Bernoulli's theorem — find velocity of efflux from a hole in a tank

Question

A large tank filled with water has a small hole at a depth $h$ below the water surface. Using Bernoulli's theorem, find the velocity of efflux from the hole (Torricelli's theorem).

(JEE Main 2022, similar pattern)

Solution — Step by Step

Identify the two points for Bernoulli's equation

Point 1: The free surface of the water (top of the tank). Point 2: The hole at depth $h$ .

Assumptions: the tank is large (so the surface drops very slowly, $v_1 \approx 0$ ), the hole is small compared to the tank, and both points are open to the atmosphere ( $P_1 = P_2 = P_{\text{atm}}$ ).

Apply Bernoulli's equation

Bernoulli's Equation

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

Taking the hole as reference level ( $h_2 = 0$ , $h_1 = h$ ):

$P_{\text{atm}} + 0 + \rho g h = P_{\text{atm}} + \frac{1}{2}\rho v^2 + 0$

Solve for v

$\rho g h = \frac{1}{2}\rho v^2$

$\boxed{v = \sqrt{2gh}}$

This is Torricelli's theorem: the velocity of efflux equals the speed a body would acquire falling freely through the same height $h$ .

Why This Works

Bernoulli's equation is energy conservation per unit volume for a fluid. The pressure energy and gravitational PE at the top convert into kinetic energy at the hole. Since both points are open to atmosphere, the pressure terms cancel.

The result $v = \sqrt{2gh}$ is identical to the free-fall formula — and this is not a coincidence. The water at the hole has "fallen" through height $h$ , converting PE to KE. Bernoulli's equation simply formalises this for fluids.

For a tank of cross-section $A$ with a hole of area $a$ : if $a \ll A$ , our approximation $v_1 \approx 0$ holds. For comparable sizes, you'd need to use the continuity equation $Av_1 = av_2$ and the result modifies to $v = \sqrt{\frac{2gh}{1 - (a/A)^2}}$ .

Alternative Method — Using energy conservation directly

Consider a small volume $\Delta V$ of water at the surface that eventually exits through the hole.

Energy at top: $PE = \rho g h \cdot \Delta V$ , $KE \approx 0$

Energy at hole: $PE = 0$ , $KE = \frac{1}{2}\rho \Delta V \cdot v^2$

Equating: $v = \sqrt{2gh}$ . Same result without invoking Bernoulli explicitly.

💡 Expert Tip

Torricelli's theorem is the starting point for many JEE problems: "time to empty a tank," "range of the water jet," "at what height should the hole be for maximum range." For maximum horizontal range of the jet, the hole should be at $h/2$ from the surface (or equivalently, at $h/2$ from the bottom). This is a classic MCQ result.

Common Mistake

⚠️ Common Mistake

Students sometimes forget to cancel atmospheric pressure. If the hole is open to the atmosphere, $P_2 = P_{\text{atm}}$ , and if the tank is also open, $P_1 = P_{\text{atm}}$ . These cancel. But if the tank is sealed with pressure $P_0$ above the water, the velocity becomes $v = \sqrt{2gh + 2(P_0 - P_{\text{atm}})/\rho}$ . Read the problem carefully to check whether the tank is open or sealed.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using energy conservation directly

Common Mistake

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