Pascal's law — pressure transmits equally: Worked examples

2.2.6 · D3 · Physics › Fluid Mechanics › Pascal's law — pressure transmits equally

Scenario matrix

Pascal ke problems ek chhote grid mein rehte hain. Do knobs ghoomte hain: area ratio $k = A_{2} / A_{1}$ aur woh kya poochhte hain (output force, input force, distance, pressure change, energy). Neech diya hua sab iss table ka ek cell fill karta hai.

Cell	Situation	Kya special hai	Example
A	$k > 1$ (bada output piston)	force multiply hoti hai	Ex 1
B	Distance trade-off	chhota piston door tak jaata hai	Ex 2
C	Energy check	work in $=$ work out	Ex 3
D	$k < 1$ (reversed)	force kam hoti hai, distance milti hai	Ex 4
E	$k = 1$ (equal pistons)	degenerate — kuch bhi multiply nahi	Ex 5
F	Pure $Δ p$ transmission	koi piston nahi, sirf depth	Ex 6
G	Word problem (car jack)	real numbers, unit ka dhyan	Ex 7
H	Exam twist (piston ka weight hai)	extra $ρ g h$ + weight bias	Ex 8
I	Limiting case ( $A_{1} \to 0$ )	infinite advantage myth	Ex 9

Neech diya hua har numeric answer verify block mein machine-checked hai.

Woh ek picture jo poora page chalati hai

Figure — Pascal's law — pressure transmits equally

Figure dekho. Red chhota piston (area $A_{1}$ ) ko $F_{1}$ force se neeche push kiya ja raha hai. Confined fluid same pressure change $Δ p = F_{1} / A_{1}$ ko bade piston (area $A_{2}$ ) tak le jaati hai, jo $F_{2} = Δ p A_{2}$ se wापas push karta hai. Kyunki fluid incompressible hai, chhoti side se jaane wala volume $A_{1} d_{1}$ badi side par aane wale volume $A_{2} d_{2}$ ke barabar hona chahiye — yeh ek fact cells B, C, D, aur I ko chalata hai.

Formula Chaar equations jo hum baar baar use karenge

$Pascal: equal p \frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}}$ Yeh step kyun? Pascal kehta hai applied pressure har jagah identical hai, isliye chhoti side par force-over-area badi side par force-over-area ke barabar hai. Yeh jawab deta hai "pistons ke beech kya share hota hai?" — pressure. $incompressible A_{1} d_{1} = A_{2} d_{2}$ Yeh step kyun? Fluid ko squash nahi kiya ja sakta, isliye ek cylinder se bahar push kiya gaya volume ( $A_{1} d_{1}$ ) doosre mein reappear hona chahiye ( $A_{2} d_{2}$ ). Yeh jawab deta hai "agar ek piston move kare, to doosra kitna door jaata hai?" $force mult. F_{2} = F_{1} \frac{A _{2}}{A _{1}}$ Yeh step kyun? Pehli equation ko $F_{2}$ ke liye rearrange karne se directly output force milti hai. Yeh jawab deta hai "mujhe kitni force bahar milegi?" $energy conserved F_{1} d_{1} = F_{2} d_{2}$ Yeh step kyun? Dono sides par force ko distance se multiply karo aur area ratios cancel ho jaate hain — prove karta hai ki work in equals work out. Yeh jawab deta hai "kya yeh free-energy machine hai?" (Nahi.)

Circular areas: $A = π r^{2}$ , isliye $\frac{A _{2}}{A _{1}} = (\frac{r _{2}}{r _{1}})^{2}$ . Squared kyun? Area radius ke square ke saath badhta hai — radius double karo, area chaar guna ho jaata hai. Yahi "huge boost" ka poora source hai.

Cell A — bada output piston ( $k > 1$ )

Figure mein, red patla input piston $F_{1}$ se push kiya ja raha hai; chauda kala output piston multiplied $F_{2}$ deliver karta hai. Dekho kaise same $Δ p$ ek bahut bade kale face par spread hoti hai.

Worked example Ex 1 — force multiplication

Chhote piston ki radius $r_{1} = 3 cm$ , bade piston ki radius $r_{2} = 15 cm$ . Tum $F_{1} = 200 N$ se press karte ho. Output force $F_{2}$ kya milegi?

Forecast: radius ratio $5$ hai, to guess karo: kya force $\times 5$ hai, ya kuch bada?

Areas ka ratio: $\frac{A _{2}}{A _{1}} = (\frac{r _{2}}{r _{1}})^{2} = (\frac{15}{3})^{2} = 25$ . Yeh step kyun? Force area ratio se multiply hoti hai (yahi "equal pressure over bigger area" ka matlab hai), aur area radius squared use karta hai — isliye boost $25$ hai, $5$ nahi.
$F_{2} = F_{1} \frac{A _{2}}{A _{1}} = 200 \times 25 = 5000 N$ . Yeh step kyun? Directly $F_{2} = F_{1} A_{2} / A_{1}$ , Pascal press formula se.

Verify: Check karo ki pressure dono sides par equal hai. $p = F_{1} / A_{1} = 200/ (π \cdot 0.0 3^{2}) \approx 70736 Pa$ ; aur $F_{2} / A_{2} = 5000/ (π \cdot 0.1 5^{2}) \approx 70736 Pa$ . ✓ Same pressure, exactly jaisa Pascal maangta hai.

Cell B — woh distance jo tum chukate ho

Figure dono travel distances mark karta hai: chhoti side par red double-arrow ( $d_{1} = 50 cm$ ) bahut bada hai, badi side par kala wala ( $d_{2} = 2 cm$ ) tiny hai. Woh visual mismatch hi force boost ki kimat hai.

Cell C — energy conserved hai (koi free lunch nahi)

Figure work ko area ki tarah draw karta hai: red input bar tall-and-thin hai (chhoti force, lambi distance), kala output bar short-and-wide hai (badi force, chhoti distance). Dono rectangles ki same area hai — woh equal area conserved $100 J$ hai.

Cell D — reversed press ( $k < 1$ )

Yahan red arrow bade piston par hai — is baar hum wide side push karte hain. Figure chhote kale piston ko reduced output force deliver karte dikhata hai. Cell A ka sab kuch ulta chal raha hai.

Worked example Ex 4 —

bade piston ko push karna Ab large piston ( $r = 15 cm$ ) ko $F_{1} = 5000 N$ se push karo aur small piston ( $r = 3 cm$ ) par force padhlo. Yahan input area bada hai, isliye $k = A_{2} / A_{1} < 1$ .

Forecast: bade piston ko input ki tarah use karne se, kya zyada ya kam force bahar milti hai?

Area ratio (output/input) $= (\frac{3}{15})^{2} = \frac{1}{25} = 0.04$ . Yeh step kyun? Formula hamesha output area ko input area se divide karta hai — hum kaun sa piston push karte hain yeh swap karna sirf ratio ko flip karta hai.
$F_{2} = F_{1} \times 0.04 = 5000 \times 0.04 = 200 N$ . Yeh step kyun? Same Pascal formula; chhoti output area chhoti force deti hai.

Verify: pressures equal hain? $5000/ (π \cdot 0.1 5^{2}) \approx 70736 Pa$ aur $200/ (π \cdot 0.0 3^{2}) \approx 70736 Pa$ . ✓ Force $25 \times$ kam hui, lekin chhota piston $25 \times$ door jaata — Ex 1–2 ka exact mirror.

Cell E — degenerate case, equal pistons ( $k = 1$ )

Figure identical width ke do pistons dikhata hai. Input (red) aur output (black) arrows same length ke hain — koi multiplication nahi, koi reduction nahi. Yeh Cell A aur Cell D ke beech exact boundary hai.

Cell F — pure transmission, koi piston nahi

Figure ek sealed tank dikhata hai; red top piston $Δ p = 6000 Pa$ se squeeze kiya gaya hai. Bottom par label follow karo: same $6000 Pa$ wahan pahuncha, left par marked $h = 4 m$ depth se untouched.

Worked example Ex 6 — ek sealed tank press karna

Ek sealed tank of water, $4 m$ deep, upar ek piston hai. Tum surface pressure $Δ p = 6000 Pa$ badhate ho. Ekdum bottom par pressure kitni badhegi? ( $ρ = 1000 kg/m^{3}$ , $g = 9.8$ lo.)

Forecast: deeper water pehle se zyada pressure carry karta hai. Kya extra push bottom tak pahunchte pahunchte weak ho jaata hai?

Absolute bottom pressure: $p_{bottom} = p_{surface} + ρ g h$ (hydrostatics se). Yahan plain $p$ ka matlab bottom par total pressure hai, $p_{surface}$ top par total pressure hai. Yeh step kyun? Yeh total bottom pressure ko ek surface part (jise hum change karte hain) aur ek depth part $ρ g h$ (jise hum touch nahi karte) mein split karta hai.
Surface ki total pressure ko applied change $Δ p$ se badhao: new bottom $= (p_{surface} + Δ p) + ρ g h = p_{bottom,old} + Δ p$ . Yeh step kyun? $ρ g h$ term pehle aur baad mein identical hai (same fluid, same depth), isliye cancel ho jaata hai — sirf applied change $Δ p$ bachta hai.
Isliye $Δ p_{bottom} = 6000 Pa$ — depth irrelevant hai. Note: absolute $p_{bottom}$ bahut bada hai; sirf change $6000 Pa$ ke barabar hai. Yeh step kyun? Old bottom pressure ko new se subtract karne par sirf $Δ p$ bachta hai; fixed $ρ g h$ baseline difference se bahar nikal jaata hai, isliye change depth par depend nahi kar sakta.

Verify: Unchanged baseline $ρ g h = 1000 \times 9.8 \times 4 = 39200 Pa$ old aur new dono bottom pressures mein appear karta hai, isliye change ko affect nahi kar sakta. $Δ p_{bottom} = 6000 Pa$ exactly. ✓ Transmission undiminished hai — Pascal apne purest form mein.

Cell G — real-world word problem

Figure jack setup karta hai: red narrow handle piston, upar car block ke saath wide lifting piston $8820 N$ labelled, aur output force $6250 N$ . Isko padhte hue, tum dekh sakte ho ki output load se chhota hai — one-stroke failure.

Worked example Ex 7 — car jack

Ek hydraulic jack mein pump handle piston ka diameter $1.2 cm$ aur lifting piston ka diameter $6 cm$ hai. Ek person handle piston ko $F_{1} = 250 N$ se push karta hai. (a) Woh kitna weight utha sakta hai? (b) Car mass $900 kg$ hai — kya woh uthaa sakta hai?

Forecast: diameters ka ratio $5$ hai. $900 kg$ car ke liye kaafi hai?

Diameters directly use karo: $\frac{A _{2}}{A _{1}} = (\frac{d _{2}}{d _{1}})^{2} = (\frac{6}{1.2})^{2} = 25$ . Yeh step kyun? Kyunki $A = π (D /2)^{2}$ , ratio mein $π$ aur $\frac{1}{4}$ cancel ho jaate hain — tum diameters use kar sakte ho, halve karne ki zaroorat nahi.
Output force: $F_{2} = 250 \times 25 = 6250 N$ . Yeh step kyun? Pascal press formula.
Car ko jitna weight uthana hai: $W = m g = 900 \times 9.8 = 8820 N$ . Yeh step kyun? Weight = mass times gravity — yeh woh load hai jise beat karna hai.
Compare: $6250 N < 8820 N$ → ek press mein nahi utha sakta. Yeh step kyun? Output force kam se kam load ke barabar honi chahiye. Real jacks ek ratchet use karte hain: kai pump strokes, har ek fluid ka ek slug transfer karta hai, isliye kai strokes mein effective advantage bahut zyada hota hai.

Verify: $F_{2} = 6250 N$ , load $= 8820 N$ , aur $6250 < 8820$ true hai → single stroke fails. ✓ Units: N vs N, consistent.

Cell H — exam twist: piston ka weight hai

Figure load aur platform ka apna weight bade piston par stacked dikhata hai, plus ek red input arrow $\approx 30 N$ aur dono piston levels ke beech height marker $h = 0.5 m$ . Woh do extras — grey platform weight aur height gap — exactly woh cheezein hain jo input ko naive value se upar push karti hain.

Worked example Ex 8 — badi side par heavy platform

Chhote piston ka area $A_{1} = 5 cm^{2}$ , bade piston ka area $A_{2} = 250 cm^{2}$ . Bada piston khud (platform) $W_{p} = 400 N$ weighs karta hai aur chhote wale se $h = 0.5 m$ upar baithta hai. Tum bade piston par $L = 1000 N$ load ko just start uthana chahte ho. Input force $F_{1}$ kitni chahiye? (fluid $ρ = 800 kg/m^{3}$ oil, $g = 9.8$ .)

Forecast: naive answer platform weight aur height ignore karta hai. Guess low karo, phir dekho corrections ise upar kaise push karti hain.

Yahan plain $p_{1}, p_{2}$ har piston ke neeche absolute fluid pressures hain (wahan total pressure), applied changes nahi.

Badi side total downward force $= L + W_{p} = 1000 + 400 = 1400 N$ , area $A_{2} = 250 cm^{2} = 0.025 m^{2}$ par. Yeh step kyun? Bade piston ke neeche fluid pressure load aur platform ka apna weight dono hold karna chahiye.
Bade piston ke neeche required pressure: $p_{2} = \frac{1400}{0.025} = 56000 Pa$ . Yeh step kyun? Pressure $=$ force/area; yeh total pressure hai jo fluid ko wahan supply karni hai.
Chhota piston $h = 0.5 m$ neeche hai, isliye hydrostatics se uski fluid pressure $ρ g h$ se zyada hai: $p_{1} = p_{2} + ρ g h = 56000 + 800 \times 9.8 \times 0.5 = 56000 + 3920 = 59920 Pa$ . Yeh step kyun? Pascal applied change ko equally transmit karta hai, lekin dono pistons alag heights par hain, isliye unke absolute pressures $ρ g h$ se differ karte hain — ise wापas add karna padega.
Input force: $F_{1} = p_{1} A_{1} = 59920 \times 5 \times 1 0^{- 4} = 29.96 N$ . Yeh step kyun? Force $=$ pressure $\times$ area chhote piston par.

Verify: $F_{1} \approx 29.96 N$ . Sanity: ek naive "weight & height ignore karo" answer hoga $\frac{L}{A _{2}} \cdot A_{1} = \frac{1000}{0.025} \times 5 \times 1 0^{- 4} = 20 N$ — platform weight aur height difference ne correctly requirement ko $\approx 30 N$ tak badha diya. ✓

Cell I — limiting case ( $A_{1} \to 0$ )

Figure input piston ko ek red sliver tak shrunk dikhata hai, side mein wide kala output piston, aur input stroke far upar khichta hua — ek visual reminder ki jab $A_{1}$ shrink hota hai, chhote piston ko travel karni padhne wali distance utni hi fast blow up hoti hai jitni force hoti hai.

Worked example Ex 9 — infinite advantage?

Jab input piston shrink hota hai, $A_{1} \to 0$ , ratio $A_{2} / A_{1} \to \infty$ . Kya force ki ek whisper kuch bhi utha sakti hai, free mein?

Forecast: multiplier blow up ho raha hai — kya yeh free-energy machine hai?

Force out: $F_{2} = F_{1} \frac{A _{2}}{A _{1}} \to \infty$ jab $A_{1} \to 0$ (fixed $F_{1}$ ke liye). Yeh step kyun? Ratio diverge karta hai, isliye koi bhi fixed input force unbounded output force deti hai — mathematically true.
Distance in: $d_{1} = d_{2} \frac{A _{2}}{A _{1}} \to \infty$ bhi. Yeh step kyun? Volume conservation tiny piston ko same volume sweep karne par majboor karta hai ek chhote face se, isliye uski travel distance usi diverging ratio se badhti hai — big piston ko finite amount move karne ke liye infinite stroke chahiye.
Energy: $W = F_{1} d_{1}$ finite rehti hai aur $F_{2} d_{2}$ ke barabar; infinities cancel ho jaati hain. Yeh step kyun? $F_{2} d_{2} = (F_{1} \cdot k) (d_{2})$ jabki $F_{1} d_{1} = F_{1} (d_{2} \cdot k)$ — dono $F_{1} k d_{2}$ ke barabar hain, isliye diverging $k$ dono sides par identically appear karta hai aur kabhi extra energy create nahi karta.

Verify: $k = 1000$ , $F_{1} = 10 N$ , $d_{2} = 0.01 m$ lo: $F_{2} = 10000 N$ , $d_{1} = 10 m$ . $W_{1} = 10 \times 10 = 100 J$ , $W_{2} = 10000 \times 0.01 = 100 J$ . ✓ Equal. Limit free energy nahi hai — yeh ek impractical stroke length maangti hai. Reality mein bhi cap lagti hai: koi bhi fluid perfectly incompressible nahi hota aur pistons ki finite size hoti hai.

Poora matrix ek map par

Upar ka decision tree ek figure ki tarah bhi draw kiya gaya hai taaki tum ise follow kar sako chahe Mermaid render na ho. "Equal pressure" se start karo, force questions ke liye area ratio $k$ par split karo, distance/energy questions ke liye volume conservation par aao, aur $ρ g h$ sirf tab add karo jab dono pistons alag heights par hon.

Recall Main kis cell mein hoon?

Pehla sawaal: kya woh force pooch rahe hain, ya distance, ya energy? ::: Force → $F_{2} = F_{1} A_{2} / A_{1}$ use karo (cells A/D/E). Distance → $A_{1} d_{1} = A_{2} d_{2}$ (cell B). Energy → $F_{1} d_{1} = F_{2} d_{2}$ (cell C). Bada output piston force ko kaunse number se multiply karta hai? ::: $A_{2} / A_{1} = (r_{2} / r_{1})^{2}$ — squared radius ratio. Pistons alag heights par hain — kya add karna padega? ::: $ρ g h$ baseline difference (cell H), transmitted change ke upar. Kya $A_{1} \to 0$ free energy deta hai? ::: Nahi — work finite rehti hai; tiny piston ko infinite travel chahiye (cell I). Agar ek question sirf surface pressure raise karta hai aur neeche change poochhe? ::: Yeh undiminished transmit hoti hai — $Δ p$ har depth par same hai (cell F); change track karo, absolute pressure nahi.

Connections

Parent: Pascal's law — woh principle jinhe yeh examples exercise karte hain.
Pressure — force per unit area — $p = F / A$ , har cell mein use hota hai.
Hydraulic systems — brakes, lifts, jacks — cells A, D, G real world mein.
Incompressibility & continuity equation — cells B, C, I ke peechhe $A_{1} d_{1} = A_{2} d_{2}$ .
Conservation of energy in machines — isliye cell C aur cell I free lunches nahi hain.
Hydrostatic pressure — p = p0 + ρgh — cells F aur H mein $ρ g h$ correction.