square metre surface par kitni force lagata hai. Units: pascals (Pa), jahan $1 Pa = 1 N/m^{2}$ .
$Δ P$ (padho "delta P") = do points ke beech pressure mein change (difference) — $Δ$ symbol ka matlab hamesha "jis cheez ke aage likha hai usmein change" hota hai. Yahan ye lower point ka pressure minus higher point ka pressure hai.
$ρ$ (Greek "rho") = density = ek cubic metre mein kitne kilograms fluid samata hai. Units $kg/m^{3}$ .
$g$ = gravity ki strength = $9.8 m/s^{2}$ . Ye mass ko weight mein convert karta hai (ek neeche ki taraf pull).
$h$ = fluid mein do points ke beech vertical height difference, metres mein. Aakhri teen ko multiply karo aur tumhe $ρ g h$ milta hai = vertical distance $h$ neeche jaane par jo extra pressure milta hai. Neeche picture mein dekho ki "vertical only" ka kya matlab hai.

Figure 1 (description): ek fluid tube jo lower-left se upper-right tak slant hai. Kali hypotenuse par likha hai "tube length $L$ "; dayi taraf ek lal vertical segment par likha hai "vertical $h$ (yahi count hota hai)"; ek dashed horizontal base par likha hai "horizontal (pressure add nahi karta)". Ek arc incline angle $θ$ dikhata hai. Message: tube moti ho, patli ho, ya slanted ho, $ρ g h$ mein sirf vertical drop (lal line) enter karta hai.

Level 1 — Recognition

Kya tum sahi formula pick kar sakte ho aur diagram se height padh sakte ho?

L1.1

Ek mercury barometer mein column height $h = 0.76 m$ hai. Ye kaun sa atmospheric pressure report karta hai?

Recall Solution

DEVICE kya hai: ek barometer ke upar vacuum hota hai, isliye column poori atmospheric push ko balance karta hai. Matlab ye seedha absolute pressure read karta hai. YE FORMULA KYUN: vacuum ( $P \approx 0$ ) upar aur $P_{a t m}$ dish par press karte waqt, dish level par pressures match karne par $P_{a t m} = 0 + ρ_{H g} g h$ milta hai. $P_{a t m} = ρ_{H g} g h = 13600 \times 9.8 \times 0.76 = 1.013 \times 1 0^{5} Pa$ Ye exactly $760$ mmHg $= 1$ atm hai. ✓

L1.2

Ek gas se connected open-tube mercury manometer mein, open arm ki mercury surface, gas arm ki surface se $h = 0.12 m$ upar hai. Kya gas pressure atmospheric se zyada hai ya kam, aur kitne se (gauge)?

Recall Solution

HEIGHT kya batata hai: open arm ka upar hona matlab gas ne apni side ko neeche push kiya aur doosri side ko upar dhakel diya. Toh gas zyada force se push kar raha hai atmosphere se. ADD KYUN: gas atmosphere se zyada hai, isliye $P_{g a s} = P_{a t m} + ρ g h$ ; gauge part (atmosphere se excess) woh $ρ g h$ term hai. $P_{g a ug e} = ρ_{H g} g h = 13600 \times 9.8 \times 0.12 = 15993.6 Pa \approx 1.60 \times 1 0^{4} Pa$ Gas atmospheric se upar hai lagbhag $1.6 \times 1 0^{4}$ Pa se.

Level 2 — Application

$P = P_{a t m} \pm ρ g h$ mein sahi sign ke saath numbers plug karo.

L2.1

Ek gas vessel se attached open-tube water manometer mein gas arm ka water level open arm se $0.35 m$ upar hai. Gas ka absolute pressure nikalo.

Recall Solution

GEOMETRY kya kehta hai: gas arm upar hai, isliye atmosphere ne gas side ko kam neeche push kiya — actually gas atmosphere se kamzor hai, open side ko neeche gir jaane deta hai. Gas atmospheric se neeche hai ⇒ subtract. $P_{ab s} = P_{a t m} - ρ_{w a t er} g h = 1.013 \times 1 0^{5} - 1000 (9.8) (0.35)$ $= 1.013 \times 1 0^{5} - 3430 = 9.787 \times 1 0^{4} Pa$ Gauge pressure $- 3430$ Pa hai (thoda vacuum).

L2.2

Ek barometer ko pahaad par le jaaya jaata hai jahan mercury column $h = 0.65 m$ tak gir jaata hai. Wahan atmospheric pressure kya hai, aur ye sea-level pressure ka kitna fraction hai?

Recall Solution

$P = ρ_{H g} g h = 13600 \times 9.8 \times 0.65 = 86632 Pa \approx 8.66 \times 1 0^{4} Pa$ Sea level ka fraction: $\frac{86632}{1.013 \times 1 0 ^{5}} = 0.855 \approx 85.5%$ KYUN gira: upar jaane par, tumhare upar kam hawa hoti hai, toh atmosphere kam force se push karta hai aur sirf chhota column hi hold kar pata hai.

L2.3

Standard atmospheric pressure ( $1.013 \times 1 0^{5}$ Pa) read karne ke liye ek water barometer kitna lamba hona chahiye?

Recall Solution

FORMULA INVERT KYUN: hum $P$ jaante hain aur $h$ chahiye, toh $P = ρ g h$ ko $h$ ke liye solve karo. $h = \frac{P}{ρ _{w a t er} g} = \frac{1.013 \times 1 0 ^{5}}{1000 \times 9.8} = 10.34 m$ Isliye hum mercury use karte hain: $13.6 \times$ zyada dense fluid ko $13.6 \times$ chhota column chahiye ( $10.34/13.6 \approx 0.76$ m). ✓

Level 3 — Analysis

Ab tube tilted hai, ya do fluids stack hain — vertical height nikalni padegi.

Figure 2 (description): ek right triangle. Kali hypotenuse fluid path hai jis par " $L = 0.20$ m" likha hai, aur ek lal dot mercury slug dikhata hai. Lal vertical side par " $h = L sin θ = 0.10$ m" likha hai. Horizontal side kali hai, aur base par right angle ka chhota square hai. Ek arc incline " $θ = 3 0^{\circ}$ " label karta hai. Message: fluid hypotenuse $L$ travel karta hai, lekin sirf vertical side $h$ hi $ρ g h$ mein jaata hai.

L3.1

Ek manometer tube horizontal se $θ = 3 0^{\circ}$ par inclined hai. Mercury slug tube ke along length $L = 0.20 m$ move karta hai. Ye kaun si vertical height $h$ represent karta hai, aur ye kaunsa gauge pressure (Pa mein) correspond karta hai?

Recall Solution

$sin$ KYUN, poora length nahi: sirf vertical rise gravity se ladta hai. Upar figure mein right triangle dekho, tube length $L$ hypotenuse hai, aur vertical side angle $θ$ ki "opposite" side hai. Toh $h = L sin θ = 0.20 \times sin 3 0^{\circ} = 0.20 \times 0.5 = 0.10 m$ INCLINED TUBES KYUN USE KARTE HAIN: ye ek chhote vertical change ko tube ke along ek bade, aasaani se padhne wale length mein magnify karte hain — better sensitivity. $P_{g a ug e} = ρ_{H g} g h = 13600 \times 9.8 \times 0.10 = 13328 Pa \approx 1.33 \times 1 0^{4} Pa$

L3.2

Ek U-tube mein paani hai. Left arm mein oil ( $ρ_{o i l} = 800 kg/m^{3}$ ) daala jaata hai aur paani par tairta hua $0.10 m$ lamba column banta hai. Equilibrium par right (open) arm mein water level, left arm ke water level se kitna upar uthta hai?

Recall Solution

BALANCE KYUN: reference line oil–water interface par lo (woh lowest point jahan dono sides pure water/continuous hain). Same fluid, same level dono sides par ⇒ equal pressure. Interface par left arm: oil column neeche press karta hai: $P_{a t m} + ρ_{o i l} g (0.10)$ . Same level par right arm: paani us interface level se $h_{w}$ upar utha, deta hai $P_{a t m} + ρ_{w a t er} g h_{w}$ . Equal karo: $ρ_{o i l} g (0.10) = ρ_{w a t er} g h_{w}$ $h_{w} = \frac{ρ _{o i l}}{ρ _{w a t er}} \times 0.10 = \frac{800}{1000} \times 0.10 = 0.08 m$ $0.10$ m SE KAM KYUN: oil paani se halka hai, isliye $0.10$ m oil column ka weight sirf $0.08$ m paani ke column ke barabar hota hai. Water surfaces ka difference $0.08$ m hai.

Level 4 — Synthesis

Hydrostatics ko ek doosre principle ke saath combine karo: sealed gas, ya two-fluid barometer.

L4.1

Ek closed-tube manometer (ek arm par mercury ke upar vacuum sealed) mein mercury height difference $h = 0.68 m$ read hota hai, sealed arm upar hai. Connected gas ka kaun sa absolute pressure ye indicate karta hai?

Recall Solution

OPEN TUBE SE KYA ALAG HAI: sealed arm ka top vacuum hai, $P \approx 0$ — bilkul barometer ki tarah. Toh mercury ka difference pura gas pressure balance karta hai, seedha absolute deta hai. BOTTOM LEVEL EQUAL KYUN SET KARTE HAIN: dono arms connected mercury se jude hain, isliye lowest common horizontal level par dono sides mein pressure identical hona chahiye (warna fluid sideways flow karta rehta jab tak nahi hota). Us shared level se har surface tak upar trace karna dono arms ko equate karne deta hai. Gas side par, gas se reference tak trace karte waqt $P_{g a s}$ se upar kuch extra nahi add hota; sealed side par, reference pressure vacuum ( $0$ ) plus lambe mercury column ka weight hai, $ρ_{H g} g h$ . Dono ko equate karo: $P_{g a s} = 0 + ρ_{H g} g h = 13600 \times 9.8 \times 0.68 = 90630.4 Pa \approx 9.06 \times 1 0^{4} Pa$ Ye ek absolute reading hai — koi $P_{a t m}$ add nahi karna, kyunki reference arm vacuum hai, atmosphere nahi.

L4.2

Ek barometer tube mein accidentally upar mercury ke upar ek chhota air pocket trap ho jaata hai, jo $2000$ Pa pressure exert karta hai. True atmospheric pressure $1.013 \times 1 0^{5}$ Pa hai. Ye faulty barometer kaun si mercury column height dikhayega, aur agar tum (galti se) upar vacuum assume karo toh $P_{a t m}$ mein kitna percent error hoga?

Recall Solution

TRAPPED AIR KYA KARTA HAI: ye upar se mercury par neeche push karta hai, isliye column utna upar nahi uth pata. Dish level par balance: $P_{a t m} = P_{ai r} + ρ_{H g} g h$ $h = \frac{P _{a t m} - P _{ai r}}{ρ _{H g} g} = \frac{1.013 \times 1 0 ^{5} - 2000}{13600 \times 9.8} = \frac{99300}{133280} = 0.74504 m$ Agar tum vacuum assume karo aur $P_{a t m} = ρ_{H g} g h = 133280 \times 0.74504 = 99300 Pa$ read karo, tum true value se kam estimate karoge. $error = \frac{1.013 \times 1 0 ^{5} - 99300}{1.013 \times 1 0 ^{5}} = \frac{2000}{101300} = 0.01974 = 1.97%$ KYUN: reading exactly trapped-air pressure ( $2000$ Pa) se kam hai, jo true $P_{a t m}$ ka $\approx 1.97%$ hai.

Level 5 — Mastery

Sab kuch ek saath — two immiscible fluids, sign bookkeeping, aur ek sanity limit.

L5.1

Ek U-tube dono taraf se open hai aur usme mercury hai. Left arm mein paani daala jaata hai $0.40 m$ ki height tak (mercury–water interface se measure kiya gaya). Right arm mein oil ( $ρ_{o i l} = 850 kg/m^{3}$ ) daala jaata hai. Agar dono arms mein mercury levels equal ho jaayein, toh right arm mein oil column ki height kitni hogi?

Recall Solution

EQUAL MERCURY LEVELS SAB SIMPLIFY KYUN KARTE HAIN: agar mercury surfaces same height par hain, toh us common level par mercury dono sides pe equally contribute karta hai aur cancel ho jaata hai. Dono arms se mercury par neeche push karne wale pressures isliye equal hone chahiye. Left arm (water) mercury surface par neeche push karta hai: $P_{a t m} + ρ_{w a t er} g (0.40)$ . Right arm (oil) mercury surface par neeche push karta hai: $P_{a t m} + ρ_{o i l} g h_{o i l}$ . Equal karo ( $P_{a t m}$ aur $g$ cancel ho jaate hain): $ρ_{w a t er} (0.40) = ρ_{o i l} h_{o i l}$ $h_{o i l} = \frac{ρ _{w a t er}}{ρ _{o i l}} \times 0.40 = \frac{1000}{850} \times 0.40 = 0.4706 m$ Sanity limit: oil paani se halka hai, isliye same weight se press karne ke liye lamba oil column chahiye ( $0.47 > 0.40$ m). ✓ Agar oil paani jitna dense hota, toh exactly $0.40$ m milta — formula us limit mein agree karta hai.

L5.2

Ek single manometer gas ka pressure measure karta hai, lekin connecting fluid paani hai jabki U-tube ka measuring fluid mercury hai, dono bina mix hue stacked hain. Gas ek $0.30 m$ high water column ko mercury par neeche push karta hai, aur mercury $0.05 m$ ka height difference dikhata hai (open arm upar). Gas ka absolute pressure nikalo. ( $ρ_{w a t er} = 1000$ , $ρ_{H g} = 13600$ lo.)

Recall Solution

TRACK KYUN KARO: gas se shuru karo, paani ke column se neeche jaao (neeche jaate waqt pressure barhta hai), bottom cross karo, phir open side par mercury difference upar jaao atmosphere tak pahuncho. Gas se shuru: $P_{g a s}$ . $0.30$ m paani neeche jaao: $ρ_{w a t er} g (0.30)$ add karo. Ab open side par $0.05$ m mercury upar jaao atmosphere tak: $ρ_{H g} g (0.05)$ subtract karo. $P_{a t m}$ par khatam: $P_{g a s} + ρ_{w a t er} g (0.30) - ρ_{H g} g (0.05) = P_{a t m}$ $P_{g a s} = P_{a t m} - ρ_{w a t er} g (0.30) + ρ_{H g} g (0.05)$ $= 1.013 \times 1 0^{5} - 1000 (9.8) (0.30) + 13600 (9.8) (0.05)$ $= 101300 - 2940 + 6664 = 105024 Pa \approx 1.050 \times 1 0^{5} Pa$ SIGNS FLIP KYUN: reference ki taraf neeche jaana pressure add karta hai; usse upar jaana subtract karta hai. Har leg ki direction ka bookkeeping karna hi poora skill hai.

Connections

Hydrostatic Pressure — $ρ g h$ engine jo yahan har problem ke peeche hai.
Pascal's Law — "same fluid, same level ⇒ same pressure" ko justify karta hai.
Atmospheric Pressure — barometer problems (L1.1, L2.2) yahi measure karte hain.
Density and Specific Gravity — two-fluid problems (L3.2, L5.1) $ρ$ compare karne par hinge karte hain.
Buoyancy and Archimedes Principle — floating oil column same depth-pressure idea par tika hai.
Bernoulli's Equation — jahan ye static results moving fluids tak generalise hote hain.