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Fluid Mechanics formula sheet

Every formula in the Fluid Mechanics chapter for JEE Main, with one line on when it applies. Topic 1 is open to everyone; sign in free to read all 32 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 formula, and "last" is the latest year one appeared. ★ marks formulas asked 8 or more times.

Pressure and Pascal's law

Pressure
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; acts equally in all directions at a point; force always normal to a surface. 1 bar =105=10^5 Pa.
Hydrostatic pressure
5× asked, last 2024
P=P0+ρgh,Pgauge=ρghP=P_0+\rho gh,\qquad P_{\text{gauge}}=\rho gh
hh = depth; same on one level of a connected liquid. Independent of vessel shape. 10 m water ≈ 1 atm.
Bubble rising (constant TT)
Vtop=Vbottom P0+ρghP0V_{\text{top}}=V_{\text{bottom}}\,\frac{P_0+\rho gh}{P_0}
PVPV fixed.
Pascal's law: 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 liquid, pistons level. Multiplies force, never work.
U-tube with two liquids
ρ1h1=ρ2h2\rho_1h_1=\rho_2h_2
Heights above the interface; the denser column is shorter.
Barometer and manometer
P0=ρHgghHg,P−P0=ρmghP_0=\rho_{\text{Hg}}gh_{\text{Hg}},\qquad P-P_0=\rho_mgh
hHg≈76h_{\text{Hg}}\approx76 cm (water 10.3 m). Open manometer reads gauge pressure.
Liquid in an accelerating container
tan⁡θ=ag (horizontal a)P=P0+ρ(g+a)h (upward a)\begin{aligned}&\tan\theta=\frac ag\ (\text{horizontal }a)\\&P=P_0+\rho(g+a)h\ (\text{upward }a)\end{aligned}
Surface rises at the back. Downward aa: g−ag-a; free fall: no pressure difference.
Rotating liquid
1× asked, last 2020
h=ω2r22g,ΔP=12ρω2r2h=\frac{\omega^2r^2}{2g},\qquad \Delta P=\tfrac12\rho\omega^2r^2
Height at radius rr above the lowest point; paraboloid surface.
Force on a wall or dam
2× asked, last 2025
F=12ρgh2w,at h3 above the baseF=\tfrac12\rho gh^2w,\qquad \text{at }\tfrac h3\text{ above the base}
Vertical wall of width ww, depth hh (gauge). Average pressure × wet area.
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 gS(h_1-h_2)^2
Equal base areas SS; ΔU\Delta U lost as heat.

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22 more formulas in 4 topics

Buoyancy and Archimedes' principle, Continuity and Bernoulli's theorem, Applications of Bernoulli's theorem and Viscosity and Stokes' law. 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×
  • Mixing two liquids
  • Body between two liquids1×
  • Hollow sphere that just floats2×
  • Buoyancy in an accelerating lift
  • Ice melting

Continuity and Bernoulli's theorem

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

Applications of Bernoulli's theorem

  • Venturimeter2×
  • Torricelli's law6×
  • 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×
  • Power at terminal velocity
  • Drops merging4×
  • Reynolds number3×

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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.

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