Pressure due to fluid column
P = P_0 + ρ g h, where P_0 is atmospheric pressure, ρ is fluid density, h is depth. Pressure increases linearly with depth in a static fluid.
-- NCERT, p. 3Pressure at depth: the one formula NEET keeps testing in disguise.
The static fluid pressure formula P = P₀ + ρgh looks simple, but NEET recycles it in setups that confuse well-prepared students. The common trap is not the formula itself — it is misidentifying what h, ρ, or P₀ represent in a given scenario.
The core idea. In a fluid at rest, gravity pulls every layer downward. Each layer must support the weight of the fluid above it. At depth h below the free surface of a fluid with uniform density ρ, the pressure exceeds the surface pressure P₀ by exactly ρgh. This is gauge pressure — the extra pressure due to the fluid column alone (NCERT Class 11 Physics, Chapter 10 — Mechanical Properties of Fluids, page 3).
Three conditions that must hold:
Where students lose marks:
NEET bridge. Questions on this topic appear as straightforward depth-pressure calculations, as setups involving manometers and U-tubes, or as conceptual questions about pressure equality at the same horizontal level. The arithmetic is light; the marks are lost to misreading which pressure (gauge vs. absolute) the question demands.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The pressure at a point inside a static fluid depends on which of the following?
What is the SI unit of pressure?
In the formula P = P₀ + ρgh for static fluid pressure, what does P₀ represent?
A tank is filled with water (density 1.0 × 10³ kg/m³) to a depth of 5.0 m. Taking atmospheric pressure as 1.013 × 10⁵ Pa and g = 9.8 m/s² (exact for this problem), what is the absolute pressure at the bottom of the tank?
A U-tube manometer open to the atmosphere on both sides contains mercury (density 1.36 × 10⁴ kg/m³). If the mercury level in one arm is 8.0 cm higher than in the other, what is the gauge pressure difference between the two arms? (Take g = 9.8 m/s², exact for this problem.)
A sealed container has gas at pressure P_gas trapped above a liquid column of height h and density ρ. The bottom of the container is open to the atmosphere (P₀). At the bottom, the pressure is P₀. What is the gas pressure P_gas?
Three open containers — a narrow cylinder, a wide cylinder, and a cone (wider at top) — are filled with the same liquid to the same vertical height h. How do the pressures at the bottom compare?
A vertical U-tube has water (density 1.0 × 10³ kg/m³) in one arm and oil (density 8.0 × 10² kg/m³) in the other. The oil column is 20.0 cm above the water-oil interface. If the water surface in the other arm is open to the atmosphere, how much higher is the water level in the water arm than the interface level? (Take g = 9.8 m/s², exact for this problem.)
Given
A lake has fresh water with density ρ = 1.00 × 10³ kg/m³. A diver descends to a depth of h = 15.0 m. Atmospheric pressure at the surface is P₀ = 1.013 × 10⁵ Pa. Take g = 9.80 m/s².
Required
Find (a) the gauge pressure at the diver's depth, and (b) the absolute pressure.
Concept
In a static incompressible fluid, the pressure increases linearly with depth. The hydrostatic formula P = P₀ + ρgh gives the absolute pressure. The gauge pressure is the excess above atmospheric: P_gauge = ρgh.
Formula
- Gauge pressure: P_gauge = ρgh - Absolute pressure: P = P₀ + ρgh
Substitution
P_gauge = (1.00 × 10³)(9.80)(15.0) P = 1.013 × 10⁵ + P_gauge
Calculation
P_gauge = 1.00 × 10³ × 9.80 × 15.0 = 1.470 × 10⁵ Pa Note on exact constants: g = 9.80 m/s² is given as a three-significant-figure value (not an exact constant here). All given quantities have 3 significant figures, so the result is reported to 3 significant figures. P = 1.013 × 10⁵ + 1.470 × 10⁵ = 2.483 × 10⁵ Pa
Final answer
(a) Gauge pressure at 15.0 m depth = 1.47 × 10⁵ Pa (b) Absolute pressure at 15.0 m depth = 2.48 × 10⁵ Pa (≈ 2.5 atm) Both reported to 3 significant figures, consistent with the given data.
Common trap
The most frequent mark-losing error on this type of question: computing ρgh (gauge pressure) and writing it as the final answer when the question asks for absolute pressure — or vice versa. Always re-read the stem: "pressure at depth" without qualification usually means absolute pressure; "excess pressure" or "gauge pressure" means ρgh alone. A second trap: using h as the distance along a slanted path rather than the vertical depth. Only the vertical component matters for hydrostatic pressure.
Similar NEET-style question
A submarine is at a depth of 200 m in sea water (density 1.03 × 10³ kg/m³). Taking atmospheric pressure as 1.01 × 10⁵ Pa and g = 9.80 m/s², find the absolute pressure on the hull. (Answer: ≈ 2.12 × 10⁶ Pa ≈ 21 atm.) ---
P = P_0 + ρ g h, where P_0 is atmospheric pressure, ρ is fluid density, h is depth. Pressure increases linearly with depth in a static fluid.
-- NCERT, p. 3Conservation of energy along a streamline of incompressible non-viscous flow.
| Symbol | Quantity | SI Unit |
|---|---|---|
| P | pressure | Pa |
| rho | density | kg/m^3 |
| v | speed | m/s |
| g | gravity | m/s^2 |
| h | height | m |
Resistance of a material to uniform compression. Inverse: compressibility.
| Symbol | Quantity | SI Unit |
|---|---|---|
| K | bulk modulus | Pa |
| V | volume | m^3 |
| P | pressure | Pa |
Height a liquid rises (or falls) in a capillary tube. cos(theta) > 0: rises (wetting); < 0: depresses.
| Symbol | Quantity | SI Unit |
|---|---|---|
| h | capillary height | m |
| T | surface tension | N/m |
| theta | contact angle | rad |
| rho | density | kg/m^3 |
| r | tube radius | m |
Pressure at depth h below free surface of fluid of density rho.
| Symbol | Quantity | SI Unit |
|---|---|---|
| P | total pressure | Pa |
| P0 | atmospheric/surface pressure | Pa |
| rho | density | kg/m^3 |
| h | depth | m |
Heat absorbed/released during phase change at constant T. L_fusion or L_vaporisation.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Q | heat | J |
| m | mass | kg |
| L | latent heat | J/kg |
Heat required to raise mass m by temperature Delta_T. Specific heat c is material property.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Q | heat | J |
| m | mass | kg |
| c | specific heat | J/kg/K |
| Delta_T | temp change | K |
Radiation power from a body. Black body epsilon=1; net to surroundings P = sigma*epsilon*A*(T^4 - T_s^4).
| Symbol | Quantity | SI Unit |
|---|---|---|
| sigma | Stefan-Boltzmann = 5.67e-8 | W/m^2/K^4 |
| epsilon | emissivity (0-1) | - |
| A | surface area | m^2 |
| T | absolute temp | K |
Drag force on a sphere of radius r moving with velocity v through viscous fluid (low Reynolds number).
| Symbol | Quantity | SI Unit |
|---|---|---|
| F | drag force | N |
| eta | viscosity | Pa*s |
| r | sphere radius | m |
| v | velocity | m/s |
Excess internal pressure due to surface tension. Bubble has 2 surfaces, hence factor 4.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Delta_P | excess pressure | Pa |
| T | surface tension | N/m |
| r | radius | m |
Constant velocity reached when net force is zero (gravity balanced by buoyancy + viscous drag).
| Symbol | Quantity | SI Unit |
|---|---|---|
| v_t | terminal velocity | m/s |
| r | sphere radius | m |
| rho_s | sphere density | kg/m^3 |
| rho_f | fluid density | kg/m^3 |
| eta | viscosity | Pa*s |
Fractional change in length, area, volume per degree temperature change.
| Symbol | Quantity | SI Unit |
|---|---|---|
| alpha | linear coefficient | 1/K |
| beta | volume coefficient | 1/K |
| Delta_T | temperature change | K |
Ratio of longitudinal stress to longitudinal strain in a stretched wire/rod within elastic limit.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Y | Young's modulus | Pa |
| F | applied force | N |
| A | cross-section area | m^2 |
| L | original length | m |
| Delta_L | extension | m |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
Student uses 2T/r for soap bubble (drop formula). Bubble has 2 surfaces → 4T/r.
Question mentions soap bubble OR liquid drop OR air bubble in liquid.
Drop in air: 1 surface → 2T/r. Soap bubble in air: 2 surfaces → 4T/r. Air bubble in liquid: 1 surface → 2T/r.
Category: Similar Terms
Student uses Y formula when problem is about volumetric compression (use K) or vice versa.
Problem describes longitudinal stretching (use Y), volumetric pressure (use K), or shear (use G).
Y: longitudinal stress/strain. K: volumetric. G: shear. Match modulus to deformation type.
Root cause: concept gap
Drop in air: 1 surface → ΔP = 2T/r. Soap bubble: 2 surfaces (inner + outer) → ΔP = 4T/r. Air bubble inside liquid: 1 surface → 2T/r.
Root cause: formula misuse
v_t ∝ r² (because Stokes drag ∝ r v, gravity ∝ r³ - r³ = r³_diff). Doubling radius quadruples terminal velocity, not doubles.
Root cause: formula misuse
Y for longitudinal stretch (FL/A·ΔL); K for volumetric compression (-V·dP/dV); G for shear. Match modulus type to deformation type before computing.
If a soap bubble expands, the pressure inside the bubble
forgets height term
Drops rho*g*h term
equates pressures incorrectly
Picks wrong reference points
ignores mass difference
Compares only specific heats, not masses
uses 2T r instead of 4T r for bubble
Treats soap bubble like a drop
expects linear acceleration
Default to constant acceleration without recognising drag-induced terminal v
uses bulk modulus formula
Confuses Y with K
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