Intuition The big pictureA fluid at rest still has weight . The deeper you go, the more fluid sits on top of you , pressing down. So pressure grows with depth . The relation P = ρ g h P = \rho g h P = ρ g h “weight of the column above, spread over an area.”
==Deeper ⇒ \Rightarrow ⇒ ⇒ \Rightarrow ⇒
Pressure P P P P = F A P = \dfrac{F}{A} P = A F pascal Pa = N/m 2 \text{Pa} = \text{N/m}^2 Pa = N/m 2 Density ρ \rho ρ ρ = m V \rho = \dfrac{m}{V} ρ = V m kg/m 3 \text{kg/m}^3 kg/m 3 Hydrostatic = fluid at rest (no flow, no acceleration).Gauge pressure : pressure above atmospheric . Absolute pressure : total, including atmosphere.
WHY does it matter? Dams, submarines, blood pressure, barometers, buoyancy — every one of them lives or dies by how pressure changes with depth.
We use only Newton's first law: a fluid at rest has zero net force on every piece of it.
Pick an imaginary column of fluid of cross-section area A A A h h h
Step 1 — Choose the body.
A vertical cylinder of fluid: area A A A h h h h h h
Why this step? A column makes the geometry simple — only top and bottom faces matter for vertical balance; the sides push horizontally and cancel.
Step 2 — Find the mass of the column.
V = A h , m = ρ V = ρ A h V = A h, \qquad m = \rho V = \rho A h V = A h , m = ρ V = ρ A h
Why this step? Pressure is about weight, and weight needs mass; mass comes from density × volume.
Step 3 — Forces on the column (vertical only).
Weight pulling down : W = m g = ρ A h g W = mg = \rho A h\, g W = m g = ρ A h g
Atmosphere pushing down on the top: F top = P 0 A F_{\text{top}} = P_0 A F top = P 0 A
Fluid below pushing up on the bottom: F bot = P A F_{\text{bot}} = P A F bot = P A
Why this step? Each face feels pressure × \times ×
Step 4 — Apply equilibrium (net force = 0 =0 = 0
F bot = F top + W F_{\text{bot}} = F_{\text{top}} + W F bot = F top + W P A = P 0 A + ρ A h g P A = P_0 A + \rho A h g P A = P 0 A + ρ A h g
Why this step? The column is at rest ⇒ \Rightarrow ⇒
Step 5 — Divide by A A A
P = P 0 + ρ g h \boxed{P = P_0 + \rho g h} P = P 0 + ρ g h
The gauge (extra) pressure due to the fluid alone is:
P gauge = P − P 0 = ρ g h \boxed{P_{\text{gauge}} = P - P_0 = \rho g h} P gauge = P − P 0 = ρ g h
Intuition Hydrostatic paradox (WHY shape doesn't matter)A wide tank and a thin tube of the same depth have the same bottom pressure, even though the wide tank holds far more water. Pressure cares about depth , not total volume — because the extra side walls hold up the extra weight.
Worked example 1) Pressure at the bottom of a 3 m poolWater ρ = 1000 kg/m 3 \rho = 1000\,\text{kg/m}^3 ρ = 1000 kg/m 3 g = 9.8 m/s 2 g = 9.8\,\text{m/s}^2 g = 9.8 m/s 2 h = 3 m h = 3\,\text{m} h = 3 m Gauge: P = ρ g h = 1000 × 9.8 × 3 = 2.94 × 10 4 Pa P = \rho g h = 1000 \times 9.8 \times 3 = 2.94\times10^4\,\text{Pa} P = ρ g h = 1000 × 9.8 × 3 = 2.94 × 1 0 4 Pa Why this step? We want the extra pressure from water, so use ρ g h \rho g h ρ g h Absolute: add atmosphere: P a b s = 1.01 × 10 5 + 2.94 × 10 4 ≈ 1.30 × 10 5 Pa P_{abs} = 1.01\times10^5 + 2.94\times10^4 \approx 1.30\times10^5\,\text{Pa} P ab s = 1.01 × 1 0 5 + 2.94 × 1 0 4 ≈ 1.30 × 1 0 5 Pa
Worked example 2) How deep for pressure to double atmospheric?Want ρ g h = P 0 = 1.01 × 10 5 \rho g h = P_0 = 1.01\times10^5 ρ g h = P 0 = 1.01 × 1 0 5 h = P 0 ρ g = 1.01 × 10 5 1000 × 9.8 ≈ 10.3 m h = \dfrac{P_0}{\rho g} = \dfrac{1.01\times10^5}{1000\times 9.8} \approx 10.3\,\text{m} h = ρ g P 0 = 1000 × 9.8 1.01 × 1 0 5 ≈ 10.3 m Why this step? Set the water's gauge pressure equal to one atmosphere, then solve for h h h ∼ 10 \sim10 ∼ 10
Worked example 3) Mercury barometerAtmosphere holds up a mercury column of height h h h ρ H g = 13600 kg/m 3 \rho_{Hg}=13600\,\text{kg/m}^3 ρ H g = 13600 kg/m 3 P 0 = ρ g h ⇒ h = 1.01 × 10 5 13600 × 9.8 ≈ 0.758 m = 758 mm P_0 = \rho g h \Rightarrow h = \dfrac{1.01\times10^5}{13600 \times 9.8} \approx 0.758\,\text{m} = 758\,\text{mm} P 0 = ρ g h ⇒ h = 13600 × 9.8 1.01 × 1 0 5 ≈ 0.758 m = 758 mm Why this step? The vacuum above has P = 0 P=0 P = 0 ρ g h \rho g h ρ g h why pressure is quoted in "mm of Hg."
Common mistake "Bigger tank⇒ \Rightarrow ⇒ Why it feels right: more water = more total weight, surely more push. Fix: the extra width is held up by the bottom and walls , not stacked over one point. Only the column directly above a point matters; pressure = ρ g h =\rho g h = ρ g h depth , not volume.
Common mistake ForgettingP 0 P_0 P 0 Why it feels right: ρ g h \rho g h ρ g h the answer. Fix: ρ g h \rho g h ρ g h gauge (extra) pressure. Absolute = P 0 + ρ g h = P_0 + \rho g h = P 0 + ρ g h
h h h Why it feels right: "distance from the edge." Fix: h h h vertical depth below the free surface , measured straight down , regardless of how the container slants.
Common mistake Mixing density units.Why it feels right: 1 g/cm 3 1\,\text{g/cm}^3 1 g/cm 3 Fix: convert to SI: 1 g/cm 3 = 1000 kg/m 3 1\,\text{g/cm}^3 = 1000\,\text{kg/m}^3 1 g/cm 3 = 1000 kg/m 3
Recall Feynman: explain to a 12-year-oldImagine a tall stack of pillows. If you lie at the bottom of the stack, lots of pillows squash you. Near the top , only a few do. Water is the same: the deeper you dive, the more water sits above you and squeezes you. Twice as deep ⇒ \Rightarrow ⇒ deep you go counts, because only the water right above your head is pressing on you.
“RGH = Really Gets Heavier (with depth).”
P = ρ g h P = \rho g h P = ρ g h R ho, G ravity, H eight-of-column. Drop the h h h from .
Hydrostatic pressure formula (gauge) What does each symbol mean in ρ g h \rho g h ρ g h ρ \rho ρ = fluid density,
g g g = gravity,
h h h = vertical depth below surface
Absolute vs gauge pressure relation P a b s = P 0 + ρ g h P_{abs} = P_0 + \rho g h P ab s = P 0 + ρ g h ; gauge
= ρ g h = \rho g h = ρ g h Does pressure depend on container shape/area? No — only on vertical depth
h h h (hydrostatic paradox)
Starting principle used to derive P = ρ g h P=\rho g h P = ρ g h Fluid at rest
⇒ \Rightarrow ⇒ net force on any column is zero (force balance)
Why do side forces drop out of the derivation? They are horizontal and cancel by symmetry; only top/bottom faces give vertical force
Roughly how much depth of water adds 1 atmosphere? About 10 m
SI unit of pressure pascal,
Pa = N/m 2 \text{Pa} = \text{N/m}^2 Pa = N/m 2 Height of mercury barometer at 1 atm ≈ 0.76 m \approx 0.76\,\text{m} ≈ 0.76 m (760 mm Hg)
Convert 1 g/cm 3 1\,\text{g/cm}^3 1 g/cm 3 1000 kg/m 3 1000\,\text{kg/m}^3 1000 kg/m 3 Why does pressure increase with depth (one line)? More fluid weight sits above, spread over the same area
Pascal's Law — pressure applied to enclosed fluid transmits equallyAtmospheric Pressure & Barometer — measuring P 0 P_0 P 0 ρ g h \rho g h ρ g h Buoyancy & Archimedes' Principle — net upward force comes from pressure difference with depthManometers — read pressure differences as height differencesBernoulli's Equation — adds motion; reduces to ρ g h \rho g h ρ g h Density and Specific Gravity — supplies ρ \rho ρ
deeper means more fluid above
Pressure grows with depth
Imaginary fluid column area A depth h
Atmosphere pushes down P0 A
Hydrostatic paradox shape irrelevant
Dams submarines barometers buoyancy
Intuition Hinglish mein samjhoDekho yaar, hydrostatics ka core idea bilkul simple hai: fluid jab rest pe hota hai, tab bhi uska weight hota hai. Jitna neeche jaoge, utna zyada paani tumhare upar baitha hota hai, aur wo dabaata hai. Isi liye depth badhne se pressure badhta hai . Formula P = ρ g h P = \rho g h P = ρ g h
Derivation samajhne ke liye ek imaginary column lo — area A A A h h h P A PA P A P 0 A P_0 A P 0 A ρ A h g \rho A h g ρ A h g P A = P 0 A + ρ A h g PA = P_0 A + \rho A h g P A = P 0 A + ρ A h g A A A P = P 0 + ρ g h P = P_0 + \rho g h P = P 0 + ρ g h
Ek important baat: pressure sirf depth pe depend karta hai, container ke size ya shape pe nahi. Isko hydrostatic paradox kehte hain — chhoti patli tube aur badi tank, dono ka same depth pe bottom pressure same. Reason: jo extra paani hai badi tank mein, uska weight walls aur base sambhalte hain, ek point ke upar nahi aata.
Exam tip: agar "gauge" pressure pucha hai to sirf ρ g h \rho g h ρ g h P 0 P_0 P 0 h h h 1 g/cm 3 = 1000 kg/m 3 1\,\text{g/cm}^3 = 1000\,\text{kg/m}^3 1 g/cm 3 = 1000 kg/m 3
