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Fluid Mechanics revision notes

Every idea in the Fluid Mechanics chapter for JEE Main, with the usual mistakes and the fast methods. Topic 1 is open to everyone; sign in free to read all 30 and download the PDF.

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How we count: 7× asked means 7 past JEE Main questions in the PrepWiser bank needed that concept, and "last" is the latest year one appeared. ★ marks ideas asked 8 or more times.

Pressure and Pascal's law

Pressure in a fluid
2× asked, last 2026
P=F⊥A,1 atm=1.013×105 Pa=76 cm of HgP=\frac{F_\perp}{A},\qquad 1\ \text{atm}=1.013\times10^5\ \text{Pa}=76\ \text{cm of Hg}
Scalar; unit Pa = N m⁻², 1 bar = 10⁵ Pa. The force it gives is always normal to the surface.
  • Pressure exists at every point inside a still fluid, not only at the walls, and acts equally in all directions there. A fluid at rest exerts no shear force.
Hydrostatic pressure
5× asked, last 2024
P=P0+ρgh,Pgauge=P−P0=ρghP=P_0+\rho g h,\qquad P_{\text{gauge}}=P-P_0=\rho g h
hh = depth below the free surface. Same at every point on one level of a connected liquid at rest.
hh (m)PP (atm)1020P0P_023
Water: every 10 m adds about 1 atm; the line starts at P0P_0, not at 0
  • Depends only on depth, not on the vessel's shape or the amount of liquid. With no flow, P1−P2=ρg(h2−h1)P_1-P_2=\rho g(h_2-h_1) for heights h1,h2h_1,h_2. About 10 m of water adds 1 atm.
  • Bubble rising at constant temperature: PVPV stays fixed, so Vtop=Vbottom(P0+ρgh)/P0V_{\text{top}}=V_{\text{bottom}}(P_0+\rho gh)/P_0.
  • Not double the total pressure when the depth doubles. Only ρgh\rho gh doubles; P0P_0 stays. Percentage rise =Δ(ρgh)/(P0+ρgh)×100=\Delta(\rho gh)/(P_0+\rho gh)\times100.
  • Not ρghA\rho ghA alone for the force on the base of a liquid open to air. The liquid passes P0P_0 on too, so F=(P0+ρgh)AF=(P_0+\rho gh)A unless gauge values are asked.
Pascal's law and hydraulic lift
6× asked, last 2025
F2F1=A2A1=(d2d1)2,F1x1=F2x2\frac{F_2}{F_1}=\frac{A_2}{A_1}=\left(\frac{d_2}{d_1}\right)^2,\qquad F_1x_1=F_2x_2
Enclosed incompressible liquid, pistons at one level; dd = piston diameter, xx = piston's travel. F/AF/A is equal under both.
F1F_1 on A1A_1F2=F1A2A1F_2=F_1\dfrac{A_2}{A_1}
Same pressure under both pistons: area ×4\times4 gives force ×4\times4
  • A pressure change applied to an enclosed liquid reaches every part of it and the walls undiminished. Toothpaste tubes, hydraulic brakes and lifts all use it.
  • Not more work out. A lift multiplies force, never work. The big piston moves A1/A2A_1/A_2 times as far, so pair each force with its own piston's displacement.
  • Quick method. Diameters given: square their ratio. Changing both diameters multiplies the two squared factors: large ×2 and small ÷2 gives 22×22=162^2\times2^2=16 times the load.
U-tube with two liquids
ρ1h1=ρ2h2\rho_1 h_1=\rho_2 h_2
Immiscible liquids at rest; heights measured above the interface level. The denser column is shorter.
h1h_1 (oil)h2h_2 (water)
Above the interface: ρ1h1=ρ2h2\rho_1h_1=\rho_2h_2; the lighter oil stands taller
Barometer and manometer
P0=ρHg g hHg,P−P0=ρmghP_0=\rho_{\text{Hg}}\,g\,h_{\text{Hg}},\qquad P-P_0=\rho_m g h
hHg≈76h_{\text{Hg}}\approx76 cm at sea level (water 10.3 m). Open manometer: gauge PP; ρm\rho_m = liquid's density, hh = level gap.
Liquid in an accelerating container
tan⁡θ=ag (horizontal a),P=P0+ρ(g+a)h (upward a)\tan\theta=\frac{a}{g}\ (\text{horizontal } a),\qquad P=P_0+\rho(g+a)h\ (\text{upward } a)
Surface tilts, rising at the back. Downward aa: use g−ag-a; in free fall there is no pressure difference.
θ\thetaaa
Accelerating forward: the surface tilts by tan⁡θ=a/g\tan\theta=a/g, higher at the back
Rotating liquid
1× asked, last 2020
h=ω2r22g,ΔP=12ρ ω2r2h=\frac{\omega^2 r^2}{2g},\qquad \Delta P=\tfrac12\rho\,\omega^2 r^2
Height at radius rr above the lowest point, on the axis. The free surface is a paraboloid.
h=ω2r22gh=\dfrac{\omega^2r^2}{2g}
Spinning at ω\omega: the surface is a paraboloid, lowest on the axis
Force on a wall or dam
2× asked, last 2025
F=12ρgh2w,acting at h3 above the baseF=\tfrac12\rho g h^2 w,\qquad \text{acting at } \tfrac{h}{3}\ \text{above the base}
Vertical rectangular wall of width ww, liquid depth hh, gauge pressure. FF = average pressure × wet area.
F=12ρgh2wF=\frac12\rho gh^2wh/3h/3
Pressure grows with depth (triangle); the total acts h/3h/3 above the base
  • Quick method. Layers or two sides: average pressure × area per part; a lower layer also carries the full pressure of the layer above. Door wet all over on both sides: ΔP A\Delta P\,A, ΔP\Delta P at its centre; else subtract each side's force.
Joining two vessels
2× asked, last 2025
hf=h1+h22,ΔU=14ρgS (h1−h2)2h_f=\frac{h_1+h_2}{2},\qquad \Delta U=\tfrac14\rho g S\,(h_1-h_2)^2
Equal base areas SS, joined at the bottom. ΔU\Delta U = work done by gravity, lost as heat.

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Practise Fluid Mechanics

21 more concepts in 5 topics

Buoyancy and Archimedes' principle, Continuity and Bernoulli's theorem, Applications of Bernoulli's theorem and Viscosity and Stokes' law, plus ideas from surface tension, elasticity, heat. Read the first part free. Sign in free with Google to read the rest and download the PDF.

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Buoyancy and Archimedes' principle

  • Archimedes' principle3×
  • Floating body4×
  • Body between two liquids1×
  • Hollow sphere that just floats2×
  • Buoyancy in an accelerating lift
  • Ice melting and the water level

Continuity and Bernoulli's theorem

  • Equation of continuity5×
  • Bernoulli's theorem6×
  • Horizontal pipe of two sections7×

Applications of Bernoulli's theorem

  • Venturimeter2×
  • Torricelli's law (speed of efflux)6×
  • Range of a jet1×
  • Time to empty a tank
  • Force of a jet2×
  • Dynamic lift: wing and roof3×

Viscosity and Stokes' law

  • Newton's law of viscosity5×
  • Stokes' law4×
  • Terminal velocity22×
  • Drops merging or splitting4×
  • Measuring viscosity by terminal velocity2×
  • Streamline flow and Reynolds number3×

Also asked here: surface tension, elasticity, heat

  • Key points3

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About 85 past JEE Main questions from this chapter, with instant answer checking and step-by-step solutions where available. Free, and you can start without signing up.

Practise Fluid Mechanics

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