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. 183Fluid Pressure
8 MCQs4 revision cards9-step worked example
Source: NCERT Properties of Solids and LiquidsOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Fluid Pressure, explained for NEET
The pressure at any point inside a static fluid increases with depth. This is one of those topics where the formula looks simple — P = P₀ + ρgh — but NEET questions test whether you truly understand what each symbol means and what conditions make the formula valid.
The core idea. Consider a horizontal surface at depth h below the free surface of a fluid at rest. The weight of the fluid column above that surface creates additional pressure beyond whatever pressure acts on the free surface. NCERT Class 11 Physics Chapter 9 (Mechanical Properties of Fluids), page 182, derives this by balancing forces on a thin fluid element: the pressure difference between top and bottom faces equals the weight of the fluid slab per unit area.
The result: P = P₀ + ρgh, where P₀ is the pressure at the free surface (usually atmospheric pressure), ρ is the fluid density, g is gravitational acceleration, and h is the vertical depth below the free surface.
Three conditions that must hold:
- The fluid is static — no flow.
- The fluid is incompressible — ρ is constant throughout.
- g is constant over the depth range considered.
What to watch for in NEET questions:
- Pressure depends on vertical depth only, not on the shape of the container. Two containers with different shapes but the same fluid and same depth have the same pressure at that depth — this is the hydrostatic paradox.
- Gauge pressure is ρgh alone (excluding atmospheric pressure). When a problem asks for "pressure due to the liquid column," it wants gauge pressure, not absolute pressure.
- The formula fails if the fluid is flowing (use Bernoulli instead) or if density varies with depth (e.g., compressible gas columns — outside NEET scope for this formula).
- h is measured vertically downward from the free surface, not along a slanted tube wall.
Can you answer these Fluid 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 liquid depends on which of the following?
Show answer and why every option is right or wrong
Answer: B. Pressure in a static incompressible fluid is P = P₀ + ρgh, which depends on depth h, fluid density ρ, and g — not on the container's shape, total volume, or total mass. This is stated directly in NCERT Class 11 Physics Chapter 9, page 182.
Why A is wrong: A is wrong because pressure at a given depth is independent of container shape — this is the hydrostatic paradox. Two containers of different shapes filled to the same height with the same fluid have identical pressure at the same depth.
Why C is wrong: C is wrong because the total volume of liquid does not appear in P = P₀ + ρgh. Only depth, density, and g matter.
Why D is wrong: D is wrong because total mass is irrelevant; pressure depends on the density of the fluid and the vertical depth, not on how much fluid the container holds overall.
MCQ 2Easy RecallPractice
What is the SI unit of pressure?
Show answer and why every option is right or wrong
Answer: C. Pressure is force per unit area. The SI unit is the pascal (Pa), where 1 Pa = 1 N/m². This is a fundamental definition from NCERT Class 11 Physics Chapter 9, page 182.
Why A is wrong: A is wrong because N/m is the unit of force per unit length (surface tension), not force per unit area.
Why B is wrong: B is wrong because J/m² = (N·m)/m² = N/m, which is surface tension or energy per unit area, not pressure.
Why D is wrong: D is wrong because kg/(m·s) is the unit of dynamic viscosity, not pressure. Pressure in base SI units is kg/(m·s²).
MCQ 3Easy RecallPractice
Gauge pressure at a depth h in a liquid of density ρ is:
Show answer and why every option is right or wrong
Answer: C. Gauge pressure is the pressure above atmospheric, i.e., the ρgh term alone. Absolute pressure is P₀ + ρgh; gauge pressure subtracts the atmospheric contribution. NCERT Class 11 Physics Chapter 9, page 182.
Why A is wrong: A is wrong because P₀ + ρgh is the absolute pressure, not gauge pressure. Gauge pressure excludes the atmospheric term P₀.
Why B is wrong: B is wrong because subtracting ρgh from atmospheric pressure would give a value less than atmospheric, which is not what gauge pressure means.
Why D is wrong: D is wrong because ρg/h has dimensions of Pa/m², which is not dimensionally consistent with pressure. Pressure requires the product ρgh, not ρg divided by h.
MCQ 4Direct ApplicationPractice
A tank contains water (ρ = 1.0 × 10³ kg/m³) open to the atmosphere. What is the gauge pressure at a depth of 5.0 m? (Take g = 10 m/s², exact.)
Show answer and why every option is right or wrong
Answer: B. Gauge pressure = ρgh = (1.0 × 10³)(10)(5.0) = 5.0 × 10⁴ Pa. Here g = 10 m/s² is problem-defined exact and does not limit significant figures. NCERT Class 11 Physics Chapter 9, page 182.
Why A is wrong: A is wrong because this value (5.0 × 10³ Pa) is off by a factor of 10 — likely from a power-of-ten error in multiplying 1.0 × 10³ × 10 × 5.0.
Why C is wrong: C is wrong because 1.0 × 10⁵ Pa is approximately standard atmospheric pressure (1.013 × 10⁵ Pa). This would be the answer only if the question asked for absolute pressure AND used atmospheric pressure ≈ 1.0 × 10⁵ Pa, then added ρgh — but gauge pressure is ρgh alone.
Why D is wrong: D is wrong because 1.5 × 10⁵ Pa would correspond to absolute pressure (atmospheric + gauge) with P₀ ≈ 1.0 × 10⁵ Pa. The question asks for gauge pressure, so atmospheric pressure should not be added.
MCQ 5Direct ApplicationPractice
A U-tube manometer contains mercury (ρ = 13.6 × 10³ kg/m³). The difference in mercury levels between the two arms is 20.0 cm. What is the pressure difference being measured? (Take g = 9.8 m/s².)
Show answer and why every option is right or wrong
Answer: D. ΔP = ρgh = (13.6 × 10³)(9.8)(0.200) = 2.6656 × 10⁴ Pa ≈ 2.67 × 10⁴ Pa. Note h must be converted to metres: 20.0 cm = 0.200 m. NCERT Class 11 Physics Chapter 9, page 182.
Why A is wrong: A is wrong because 2.67 × 10⁵ Pa is ten times too large. It comes from using h = 2.00 m instead of 0.200 m — a slip in the cm-to-m conversion.
Why B is wrong: B is wrong because 2.67 × 10³ Pa is off by a factor of 10, likely from failing to convert 20.0 cm to 0.200 m and instead using 0.0200 m or from a decimal error.
Why C is wrong: C is wrong because 1.33 × 10⁴ Pa uses h = 10.0 cm — the rise in one arm above the original level — instead of the full 20.0 cm difference between the arms.
MCQ 6Direct ApplicationPractice
Two connected vessels contain the same liquid at rest. Vessel A has a narrow neck and vessel B has a wide base. The liquid level in both vessels is the same. Which statement is correct about the pressure at the bottom of each vessel?
Show answer and why every option is right or wrong
Answer: A. Pressure at the bottom depends only on ρ, g, and the vertical height of the liquid column — not on the container's shape or the total amount of liquid. Since both have the same liquid and the same level, the pressure at the bottom is identical. This is the hydrostatic paradox, discussed in NCERT Class 11 Physics Chapter 9, page 182.
Why B is wrong: B is wrong because the total amount of liquid is irrelevant to the pressure at a given depth. Pressure depends on depth, density, and g — not volume or mass of liquid in the container.
Why C is wrong: C is wrong because cross-sectional area does not appear in P = P₀ + ρgh. Narrower containers do not produce higher pressure at the same depth.
Why D is wrong: D is wrong because the volume information is unnecessary. The hydrostatic pressure formula P = P₀ + ρgh contains no volume term, so the pressure can be fully determined from depth and density alone.
MCQ 7Concept TrapPractice
A closed tank is completely filled with water and pressurised so that the pressure at the top surface is 2.0 × 10⁵ Pa. The tank is 3.0 m deep. What is the absolute pressure at the bottom? (ρ_water = 1.0 × 10³ kg/m³, g = 9.8 m/s².)
Show answer and why every option is right or wrong
Answer: D. P = P₀ + ρgh = 2.0 × 10⁵ + (1.0 × 10³)(9.8)(3.0) = 2.0 × 10⁵ + 2.94 × 10⁴ = 2.294 × 10⁵ Pa. The surface pressure here is not atmospheric but the given pressurised value. NCERT Class 11 Physics Chapter 9, page 182.
Why A is wrong: A is wrong because this ignores the ρgh contribution entirely. Even though the tank is closed, the fluid column still adds pressure with depth.
Why B is wrong: B is wrong because subtracting ρgh from P₀ would imply pressure decreases with depth, which contradicts the hydrostatic formula. Pressure always increases with depth in a static fluid.
Why C is wrong: C is wrong because 2.94 × 10⁴ Pa is only the gauge pressure (ρgh). The question asks for absolute pressure, which requires adding the surface pressure P₀ = 2.0 × 10⁵ Pa.
MCQ 8CalculationPractice
A swimming pool has fresh water (ρ = 1.00 × 10³ kg/m³) to a depth of 4.0 m. A layer of oil (ρ = 8.0 × 10² kg/m³) of thickness 1.0 m floats on top. What is the gauge pressure at the bottom of the pool? (Take g = 10 m/s², exact.)
Show answer and why every option is right or wrong
Answer: A. Gauge pressure at the bottom = ρ_oil × g × h_oil + ρ_water × g × h_water = (8.0 × 10²)(10)(1.0) + (1.00 × 10³)(10)(4.0) = 8.0 × 10³ + 4.0 × 10⁴ = 4.8 × 10⁴ Pa. Two separate fluid layers require summing their individual ρgh contributions. NCERT Class 11 Physics Chapter 9, page 182.
Why B is wrong: B is wrong because this accounts only for the water layer (1.00 × 10³ × 10 × 4.0 = 4.0 × 10⁴ Pa) and ignores the oil layer on top. Both layers contribute to the pressure at the bottom.
Why C is wrong: C is wrong because 5.0 × 10⁴ Pa would result from treating both layers as having water's density: (1.00 × 10³)(10)(5.0) = 5.0 × 10⁴. The oil has a lower density (8.0 × 10² kg/m³), so its contribution is less than water's.
Why D is wrong: D is wrong because this could result from a miscalculation such as using an incorrect oil density or thickness. The correct oil contribution is 8.0 × 10³ Pa, not 4.0 × 10³ Pa.
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Fluid Pressure: quick recall before you leave
How do you solve a Fluid Pressure question? A worked example
- 1
Given
A cylindrical vessel contains two immiscible liquids. The bottom layer is mercury (ρ₁ = 13.6 × 10³ kg/m³) of height h₁ = 5.00 cm. The top layer is water (ρ₂ = 1.00 × 10³ kg/m³) of height h₂ = 40.0 cm. The vessel is open to the atmosphere (P₀ = 1.013 × 10⁵ Pa). Take g = 9.80 m/s².
- 2
Required
Find the absolute pressure at the bottom of the vessel.
- 3
Concept
For two immiscible fluid layers stacked vertically, the pressure at the bottom is the sum of atmospheric pressure and the ρgh contributions of each layer: P = P₀ + ρ₂gh₂ + ρ₁gh₁.
- 4
Formula
P = P₀ + ρ₂gh₂ + ρ₁gh₁
- 5
Substitution
Convert heights to metres: h₁ = 5.00 cm = 5.00 × 10⁻² m; h₂ = 40.0 cm = 4.00 × 10⁻¹ m.
P = 1.013 × 10⁵ + (1.00 × 10³)(9.80)(4.00 × 10⁻¹) + (13.6 × 10³)(9.80)(5.00 × 10⁻²) - 6
Calculation
Water contribution: (1.00 × 10³)(9.80)(4.00 × 10⁻¹) = 3.920 × 10³ Pa
Mercury contribution: (13.6 × 10³)(9.80)(5.00 × 10⁻²) = 6.664 × 10³ Pa
Total gauge pressure: 3.920 × 10³ + 6.664 × 10³ = 1.0584 × 10⁴ Pa
Absolute pressure: 1.013 × 10⁵ + 1.058 × 10⁴ = 1.119 × 10⁵ Pa
Note on significant figures: g = 9.80 m/s² is given to 3 significant figures. All given quantities have 3 significant figures, so the final answer is reported to 3 significant figures. - 7
Final answer
P ≈ 1.12 × 10⁵ Pa (3 significant figures).
- 8
Common trap
A frequent error is forgetting to convert centimetres to metres before substituting into ρgh. Using h = 40.0 (without converting) gives an answer 100 times too large. Another common error: treating the two-layer system as a single fluid by using only one density — each layer must have its own ρgh term.
- 9
Similar NEET-style question
A vessel open to the atmosphere contains oil (ρ = 9.00 × 10² kg/m³, height 30.0 cm) on top of water (ρ = 1.00 × 10³ kg/m³, height 20.0 cm). Find the gauge pressure at the bottom.
Answer: ρ_oil × g × h_oil + ρ_water × g × h_water = (9.00 × 10²)(9.80)(0.300) + (1.00 × 10³)(9.80)(0.200) = 2.646 × 10³ + 1.960 × 10³ = 4.61 × 10³ Pa.
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What to remember before solving Fluid Pressure questions
Which Fluid Pressure formulas do you need for NEET?
1 formula — click to collapse
Pressure in static fluid
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 |
Valid when
- Static fluid (no flow)
- Constant g
- Constant rho (incompressible)
More in Properties of Bulk Matter: 5 exam traps and mistakes · 11 formulas · 5 question patterns from its other lessons.
Fluid 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
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