Formula
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 Class 11 Physics, Ch. 9, p. 183Effect of Gravity on Pressure
8 MCQs4 revision cards9-step worked example
Source: NCERT Properties of Solids and LiquidsOfficial key: NTA-verifiedLast updated: 24 Sep 2026
Effect of Gravity on Pressure, explained for NEET
Pressure 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 9 — Mechanical Properties of Fluids, page 182).
Three conditions that must hold:
- The fluid is static (no flow).
- Density ρ is constant (incompressible fluid).
- Gravitational acceleration g is uniform over the depth considered.
Where students lose marks:
- Confusing depth with height above a reference. h is measured vertically downward from the free surface, not along a tilted tube or from the container bottom.
- Forgetting that pressure at the same horizontal level is equal. In connected vessels with the same fluid, the pressure depends only on vertical depth — the shape of the container is irrelevant (Pascal's vases).
- Mixing gauge pressure and absolute pressure. Gauge pressure = ρgh. Absolute pressure = P₀ + ρgh. NEET stems sometimes ask for one when you instinctively compute the other.
- Applying the formula to a flowing fluid. P = P₀ + ρgh holds only in a static fluid. If the fluid moves, you need Bernoulli's equation instead.
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.
Can you answer these Effect of Gravity on Pressure MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The pressure at a point inside a static fluid depends on which of the following?
Show answer and why every option is right or wrong
Answer: D. In a static incompressible fluid, pressure at any point is P = P₀ + ρgh, depending only on vertical depth h below the free surface (NCERT Class 11 Physics, Chapter 9, page 182). Container shape, total volume, and horizontal position do not affect the pressure at a given depth.
Why A is wrong: A is wrong because pressure in a static fluid is independent of container shape — Pascal's vases demonstrate that differently shaped containers filled to the same height produce the same pressure at the bottom.
Why B is wrong: B is wrong because the total volume of fluid does not enter the formula P = P₀ + ρgh. Only the vertical depth h and density ρ matter.
Why C is wrong: C is wrong because horizontal position does not affect static fluid pressure. All points at the same vertical depth in a connected body of the same fluid have the same pressure.
MCQ 2Direct ApplicationPractice
The atmospheric pressure at the surface of a lake is 1.0 × 10⁵ Pa. Taking the density of water as 1000 kg/m³ and g = 10 m/s², the absolute pressure at a depth of 10 m is:
Show answer and why every option is right or wrong
Answer: A. A is correct. Pressure increases with depth because each layer supports the weight of the water above it: P = P₀ + ρgh = 1.0 × 10⁵ + (1000)(10)(10) = 2.0 × 10⁵ Pa. Every 10 m of water adds about one atmosphere.
Why B is wrong: B is wrong because 1.0 × 10⁵ Pa is either the atmospheric pressure alone or the gauge pressure ρgh alone. The absolute pressure is their sum.
Why C is wrong: C is wrong because 1.1 × 10⁵ Pa leaves g out of the depth term: P₀ + ρh = 1.0 × 10⁵ + 1.0 × 10⁴. Without g, ρh is not even a pressure.
Why D is wrong: D is wrong because 5.0 × 10⁴ Pa is ρgh/2, an average over the column. The pressure at a given depth depends on the full depth of water above that point.
MCQ 3Easy RecallPractice
In the formula P = P₀ + ρgh for static fluid pressure, what does P₀ represent?
Show answer and why every option is right or wrong
Answer: C. P₀ is the pressure at the free surface of the fluid — typically atmospheric pressure when the surface is open to the atmosphere (NCERT Class 11 Physics, Chapter 9, page 182). The total (absolute) pressure at depth h is P₀ plus the hydrostatic contribution ρgh.
Why A is wrong: A is wrong because gauge pressure is ρgh alone — the excess over the surface pressure. P₀ is the surface pressure itself, not the gauge component.
Why B is wrong: B is wrong because density is represented by ρ in the formula, not P₀. P₀ has units of pressure (Pa), while density has units of kg/m³.
Why D is wrong: D is wrong because ρgh is the hydrostatic pressure component, not P₀. P₀ is the base pressure at the free surface.
MCQ 4Direct ApplicationPractice
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?
Show answer and why every option is right or wrong
Answer: B. P = P₀ + ρgh = 1.013 × 10⁵ + (1.0 × 10³)(9.8)(5.0) = 1.013 × 10⁵ + 4.9 × 10⁴ = 1.503 × 10⁵ ≈ 1.50 × 10⁵ Pa. The question asks for absolute pressure, so atmospheric pressure must be included.
Why A is wrong: A is wrong because 4.9 × 10⁴ Pa is only the gauge pressure (ρgh). The question asks for absolute pressure, which requires adding atmospheric pressure P₀. This is a common trap — misreading 'absolute' as 'gauge'.
Why C is wrong: C is wrong because 1.013 × 10⁵ Pa is atmospheric pressure alone, with no contribution from the water column. This would only be correct at the surface, not at the bottom.
Why D is wrong: D is wrong because 2.0 × 10⁵ Pa does not match the correct arithmetic. It likely results from rounding errors or using incorrect values for g or h.
MCQ 5Direct ApplicationPractice
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.)
Show answer and why every option is right or wrong
Answer: A. Gauge pressure difference = ρgh = (1.36 × 10⁴)(9.8)(0.080) = 1.066 × 10⁴ ≈ 1.07 × 10⁴ Pa. Note: h must be converted from cm to m (8.0 cm = 0.080 m).
Why B is wrong: B is wrong because 1.07 × 10³ Pa is off by a factor of 10. This error arises from using h = 0.008 m (forgetting that 8.0 cm = 0.080 m, not 0.008 m) — a unit-conversion trap.
Why C is wrong: C is wrong because 1.36 × 10⁴ Pa equals the density value itself, not ρgh. The density must be multiplied by both g and h to obtain pressure.
Why D is wrong: D is wrong because 7.84 × 10² Pa results from using ρ = 1.0 × 10³ kg/m³ (water density) instead of 1.36 × 10⁴ kg/m³ (mercury density). Always use the manometric fluid's density.
MCQ 6Direct ApplicationPractice
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?
Show answer and why every option is right or wrong
Answer: D. At the bottom (open to atmosphere): P_bottom = P₀. Moving upward through the liquid of height h, pressure decreases. At the gas-liquid interface: P_gas + ρgh = P_bottom = P₀, so P_gas = P₀ − ρgh. The gas pressure is less than atmospheric because it must support a column that alone would produce pressure P₀ at the bottom.
Why A is wrong: A is wrong because P₀ + ρgh would mean the gas pressure exceeds atmospheric pressure by ρgh. This reverses the relationship — the liquid column sits below the gas, so the gas must be at lower pressure, not higher, for equilibrium at the open bottom.
Why B is wrong: B is wrong because P_gas = P₀ would mean the liquid column contributes nothing to the pressure balance, which contradicts the existence of a liquid column of height h.
Why C is wrong: C is wrong because ρgh is only the hydrostatic contribution of the liquid column. It ignores P₀ entirely. The gas pressure must account for the atmosphere pressing at the bottom.
MCQ 7Concept TrapPractice
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?
Show answer and why every option is right or wrong
Answer: B. Pressure at the bottom of an open container depends only on the height of the liquid column and its density: P = P₀ + ρgh. Container shape does not affect the pressure at a given depth. This is the principle demonstrated by Pascal's vases (NCERT Class 11 Physics, Chapter 9, page 182).
Why A is wrong: A is wrong because pressure does not depend on the cross-sectional area of the container. A narrow container has less total fluid weight, but also less base area — the ratio (force/area = ρgh) is the same.
Why C is wrong: C is wrong because the conical shape does not increase bottom pressure. Although the cone holds more fluid near the top, the pressure at the bottom still depends only on vertical depth h and density ρ.
Why D is wrong: D is wrong for the same reason as A — the wide cylinder has more total weight but proportionally more base area. Pressure (force per unit area) at the bottom is still ρgh, independent of width.
MCQ 8CalculationPractice
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.)
Show answer and why every option is right or wrong
Answer: C. At the interface, pressure from both arms must be equal. Let h_w be the height of water above the interface. Pressure balance: P₀ + ρ_water × g × h_w = P₀ + ρ_oil × g × h_oil. The P₀ and g cancel: ρ_water × h_w = ρ_oil × h_oil → h_w = (ρ_oil/ρ_water) × h_oil = (8.0 × 10²/1.0 × 10³) × 20.0 = 0.80 × 20.0 = 16.0 cm. Since oil is less dense than water, the water column is shorter than the oil column for pressure balance.
Why A is wrong: A is wrong because 25.0 cm comes from inverting the ratio: h_w = (ρ_water/ρ_oil) × h_oil = (1000/800) × 20 = 25. This incorrectly places the larger density in the numerator, implying the denser fluid needs a taller column — the opposite of reality.
Why B is wrong: B is wrong because 20.0 cm assumes h_w = h_oil, which would only hold if both fluids had the same density. Since oil is less dense than water, a shorter water column balances the taller oil column.
Why D is wrong: D is wrong because 10.0 cm likely results from halving the oil height arbitrarily or misapplying the formula. The correct ratio is ρ_oil/ρ_water = 0.80, giving 16.0 cm.
Free NEET study resources
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
Effect of Gravity on Pressure: quick recall before you leave
How do you solve a Effect of Gravity on Pressure question? A worked example
- 1
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.
- 2
Required
Find (a) the gauge pressure at the diver's depth, and (b) the absolute pressure.
- 3
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.
- 4
Formula
• Gauge pressure: P_gauge = ρgh• Absolute pressure: P = P₀ + ρgh
- 5
Substitution
P_gauge = (1.00 × 10³)(9.80)(15.0)
P = 1.013 × 10⁵ + P_gauge - 6
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 - 7
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. - 8
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. - 9
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.)
---
What to remember before solving Effect of Gravity on Pressure questions
More in Properties of Bulk Matter: 5 exam traps and mistakes · 12 formulas · 5 question patterns from its other lessons.
Effect of Gravity on Pressure questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
All 17 past-paper questions from Properties of Bulk Matter →
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →