Pascal's law — pressure transmits equally: Question bank

2.2.6 · D5 · Physics › Fluid Mechanics › Pascal's law — pressure transmits equally

Shuru karne se pehle, teen anchor ideas yaad rakho — neeche ke har trap mein se ek ka twist hai:

Anchor A — isotropy: ek static fluid mein kisi ek point par, pressure har direction mein same hota hai (wedge argument).
Anchor B — undiminished transmission: ek applied change $Δ p$ har point tak equally pahunchta hai, bina kisi kami ke.
Anchor C — no free lunch: hydraulic press distance ko force ke saath trade karta hai, isliye work $F d$ conserved rehta hai.

Yahan $p$ pressure hai (force per area, dekho Pressure — force per unit area), $ρ$ fluid density hai, $g$ gravity hai, $h$ depth hai, aur $Δ p$ ka matlab hai "woh pressure change jo tumne add kiya."

True or false — justify karo

Har item: true/false decide karo, phir reason do.

"Pascal's law kehta hai ki ek tank mein pressure sab depths par same hota hai."

False. Total pressure depth ke saath badhta hai jaise

p = p_{0} + ρ g h

; Pascal sirf kehta hai ki ek added change

Δ p

sab depths tak equally pahunchta hai.

"Agar main surface pressure

200 Pa

badhata hoon, toh bottom pressure

200 Pa

se zyada badhega kyunki water column usmein add hota hai."

False.

ρ g h

term fixed hai aur unchanged rehta hai; bottom par increase exactly

200 Pa

hoga (Anchor B).

"Ek hydraulic lift force multiply karta hai, toh energy bhi multiply karta hai."

False. Output force bada hota hai lekin uska piston proportionally kam move karta hai, isliye

F_{1} d_{1} = F_{2} d_{2}

— energy conserved rehti hai (Anchor C).

"Still water mein ek point par, sideways push karne wala pressure neeche push karne wale pressure se chhota hota hai."

False. Static fluid mein ek point par pressure isotropic hota hai — har direction mein equal (Anchor A, wedge proof).

"Pascal's law sealed cylinder of air ke liye bhi utni hi achhi tarah kaam karta hai."

Practice mein False. Gas compressible hoti hai, isliye woh shrink hokar change ka kuch hissa absorb kar leti hai; transmission na undiminished hoti hai na instant.

"Pascal's law ko kaam karne ke liye gravity chahiye."

False. Isotropy ka result ek tiny element par force balance se aata hai jahan gravity negligible hoti hai; yeh free fall ya space mein bhi hold karta hai.

"Agar fluid enclosed nahi hai (ek open puddle), toh usse push karne par bhi pressure har jagah equally transmit hoga."

False. Confinement ke bina fluid simply flow/spill ho jaata hai aur applied change escape ho jaata hai instead of build up hone ke.

"Dono piston radii double karne par mechanical advantage unchanged rehta hai."

True. Advantage hai

A_{2} / A_{1} = (r_{2} / r_{1})^{2}

; dono double karne par ratio, aur isliye advantage, same rehta hai.

"Ek hydraulic press mein bina friction ke, input work exactly output work ke barabar hota hai."

True.

F_{1} d_{1} = F_{2} d_{2}

equal pressure plus incompressible volume conservation se follow karta hai.

Error pakdo

Har statement mein ek flawed step hai. Use name karo.

"Kyunki

F_{2} = F_{1} A_{2} / A_{1}

aur

A_{2} > A_{1}

, bada piston hamesha jeetta hai, isliye ek hydraulic jack koi bhi weight instantly lift kar sakta hai."

Distance ko ignore karta hai: bade piston ko thoda bhi upar uthane ke liye chhote ko kaafi zyada push karna padta hai, isliye 'instantly/any weight' kaam ki cost ko skip karta hai (Anchor C).

p_{x} = p_{n}

isliye nikla kyunki humne dono sides se

d A_{n} sin θ

cancel kiya — lekin hum

sin θ = 0

bhi set kar sakte the."

Galat:

sin θ = 0

ek degenerate wedge hai (koi slant face nahi). Valid cancellation har

θ

ke liye hold karta hai, aur yehi reason hai ki result general hai.

"Kyunki pressure depth ke saath badhta hai, agar main upar piston push karunga toh sabse gehre paani ko push sabse zyada strongly feel hogi."

Absolute pressure (

ρ g h

, depth-dependent) ko added

Δ p

ke saath confuse karta hai, jo depth-independent hai — har point ko same extra push feel hoti hai.

"Paani 'incompressible' hai, isliye uska pressure kabhi nahi badalta."

Incompressible ka matlab hai volume/density mein bahut kam change hota hai; pressure phir bhi freely change ho sakta hai — yehi toh Pascal's law transmit karta hai.

"Ek tiny wedge gravity feel karta hai, isliye isotropy prove karne ke liye humein

ρ g

weight term rakhna padega."

Jaise wedge shrink hota hai, uska volume (weight

\propto L^{3}

) uske face areas (force

\propto L^{2}

) se zyada tezi se vanish hota hai, isliye limit mein weight term drop out ho jaata hai.

"Ek press ka mechanical advantage

A_{1} / A_{2}

hai."

Ulta hai. Advantage

= A_{2} / A_{1}

: bada output area chhote input area par — yahi amplification deta hai.

"Kyunki dono pistons mein

p_{1} = p_{2}

hai, dono forces

F_{1}

aur

F_{2}

bhi equal hone chahiye."

Equal pressure hai, equal force nahi.

F = p A

, aur areas alag hain, isliye forces area ratio se alag hote hain.

Why questions

"Ek point par pressure har direction mein same kyun hota hai, words mein?"

Agar nahi hota, toh unbalanced sideways force fluid ko flow karata — lekin static fluid flow nahi karta, isliye sab directions balance hone chahiye (isotropy).

"Jab hum surface par

Δ p

add karte hain toh

ρ g h

term constant kyun rehta hai?"

Same fluid, same density, same heights —

ρ

g

h

change nahi hote, isliye sirf surface term

p_{0}

shift hota hai aur uski shift unchanged neeche pahunchti hai.

"Press ke liye volume conserve karne ke liye fluid incompressible kyun hona chahiye (

A_{1} d_{1} = A_{2} d_{2}

)?"

Agar compress karta, toh kuch pushed-in volume denser fluid mein 'disappear' ho jaata instead of dusre piston par emerge hone ke, aur volume bookkeeping toot jaati.

"Ek chhoti si force ek bade weight ko kyun lift karti hai — 'strength' kahaan se aati hai?"

Same pressure ek bade area par act karta hai, aur force

=

pressure

\times

area, isliye ek bada area modest pressure ko ek badi force mein convert kar deta hai.

"Pascal's law pressure mein change ke baare mein kyun hai, total ke baare mein nahi?"

Total mein pehle se fixed hydrostatic profile included hai; law ka content yeh hai ki koi bhi naya applied change khud ko us profile ke upar equally redistribute karta hai.

"Ek hydraulic press perpetual-motion / free-energy machine kyun nahi ho sakta?"

Kyunki amplified force proportionally shorter distance par act karta hai, isliye energy in equals energy out — dekho Conservation of energy in machines.

"Buoyancy Pascal-style isotropy par kyun depend karta hai?"

Buoyant force submerged body ke sab faces par equally andar pressure push karne se aati hai; gehre faces zyada feel karte hain, jo ek net upward push deta hai (dekho Archimedes' principle — buoyancy).

Edge cases

"Agar dono pistons ka area equal hai,

A_{1} = A_{2}

, toh mechanical advantage kya hai?"

Exactly 1 — press force ko bina amplification aur bina reduction ke transmit karta hai; yeh sirf ek pressure relay hai.

"Agar 'incompressible' fluid mein ek chhota air bubble trapped ho toh transmitted change ka kya hoga?"

Bubble compress hota hai aur applied change ka kuch hissa absorb kar leta hai, isliye transmission sluggish aur diminished ho jaati hai — yehi wajah hai ki brake lines bleed ki jaani chahiye.

"Jab wedge angle

θ \to 0

, slant face vertical ki taraf flatten hoti hai — kya proof phir bhi hold karta hai?"

Haan; ratios

d A_{x} = d A_{n} sin θ

aur

d A_{y} = d A_{n} cos θ

sab

θ

ke liye hold karte hain, aur pressures angle se regardless identically cancel ho jaate hain.

"Agar tum

Δ p

apply karo lekin container walls perfectly rigid hain aur fluid truly incompressible hai, toh kya kuch move karta hai?"

Kuch nahi move karta; pressure simply har jagah

Δ p

se rise karta hai aur walls par zyada press karta hai — pressure transmission ke liye koi flow zaroori nahi.

"Us surface par jahan depth

h = 0

hai, kya applied change phir bhi register hota hai?"

Haan —

h = 0

ke saath pressure sirf

p_{0}

hai, aur

Δ p

add karne par yeh

p_{0} + Δ p

ho jaata hai, kisi bhi depth par same rise.

"Ek sealed tank mein jahan piston upar ki jagah neeche ho, kya transmitted

Δ p

alag hota hai?"

Nahi — Pascal's law is baat se indifferent hai ki change kahaan apply kiya gaya hai; har point phir bhi same

Δ p

se rise karta hai.

"Kya hoga agar fluid density

ρ

depth ke saath vary kare (jaise layered oil over water)?"

Hydrostatic profile piecewise ho jaata hai lekin phir bhi fixed rehta hai; applied

Δ p

phir bhi us profile ke upar equally add hota hai.

Recall Har trap ki ek-line summary

Yahan lagbhag har mistake teen confusions mein se ek hai: total pressure ko applied change ke saath mix karna (Anchor B), ek point par isotropy bhool jaana (Anchor A), ya force multiplication se free energy expect karna (Anchor C).

Connections

Pascal's law — pressure transmits equally — parent topic jinhe ye traps drill karte hain.
Hydrostatic pressure — p = p0 + ρgh — woh fixed depth term jo kai traps mein $Δ p$ ke saath confuse hota hai.
Pressure — force per unit area — kyun equal pressure unequal forces deta hai.
Hydraulic systems — brakes, lifts, jacks — jahan bubble/incompressibility edge cases bite karte hain.
Conservation of energy in machines — "no free lunch" answers.
Incompressibility & continuity equation — volume-conservation edge cases.
Archimedes' principle — buoyancy — isotropy submerged bodies par apply hoti hai.