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Exam level4 past questions

Fluid dynamics and Bernoulli's theorem

Covers the motion of fluids, including Bernoulli's theorem, Torricelli's law, dynamic lift, and viscous flow.

Why this shows up in the exam

NEET often asks you to analyze moving fluids and apply conservation of energy and momentum in fluid systems.

How NEET tests this

Direct recall · 2 QsApplication · 2 QsNumerical · 2 Qs

Learn the idea

Bernoulli’s theorem tells us that in a steady, incompressible flow the sum of pressure, kinetic and potential energy per unit volume is constant; the kinetic part (½ρv²) creates a pressure drop that can be turned into a force on a surface.

🧠 Memory hook: Half the wind’s density times its speed squared is the pressure push that lifts the roof.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Bernoulli’s equation: P+½ρv²+ρgh = constant for incompressible steady flow
  • Dynamic pressure = ½ρv²
  • For two points at the same height, pressure difference ΔP = ½ρ(v₁²‑v₂²)
  • Force on a flat surface = ΔP × area, direction from high to low pressure
  • Venturi meter works on the pressure drop when fluid speed increases
  • Torricelli’s law: speed of efflux from an opening = √(2gh)

How to approach it

  1. 1Identify a point on the roof where air moves with speed v and a point inside where air is at rest (v=0) and both at the same height
  2. 2Apply Bernoulli between the two points, dropping the ρgh term, to get ΔP = ½ρv²
  3. 3Calculate ΔP using the given density and speed
  4. 4Multiply ΔP by the roof area to obtain the net force and state that it acts from the higher‑pressure interior toward the lower‑pressure exterior (upward)

Worked example — watch it click

A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m². Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be: (ρₗₐₒ = 1.2 kg/m³)

  • ✅2.4 x 10⁵ N, upwards
  • B)2.4 x 10⁵ N, downwards
  • C)4.8 x 10⁵ N, downwards
  • D)4.8 x 10⁵ N, upwards

The concept behind this problem

The example asks you to convert the Bernoulli‑derived pressure drop over the moving roof into a measurable upward force, testing the direct use of ΔP = ½ρv² and F = ΔP·A.

Step by step

  1. 1By Bernoulli's principle, pressure above roof P₁ + ½ρv² = P_atm.
  2. 2Pressure difference ΔP = ½ρv² = ½(1.2)(40)² = 960 Pa.
  3. 3Force = ΔP × A = 960 × 250 = 2.4×10⁵ N, directed upward (lower pressure above).

Watch out

Students often forget that faster flow lowers pressure and thus answer ‘downward’ instead of the correct upward force.

Common slip-ups that cost marks

  • •Missing the factor ½ in the dynamic pressure term
  • •Assuming a height difference when the two points lie at the same level, which would incorrectly keep the ρgh term
  • •Reversing the pressure difference and therefore giving the force the wrong direction

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 4NEET 2015

A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m². Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be: (ρₗₐₒ = 1.2 kg/m³)

Push further

More challenging

4 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 4

According to Stokes' law, the viscous drag force (F) on a spherical object moving through a viscous fluid is given by F = 6πηrv, where η is the coefficient of viscosity, r is the radius of the sphere, and v is its velocity. Which of the following conditions must be met for Stokes' law to be applicable?

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