Pascal's law — pressure transmits equally: Exercises

2.2.6 · D4 · Physics › Fluid Mechanics › Pascal's law — pressure transmits equally

Shuru karne se pehle, yahan poora toolkit ek card pe hai taaki koi bhi symbol unexplained na rahe.

Formula Paanch tools (har ek parent note mein earn kiye gaye hain)

Pressure $p = \frac{F}{A}$ — force $F$ (newtons mein, N) seedha area $A$ (square metres mein, $m^{2}$ ) par dabaya gaya. Dekho Pressure — force per unit area.
Pascal transmission — enclosed incompressible fluid par lagaya gaya pressure change $Δ p$ har jagah equally pahunchta hai. Dekho the parent note.
Pistons ke across same pressure $\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}} \Rightarrow F_{2} = F_{1} \frac{A _{2}}{A _{1}}$ .
Volume conservation (incompressible): $A_{1} d_{1} = A_{2} d_{2}$ — ek cylinder se bahar push kiya gaya fluid ( $A_{1} d_{1}$ ) doosre mein dikhai deta hai. Dekho Incompressibility & continuity equation.
Hydrostatic add-on $p = p_{0} + ρ g h$ — density $ρ$ , height $h$ wale fluid ke column se extra pressure, jahan $g \approx 9.8 m/s^{2}$ . Dekho Hydrostatic pressure — p = p0 + ρgh.

Yahan $A$ = area, $d$ = distance jitna ek piston move karta hai, $ρ$ (Greek "rho") = density $kg/m^{3}$ mein.

Figure — Pascal's law — pressure transmits equally

Figure dekho: chhota piston (left, lavender) ka area $A_{1}$ hai; tum isko force $F_{1}$ se push karte ho aur yeh kaafi door $d_{1}$ tak sink karta hai. Bada piston (right, coral) ka area $A_{2}$ hai; yeh force $F_{2}$ se uthta hai lekin sirf thodi door $d_{2}$ tak. Unke beech ka mint fluid same pressure dono tak pohunchata hai. Har problem ke liye yeh picture dimag mein rakho.

Level 1 — Recognition

Recall Solution L1.1

Pressure change dono pistons par equal hoti hai — yahi Pascal's law hai exactly. Force bade piston par zyada hoti hai, aur move ki gayi distance bade piston par kam hoti hai. Toh na force equal hai aur na distance; sirf pressure equal hai.

Recall Solution L1.2

Exactly $3000 Pa$ . Likho $p = p_{0} + ρ g h$ . $p_{0}$ ko $Δ p$ se badhane par $p_{new} = (p_{0} + Δ p) + ρ g h$ milta hai. $ρ g h$ term (jo depth par depend karta hai) unchanged rehta hai, toh rise har depth par $Δ p$ hi hoga — $4 m$ par bhi. Depth ek distractor hai.

Level 2 — Application

Recall Solution L2.1

Area radius ke square ke saath scale karta hai, isliye humhe kabhi $π$ ki zaroorat nahi — yeh cancel ho jaata hai: $\frac{A _{2}}{A _{1}} = \frac{π r _{2}^{2}}{π r _{1}^{2}} = (\frac{r _{2}}{r _{1}})^{2} = (\frac{30}{3})^{2} = 1 0^{2} = 100.$ Square kyun? Figure dekho: radius double karne par disc ka area chaar guna ho jaata hai, toh force area ratio se multiply hoti hai, radius ratio se nahi. $F_{2} = F_{1} \frac{A _{2}}{A _{1}} = 200 \times 100 = 20, 000 N .$

Recall Solution L2.2

Fluid incompressible hai, toh volume out = volume in: $A_{1} d_{1} = A_{2} d_{2}$ . $d_{1} = d_{2} \frac{A _{2}}{A _{1}} = 2 cm \times 100 = 200 cm = 2 m .$ Chhota piston bahut badi distance travel karta hai bade piston ko thoda sa move karane ke liye — yahi force boost ki keemat hai.

Recall Solution L2.3

Areas seedhe diye gaye hain, toh yahan koi squaring ki zaroorat nahi: $F_{2} = F_{1} \frac{A _{2}}{A _{1}} = 150 \times \frac{20}{5} = 150 \times 4 = 600 N .$

Level 3 — Analysis

Recall Solution L3.1

Work = force $\times$ force ki direction mein move ki gayi distance.

Input work: $W_{1} = F_{1} d_{1} = 200 \times 2 = 400 J$ .
Output work: $W_{2} = F_{2} d_{2} = 20, 000 \times 0.02 = 400 J$ . Dono equal hain: $W_{1} = W_{2} = 400 J$ . Press force ko 100 se multiply karti hai lekin distance ko 100 se shrink karti hai, toh unka product — energy — unchanged rehta hai. Dekho Conservation of energy in machines.

Recall Solution L3.2

Bade piston par pressure = tumhare push ka pressure plus uske upar khade oil column ka weight jo height $h$ par hai (from Hydrostatic pressure — p = p0 + ρgh): $p_{2} = \frac{F _{1}}{A _{1}} + ρ g h .$ Step 1 — applied pressure: $\frac{F _{1}}{A _{1}} = \frac{300}{4 \times 1 0 ^{- 4}} = 750, 000 Pa$ . Step 2 — hydrostatic add-on: $ρ g h = 850 \times 9.8 \times 1.5 = 12, 495 Pa$ . Step 3 — bade piston par total: $p_{2} = 750, 000 + 12, 495 = 762, 495 Pa$ . Step 4 — force: $F_{2} = p_{2} A_{2} = p_{2} \times 50 A_{1} = 762, 495 \times 50 \times 4 \times 1 0^{- 4}$ $F_{2} = 762, 495 \times 0.02 = 15, 249.9 N \approx 1.52 \times 1 0^{4} N .$ Pure-Pascal estimate hoga $300 \times 50 = 15, 000 N$ ; oil column lagbhag $250 N$ add karta hai.

Level 4 — Synthesis

Recall Solution L4.1

(a) Zaroori output force = car ka weight: $F_{2} = m g = 1500 \times 9.8 = 14, 700 N$ . Area ratio: $\frac{A _{2}}{A _{1}} = (\frac{r _{2}}{r _{1}})^{2} = (\frac{25}{2.5})^{2} = 1 0^{2} = 100$ . $F_{1} = F_{2} \frac{A _{1}}{A _{2}} = 14, 700 \times \frac{1}{100} = 147 N .$ (b) Volume conserved: $A_{1} d_{1} = A_{2} d_{2} \Rightarrow d_{1} = d_{2} \frac{A _{2}}{A _{1}} = 0.5 \times 100 = 50 m$ . (c) Input work $W_{1} = F_{1} d_{1} = 147 \times 50 = 7350 J$ . Car ki gravitational energy gain $= m g d_{2} = 14, 700 \times 0.5 = 7350 J$ . ✓ Dono match karte hain: tumhara har joule push karna car ki height energy mein wapas aata hai. Koi free lunch nahi — tumne bas ek chhote piston ko lambi $50 m$ door push kiya.

Recall Solution L4.2

Step 1 — press karne se pehle bottom pressure: $p_{before} = p_{0} + ρ g h = 101, 325 + 1000 \times 9.8 \times 3 = 101, 325 + 29, 400 = 130, 725 Pa$ . Step 2 — Pascal applied change ko bina ghate add karta hai: $p_{after} = p_{before} + Δ p = 130, 725 + 20, 000 = 150, 725 Pa$ . Step 3 — hatch par force: $F = p_{after} A = 150, 725 \times 0.02 = 3014.5 N$ .

Level 5 — Mastery

Recall Solution L5.1

(a) Same height ⇒ same pressure $p = F_{1} / A_{1} = 50/ (1 \times 1 0^{- 4}) = 500, 000 Pa$ har jagah. $F_{2} = p A_{2} = 500, 000 \times 4 \times 1 0^{- 4} = 200 N, F_{3} = p A_{3} = 500, 000 \times 9 \times 1 0^{- 4} = 450 N .$ (b) Fluid incompressible hai: chhota piston jitna volume andar push karta hai woh baaki pistons ke liye total volume ke barabar hai: $A_{1} d_{1} = A_{2} d_{2} + A_{3} d_{3} .$ $A_{1} = 1 cm^{2}$ se divide karo (toh ratios $A_{2} / A_{1} = 4$ , $A_{3} / A_{1} = 9$ ): $d_{1} = \frac{A _{2}}{A _{1}} d_{2} + \frac{A _{3}}{A _{1}} d_{3} = 4 d_{2} + 9 d_{3} . ■$ (c) Real fluids aur seals friction cause karte hain, aur koi bhi trapped air compressible hoti hai, isliye kuch input work heat ke roop mein kho jaata hai / compression mein absorb ho jaata hai — output forces thodi kam rehti hain aur transmission perfectly instant nahi hoti. (Parent note mein "any fluid" mistake aur Conservation of energy in machines dekho.)

Recall Ek-line self-test

Force ka sawaal ::: $F_{2} = F_{1} A_{2} / A_{1}$ use karo (equal height par pressures equal). Distance ka sawaal ::: $A_{1} d_{1} = A_{2} d_{2}$ use karo (volume conserved). Alag heights ::: pistons ke beech ke column ke liye $ρ g h$ add karo. Kya yeh kabhi free energy deta hai? ::: Nahi — $F_{1} d_{1} = F_{2} d_{2}$ hamesha.

Connections

Parent: Pascal's law — woh theory jise yeh drills test karti hain.
Pressure — force per unit area — $p = F / A$ jo poori jagah use hoti hai.
Incompressibility & continuity equation — $A_{1} d_{1} = A_{2} d_{2}$ volume rule.
Hydrostatic pressure — p = p0 + ρgh — L3.2 aur L4.2 mein $ρ g h$ term.
Hydraulic systems — brakes, lifts, jacks — L2.3 brake aur L4.1 lift.
Conservation of energy in machines — L3.1, L4.1, L5.1 mein "no free lunch" checks.