Bernoulli Equation — When and How to Apply

Bernoulli’s Equation — When and How to Apply

Most students learn Bernoulli’s equation as a one-liner and then mis-apply it everywhere. The equation is powerful but conditional. We will go through when it actually works, how to set it up, and the four classic JEE/NEET applications that examiners love. Once you internalize the conditions, you stop making the silly errors that cost easy marks.

The equation itself is just energy conservation per unit volume for an ideal fluid. Pressure energy + kinetic energy + gravitational potential energy = constant along a streamline.

Key Terms & Definitions

Streamline. The path traced by a fluid particle. In steady flow, streamlines don’t cross, and Bernoulli’s equation is valid along each streamline.

Steady flow. Flow where velocity at every point doesn’t change with time. The flow pattern looks the same at $t = 1$ s and $t = 100$ s.

Incompressible fluid. Density $\rho$ is constant. Most JEE problems involve liquids (water, oil) or low-speed air, both treated as incompressible.

Non-viscous fluid. No internal friction. We ignore viscosity in Bernoulli problems unless the problem explicitly mentions it.

The Equation

$P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}$

along a streamline, where $P$ is pressure, $v$ is fluid speed, $h$ is height above a reference, and $\rho$ is density.

Three conditions must hold to apply this:

The fluid is incompressible.
The fluid is non-viscous.
The flow is steady and the points are on the same streamline.

If any condition fails, Bernoulli doesn’t apply directly. You need a modified form (with viscous losses) or a completely different approach.

How to Apply Bernoulli — The 4-Step Method

Almost always: one point where you know everything (e.g., the surface of a tank) and one point where you want to find something (e.g., the exit hole). The two points must be connected by a streamline.

Set $h = 0$ at one of the two points or at any convenient horizontal level. The reference must be the same for both points in a single equation.

Make a small table: $P_1, v_1, h_1$ on one side; $P_2, v_2, h_2$ on the other. Mark unknowns clearly.

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

Plug in, solve for the unknown.

Application 1 — Torricelli’s Theorem

A tank of water has a small hole at depth $h$ below the surface. Find the exit speed.

Take point 1 at the surface and point 2 at the hole. At the surface: $P_1 = P_{\text{atm}}$ , $v_1 \approx 0$ (large tank), $h_1 = h$ . At the hole: $P_2 = P_{\text{atm}}$ , $v_2 = ?$ , $h_2 = 0$ .

Bernoulli: $P_{\text{atm}} + 0 + \rho g h = P_{\text{atm}} + \tfrac{1}{2}\rho v_2^2 + 0$

$v_2 = \sqrt{2 g h}$

This is just the speed of a freely falling body dropped from height $h$ — fluid pressure at depth converts entirely to kinetic energy at the exit.

Application 2 — Venturi Meter

A pipe narrows from area $A_1$ to area $A_2$ . Pressure difference between wide and narrow sections allows us to compute flow rate.

By continuity: $A_1 v_1 = A_2 v_2$ , so $v_2 = v_1 (A_1/A_2)$ .

By Bernoulli (same height): $P_1 + \tfrac{1}{2}\rho v_1^2 = P_2 + \tfrac{1}{2}\rho v_2^2$

$P_1 - P_2 = \tfrac{1}{2}\rho (v_2^2 - v_1^2) = \tfrac{1}{2}\rho v_1^2 \left(\frac{A_1^2}{A_2^2} - 1\right)$

This is how we measure flow rate in industrial pipes — measure $\Delta P$ with a manometer, solve for $v_1$ .

Application 3 — Aeroplane Wing Lift

The aerofoil is shaped so air flows faster over the top surface than the bottom (the streamlines on top are forced into a longer, faster path). Higher speed on top means lower pressure on top (Bernoulli). The pressure difference across the wing × area = lift force.

Order of magnitude: a 100 m/s vs 95 m/s difference at $\rho = 1.2$ kg/m³ gives $\Delta P \approx 580$ Pa. For a 30 m² wing area, lift $\approx 17000$ N — enough to support a small aircraft.

Application 4 — Atomizer (Spray Bottle)

When you squeeze the bulb of a spray bottle, air rushes through a narrow nozzle at high speed. By Bernoulli, the pressure in this fast-moving air drops below atmospheric. The atmospheric pressure pushes liquid up a vertical tube into the airstream, where it gets dispersed as a fine mist.

This is the same physics that lets a perfume bottle, a paint spray gun, and a Bunsen burner all work.

Solved Examples

Example 1 (Easy, NEET)

Water flows through a horizontal pipe of varying cross-section. At the wider section ( $A_1 = 4$ cm²), the speed is $v_1 = 2$ m/s and pressure is $P_1 = 2 \times 10^5$ Pa. Find the pressure at the narrower section ( $A_2 = 1$ cm²).

By continuity: $v_2 = v_1 (A_1/A_2) = 2 \times 4 = 8$ m/s.

By Bernoulli: $P_2 = P_1 + \tfrac{1}{2}\rho(v_1^2 - v_2^2) = 2 \times 10^5 + \tfrac{1}{2}(1000)(4 - 64) = 2 \times 10^5 - 30000 = 1.7 \times 10^5$ Pa.

Example 2 (Medium, JEE Main)

A tank of height 5 m is full of water. A hole is drilled at the bottom. Find the time taken for the water level to drop from 5 m to 1 m. The hole has area $a$ and the tank has cross-section $A$ , with $A \gg a$ .

Exit speed at depth $h$ : $v = \sqrt{2gh}$ . Flow rate $= a\sqrt{2gh}$ . Volume balance: $-A \dfrac{dh}{dt} = a\sqrt{2gh}$ .

Integrate from $h = 5$ to $h = 1$ :

$t = \frac{A}{a}\sqrt{\frac{2}{g}}\left(\sqrt{H_0} - \sqrt{H_f}\right) = \frac{A}{a}\sqrt{\frac{2}{g}}(\sqrt{5} - 1)$

Example 3 (Hard, JEE Advanced)

A horizontal pipe of cross-section $A$ branches into two pipes of cross-sections $A_1$ and $A_2$ . Flow speed in the main pipe is $v_0$ , and in the second branch is $v_2$ . Find the speed in the first branch and pressure relations.

By continuity: $A v_0 = A_1 v_1 + A_2 v_2 \Rightarrow v_1 = (A v_0 - A_2 v_2)/A_1$ .

Both branches have the same outlet pressure, so by Bernoulli applied separately to each branch from the main, we get $P_{\text{main}} - P_{\text{out}} = \tfrac{1}{2}\rho(v_1^2 - v_0^2) = \tfrac{1}{2}\rho(v_2^2 - v_0^2)$ , which forces $v_1 = v_2$ — but that contradicts continuity unless $A_1 = A_2$ . So the problem only makes sense if the branches differ in pressure — i.e., the outlets aren’t both at atmospheric.

Exam-Specific Tips

JEE Main weightage. Bernoulli appears in roughly 1 question per shift, usually as a 4-marker. Direct Torricelli or Venturi questions are easy targets — practice 10 of each.

NEET style. NEET often combines Bernoulli with continuity equation in a 1-mark MCQ. Get continuity ( $A_1 v_1 = A_2 v_2$ ) cold.

CBSE Class 11 boards. Boards usually ask one numerical from Bernoulli or Torricelli — direct application, no twists. Free marks.

Common Mistakes to Avoid

Mistake 1: Applying Bernoulli when streamlines aren’t connected. You can’t apply Bernoulli between two random points in a fluid. They must lie on the same streamline.

Mistake 2: Forgetting continuity. Bernoulli alone has two unknowns ( $v_2$ and $P_2$ ). You need continuity ( $A_1 v_1 = A_2 v_2$ ) to solve. Always write both equations.

Mistake 3: Wrong sign for $h$ . $h$ in Bernoulli is height above a reference. Going down decreases $h$ — students sometimes mix this up.

Mistake 4: Ignoring atmospheric pressure when both points are open. If both points are exposed to atmosphere, the $P$ terms cancel, but you must still write them — students sometimes forget and miss a sign.

Mistake 5: Applying Bernoulli to viscous flows. If the problem mentions viscosity, friction, or terminal velocity, Bernoulli alone won’t work. You need Stokes’ law or modified energy equations.

Practice Questions

Q1. A tank holds water 4 m deep. A small hole is at the bottom. Find the exit speed. (Take $g = 10$ m/s².)

$v = \sqrt{2 \times 10 \times 4} = \sqrt{80} \approx 8.94$ m/s.

Q2. Water flows in a pipe at 5 m/s with pressure $1.5 \times 10^5$ Pa. The pipe narrows so velocity becomes 10 m/s. Find the new pressure (horizontal pipe).

$P_2 = P_1 + \tfrac{1}{2}\rho(v_1^2 - v_2^2) = 1.5 \times 10^5 + 500(25 - 100) = 1.5 \times 10^5 - 37500 = 1.125 \times 10^5$ Pa.

Q3. Air flows over the top of a wing at 60 m/s and below at 40 m/s. Find the pressure difference across the wing. ( $\rho_{\text{air}} = 1.2$ kg/m³.)

$\Delta P = \tfrac{1}{2}(1.2)(60^2 - 40^2) = 0.6 \times 2000 = 1200$ Pa.

Q4. A horizontal pipe carries water at 2 m/s where the cross-section is 50 cm². Find the speed where the cross-section is 10 cm².

By continuity: $v_2 = 2 \times 5 = 10$ m/s.

Q5. Find the maximum range of water emerging from a hole in a tank of water depth $H$ , when the hole is at depth $h$ below the surface.

Range $x = 2\sqrt{h(H-h)}$ , maximum when $h = H/2$ , giving $x_{\max} = H$ .

Q6. A Venturi meter has $A_1 = 10$ cm² and $A_2 = 5$ cm². Pressure difference $P_1 - P_2 = 600$ Pa. Find $v_1$ for water.

$v_2 = 2 v_1$ . $\tfrac{1}{2}(1000)(4v_1^2 - v_1^2) = 600 \Rightarrow 1500 v_1^2 = 600 \Rightarrow v_1 \approx 0.632$ m/s.

Q7. A tank of cross-section $A$ has a small hole of area $a$ at depth $h$ . Find the rate at which the water level drops.

By continuity: $A \dfrac{dh}{dt} = -av = -a\sqrt{2gh}$ , so $dh/dt = -(a/A)\sqrt{2gh}$ .

Q8. Water enters a horizontal pipe at pressure $2 \times 10^5$ Pa and exits at atmospheric pressure ( $10^5$ Pa). Inlet area is twice the outlet area. Find the inlet speed.

$v_2 = 2v_1$ . $2 \times 10^5 + \tfrac{1}{2}(1000)v_1^2 = 10^5 + \tfrac{1}{2}(1000)(2v_1)^2$ . $10^5 = 1500 v_1^2 \Rightarrow v_1 \approx 8.16$ m/s.

FAQs

Q: Can Bernoulli be applied to gases?

Yes, but only at low speeds (Mach number < 0.3) where the gas is essentially incompressible. For high-speed flows, use compressible flow equations.

Q: What if the flow is unsteady?

Bernoulli requires steady flow. For unsteady flow, you need the unsteady form (with $\partial v/\partial t$ terms) — beyond JEE/NEET syllabus.

Q: Why does the Magnus effect (curving cricket ball) work?

A spinning ball drags air with it. On one side, air speeds up; on the other, slows down. Bernoulli says pressure differs across the ball, producing a sideways force.

Q: Do I need to memorize the Venturi formula?

No — derive it on the spot from Bernoulli + continuity. Memorizing the formula without derivation traps students when the geometry changes slightly.

Q: Is Torricelli’s theorem the same as energy conservation?

Yes, exactly. Bernoulli is energy conservation per unit volume. Torricelli is a special case where the only relevant terms are pressure (cancels) and gravity vs. kinetic energy.

Q: Why don’t roofs blow off in wind even though Bernoulli predicts low pressure above?

They sometimes do. In high winds, the pressure difference can be huge — that’s how cyclones lift roofs. Standard buildings are reinforced to resist this force.

Q: Can I use Bernoulli for viscous flows like blood through arteries?

Approximately yes for short distances or large arteries, but real blood flow needs the Hagen-Poiseuille equation that includes viscosity. Beyond Class 11.