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Bernoulli’s principle and applications: NEET MCQs with answers

50 questions on this topic of Properties of Solids and Liquids (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. In a car carburetor, petrol is drawn into the air stream primarily because:

  1. The high-speed air creates a region of reduced pressure near the petrol nozzle
  2. The high-speed air creates a region of increased pressure forcing petrol into it
  3. Capillary action pulls the petrol into the chamber against gravity
  4. Viscous drag forces of air carry the heavy petrol droplets upward
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Answer: (A) The high-speed air creates a region of reduced pressure near the petrol nozzle

In the carburetor, air passes through a constriction (venturi). The narrowing of the channel increases the velocity of the air, which according to Bernoulli's principle lowers the local pressure below atmospheric pressure, causing petrol to be sucked into the stream.

medium

2. Water flows smoothly through a horizontal pipe. At section X, the static pressure is 3.0 × 10⁴ Pa and flow speed is 2 m s⁻¹. At section Y, the static pressure is 2.55 × 10⁴ Pa. Assuming density of water is 1000 kg m⁻³, the flow speed at section Y is:

  1. 3.0 m s⁻¹
  2. 4.5 m s⁻¹
  3. 3.6 m s⁻¹
  4. 5.0 m s⁻¹
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Answer: (C) 3.6 m s⁻¹

Applying Bernoulli's equation: P_X + (1/2)ρv_X² = P_Y + (1/2)ρv_Y². Substituting the values: 3.0 × 10⁴ + (1/2)(1000)(4) = 2.55 × 10⁴ + 500 v_Y². This gives 32000 - 25500 = 500 v_Y², so 6500 = 500 v_Y², which yields v_Y² = 13 and v_Y = √13 ≈ 3.6 m s⁻¹.

hard

3. An enclosed large storage tank contains water of density 1000 kg m⁻³. The space above the water is filled with compressed air at an absolute pressure of 3.0 × 10⁵ Pa. A small hole develops at a depth of 5.0 m below the water level. Taking atmospheric pressure as 1.0 × 10⁵ Pa and g = 10 m s⁻¹, the speed of efflux from the hole is:

  1. 20 m s⁻¹
  2. 10√3 m s⁻¹
  3. 10√5 m s⁻¹
  4. 25 m s⁻¹
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Answer: (C) 10√5 m s⁻¹

Applying Bernoulli's theorem between the surface and the hole: P_inside + ρgh = P_atm + (1/2)ρv². Here, P_inside - P_atm = 2.0 × 10⁵ Pa, and ρgh = 1000 × 10 × 5 = 0.5 × 10⁵ Pa. Thus, (1/2)ρv² = 2.5 × 10⁵ Pa, giving v² = 2 × 2.5 × 10⁵ / 1000 = 500. Therefore, v = √500 = 10√5 m s⁻¹.

medium

4. The dynamic lift on an aeroplane wing arises because air flows:

  1. Slower over the upper surface
  2. Equally on both surfaces
  3. Only below the wing
  4. Faster over the upper surface, reducing pressure there
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Answer: (D) Faster over the upper surface, reducing pressure there

The wing shape makes air move faster above it than below. By Bernoulli's principle, pressure above is lower, giving an upward lift.

NCERT Class 11

easy

5. A wide open container filled with water to a height H is placed on a horizontal table. A small hole is made in the side wall at a depth y below the free surface of water. The emerging water jet strikes the table at a horizontal distance R from the base of the tank. For what value of y is the horizontal range R maximum?

  1. H/4
  2. H/2
  3. H/3
  4. 2H/3
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Answer: (B) H/2

The velocity of efflux is v = √(2gy). The time taken for the liquid to fall through the remaining height (H - y) to the table level is t = √[2(H - y)/g]. The horizontal range is R = v × t = √(2gy) × √[2(H - y)/g] = 2√[y(H - y)]. This product y(H - y) is maximum when y = H - y, which gives y = H/2.

hard

6. An aeroplane of total mass 4000 kg is in level flight at constant velocity. The total wing area is 20 m². The air speed over the lower wing surface is 180 m s⁻¹. Taking the density of air as 1.0 kg m⁻³ and g = 10 m s⁻², the air speed over the upper wing surface must be:

  1. 220 m s⁻¹
  2. 200 m s⁻¹
  3. 210.5 m s⁻¹
  4. 190.8 m s⁻¹
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Answer: (D) 190.8 m s⁻¹

For level flight, lift force equals weight: ΔP × A = mg => ΔP = (4000 × 10) / 20 = 2000 Pa. By Bernoulli's equation, ΔP = (1/2)ρ(v₂² - v₁²). Thus, 2000 = (1/2)(1.0)(v₂² - 180²), which gives v₂² - 32400 = 4000, so v₂² = 36400, leading to v₂ = √36400 ≈ 190.8 m s⁻¹.

easy

7. Which of the following physical quantities is directly conserved in the derivation of Bernoulli's equation for fluid flow?

  1. Linear momentum of the fluid
  2. Mechanical energy of the fluid
  3. Angular momentum of the fluid
  4. Mass flow rate across streamlines
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Answer: (B) Mechanical energy of the fluid

Bernoulli's equation is essentially a statement of the work-energy theorem (conservation of total mechanical energy) applied to the steady streamline flow of an incompressible, non-viscous fluid.

easy

8. A liquid flows through a horizontal pipe of non-uniform diameter. At a point where the diameter is 6 cm, the fluid velocity is 1 m s⁻¹. What is the velocity at another point in the same pipe where the diameter is 2 cm?

  1. 9 m s⁻¹
  2. 6 m s⁻¹
  3. 3 m s⁻¹
  4. 12 m s⁻¹
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Answer: (A) 9 m s⁻¹

By the equation of continuity, A₁v₁ = A₂v₂. Therefore, v₂ = v₁ × (d₁ / d₂)² = 1 × (6 / 2)² = 1 × 3² = 9 m s⁻¹.

easy

9. An open water tank has a small hole located at depth h below the water surface. If the entire tank is placed inside an elevator accelerating downwards with an acceleration a (where a < g), the speed of efflux from the hole is given by:

  1. √(2gh)
  2. √(2(g + a)h)
  3. √(2(g² - a²)h / g)
  4. √(2(g - a)h)
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Answer: (D) √(2(g - a)h)

In an elevator accelerating downward with acceleration a, the effective acceleration due to gravity is g_eff = g - a. By Torricelli's law in an accelerated frame, the speed of efflux becomes v = √(2 g_eff h) = √(2(g - a)h).

easy

10. According to the equation of continuity, when water flows from a wide pipe into a narrow pipe, its speed:

  1. Becomes zero
  2. Increases
  3. Decreases
  4. Remains unchanged
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Answer: (B) Increases

For an incompressible fluid, Av = constant. A smaller cross-sectional area means a larger flow speed.

NCERT Class 11

easy

11. A large open tank has two small holes on its side wall: hole 1 at depth h₁ = 0.2 m and hole 2 at depth h₂ = 1.8 m below the liquid surface. The ratio of the speed of efflux from hole 1 to that from hole 2 (v₁ / v₂) is:

  1. 1 : 9
  2. 1 : 3
  3. 3 : 1
  4. 1 : √3
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Answer: (B) 1 : 3

According to Torricelli's law, efflux speed v = √(2gh). Therefore, v₁ / v₂ = √(h₁ / h₂) = √(0.2 / 1.8) = √(1/9) = 1/3.

easy

12. In the steady streamline flow of an incompressible, non-viscous fluid, the total mechanical energy per unit volume is constant throughout the fluid. The dimension of the ratio (Total Energy per unit volume) / (Total Energy per unit mass) is equivalent to that of:

  1. Density
  2. Velocity
  3. Pressure
  4. Acceleration
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Answer: (A) Density

(Energy per unit volume) / (Energy per unit mass) = (E/V) / (E/m) = m/V = Density (ρ). Its SI unit is kg m⁻³ and dimensional formula is [M L⁻³ T⁰].

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