ARCHETYPE: Draining Tanks & Venturi Tubes

Punch a small hole in the side of a full water tank and the jet that escapes carries a secret: it leaves at exactly the speed a stone would have if you dropped it from the water's surface down to the hole. Evangelista Torricelli announced this in 1644 — $v = \sqrt{2gh}$ , a free-fall speed smuggled out of a pressure problem — nearly a century before Bernoulli's equation existed to explain why. The tank doesn't push the water out so much as let it fall, with pressure acting as gravity's accountant. That single result seeds a full problem archetype. Where does the escaping jet land? (A projectile problem grafted onto a fluids problem — examiners' favourite marriage.) Which of two holes throws water farther? (There's a gorgeous symmetry hiding in the answer.) What if the tank is pressurized, or the hole is large, or the jet must fill a bucket? And the same Bernoulli-plus-continuity toolkit, turned horizontal, becomes the venturimeter — the throat-and-gauge instrument that measures flow in pipes from refineries to hospital oxygen lines. This archetype is the spending of the concept module before it: continuity and Bernoulli, now at exam speed. CBSE and NEET ask Torricelli speed and flow-rate fills; JEE-Main runs the jet-range family and venturimeter numericals; JEE-Advanced pressurizes the tank and asks for the symmetry argument; AP's fluids unit names tank-draining as a core context. Board fills first, then the escalations, with timing notes.

Key takeaways

Torricelli's law: efflux speed from a small hole a depth $h$ below the free surface is $v = \sqrt{2gh}$ — the free-fall speed from the surface to the hole.
It follows from Bernoulli between surface and jet: both sit at atmospheric pressure, so the entire $\rho gh$ converts to $\tfrac12\rho v^2$ (small hole ⇒ surface barely moves).
Volume flow from a hole of area $a$ : $Q = a\sqrt{2gh}$ — hole size sets the flow, not the speed.
Jet range (hole in a side wall, tank base at ground): $R = 2\sqrt{h(H-h)}$ — maximum at $h = H/2$ ; holes at depths $h$ and $H-h$ throw equally far.
Pressurized tank (gauge pressure $P_g$ over the liquid): $v = \sqrt{2gh + 2P_g/\rho}$ .
Venturimeter: flow rate from the throat's pressure drop, $Q = A_1A_2\sqrt{\dfrac{2\,\Delta P}{\rho\,(A_1^2 - A_2^2)}}$ .
Key idea: the surface-to-jet Bernoulli ledger converts height (or applied pressure) into efflux speed — then kinematics takes over.

The core insight

For a small hole, Bernoulli's equation between the free surface and the emerging jet — both at atmospheric pressure — gives Torricelli's efflux speed $v = \sqrt{2gh}$ , after which the jet is pure projectile motion; the same two-station ledger with a manometer at the throat turns a constricted pipe into a flow meter.

Where this lives today

Herschel's billing problem never went away: venturi meters still meter city water mains, refinery feeds, and natural-gas pipelines, prized because they sip pressure (the smooth taper recovers most of the drop) and have no moving parts to wear. Medicine runs on the same throat: a Venturi mask clips over a patient's face and uses an oxygen jet through a calibrated constriction to entrain a precise fraction of room air — the constriction's geometry, not a valve, sets the oxygen percentage. Dam engineers size spillways with Torricelli-family calculations, and the "orifice equation" ( $Q = C_d\,a\sqrt{2gh}$ , with the discharge coefficient $C_d \approx 0.6$ carrying the vena-contracta honesty) is standard in every hydraulics handbook, fire-sprinkler code, and irrigation design. Even play obeys the archetype: a water-bottle rocket is worked example 3 inverted — pressurize, invert, and the $\sqrt{2P_g/\rho}$ jet provides the thrust. The draining tank's slowing trickle, meanwhile, is the water clock: from Egyptian clepsydrae to Galileo timing his inclined planes, the first precision timers in history were exactly this article's physics.

The setup

A tank holds liquid to height $H$ above its base; a small hole sits at depth $h$ below the free surface (so at height $H - h$ above the base). "Small" carries two paid duties: the hole's area $a$ is tiny compared to the tank's cross-section $A$ , so the surface descends negligibly slowly ( $v_{\text{surface}} = (a/A)v \approx 0$ by continuity); and the jet leaves horizontally as a clean stream at atmospheric pressure.

Figure

The draining tank — surface to jet, then projectile; the range's hidden symmetry

Recognize the costumes on sight. "Speed of efflux / how fast does it leave" — Torricelli direct. "How far from the base does the jet land" — Torricelli times projectile. "Two holes / where should the hole be for maximum range" — the symmetry family. "Closed tank with compressed air above" — pressurized Torricelli. "Manometer across a constriction, find the flow" — venturimeter. The first move in all of them is the same two-station Bernoulli ledger.

The master solution pattern

Derivation ATorricelli's law — Bernoulli from surface to jet

We derive the efflux speed by running Bernoulli's equation along a streamline from the free surface (station 1) to the emerging jet just outside the hole (station 2). The chosen stations are the whole trick: both are exposed to the atmosphere, so pressure drops out entirely, and the small-hole condition kills the surface's kinetic term. Watch how the result reproduces free fall — that is not a coincidence but energy conservation wearing fluid clothes.

Step 1. Stations and data. Surface: $P_1 = P_0$ , $v_1 \approx 0$ (since $a \ll A$ , continuity gives $v_1 = (a/A)v_2 \approx 0$ ), height $h$ above the hole. Jet: $P_2 = P_0$ (a free jet in open air is at atmospheric pressure), $v_2 = v$ , height $0$ .

Step 2. Bernoulli between the stations:

P_0 + 0 + \rho g h = P_0 + \tfrac12\rho v^2 + 0

Step 3. Cancel $P_0$ and solve:

v = \sqrt{2gh}

Why the free-fall echo: with pressures equal at both ends, Bernoulli is exactly "PE per volume → KE per volume" — each parcel arrives at the hole with the kinetic energy it would have gained falling freely from the surface. Pressure inside the tank merely brokers the exchange. *Why the jet is at atmospheric pressure: outside the hole nothing confines the stream sideways; any pressure excess would make it bulge until balanced. (Fine print: streamlines converge just outside to a narrower section — the vena contracta — so the effective hole area is ~0.6–0.65 of the geometric one; exams ignore this unless they say otherwise.) Verified: units $\sqrt{\mathrm{m/s^2 \cdot m}} = \mathrm{m/s}$ ✓; limits — $h \to 0$ : no jet; matches the UG Bernoulli article's worked example ✓. Anchor: $h = 5$ m, $g = 10$ : $v = 10$ m/s — worked example 1. Standard: NCERT Ch. 9; Torricelli, De Motu Aquarum (1644).*

Once the jet leaves, fluids hands the baton to kinematics. A horizontal jet from a hole at height $y$ above the ground falls for time $t = \sqrt{2y/g}$ and lands at $R = vt$ . For the classic geometry (tank of depth $H$ standing on the ground, hole at depth $h$ below the surface, i.e. height $H-h$ above the ground):

R = \sqrt{2gh}\,\sqrt{\frac{2(H-h)}{g}} = 2\sqrt{h\,(H-h)} \qquad (1)

where: $g$ has cancelled — the range depends only on the two lengths, and their product exposes the symmetry: swapping $h \leftrightarrow H-h$ leaves $R$ untouched, so holes equidistant from the top and the bottom throw equally far (deep hole: fast but low; shallow hole: slow but with a long fall — the trades cancel exactly). The product $h(H-h)$ is maximized when the factors are equal: $h = H/2$ , giving $R_{\max} = H$ . A hole at mid-depth throws the jet a distance equal to the tank's full height — a clean, memorable pair of facts examiners never tire of. Two upgrades complete the tank family. Pressurized tank: gas at gauge pressure $P_g$ sits above the liquid. The surface station now carries $P_0 + P_g$ , and the ledger yields

v = \sqrt{2gh + \frac{2P_g}{\rho}} \qquad (2)

where: the applied pressure acts like extra depth $P_g/\rho g$ — a rocket-water-bottle in equation form (and with $h \to 0$ it is the pure pressure-jet: $v = \sqrt{2P_g/\rho}$ ). Draining time: as the level falls, $v = \sqrt{2gh}$ weakens, so draining slows down — the last centimetres take the longest. The exact time needs calculus (it comes out twice the constant-initial-speed estimate); at HS level state the qualitative slowdown and, if asked "estimate", flag which assumption you froze.

The venturimeter is the same ledger turned horizontal and instrumented. Fluid of density $\rho$ flows through a pipe of area $A_1$ that narrows to a throat $A_2$ ; a manometer reads the pressure drop $\Delta P = P_1 - P_2$ between pipe and throat. Continuity gives $v_2 = (A_1/A_2)v_1$ ; Bernoulli at constant height gives $\Delta P = \tfrac12\rho(v_2^2 - v_1^2)$ . Eliminate $v_2$ :

Q = A_1 v_1 = A_1 A_2\,\sqrt{\frac{2\,\Delta P}{\rho\,(A_1^2 - A_2^2)}} \qquad (3)

where: $\Delta P$ usually arrives as a manometer reading ( $\rho_m g\,h_m$ with the gauge liquid's density — a mercury column under water strictly reads $(\rho_{\text{Hg}} - \rho_w)gh_m$ , though many exam items simplify; use what the paper defines). Read the structure: flow scales as $\sqrt{\Delta P}$ — to double the measured flow, the gauge must read four times higher. That square root is the working principle of most flow meters on Earth.

Figure

The venturimeter — the throat's pressure drop is the flow's signature

Limiting cases

case	result	meaning
$h \to 0$ (hole at surface)	$v \to 0$ , $R \to 0$	no head, no jet
hole at the base ( $h = H$ )	fastest jet, zero range	it lands immediately — speed isn't range
$h = H/2$	$R_{\max} = H$	the product $h(H-h)$ peaks at equal factors
$h \leftrightarrow H - h$	same range	the mirror-hole symmetry
$a \to$ comparable to $A$	$v_1$ not negligible	use full Bernoulli+continuity; Torricelli under-predicts
$P_g \gg \rho gh$	$v \to \sqrt{2P_g/\rho}$	pressure-driven jet; depth irrelevant
$A_2 \to A_1$ (no throat)	$\Delta P \to 0$	a venturimeter needs a real constriction to read anything

Worked examples

1 · Board level — speed and flow rate (CBSE/AP pattern, 3 marks)

A large open tank holds water to a depth of $5.0\ \mathrm{m}$ . A small hole of area $2.0\ \mathrm{cm^2}$ opens in its base. Take $g = 10\ \mathrm{m/s^2}$ . Find (a) the efflux speed and (b) the initial volume flow rate.

Step 1. (a) Torricelli with $h = 5.0$ m: $v = \sqrt{2\times10\times5.0} = \boxed{10\ \mathrm{m/s}}$ — the speed of a stone dropped 5 m.

Step 2. (b) Convert the hole: $a = 2.0\ \mathrm{cm^2} = 2.0\times10^{-4}\ \mathrm{m^2}$ .

Step 3. $Q = a\,v = 2.0\times10^{-4}\times10 = \boxed{2.0\times10^{-3}\ \mathrm{m^3/s}}$ — two litres per second.

Step 4. Sanity: hole size touched the flow, not the speed — a pinhole and a coin-sized hole at 5 m depth eject at the same 10 m/s. (And the rate will decay as the tank drains: this is the initial value.)

2 · Board level — where does the jet land? (ISC/JEE-Main entry pattern, ~2 minutes)

A tank of height $2.0\ \mathrm{m}$ stands on the ground, full of water. A small hole is punched $1.5\ \mathrm{m}$ below the surface. Find (a) the efflux speed, (b) the horizontal distance from the tank at which the jet strikes the ground ( $g = 10\ \mathrm{m/s^2}$ ).

Step 1. (a) $v = \sqrt{2\times10\times1.5} = \sqrt{30} = 5.48\ \mathrm{m/s}$ .

Step 2. (b) The hole sits $H - h = 0.5$ m above the ground; fall time $t = \sqrt{2\times0.5/10} = 0.316\ \mathrm{s}$ .

Step 3. Range: $R = vt = 5.48\times0.316 = \boxed{1.73\ \mathrm{m}}$ . Formula check: $R = 2\sqrt{h(H-h)} = 2\sqrt{1.5\times0.5} = 2\sqrt{0.75} = 1.73$ ✓.

Step 4. Symmetry bonus (often part c): a hole at depth $0.5$ m — the mirror of 1.5 — lands at the same 1.73 m; the farthest-throwing hole would sit at $h = 1.0$ m, reaching $R = H = 2.0$ m.

3 · Competitive level — the pressurized tank (JEE-Adv flavour, ~3 minutes)

A sealed tank contains water to depth $h = 1.8\ \mathrm{m}$ ; compressed air above the water is held at gauge pressure $P_g = 3.2\times10^4\ \mathrm{Pa}$ . A small hole opens at the tank's base. ( $\rho = 1000\ \mathrm{kg/m^3}$ , $g = 10\ \mathrm{m/s^2}$ .) (a) Find the efflux speed. (b) What fraction of the exit kinetic energy came from the compressed air? (c) What equivalent extra water depth would replace the air pressure?

Step 1. (a) The generalized ledger, equation (2): $v = \sqrt{2gh + 2P_g/\rho} = \sqrt{2(10)(1.8) + 2(3.2\times10^4)/1000} = \sqrt{36 + 64} = \boxed{10\ \mathrm{m/s}}$ .

Step 2. (b) The two terms under the root are the two energy sources per unit mass: gravity contributed $36\ \mathrm{J/kg}$ of the total $100$ — so the air supplied $64/100 = \boxed{64\%}$ of the jet's kinetic energy.

Step 3. (c) Equivalent depth: $h_{\text{eq}} = P_g/\rho g = 3.2\times10^4/10^4 = \boxed{3.2\ \mathrm{m}}$ — the sealed tank behaves like an open tank 5.0 m deep, consistent with $v = \sqrt{2(10)(5.0)} = 10$ m/s ✓. Timing note: pressurized-tank items are 90 seconds once you treat $P_g/\rho g$ as bonus depth — convert first, then run plain Torricelli. The energy-split part (b) is read directly off the two terms under the root; computing kinetic energies from scratch is a two-minute detour for the same mark.

4 · Competitive level — venturimeter numerical (JEE-Main/NEET pattern, ~3 minutes)

Water flows through a horizontal pipe of cross-section $8.0\ \mathrm{cm^2}$ that narrows to a throat of $4.0\ \mathrm{cm^2}$ . A differential gauge reads $\Delta P = 4500\ \mathrm{Pa}$ between pipe and throat. ( $\rho = 1000\ \mathrm{kg/m^3}$ .) Find (a) the volume flow rate and (b) the speed in the pipe.

Step 1. Convert areas: $A_1 = 8.0\times10^{-4}$ , $A_2 = 4.0\times10^{-4}\ \mathrm{m^2}$ ; so $A_1^2 - A_2^2 = (64 - 16)\times10^{-8} = 4.8\times10^{-7}\ \mathrm{m^4}$ .

Step 2. (a) Equation (3): $Q = A_1A_2\sqrt{\dfrac{2\Delta P}{\rho(A_1^2 - A_2^2)}} = (3.2\times10^{-7})\sqrt{\dfrac{9000}{1000\times4.8\times10^{-7}}}$ .

Step 3. Under the root: $9000/(4.8\times10^{-4}) = 1.875\times10^{7}$ ; root $= 4.33\times10^{3}$ . So $Q = 3.2\times10^{-7}\times4330 = \boxed{1.4\times10^{-3}\ \mathrm{m^3/s}}$ — 1.4 litres per second.

Step 4. (b) $v_1 = Q/A_1 = 1.4\times10^{-3}/8.0\times10^{-4} = \boxed{1.7\ \mathrm{m/s}}$ . Sanity loop: throat speed $= 2v_1 = 3.5$ m/s; $\tfrac12\rho(v_2^2 - v_1^2) = \tfrac12(1000)(12.0 - 3.0) = 4500$ Pa ✓ — the ledger closes. Timing note: the formula is heavy — but MCQs usually rig a 2:1 area ratio, and then a shortcut flies: $v_2 = 2v_1 \Rightarrow \Delta P = \tfrac{3}{2}\rho v_1^2$ , solve for $v_1$ in one line ( $v_1 = \sqrt{2\times4500/3000} = 1.73$ m/s), then $Q = A_1v_1$ . Learn the shortcut, keep equation (3) for odd ratios. The Bernoulli-closure check in step 4 costs 15 seconds and has caught a thousand slipped square roots.

Exam traps

Two different lengths. Speed uses depth below the surface ( $h$ ); fall-time uses height above the floor ( $H-h$ ). Draw the tank, label both, and the mislabeling trap dies.

Hole size ≠ speed. $v = \sqrt{2gh}$ is area-free (small-hole regime); area enters only in $Q = av$ . "Bigger hole, faster jet" options are eliminable on sight.

The symmetry pair. Depths $h$ and $H-h$ : same range. Maximum at mid-depth, $R_{\max} = H$ . Quoting these without derivation is legitimate and expected in MCQs.

Not-so-small holes. If the hole's area is a stated fraction of the tank's, keep $v_1$ : continuity + Bernoulli give $v_2 = \sqrt{2gh/(1 - a^2/A^2)}$ — slightly faster than Torricelli. The question signals this by giving you both areas.

Pressurized tanks: gauge in, gauge out. Equation (2) wants the gauge pressure of the gas. If the problem quotes absolute, subtract $P_0$ first. (And a vacuum above the liquid subtracts: $v = \sqrt{2gh - 2P_0'/\rho}$ -style items are JEE-Adv bait.)

Draining slows down. Flow decays as the level falls; "time to empty = volume / initial flow" underestimates by exactly a factor of 2 (the exact result, from calculus). State the direction of the error if you estimate.

Venturi bookkeeping. $A_1^2 - A_2^2$ in the denominator — not $(A_1 - A_2)^2$ . Manometer $\Delta P$ uses the gauge liquid's (net) density. And $Q \propto \sqrt{\Delta P}$ : doubling flow quadruples the reading.

Which $g$ , which regime. JEE/NEET: usually $g = 10$ ; jets are ideal (no viscosity, no vena-contracta correction) unless the paper says otherwise. Mentioning "ideal fluid assumed" where a real-number answer looks too big is a free rigour mark on long-form papers.

Origins · 1 minGiovanni Battista Venturi and how this was discovered

Giovanni Battista Venturi

1746–1822 · Modena & Paris

Giovanni Battista Venturi — ordained priest, professor of geometry and philosophy at Modena, diplomat in Paris — published in 1797 the treatise on "the lateral communication of motion in fluids" that mapped how flow through a constriction trades pressure for speed, extending Bernoulli's and Torricelli's results into careful experiments with tapered glass tubes. He never built the flow meter; that step waited nearly a century for the American engineer Clemens Herschel, who in 1888 needed to bill hydraulic power customers and named his throat-and-gauge instrument the Venturi meter in scrupulous homage. Venturi also has a second legacy entirely: as a historian of science he was the first to systematically study Leonardo da Vinci's notebooks — announcing to a startled Paris that Leonardo had been a scientist all along — and he recovered and published the works of Galileo's circle.

History

1644

— Torricelli's De Motu Aquarum: efflux speed equals the free-fall speed from the surface — before Newton, before Bernoulli.

1738

— Daniel Bernoulli's Hydrodynamica supplies the energy ledger that explains Torricelli's rule.

1797

— Venturi's experiments on constricted tubes: the pressure-speed trade measured systematically.

1888

— Clemens Herschel patents the Venturi meter to bill water-power customers — the first practical flow meter.

1930

s — Venturi carburettors meter fuel in nearly every petrol engine; aviation adopts venturi-driven instruments.

Going further

Texts. NCERT Physics Class 11, Ch. 9 · HC Verma Vol. 1, Ch. 13 · Resnick–Halliday–Walker Ch. 14. The UG deep dive on Euler's equations & Bernoulli carries the venturi flowmeter and draining-tank calculus in full.

Provenance. All results standard. A verifier should recompute: example 1's 10 m/s and 2.0 L/s; example 2's $\sqrt{30}\times0.316 = 1.73$ m and the formula check; example 3's $\sqrt{36+64} = 10$ m/s, 64%, 3.2 m; example 4's $Q = 1.4\times10^{-3}$ m³/s, $v_1 = 1.7$ m/s, and the 4500 Pa closure.

Prerequisites. Continuity Equation & Bernoulli's Principle. Leads to. Option Elimination & Answer Checking (track capstone).

Why is the jet's pressure atmospheric the instant it leaves the hole — the water just came from a high-pressure region?

Because pressure needs confinement. Inside the tank, walls and surrounding fluid squeeze each parcel; the moment the stream is in open air, only the atmosphere touches its sides, and any internal excess would push the jet wider until it equilibrated — which happens within a jet-diameter of the exit. That's why station 2 in the derivation sits just outside the hole: pressure $P_0$ , speed $v$ . The pressure didn't vanish; it was spent, en route through the hole, buying the kinetic energy — exactly the ledger Bernoulli audits.

Torricelli says the water exits like a dropped stone. Can the jet then rise back to the surface level if aimed upward?

Almost — and the shortfall is instructive. Aim the hole upward and the ideal jet climbs to exactly the free surface's height (energy conservation round trip: $\tfrac12v^2 = gh$ ). In reality it falls short by a few percent: viscosity taxes the transit through the hole, and air drag taxes the flight. The demonstration — a fountain from a side spout arcing just under the tank's water line — is a beautiful visual proof that Torricelli's law is energy conservation, with the losses visible as the gap.

What exactly is the vena contracta, and why does my measured flow come out ~40% below Q = a√(2gh)?

Streamlines inside the tank converge toward the hole from all directions; they cannot turn a sharp corner instantly, so the jet keeps narrowing for a short distance outside, settling at a cross-section about 0.61–0.65 of the hole's geometric area (the vena contracta). The speed is genuinely $\sqrt{2gh}$ , but through the contracted area: $Q = C_d\,a\sqrt{2gh}$ with $C_d \approx 0.6$ for a sharp-edged orifice. Rounded, nozzle-shaped exits guide the streamlines and push $C_d$ toward 1. Exams use the ideal $C_d = 1$ unless they hand you a coefficient — but lab and viva questions love this honesty clause.

Why does the tank drain slower and slower — and why is the exact time exactly twice the naive estimate?

The driving depth $h$ shrinks as water leaves, so $v = \sqrt{2gh}$ decays; the last centimetre trickles. The factor of 2: flow scales as $\sqrt{h}$ , and the average of $\sqrt{h}$ over the draining process works out to half the initial value — the calculus (integrating $A\,dh/dt = -a\sqrt{2gh}$ ) gives $t = (A/a)\sqrt{2H/g}$ , precisely double the "initial-rate × volume" guess. For HS purposes: state the slowdown, know the factor-of-2 fact, and leave the integral to the UG deep dive.

The venturimeter formula looks unmemorizable. What's the minimum I should actually carry into the exam?

Carry the method, not the formula: continuity ( $v_2 = A_1v_1/A_2$ ) plus level Bernoulli ( $\Delta P = \tfrac12\rho(v_2^2 - v_1^2)$ ), two lines, always reconstructible. For the common 2:1 area ratio this collapses to $\Delta P = \tfrac32\rho v_1^2$ — solvable in your head. Equation (3) is what those two lines become after algebra; use it when ratios are ugly or the question is a pure plug-in. And keep the one structural fact: $Q \propto \sqrt{\Delta P}$ , which answers all "what happens if the reading quadruples" items instantly (flow doubles).

Is a rocket just Torricelli's law pointed downward?

A water rocket genuinely is: pressurized air drives water through the nozzle at $v \approx \sqrt{2P_g/\rho}$ (equation 2 with the depth term negligible), and the reaction to that expelled momentum is the thrust — Newton's third law cashing Bernoulli's cheque. Real chemical rockets swap the energy source (combustion, plus a converging–diverging nozzle that goes supersonic — physics beyond Bernoulli's incompressible warranty), but the accounting template — pressure energy → jet kinetic energy → momentum flux → thrust — is the same ledger this archetype teaches. The momentum half of that story lives back in the rocket-propulsion module.