On this page
A water tank springs a leak, and the question asks where the jet lands. Students often get the speed right and the distance wrong, because the two use different heights. This post shows where Torricelli's law comes from, how to find the range of the jet, why the farthest jet comes from the middle, and how long a tank takes to drain, with worked numbers.
Where is Torricelli's law in the syllabus?
The CBSE Physics curriculum for 2026-27 lists "Bernoulli's theorem and its simple applications (Torricelli's law and Dynamic lift)" in Class 11. The Syllabus for JEE (Main) 2026 includes "Bernoulli's principle and its applications", and the JEE (Advanced) 2026 Information Brochure lists "Bernoulli's theorem and its applications". Check the current year's documents before your exam, as each is reissued.
Where does v = √(2gh) come from?
NCERT's chapter Mechanical Properties of Fluids, section 9.4.1, applies Bernoulli's equation between the free surface and a small hole in the side of a tank. If the tank is much wider than the hole, the surface barely moves. If the tank is open, both points are at atmospheric pressure. The pressure terms cancel and you get v = √(2gh), where h is the depth of the hole below the water surface. That is the speed of a body falling freely through h, which is why NCERT notes the two formulas are identical.
If the tank is closed and the air above the water is at pressure P, NCERT's equation 9.14 gives v = √(2gh + 2(P − Pa)/ρ). A pressurised tank shoots water faster than an open one.
How far does the jet land?
A hole in a vertical wall sends water out horizontally, so each drop becomes a horizontal projectile. Let the surface be H above the ground and the hole y above the ground, so h = H − y.
Speed: v = √(2g(H − y))
Time to fall: t = √(2y/g)
Range: x = vt = 2√((H − y)y)
The g cancels, so the range depends only on the two heights.
Worked example 1. Water stands 2.0 m deep in an open tank on the floor. A small hole is 0.50 m above the floor. Take g = 9.8 m/s².
Depth of the hole below the surface: h = 2.0 − 0.50 = 1.5 m.
Speed: v = √(2 × 9.8 × 1.5) = √29.4 ≈ 5.42 m/s.
Falling time: t = √(2 × 0.50 / 9.8) ≈ 0.319 s.
Range: x = 5.42 × 0.319 ≈ 1.73 m. Check with the formula: 2√(1.5 × 0.50) = 2√0.75 ≈ 1.73 m.
Why does the farthest jet come from the middle?
The product (H − y)y is largest when y = H/2, which gives a maximum range equal to H. In the example, a hole at 1.0 m throws water 2√(1.0 × 1.0) = 2.0 m. Now swap the heights. A hole at 1.5 m has h = 0.50 m and y = 1.5 m, the same product, so it also lands 1.73 m away. Two holes at heights y and H − y always hit the same spot.
Put the same tank on a 0.80 m stand. The hole is now 1.3 m above the ground but still 1.5 m below the surface, so x = 2√(1.5 × 1.3) = 2√1.95 ≈ 2.79 m. The speed did not change; only the falling time did.
How long does a tank take to drain?
This JEE Advanced-style extension needs one integral. Let the tank have cross-section A and the hole area a. The outflow a√(2gh) equals the rate at which the volume drops, −A dh/dt. Separating and integrating from depth h₁ to h₂ gives
t = (A/a) √(2/g) (√h₁ − √h₂)
Worked example 2. A tank of cross-section 0.50 m² holds water 1.8 m deep, with a hole of area 1.0 cm² (1.0 × 10⁻⁴ m²) in its base. Time to empty: t = 5000 × √(2/9.8) × √1.8 ≈ 5000 × 0.452 × 1.342 ≈ 3030 s, about 50 minutes.
Because the time depends on √h, the level falls fast at first and slowly later. Draining from 1.8 m to 0.45 m, a quarter of the depth, takes half the total time, about 1515 s.
Which mistakes cost marks?
Using the hole's height above the ground in v = √(2gh). The h in the speed is always measured down from the free surface.
Forgetting a stand or table under the tank when working out the falling time.
Using √(2gh) for a sealed, pressurised tank. Add the pressure term from NCERT's equation 9.14.
Assuming the speed stays constant while the tank drains. It falls as √h, which is why the draining time needs calculus.
Ignoring the area ratio when the hole is not small. Continuity then gives v = √(2gh / (1 − a²/A²)), and the simple formula is the limit where a is much smaller than A.
For other traps in this chapter, see our post on pressure, buoyancy and Bernoulli mistakes.
A practice routine
Rohrer and Taylor (2007) found that students who practised mixed problem types scored 63% on a test a week later, against 20% for students who had practised one type at a time. Use that idea here.
Day 1: three range problems, each with a sketch marking h and y separately.
Day 2: two equal-range and maximum-range problems, proving each result from the product (H − y)y.
Day 3: one draining-time problem, deriving the formula rather than quoting it.
Day 4: a mixed set that adds one problem from our post on liquid in an accelerating container, so you must choose the method yourself.
If the shape of the jet is hard to picture, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than to read. When a problem stalls at the first step, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
Tonight, sketch a tank with holes at heights y and H − y and show, in three lines, that both jets land at the same spot.
