Fluid Statics & Dynamics: Master Notes

Chapter Overview

Fluid mechanics tests both static equilibrium and non-linear dynamic flows. Key JEE (Advanced) and Olympiad domains include accelerated and rotating fluids, Bernoulli flow with Torricelli efflux, viscous terminal velocity (Stokes' Law), and surface tension capillary dynamics.

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1. Fluid Statics & Accelerated Fluids

1.1 Hydrostatic Pressure Gradient

$$\mathrm{\nabla }P=\rho (\overrightarrow{g}-{\overrightarrow{a}}_{\text{frame}})$$

• In an accelerating container ($\overrightarrow{a}=a_x\hat{i}+a_y\hat{j}$): The free liquid surface tilts at angle $\theta$ to the horizontal:$$\boxed{{\displaystyle \tan \theta =\frac{a_x}{g+a_y}}}$$
• In a rotating cylinder with angular velocity $\omega$: The liquid surface forms a paraboloid of revolution:$$\boxed{{\displaystyle z(r)=z_0+\frac{{\omega }^2{r}^2}{2g}}}$$

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2. Ideal Fluid Dynamics: Continuity & Bernoulli's Equation

$$\text{Equation of Continuity: }{A}_1{v}_1=A_2{v}_2(\text{Mass Invariance})$$$$\text{Bernoulli’s Theorem: }\boxed{{\displaystyle P+\frac{1}{2}\rho v^2+\rho gh=\text{constant}}}$$

2.1 Torricelli's Law of Efflux & Emptying Time

Multi-Mode DiagramFluid Mechanics: Torricelli Efflux & Bernoulli Trajectory

Option 1: Publication-Grade Scientific Vector SVG

Draining fluid tank with orifice at depth h, demonstrating efflux velocity v = sqrt(2gh) and parabolic water jet range.

Torricelli Speed: $v = \sqrt{2 g h}$

Jet Range: $R = 2\sqrt{h(H - h)} \implies R_{\max} = H \text{ at } h = H/2$

Option 2: Interactive 60 FPS Canvas Physics Engine

Interactive draining tank with variable hole depth h.

Hole Depth h: 5 m

Efflux Speed ($v$):9.90 m/s

Option 3: Interactive 3D Spatial Viewport

3D cylindrical water tank reservoir.

+X

+Y

+Z

👆 Click and drag anywhere to rotate in 3D space

Azimuth Angle (ψ):35°

Elevation Angle (θ):-22°

Bernoulli Invariant:$P + \frac{1}{2}\rho v^2 + \rho g h = \text{const}$

Option 5: Reactive Parametric Physics with Live Calculus

Maximum water jet range solver.

Orifice depth h: 5 m

: 45

Speed v ($v$):$9.9 m/s$

Active Mode:Option 1 (Print & Derivation Vector SVG)Switch modes using the tabs above to explore different visualizations.

• Efflux Speed: $v=\sqrt{\frac{2gh}{1-(a/A{)}^2}}\approx \sqrt{2gh}$ (when orifice area $a\ll A$).
• Time to Empty a Tank from height $H$ to $0$:$$-\frac{dh}{dt}A=a\sqrt{2gh}⟹\boxed{{\displaystyle t_{\text{empty}}=\frac{A}{a}\sqrt{\frac{2H}{g}}}}$$
• Horizontal Range of Efflux Jet:$$R=v\cdot t_{\text{fall}}=\sqrt{2gh}\cdot \sqrt{\frac{2(H-h)}{g}}=2\sqrt{h(H-h)}$$(Max range $R_{max}=H$ occurs when the hole is drilled at the exact midpoint $h=H/2$).

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3. Surface Tension & Capillary Action

$$\text{Surface Energy: }{U}_s=T\cdot \mathrm{\Delta }A$$

3.1 Excess Pressure ($\mathrm{\Delta }P=P_{\text{in}}-P_{\text{out}}$)

• Liquid Droplet / Air Bubble in Liquid (1 surface): $\mathrm{\Delta }P=\frac{2T}{R}$
• Soap Bubble in Air (2 surfaces): $\mathrm{\Delta }P=\frac{4T}{R}$

3.2 Capillary Rise (Jurin's Law)

In a narrow tube of radius $r$ and contact angle ${\theta }_c$:

$$\boxed{{\displaystyle h=\frac{2T\cos {\theta }_c}{\rho gr}}}$$

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4. Viscosity & Stokes' Law

4.1 Viscous Drag on a Sphere (Stokes' Law)

$$F_v=6\pi \eta rv$$

4.2 Terminal Velocity ($v_t$)

When gravity, buoyancy, and viscous drag balance on a falling sphere of density $\rho$ in a fluid of density $\sigma$:

$$Mg=F_B+F_v⟹\frac{4}{3}\pi r^3\rho g=\frac{4}{3}\pi r^3\sigma g+6\pi \eta r{v}_t$$$$\boxed{{\displaystyle v_t=\frac{2r^2(\rho -\sigma )g}{9\eta }}}$$

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5. Authentic Previous Years Questions (PYQs)

PYQ 1: JEE Advanced 2021 (Paper 2) — Draining Cylindrical Tank with Reaction

Question:
A wide cylindrical vessel of cross-sectional area $A$ and height $H$ is filled with water of density $\rho$. A small orifice of area $a\ll A$ is punched in the vertical wall at the bottom. The vessel rests on a frictionless horizontal floor. Find:

1. The initial horizontal reaction force $F_{\text{thrust}}$ exerted by the exiting jet on the vessel.
2. The total time required to drain the vessel completely.

Step-by-Step Solution:

• 1. Thrust Force: The rate of mass exiting is $\frac{dm}{dt}=\rho av$.
  By momentum conservation:

  $$F_{\text{thrust}}=v\frac{dm}{dt}=v(\rho av)=\rho a{v}^2$$

  Since $v=\sqrt{2gH}$:

  $$F_{\text{thrust}}=\rho a(2gH)=\boxed{{\displaystyle 2\rho gaH=2\cdot (\text{Hydrostatic force on hole area})}}$$

  (Notice that dynamic reaction thrust is exactly DOUBLE the static hydrostatic force $\rho gHa$!)

• 2. Emptying Time:

  $$-\frac{dh}{dt}A=a\sqrt{2gh}⟹{\int }_H^0\frac{dh}{\sqrt{h}}=-\frac{a\sqrt{2g}}{A}{\int }_0^{T}dt$$$$2\sqrt{H}=\frac{a\sqrt{2g}}{A}T⟹\boxed{{\displaystyle T=\frac{A}{a}\sqrt{\frac{2H}{g}}}}$$

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PYQ 2: JEE Advanced 2020 — Surface Energy & Droplet Coalescence

Question:
$N=1000$ identical spherical water droplets, each of radius $r=1\mu \text{m}$, coalesce isothermally into a single large drop of radius $R$. If the surface tension of water is $T=0.075\text{ N/m}$, calculate the heat energy $Q$ released during this process.

Step-by-Step Solution:

1. Conservation of Volume:

   $$V_{\text{final}}=N\cdot V_{\text{initial}}⟹\frac{4}{3}\pi R^3=N\left(\frac{4}{3}\pi r^3\right)⟹R=N^{1/3}r=(1000{)}^{1/3}r=10r$$

2. Change in Surface Area:

   • Initial Area: $A_i=N(4\pi r^2)=1000\times 4\pi r^2=4000\pi r^2$
   • Final Area: $A_f=4\pi R^2=4\pi (10r{)}^2=400\pi r^2$
   $$\mathrm{\Delta }A=A_f-A_i=400\pi r^2-4000\pi r^2=-3600\pi r^2$$

3. Heat Energy Released:

   $$Q=-\mathrm{\Delta }{U}_s=-T\mathrm{\Delta }A=T(3600\pi r^2)=0.075\times 3600\times \pi \times ({10}^{-6}{)}^2$$$$Q=270\pi \times {10}^{-12}\text{ J}\approx \boxed{{\displaystyle 8.48\times {10}^{-10}\text{ J}}}$$

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6. High-Yield Formula Sheet

Concept	Formula	Key Notes
Accelerated Surface Tilt	$\tan \theta =\frac{a_x}{g+a_y}$	Plane perpendicular to effective gravity
Rotating Liquid Paraboloid	$z(r)=\frac{{\omega }^2{r}^2}{2g}$	Lowest point at axis of rotation
Torricelli Efflux	$v=\sqrt{2gh}$	Orifice area $a\ll A$
Capillary Rise	$h=\frac{2T\cos \theta }{\rho gr}$	Jurin's Law
Stokes' Terminal Speed	$v_t=\frac{2r^2(\rho -\sigma )g}{9\eta }$	Low Reynolds number viscous flow