Speed of efflux (Torricelli's law)

Punch a small hole in the side of a water tank and watch the jet shoot out sideways. Push the hole lower, near the bottom, and the jet suddenly sprays much farther across the floor — push it near the top and it barely dribbles out. Something about depth is controlling the speed of the escaping water, and the relationship turns out to be startlingly simple: the water leaves the hole exactly as fast as a stone would be moving if you dropped it from the surface and let it fall freely down to the level of the hole. No stone is actually falling, and no pipe carries it there, yet the arithmetic of free fall reappears inside a completely different situation — flowing liquid escaping under its own weight. This is Torricelli's law, one of the cleanest applications of Bernoulli's theorem in the whole "Mechanical Properties of Fluids" chapter. It needs no new physics beyond Bernoulli's equation, which is itself just energy conservation written for a moving fluid, and a touch of the equation of continuity to justify one simplifying assumption. The result, $v = \sqrt{2gh}$ , is compact enough to memorise instantly — but the CBSE board wants more than memorisation. It wants the derivation done cleanly with the assumptions stated, and it wants that speed fed straight into ordinary projectile motion to find how far the jet travels before it lands, since a leaking tank is really two physics problems stacked on top of each other. This article rebuilds the derivation from Bernoulli's equation step by step, works two numerical examples — one board-style, one competitive-style with a projectile twist — flags the traps that quietly cost marks, and closes with a compact recipe you can run in an exam in well under two minutes.

Key takeaways

Torricelli's law: a liquid escaping through a small hole at depth $h$ below the free surface leaves with speed $v = \sqrt{2gh}$ — exactly the free-fall speed through height $h$ .
It is derived by applying Bernoulli's theorem between the open top surface and the hole, both exposed to atmospheric pressure, so the pressure terms cancel.
The "large tank, small hole" assumption sets the surface's own sinking speed $v_1 \approx 0$ , using the equation of continuity $A_1 v_1 = A_2 v_2$ with $A_2 \ll A_1$ .
Once outside, the jet becomes an ordinary projectile falling under gravity — combine $v = \sqrt{2gh}$ with $H = \tfrac12 g t^2$ and $R = vt$ to find range and time of flight.
$h$ in the formula is always the depth below the free surface, never the height of the hole above the ground — mixing these up is the single most common slip.
The law fails if the tank is small (its own surface speed is not negligible) or if the hole is not open to the same atmosphere as the top surface.
Key idea: speed of efflux v = √(2gh) follows from Bernoulli's theorem applied between the open surface and the hole with equal atmospheric pressure and negligible surface-sinking speed, so the jet leaves as if it had freely fallen through depth h and then continues as a projectile.

The core insight

Applying Bernoulli's theorem between a tank's open surface and a small side hole at depth $h$ , with $P_1 = P_2 = P_0$ and $v_1 \approx 0$ , gives Torricelli's law $v = \sqrt{2gh}$ — the exit speed equals the free-fall speed through the same depth, after which the jet moves as a projectile under gravity.

Where it fits in the syllabus

This topic sits at the end of the Bernoulli's-theorem discussion inside CBSE Class 11 "Mechanical Properties of Fluids," immediately after Bernoulli's equation and the equation of continuity, and just before dynamic lift — both are listed as "simple applications" of Bernoulli's theorem in the prescribed scope. The board expects a clean derivation of $v = \sqrt{2gh}$ from Bernoulli's equation with the assumptions stated explicitly, followed by numerical use — often combined with basic projectile-motion equations from mechanics to find the range of the emerging jet. Nothing beyond this (no Poiseuille's flow-rate law, no viscous corrections to the jet) is in scope here; the hole is always treated as small, frictionless, and open to the atmosphere.

The physical picture

Picture a large tank, open to the atmosphere at the top, filled with a liquid of density $\rho$ . A small hole is punched in the side wall at a depth $h$ below the free surface. Liquid streams out of the hole, and we want its exit speed. Because the flow is steady and (ideally) non-viscous, Bernoulli's theorem applies along a streamline connecting a point at the free surface to a point right at the hole:

P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2

where: $P_1, v_1, h_1$ are the pressure, speed, and height of the surface point, and $P_2, v_2, h_2$ are the same quantities at the hole; $\rho$ is the liquid's density and $g$ is the acceleration due to gravity. The physical reading of this equation is that it is energy conservation per unit volume — pressure energy plus kinetic energy plus gravitational potential energy stays constant along the flow, exactly the way $\tfrac12 mv^2 + mgh = \text{constant}$ works for a falling body, but written per unit volume of fluid with a pressure term added for the work done by neighbouring fluid pushing it along. Two physical simplifications turn this general equation into Torricelli's law. First, both the surface and the hole are open to the same atmosphere, so $P_1 = P_2 = P_0$ (atmospheric pressure) and these terms cancel outright. Second, because the tank's cross-sectional area $A_1$ is much larger than the hole's area $A_2$ , the equation of continuity $A_1 v_1 = A_2 v_2$ forces $v_1 = v_2 (A_2/A_1)$ to be tiny compared with $v_2$ — the surface sinks so slowly it is treated as essentially still, $v_1 \approx 0$ . What survives is a direct trade between the fluid's potential energy at the surface and its kinetic energy at the hole.

Figure

A tall open tank of liquid with a small hole in its side wall at depth h below the surface, the exit jet curving downward as a projectile and landing a horizontal distance R from the tank, with an inset arrow showing a stone falling freely through the same depth h to compare speeds

The core derivation

Derivation ATorricelli's law from Bernoulli's theorem

The goal is to start from Bernoulli's equation applied between the free surface of a large open tank and a small hole in its side at depth h below that surface, and to show — using only the fact that both points sit in the same atmosphere and that the hole's area is much smaller than the tank's — that the exit speed reduces to the strikingly simple form v equals the square root of two g h, identical to the speed of a body falling freely from rest through the same height h. Two physical facts do all the work: equal atmospheric pressure at both points, and a tank large enough that its own surface barely moves while liquid drains out.

Step 1. Write Bernoulli's theorem for a streamline running from a point 1 at the open surface to a point 2 at the hole: $P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$ . This is the general statement, valid for any two points on the same streamline in steady, non-viscous, incompressible flow.

Step 2. Fix the pressures. Both the top surface and the hole are open to the same atmosphere, so $P_1 = P_2 = P_0$ . These two terms are identical on both sides of the equation and cancel completely, leaving $\frac{1}{2}\rho v_1^2 + \rho g h_1 = \frac{1}{2}\rho v_2^2 + \rho g h_2$ .

Step 3. Simplify the surface speed using the equation of continuity. If the tank's cross-sectional area is $A_1$ and the hole's area is $A_2$ , continuity gives $A_1 v_1 = A_2 v_2$ , so $v_1 = v_2\,(A_2/A_1)$ . Since the tank is large and the hole is small, $A_2 \ll A_1$ , which makes $v_1$ negligible compared with $v_2$ ; we set $v_1 \approx 0$ . This is the "large tank, small hole" assumption, and it is the only approximation in the whole derivation.

Step 4. Substitute $v_1 = 0$ and collect the height terms using $h_1 - h_2 = h$ , the depth of the hole below the surface. The equation becomes $\rho g h_1 = \frac{1}{2}\rho v_2^2 + \rho g h_2$ , which rearranges to $\rho g (h_1 - h_2) = \frac{1}{2}\rho v_2^2$ , that is, $\rho g h = \frac{1}{2}\rho v_2^2$ .

Step 5. Cancel the common factor $\rho$ from both sides (density drops out entirely, which is why the law is the same for water, oil, or any liquid) and solve for $v_2$ : $g h = \frac{1}{2}v_2^2 \Rightarrow v_2^2 = 2gh \Rightarrow v_2 = \sqrt{2gh}$ . Dropping the subscript, this is Torricelli's law, $v = \sqrt{2gh}$ — precisely the speed a body reaches falling freely from rest through a height $h$ , because both situations are governed by the identical energy trade of $mgh$ converting into $\frac{1}{2}mv^2$ .

The elegance of the result is that it hides all the fluid-specific detail — density, viscosity (assumed negligible here), even the shape of the tank — inside two clean assumptions, leaving only $g$ and $h$ . That is also precisely why the formula can mislead a student who forgets those assumptions: change the tank's geometry so the surface speed is no longer negligible, or seal the tank so the pressures no longer cancel, and $v = \sqrt{2gh}$ silently stops applying even though the tank still leaks.

Worked examples

Board level - exit speed from a known depth

A large water tank has a small hole 3.2 m below the water surface. Find the speed with which water emerges from the hole. Take $g = 9.8\ \text{m/s}^2$ .

Step 1. Identify that this is a direct application of Torricelli's law, since the hole is small and open to the atmosphere, and the depth below the surface is given directly as $h = 3.2\ \text{m}$ : $v = \sqrt{2gh}$ .

Step 2. Substitute the values. $v = \sqrt{2 \times 9.8 \times 3.2} = \sqrt{62.72}$ .

Step 3. Compute the square root to get the answer. $v \approx 7.92\ \text{m/s}$ — the same speed the water would reach falling freely from rest through 3.2 m, illustrating that the formula needs nothing beyond the depth and $g$ .

Competitive level - efflux speed combined with projectile range

A cylindrical tank of large cross-sectional area is filled with water to a height of 1.8 m above a small hole located on the side of the tank at its base. Find (a) the speed of efflux, and (b) the horizontal range at which the jet strikes the ground, given the hole itself is 0.8 m above the ground. Take $g = 9.8\ \text{m/s}^2$ .

Step 1. Find the speed of efflux using Torricelli's law with $h = 1.8\ \text{m}$ , the depth of the hole below the water surface (not the height above the ground — that number belongs to the projectile part, handled next). $v = \sqrt{2 \times 9.8 \times 1.8} = \sqrt{35.28} \approx 5.94\ \text{m/s}$ .

Step 2. Treat the exit as the start of a horizontal projectile: it leaves the hole moving horizontally at $v = 5.94\ \text{m/s}$ and then falls under gravity through the height of the hole above the ground, $H = 0.8\ \text{m}$ . Find the time of fall from $H = \frac{1}{2}g t^2$ : $0.8 = \frac{1}{2}(9.8)t^2 \Rightarrow t^2 = \frac{1.6}{9.8} = 0.1633 \Rightarrow t \approx 0.404\ \text{s}$ .

Step 3. Find the horizontal range using $R = v \times t$ : $R = 5.94 \times 0.404 \approx 2.40\ \text{m}$ . Answer: the exit speed is about 5.94 m/s, and the jet lands about 2.40 m from the base of the tank. Timing note: this two-part problem should take under two minutes once the pattern is recognised — the only real decision point is identifying which given height feeds Torricelli's law (depth below the surface) and which feeds the projectile fall (height of the hole above the ground); confusing the two is the single fastest way to lose both marks in a chain problem like this.

Common traps

Forgetting that Torricelli's law assumes $v_1 \approx 0$ at the free surface — this is only valid because the tank's cross-sectional area is much larger than the hole's; it is not valid when the two areas are comparable, since then the surface itself is draining fast enough to matter.
Using the height of the hole above the ground, instead of the depth of the hole below the liquid surface, inside the $\sqrt{2gh}$ formula — these are two different lengths in a two-part problem, and swapping them gives a wrong exit speed even if every later step is correct.
Forgetting that after leaving the hole, the liquid becomes an ordinary projectile — the speed of efflux formula gives only the initial exit speed at the instant of leaving the hole, not the jet's speed or position afterward; range and time of flight need separate projectile-motion equations.
Applying Torricelli's law to a small or sealed container where atmospheric pressure does not act equally on the top surface and the hole — the cancellation of $P_1$ and $P_2$ in the derivation requires both points to be open to the same atmosphere; a pressurised or closed tank needs the full Bernoulli equation with the actual pressures kept in.
Quoting the formula without density $\rho$ and thinking it is missing by accident — $\rho$ genuinely cancels out of the derivation in Step 5, so the speed of efflux is the same for any liquid at a given depth, which is worth stating explicitly if a question asks why.

You should be able to…

Derive $v = \sqrt{2gh}$ from Bernoulli's theorem, stating both assumptions — equal atmospheric pressure and negligible surface speed — clearly and in order.
Explain physically why the speed of efflux equals the free-fall speed through the same depth, and why density cancels out of the result.
Combine Torricelli's law with basic projectile-motion equations to find the time of fall and horizontal range of an emerging jet.
Correctly distinguish, in a multi-part problem, the depth below the surface (feeds $v = \sqrt{2gh}$ ) from the height above the ground (feeds the projectile fall).
Recognise when the "large tank, small hole" assumption is, and is not, valid, and state what changes if it fails.

Ask the tutor: Want the full derivation walked through with the equation of continuity kept explicit, or a hand-worked range/time-of-flight problem for a jet leaking from a tank at a different height — ask the AI tutor for a step-by-step walkthrough.

Why does the density of the liquid not appear in v equals root two g h?

Because it cancels out during the derivation. Bernoulli's equation has a $\rho g h$ term for potential energy and a $\frac12 \rho v^2$ term for kinetic energy, and after the pressures cancel (both points open to the same atmosphere) every remaining term still carries a factor of $\rho$ . Dividing through by $\rho$ removes it completely, leaving only $g$ and $h$ . Physically this means a light oil and a dense liquid like mercury, held at the same depth in similar tanks, would leak out through a small hole at exactly the same speed — density affects how much mass leaves per second, but not how fast each bit of fluid moves.

If the tank is not very large, does v equal root two g h still work?

Not exactly. The whole simplification $v_1 \approx 0$ in the derivation relies on the hole's area being much smaller than the tank's cross-sectional area, so that the surface sinks far more slowly than the jet exits. If the tank is narrow or the hole is comparably large, the surface speed $v_1$ is no longer negligible, and you would have to keep both $v_1$ and $v_2$ in Bernoulli's equation together with the continuity relation $A_1 v_1 = A_2 v_2$ and solve the pair simultaneously — a more general result that reduces back to $v = \sqrt{2gh}$ only in the large-tank limit. CBSE Class 11 problems always use the large-tank assumption, but it is worth knowing why it is an assumption and not a universal law.

Is the speed of efflux the same as the speed the jet has when it lands on the ground?

No, and this trips students up in projectile-style problems. $v = \sqrt{2gh}$ is only the horizontal speed at the instant the liquid leaves the hole. Once outside, gravity pulls the jet downward exactly as it would any projectile, so it gains a vertical component of speed while its horizontal component stays fixed at $v$ . By the time it lands, its total speed is larger than $v$ and no longer purely horizontal — but that is standard projectile motion from mechanics, layered on top of the efflux speed, not part of Torricelli's law itself.

Why does raising the water level make the jet shoot out farther?

Two separate effects both increase with depth, and together they explain the everyday observation. First, a deeper hole (or a fuller tank, which increases the depth for a fixed hole) gives a larger $h$ in $v = \sqrt{2gh}$ , so the water leaves faster. Second, once you also account for how far the hole itself is above the ground, projectile range depends on both the exit speed and the time of fall available. Raising the water level increases $h$ and therefore the exit speed directly, which is why the classic demonstration of punching several holes at different heights down the side of a bottle shows the lowest hole's jet reaching a maximum range at some intermediate height — a full explanation of that optimum uses calculus and lies beyond this chapter, but the direct effect of increasing $h$ on speed is exactly Torricelli's law.