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Quiz Chapter 2: Pressure

10 questions · Form 5 Physics Bab 2: Pressure

10 questions · plays in fullscreen

Full Question List & Answer Key

Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.

1. An aerofoil wing creates lift because:

  1. Air moves faster over the top surface, creating lower pressure above the wing than below it.
  2. Air moves slower over the top surface, creating higher pressure above the wing than below it.
  3. Air temperature is higher beneath the wing than above it.
  4. The wing displaces air equal to its own total mass.
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Answer: A

According to Bernoulli's Principle, faster air speed over the top curved surface produces a lower pressure region, yielding an upward net force (lift).

2. In a hydraulic jack, a force of 40 N is applied to a small input piston of area 0.02 m². What force is exerted by the large output piston of area 0.5 m²?

  1. 1000 N
  2. 100 N
  3. 1.6 N
  4. 2000 N
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Answer: A

By Pascal's Principle, F1 / A1 = F2 / A2 => 400.02 = F2 / 0.5 => 2000 = F2 / 0.5 => F2 = 1000 N.

3. A diver descends to a depth of 15 m in seawater of density 1025 kg m⁻³. What is the liquid pressure experienced by the diver due to seawater alone? (Take g = 9.81 m s⁻²)

  1. 150,922 Pa
  2. 252,222 Pa
  3. 15,375 Pa
  4. 100,562 Pa
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Answer: A

Liquid pressure P = h ρ g = 15 × 1025 × 9.81 = 150,922.5 Pa ≈ 150,922 Pa.

4. In a Venturi tube through which water flows horizontally, at which section is the fluid pressure the LOWEST?

  1. At the widest section where speed is highest
  2. At the narrowest constricted section where speed is highest
  3. At the inlet section before any constriction
  4. The pressure is uniform throughout the tube
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Answer: B

In a narrow constriction, fluid velocity increases to maintain mass flow rate. By Bernoulli's Principle, higher velocity leads to lower pressure.

5. A solid block of volume 0.004 m³ is fully submerged in freshwater (density = 1000 kg m⁻³). What is the buoyant force acting on the block? (g = 9.81 m s⁻²)

  1. 39.24 N
  2. 4.00 N
  3. 3.92 N
  4. 392.40 N
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Answer: A

Buoyant force FB = ρ V g = 1000 × 0.004 × 9.81 = 39.24 N.

6. An open U-tube manometer contains water (density = 1000 kg m⁻³). When connected to a gas supply, the water level on the open arm is 20 cm higher than the arm connected to the gas. Taking Patm = 101,000 Pa and g = 9.81 m s⁻², calculate the gas pressure.

  1. 99,038 Pa
  2. 102,962 Pa
  3. 120,620 Pa
  4. 101,962 Pa
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Answer: B

Since open arm level is higher, Pgas = Patm + h ρ g = 101,000 + (0.20 × 1000 × 9.81) = 101,000 + 1,962 = 102,962 Pa.

7. If a hydraulic brake pedal force of 50 N produces a fluid pressure of 250 kPa, what is the area of the master cylinder piston?

  1. 0.0002 m²
  2. 0.005 m²
  3. 5.0 m²
  4. 0.20 m²
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Answer: A

P = FA => 250,000 Pa = 50 N / A => A = 50250,000 = 0.0002 m².

8. What happens to the atmospheric pressure as altitude above sea level increases?

  1. Increases exponentially
  2. Decreases because the air column above is shorter and air density decreases
  3. Remains constant at 101 kPa
  4. Decreases initially then rapidly increases in the stratosphere
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Answer: B

As altitude increases, the air column overhead becomes shorter and air density diminishes, resulting in lower atmospheric pressure.

9. A wooden block floats in water with 60% of its volume submerged. What is the density of the wood? (Density of water = 1000 kg m⁻³)

  1. 600 kg m⁻³
  2. 400 kg m⁻³
  3. 1000 kg m⁻³
  4. 1200 kg m⁻³
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Answer: A

Floating condition: Weight = Buoyant Force => ρwood × Vtotal × g = ρwater × (0.60 Vtotal) × g => ρwood = 0.60 × 1000 = 600 kg m⁻³.

10. A mercury barometer reads a height of 76 cmHg. What is this atmospheric pressure value in Pascals? (Density of mercury = 13,600 kg m⁻³, g = 9.81 m s⁻²)

  1. 101,396 Pa
  2. 76,000 Pa
  3. 133,416 Pa
  4. 10,140 Pa
Show answer

Answer: A

P = h ρ g = 0.76 m × 13,600 kg m⁻³ × 9.81 m s⁻² = 101,396.16 Pa ≈ 101,396 Pa.

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