PhysicsMechanical properties of fluidsClass 11

Fluid mechanics: interactive simulation

Moving fluids follow two simple rules: what flows in must flow out (continuity), and energy per unit volume is shared between pressure, motion and height (Bernoulli). Viscosity and surface tension add the friction and the “skin”.

Useful for: JEE Main, JEE Advanced, NEET, CBSE Boards, AP Physics 2 · 4 modes in one lab

What you can explore

  1. 1

    Torricelli tank

    Efflux speed, fall time and jet range for holes at any depth.
  2. 2

    Bernoulli & Venturi

    Continuity, speed-up in the throat and the pressure drop shown by water columns.
  3. 3

    Stokes' law

    Balls of steel, glass or nylon falling through glycerine, castor oil or honey.
  4. 4

    Capillary rise

    Water, ethanol and mercury in tubes of three radii with concave and convex menisci.

Try this

Find the hole that throws water farthest, squeeze a pipe and watch the pressure drop, drop steel balls through glycerine until they stop speeding up, and compare water and mercury in thin glass tubes.

  1. 1

    Tank with H = 1.6 m: put the hole halfway down.

    What you should see

    v ≈ 3.96 m/s, fall time ≈ 0.40 s, range = H = 1.6 m — the maximum possible.

  2. 2

    Compare holes at depth 0.25H and 0.75H.

    What you should see

    Both jets land at the same spot.

  3. 3

    In Bernoulli & Venturi, double the speed in the throat (A₁/A₂ = 2) with v₁ = 1 m/s.

    What you should see

    ΔP = ½ × 1000 × (4 − 1) = 1500 Pa, a 15.3 cm drop in the water column.

  4. 4

    In Stokes’ law, drop a 2 mm steel ball in glycerine.

    What you should see

    v_t ≈ 4.0 cm/s; Re ≈ 0.14 so Stokes’ law holds. Halve r and v_t falls to a quarter.

  5. 5

    In Capillary rise, compare water and mercury in a 0.5 mm tube.

    What you should see

    Water rises about 29.7 mm; mercury is pushed down about 10.7 mm.

Formulas behind it

Torricelli’s theorem

v=2ghv = \sqrt{2gh}

Range of the jet

R=2h(H−h)R = 2\sqrt{h(H - h)}

Continuity and Bernoulli

A1v1=A2v2,P+12ρv2+ρgh=constA_1v_1 = A_2v_2,\qquad P + \tfrac{1}{2}\rho v^2 + \rho gh = \text{const}

Stokes’ law and terminal velocity

F=6πηrv,vt=2r2(ρ−σ)g9ηF = 6\pi\eta r v,\qquad v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}

Capillary rise

h=2Tcos⁡θrρgh = \frac{2T\cos\theta}{r\rho g}

Common mistakes

  • Measuring h in Torricelli’s formula from the bottom of the tank. It is the depth below the free surface.
  • Assuming pressure is higher where the fluid moves faster. Bernoulli says the opposite.
  • Forgetting buoyancy (the σ term) in the terminal velocity formula.
  • Thinking a longer capillary tube makes the liquid overflow. The liquid only rises until the meniscus adjusts its curvature.
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FAQ

Why does a spinning cricket ball swing or a football curve?

The spin makes air flow faster on one side, lowering the pressure there (the Magnus effect, a Bernoulli effect), so the ball is pushed sideways.

Why do raindrops not keep accelerating?

Air drag grows with speed until it balances the weight, so drops reach a terminal velocity.

How do plants raise water?

Capillary action helps in the narrow xylem vessels, but most of the lift comes from transpiration pulling the water column up from the leaves.

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