This MCQ module is based on: Streamline Bernoulli
TOPIC 6 OF 33
Streamline Bernoulli
🎓 Class 11
Physics
CBSE
Theory
Ch 9 – Mechanical Properties of Fluids
⏱ ~14 min
🌐 Language: [gtranslate]
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Streamline Bernoulli
9.3 Streamline Flow
So far we have studied fluids at rest. Now we consider fluids in motion. Real fluid flow is complex, but under certain conditions the flow is smooth, layered and predictable — we call this streamline (or laminar) flow.
Streamline (steady) flow: Every particle that passes a given point in space has the same velocity (magnitude and direction) as any other particle that passed or will pass through that point. The flow pattern is time-independent.
Turbulent flow: At sufficiently high velocities, the flow becomes chaotic — eddies, whirlpools and vortices appear. The velocity at a given point fluctuates rapidly in time.
Turbulent flow: At sufficiently high velocities, the flow becomes chaotic — eddies, whirlpools and vortices appear. The velocity at a given point fluctuates rapidly in time.
A streamline is the path traced by a fluid particle in steady flow. No two streamlines ever cross (if they did, a particle at the intersection would have two different velocities). A bundle of adjacent streamlines forms a tube of flow.
9.3.1 Equation of Continuity
Consider a tube of flow with cross-section \(A_1\) at point 1 (velocity \(v_1\)) and \(A_2\) at point 2 (velocity \(v_2\)). In time \(\Delta t\), the mass of fluid entering section 1 is \(\rho_1 A_1 v_1\,\Delta t\) and the mass leaving section 2 is \(\rho_2 A_2 v_2\,\Delta t\). Since no fluid is created or destroyed and fluid cannot leak through the walls, these must be equal:
\[\rho_1 A_1 v_1 = \rho_2 A_2 v_2\]
For an incompressible fluid (\(\rho_1 = \rho_2\)):
\[\boxed{\,A_1 v_1 = A_2 v_2 = \text{constant}\,}\]
This is the equation of continuity. The product \(A\,v\) (the volume flow rate, m³/s) is conserved along a streamline.
Practical consequence: where a pipe narrows, the fluid must speed up. Pinch the opening of a garden hose and the jet shoots further — not because the water gained energy, but because the area decreased and so \(v\) must increase.
9.4 Bernoulli's Principle
Daniel Bernoulli (1738) discovered a beautiful relationship between the pressure, speed and height of a fluid along a streamline. It is essentially the work–energy theorem applied to a fluid element.
9.4.1 Statement and Equation
Bernoulli's Equation: For a steady, incompressible, non-viscous flow, along any streamline:
\[\boxed{\,P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{constant}\,}\]
Equivalently: \(P_1 + \tfrac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \tfrac{1}{2}\rho v_2^2 + \rho g h_2\)
Each term has the dimensions of pressure (energy per unit volume):
- \(P\) — pressure energy per unit volume (work done by pressure).
- \(\tfrac{1}{2}\rho v^2\) — kinetic energy per unit volume.
- \(\rho g h\) — gravitational potential energy per unit volume.
9.4.2 Sketch of the Derivation
Consider a fluid element of mass \(\Delta m = \rho A\,\Delta \ell\) moving through a tube of flow between sections 1 (low, wide, slow) and 2 (high, narrow, fast). In time \(\Delta t\):
- Net work by pressure forces on the element: \((P_1 - P_2)\Delta V\), where \(\Delta V\) is the volume transferred.
- Change in kinetic energy: \(\tfrac{1}{2}\Delta m(v_2^2 - v_1^2)\).
- Change in gravitational PE: \(\Delta m\,g(h_2 - h_1)\).
Setting work = change in total mechanical energy and dividing by \(\Delta V\) yields Bernoulli's equation.
Key takeaway: For flow at the same height, where the velocity is high, the pressure is low — and vice versa. This counter-intuitive result explains why cars can be pulled toward a high-speed truck, why shower curtains billow inward, and why aircraft fly.
9.4.3 Application 1 — Speed of Efflux (Torricelli's Theorem)
A large tank of water has a small hole at depth \(h\) below the free surface. What is the speed of the water jet emerging from the hole?
Apply Bernoulli's equation between the top of the tank (point 1, area huge, \(v_1 \approx 0\), pressure \(P_0\)) and the hole (point 2, pressure \(P_0\) because the jet is in open air):
\[P_0 + 0 + \rho g h = P_0 + \tfrac{1}{2}\rho v^2 + 0\] \[\Longrightarrow\quad \boxed{\,v = \sqrt{2gh}\,}\]This is identical to the speed a freely-falling body would acquire falling through the same height \(h\). This is Torricelli's theorem.
9.4.4 Application 2 — Venturi Meter
A Venturi meter measures the flow rate of a fluid through a pipe. The pipe narrows (the throat) and then widens again. By continuity \(A_1 v_1 = A_2 v_2\), so the fluid accelerates in the throat. By Bernoulli, the pressure there falls. A manometer attached across the two sections measures the pressure difference \(\Delta P\), from which the flow rate can be deduced:
\[v_1 = A_2\sqrt{\frac{2\,\Delta P}{\rho(A_1^2 - A_2^2)}}\]9.4.5 Application 3 — Dynamic Lift on an Aerofoil
An aircraft wing (aerofoil) is shaped so that air flows faster over the curved top surface than along the flatter bottom surface. By Bernoulli's principle, the pressure on top is therefore lower than on the bottom. The net upward pressure × wing area = dynamic lift.
9.4.6 Application 4 — Magnus Effect (Spinning Ball)
When a cricket or football spins while moving forward, the air drags around with the surface on one side (speeds up relative flow) and against it on the other (slows it down). The pressure is lower on the fast-moving side, higher on the slow side — so a sideways force pushes the ball toward the low-pressure side. This is why a topspin tennis shot dips quickly, why fast bowlers swing the ball, and why footballers can "bend" free-kicks.
9.4.7 Other Applications
- Atomiser/spray gun: A horizontal blast of air over the top of a vertical tube lowers the pressure there. Atmospheric pressure at the liquid surface pushes the liquid up the tube, and the jet of air breaks it into a fine spray.
- Blood flow in narrow arteries: In a narrowed artery (say due to plaque), by continuity the blood flows faster through the constriction, lowering the pressure. If the pressure drops far enough, the artery can momentarily collapse — a warning sign of cardiovascular problems.
Worked Examples
Example 9.5: Pipe narrowing (NCERT Example 9.5)
Water flows through a horizontal pipe of diameter 12 cm with a speed of 2 m/s at one section. The pipe narrows to a diameter of 6 cm at another section. Find (a) the speed of water at the narrower section, (b) the pressure difference between the two sections. \(\rho = 1000\) kg/m³.
Given: \(d_1 = 12\) cm ⇒ \(r_1 = 6\times 10^{-2}\) m; \(d_2 = 6\) cm ⇒ \(r_2 = 3\times 10^{-2}\) m; \(v_1 = 2\) m/s.
(a) Continuity: \(A_1 v_1 = A_2 v_2 \Rightarrow v_2 = v_1(A_1/A_2) = v_1(r_1/r_2)^2 = 2\times(6/3)^2 = 2\times 4 = 8\) m/s.
(b) Bernoulli (horizontal, so \(h_1 = h_2\)): \[P_1 - P_2 = \tfrac{1}{2}\rho(v_2^2 - v_1^2) = \tfrac{1}{2}\times 1000\times(64 - 4) = 30000\text{ Pa} = 3.0\times 10^4\text{ Pa}\] Answer: \(v_2 = \boxed{8\text{ m/s}}\), \(P_1 - P_2 = \boxed{3.0\times 10^4\text{ Pa}}\) (pressure is higher in the wider section).
(a) Continuity: \(A_1 v_1 = A_2 v_2 \Rightarrow v_2 = v_1(A_1/A_2) = v_1(r_1/r_2)^2 = 2\times(6/3)^2 = 2\times 4 = 8\) m/s.
(b) Bernoulli (horizontal, so \(h_1 = h_2\)): \[P_1 - P_2 = \tfrac{1}{2}\rho(v_2^2 - v_1^2) = \tfrac{1}{2}\times 1000\times(64 - 4) = 30000\text{ Pa} = 3.0\times 10^4\text{ Pa}\] Answer: \(v_2 = \boxed{8\text{ m/s}}\), \(P_1 - P_2 = \boxed{3.0\times 10^4\text{ Pa}}\) (pressure is higher in the wider section).
Example 9.6: Speed of water leaving a tank
Water in a tank stands 4.9 m above a small hole in the side. Find the speed of the emerging jet and the horizontal distance it reaches on the ground if the hole is 1.0 m above the ground. \(g = 9.8\) m/s².
Speed (Torricelli): \(v = \sqrt{2gh} = \sqrt{2\times 9.8 \times 4.9} = \sqrt{96.04} = 9.8\) m/s (horizontal).
Time to fall 1.0 m: \(y = \tfrac{1}{2}gt^2 \Rightarrow t = \sqrt{2y/g} = \sqrt{2\times 1/9.8} = 0.452\) s.
Range: \(R = v\,t = 9.8 \times 0.452 = 4.43\) m.
Answer: \(v = \boxed{9.8\text{ m/s}}\), \(R \approx \boxed{4.4\text{ m}}\).
Time to fall 1.0 m: \(y = \tfrac{1}{2}gt^2 \Rightarrow t = \sqrt{2y/g} = \sqrt{2\times 1/9.8} = 0.452\) s.
Range: \(R = v\,t = 9.8 \times 0.452 = 4.43\) m.
Answer: \(v = \boxed{9.8\text{ m/s}}\), \(R \approx \boxed{4.4\text{ m}}\).
Example 9.7: Lift on an aeroplane wing
An aircraft has wings of total area 25 m². Air flows over the top surface at 70 m/s and under the lower surface at 63 m/s. Find the lift force. Density of air = 1.3 kg/m³.
Using Bernoulli's equation (same height):
\[P_{\text{bot}} - P_{\text{top}} = \tfrac{1}{2}\rho(v_{\text{top}}^2 - v_{\text{bot}}^2) = \tfrac{1}{2}\times 1.3\times(70^2 - 63^2)\]
\[= 0.65 \times (4900 - 3969) = 0.65\times 931 = 605\text{ Pa}\]
Lift force:
\[F = \Delta P \times A = 605\times 25 = 1.51\times 10^4\text{ N}\]
Answer: Lift \(\approx \boxed{1.5\times 10^4\text{ N}}\). This is about 1.5 tonnes of lift — enough to support a small plane.
Example 9.8: Venturi meter
A horizontal Venturi meter has inlet area \(A_1 = 40\) cm² and throat area \(A_2 = 10\) cm². Water flows through it and the manometer shows a pressure drop of 1960 Pa. Find the volume flow rate. \(\rho = 1000\) kg/m³.
Convert areas: \(A_1 = 40\times 10^{-4} = 4\times 10^{-3}\) m², \(A_2 = 10^{-3}\) m².
Continuity: \(v_2 = v_1(A_1/A_2) = 4 v_1\).
Bernoulli: \(P_1 - P_2 = \tfrac{1}{2}\rho(v_2^2 - v_1^2) = \tfrac{1}{2}\rho(16v_1^2 - v_1^2) = \tfrac{15}{2}\rho v_1^2\).
\[1960 = \tfrac{15}{2}\times 1000 \times v_1^2 \Rightarrow v_1^2 = \frac{1960}{7500} = 0.2613 \Rightarrow v_1 = 0.511\text{ m/s}\] Volume flow rate \(Q = A_1 v_1 = 4\times 10^{-3}\times 0.511 = 2.04\times 10^{-3}\) m³/s ≈ 2.0 L/s.
Answer: \(Q \approx \boxed{2.0\times 10^{-3}\text{ m}^3/\text{s}}\).
Continuity: \(v_2 = v_1(A_1/A_2) = 4 v_1\).
Bernoulli: \(P_1 - P_2 = \tfrac{1}{2}\rho(v_2^2 - v_1^2) = \tfrac{1}{2}\rho(16v_1^2 - v_1^2) = \tfrac{15}{2}\rho v_1^2\).
\[1960 = \tfrac{15}{2}\times 1000 \times v_1^2 \Rightarrow v_1^2 = \frac{1960}{7500} = 0.2613 \Rightarrow v_1 = 0.511\text{ m/s}\] Volume flow rate \(Q = A_1 v_1 = 4\times 10^{-3}\times 0.511 = 2.04\times 10^{-3}\) m³/s ≈ 2.0 L/s.
Answer: \(Q \approx \boxed{2.0\times 10^{-3}\text{ m}^3/\text{s}}\).
Example 9.9: Blood in a constricted artery
Blood flows at 0.5 m/s through an artery of radius 4 mm. At a region of plaque build-up the radius narrows to 2 mm. Find (a) the speed through the constriction, and (b) the pressure drop. Density of blood = 1060 kg/m³.
(a) \(v_2 = v_1(r_1/r_2)^2 = 0.5\times (4/2)^2 = 0.5\times 4 = 2.0\) m/s.
(b) \(\Delta P = \tfrac{1}{2}\rho(v_2^2 - v_1^2) = \tfrac{1}{2}\times 1060\times(4 - 0.25) = 530\times 3.75 = 1988\) Pa ≈ 15 mm Hg.
Answer: \(v_2 = \boxed{2.0\text{ m/s}}\), \(\Delta P \approx \boxed{2000\text{ Pa}}\). This pressure drop inside the narrowed artery can cause the soft walls to momentarily collapse — a clinically important phenomenon.
(b) \(\Delta P = \tfrac{1}{2}\rho(v_2^2 - v_1^2) = \tfrac{1}{2}\times 1060\times(4 - 0.25) = 530\times 3.75 = 1988\) Pa ≈ 15 mm Hg.
Answer: \(v_2 = \boxed{2.0\text{ m/s}}\), \(\Delta P \approx \boxed{2000\text{ Pa}}\). This pressure drop inside the narrowed artery can cause the soft walls to momentarily collapse — a clinically important phenomenon.
Interactive: Bernoulli Flow Simulator L3 Apply
A horizontal pipe narrows from area A₁ to A₂. Move the sliders and see how the narrowed-section velocity and pressure drop change.
Activity — Bernoulli with Two Sheets of Paper
L3 Apply
Predict: If you blow strongly into the gap between two sheets of paper hanging close to each other, will the sheets fly apart, or move towards each other?
- Hold two A4 sheets of paper vertically, 3–4 cm apart, one in each hand.
- Blow steadily through the gap between the two sheets.
- Observe whether the sheets separate or come together.
Observation: The sheets move towards each other.
Explanation: Your breath makes the air between the sheets move fast. By Bernoulli's principle, the pressure in the fast-moving region is lower than the (still) atmospheric pressure outside each sheet. The higher outside pressure pushes the sheets together.
Explanation: Your breath makes the air between the sheets move fast. By Bernoulli's principle, the pressure in the fast-moving region is lower than the (still) atmospheric pressure outside each sheet. The higher outside pressure pushes the sheets together.
Competency-Based Questions
A garden hose has an internal diameter of 2.0 cm. Water flows through it at 0.50 m/s. At the end of the hose is fitted a nozzle with a narrow opening of diameter 0.50 cm. The gardener is watering a plant that sits on a pedestal of height 1.25 m.
Q1. L1 Remember State the equation of continuity.
For an incompressible fluid in streamline flow, \(A_1 v_1 = A_2 v_2\) along any tube of flow.
Q2. L3 Apply Find the speed of water leaving the nozzle.
\(v_2 = v_1(d_1/d_2)^2 = 0.50 \times (2.0/0.50)^2 = 0.50\times 16 = \boxed{8.0\text{ m/s}}\).
Q3. L3 Apply If the nozzle points straight up, what maximum height can the water reach above it? (Ignore air resistance; take \(g = 10\) m/s².)
Using energy conservation for the jet, \(h_{\max} = v^2/(2g) = 64/20 = \boxed{3.2\text{ m}}\). Easily reaches the 1.25 m pedestal.
Q4. L4 Analyse Explain, using Bernoulli's equation, why the pressure inside the hose (wide part) is greater than atmospheric, but the pressure at the nozzle exit is essentially atmospheric.
Bernoulli: \(P + \tfrac{1}{2}\rho v^2 = \text{const}\). Inside the wide hose \(v\) is small so \(P\) must be large to meet the sum. At the nozzle exit, the jet is in open air so \(P = P_{\text{atm}}\); by continuity \(v\) is much larger there, soaking up the pressure difference as kinetic energy. Pressure energy has been converted to kinetic energy.
Q5. L5 Evaluate A student claims that if the nozzle is made even smaller, the water will reach unlimited height. Evaluate this claim.
The claim ignores real-world limits. By continuity a smaller \(A_2\) gives higher exit \(v_2\), and Bernoulli relates \(v^2\) to the pressure energy supplied by the supply. But the pressure in the hose is set by the tap — it is finite. When \(\tfrac{1}{2}\rho v^2\) equals the available pressure difference, further narrowing cannot increase \(v\); instead the flow rate Q falls, viscous losses rise, and eventually the supply pressure simply drives no flow. So the claim is wrong.
Assertion-Reason Questions
Assertion (A): Roofs of houses in cyclonic winds are sometimes blown off from below.
Reason (R): The fast wind moving over the roof creates a low pressure above it, while the air pressure inside remains high — the pressure difference can lift the roof.
Answer: A. This is a classic real-world application of Bernoulli's principle.
Assertion (A): When water flows through a pipe whose radius decreases, its speed increases.
Reason (R): The mass flow rate of an incompressible fluid must be conserved.
Answer: A. This is exactly the continuity equation \(Av = \)const.
Assertion (A): Bernoulli's equation applies only to steady, incompressible, non-viscous flow along a streamline.
Reason (R): The derivation uses the work-energy theorem for a fluid element and neglects heat losses due to friction (viscosity).
Answer: A. R states exactly the assumptions made in the derivation.
Frequently Asked Questions - Streamline Bernoulli
What is the main concept covered in Streamline Bernoulli?
In NCERT Class 11 Physics Chapter 9 (Mechanical Properties of Fluids), "Streamline Bernoulli" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Streamline Bernoulli useful in real-life applications?
Real-life applications of Streamline Bernoulli from NCERT Class 11 Physics Chapter 9 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Streamline Bernoulli?
Key formulas in Streamline Bernoulli (NCERT Class 11 Physics Chapter 9 Mechanical Properties of Fluids) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 9?
NCERT Class 11 Physics Chapter 9 (Mechanical Properties of Fluids) is structured so each part builds on the previous one. Streamline Bernoulli connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Streamline Bernoulli?
CBSE board questions from Streamline Bernoulli typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Streamline Bernoulli lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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