Physics questions, chapter 2

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Pressure

45 marks. Use Practice to check each answer as you go, or Exam to answer everything against the clock first. Take g = 9.81 m s⁻² unless told otherwise.

Section A: objective questions

14 marks

Choose the best answer for each question.

Question 1

Which three factors affect the pressure in a liquid?

Mass, volume and depth
Depth, density and gravitational acceleration
Depth, volume and surface area
Density, mass and shape of the container

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Liquid pressure P = hρg depends on the depth, the density of the liquid and g. Mass, volume, surface area and container shape have no effect.

For the teacher: Remembering, low difficulty, standard 2.1

Textbook p. 41

Question 2

A wide tank and a narrow tube are both filled with water to a depth of 0.5 m. Compare the pressure at their bases.

Higher in the wide tank because it holds more water
Higher in the narrow tube because the water is concentrated
Higher in the wide tank because it has a larger base area
The same in both, because only depth and density matter

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P = hρg. The cross-sectional area and shape do not affect liquid pressure (Activity 2.2).

For the teacher: Understanding, low difficulty, standard 2.1

Textbook p. 44

Question 3

What is the pressure due to the water at a depth of 2.5 m in a pool? [Density of water = 1 000 kg m⁻³]

2.5 kPa
24.5 kPa
245 kPa
2 450 kPa

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P = hρg = 2.5 × 1 000 × 9.81 = 24 525 Pa ≈ 24.5 kPa.

For the teacher: Applying, low difficulty, standard 2.1

Textbook p. 45

Question 4

In a U-tube, a column of oil 15.0 cm high balances a column of water 12.0 cm high (both measured from the same boundary level). What is the density of the oil? [Density of water = 1 000 kg m⁻³]

800 kg m⁻³
1 000 kg m⁻³
1 250 kg m⁻³
1 800 kg m⁻³

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h₁ρ₁ = h₂ρ₂ gives 15.0 × ρ = 12.0 × 1 000, so ρ = 800 kg m⁻³.

For the teacher: Applying, medium difficulty, standard 2.1

Textbook p. 46

Question 5

Why is the wall of a dam built thicker at the base?

The water at the base is warmer
The water at the base has a larger mass
Water pressure increases with depth
Atmospheric pressure is greater at the base

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Pressure increases with depth (P = hρg), so the lower part of the wall must withstand a greater force.

For the teacher: Understanding, low difficulty, standard 2.1

Textbook p. 47

Question 6

A student blows across the top surface of a piece of paper held at her lips and the paper rises. Why does this happen?

Blowing air pushes the paper upwards directly
The air below the paper is moving faster than the air above it
The air above the paper moves faster, so the pressure above is lower than below
The weight of the paper decreases when air is blown

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By Bernoulli's principle, fast air has lower pressure. The higher pressure below pushes the paper up.

For the teacher: Understanding, low difficulty, standard 2.6

Textbook p. 79

Question 7

A mercury barometer is carried from sea level to the top of a mountain. What happens to the height of the mercury column?

It decreases because atmospheric pressure is lower
It increases because the mercury is lighter
It stays the same because mercury is dense
It increases because gravitational pull is less

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Air is thinner at high altitude, so the atmospheric pressure supporting the column is lower and the column is shorter.

For the teacher: Understanding, low difficulty, standard 2.2

Textbook p. 51

Question 8

A pressure of 0.50 m Hg equals how many pascals? [Density of mercury = 13 600 kg m⁻³]

6.67 × 10²
6.67 × 10³
1.36 × 10⁴
6.67 × 10⁴

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P = hρg = 0.50 × 13 600 × 9.81 = 66 708 Pa ≈ 6.67 × 10⁴ Pa.

For the teacher: Applying, low difficulty, standard 2.2

Textbook p. 53

Question 9

A water manometer has one arm connected to a gas supply and the other arm open to the air. The water level in the arm connected to the gas is 0.20 m lower than in the open arm. Atmospheric pressure is 10.3 m H₂O. What is the gas pressure?

10.1 m H₂O
10.3 m H₂O
10.5 m H₂O
20.6 m H₂O

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The gas pushes its own level down, so the gas pressure is greater than atmospheric pressure: 10.3 + 0.20 = 10.5 m H₂O.

For the teacher: Applying, medium difficulty, standard 2.3

Textbook p. 56

Question 10

In a hydraulic system, a force of 20 N acts on an input piston of area 2.0 cm². What force acts on an output piston of area 40 cm²?

1.0 N
400 N
800 N
4 000 N

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F₂ = (A₂ ÷ A₁) × F₁ = (40 ÷ 2.0) × 20 = 400 N.

For the teacher: Applying, low difficulty, standard 2.4

Textbook p. 62

Question 11

Which property must a hydraulic fluid have?

It is easily compressed
It has a low boiling point
It evaporates easily
It cannot be compressed

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An incompressible fluid transmits the pressure fully. A compressible fluid (or one with air bubbles) wastes the movement of the input piston.

For the teacher: Remembering, low difficulty, standard 2.4

Textbook p. 63

Question 12

A solid object of volume 2.0 × 10⁻³ m³ is fully immersed in water. What is the buoyant force on it? [Density of water = 1 000 kg m⁻³]

2.0 N
9.81 N
19.6 N
196 N

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F_B = ρVg = 1 000 × 2.0 × 10⁻³ × 9.81 = 19.6 N.

For the teacher: Applying, low difficulty, standard 2.5

Textbook p. 67

Question 13

A loaded ship sails from the sea into a river where the water is less dense. What happens to the ship?

It sinks lower, because it must displace a larger volume of water to get the same buoyant force
It rises, because the buoyant force is greater in the river
It stays at the same level, because its weight is unchanged
It sinks lower, because its weight increases in the river

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The ship floats, so F_B = W, which does not change. With a smaller ρ, the displaced volume V must be larger (ρVg = W), so the ship sits lower.

For the teacher: Analysing, high difficulty, standard 2.5

Textbook p. 71

Question 14

Diagram 1 shows liquid X and water drawn up into two tubes from beakers, after some air is sucked out of the top. Which statement is correct?

Two tubes joined at the top; the left holds liquid X to 36.0 cm and the right holds water to 35.0 cm — Diagram 1

Liquid X has a density greater than water
Liquid X has a density less than water
Liquid X has the same density as water
The densities cannot be compared

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The air pressure above both columns is the same and atmospheric pressure pushes both up, so h_X ρ_X = h_w ρ_w. X is higher (36.0 cm against 35.0 cm), so ρ_X is smaller: ρ_X = 35.0 × 1 000 ÷ 36.0 = 972 kg m⁻³.

For the teacher: Analysing, medium difficulty, standard 2.1

Textbook p. 86

Section B: structured questions

31 marks

Answer all questions. Show your working.

Question 15

A diver dives to a depth of 35 m in the sea. The density of sea water is 1 060 kg m⁻³ and the atmospheric pressure is 100 kPa.

[2 marks] (a)Calculate the pressure on the diver due to the sea water.

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P = hρg = 35 × 1 060 × 9.81 [1]1
= 3.64 × 10⁵ Pa (364 kPa) [1]1

Tick each point you got.

[1 mark] (b)Calculate the actual pressure acting on the diver.

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364 + 100 = 464 kPa (4.64 × 10⁵ Pa)1

Tick each point you got.

[2 marks] (c)A diver must rise to the surface slowly. Explain why.

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At depth, high pressure makes more nitrogen dissolve in the blood and body tissues [1]1
If the pressure falls too fast, nitrogen forms bubbles in the tissues and blood vessels [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 2.2

Textbook p. 55

Question 16

Diagram 2 shows a mercury manometer connected to a flask of compressed gas. The atmospheric pressure is 76 cm Hg and the density of mercury is 13 600 kg m⁻³.

Mercury manometer connected to a flask of gas; the mercury levels differ by 40 cm — Diagram 2

[1 mark] (a)What is the gas pressure in cm Hg?

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76 + 40 = 116 cm Hg1

Tick each point you got.

[2 marks] (b)Convert the gas pressure to pascals.

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P = hρg = 1.16 × 13 600 × 9.81 [1]1
= 1.55 × 10⁵ Pa [1]1

Tick each point you got.

[2 marks] (c)Why is mercury, and not water, used for measuring this pressure?

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Mercury is much denser than water [1]1
so for the same pressure the column is much shorter (a water column would be about 13.6 times higher) [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 2.3

Textbook p. 58

Question 17

Diagram 4 shows a hydraulic system. The input piston has an area of 4.0 cm² and the output piston has an area of 100 cm². A force of 30 N acts on the input piston.

Hydraulic system with a small input piston and a large output piston — Diagram 4

[1 mark] (a)State the principle applied in this system.

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Pascal's principle: pressure applied to an enclosed fluid is transmitted uniformly in all directions1

Tick each point you got.

[1 mark] (b)Calculate the multiplying factor.

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100 ÷ 4.0 = 251

Tick each point you got.

[1 mark] (c)Calculate the output force.

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25 × 30 = 750 N1

Tick each point you got.

[2 marks] (d)Calculate the pressure transmitted through the fluid, in pascals.

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P = F ÷ A = 30 ÷ (4.0 × 10⁻⁴) [1]1
= 7.5 × 10⁴ Pa [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 2.4

Textbook p. 62

Question 18

A wooden block of mass 0.60 kg and volume 8.0 × 10⁻⁴ m³ is held fully under water and then released. The density of water is 1 000 kg m⁻³.

[2 marks] (a)Calculate the buoyant force on the block when it is fully immersed.

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F_B = ρVg = 1 000 × 8.0 × 10⁻⁴ × 9.81 [1]1
= 7.85 N [1]1

Tick each point you got.

[1 mark] (b)Calculate the weight of the block.

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W = mg = 0.60 × 9.81 = 5.89 N1

Tick each point you got.

[2 marks] (c)Describe the motion of the block immediately after it is released. Give a reason.

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It moves upwards (with an acceleration) [1]1
Buoyant force (7.85 N) is greater than the weight (5.89 N), so the resultant force is upwards [1]1

Tick each point you got.

[2 marks] (d)The block comes to rest floating on the surface. Calculate the volume of the block that is below the water.

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Floating: buoyant force = weight, so ρVg = mg and V = m ÷ ρ [1]1
V = 0.60 ÷ 1 000 = 6.0 × 10⁻⁴ m³ [1]1

Tick each point you got.

For the teacher: Evaluating, medium difficulty, standard 2.5

Textbook p. 77

Question 19

An aeroplane is taking off. The air flows over its aerofoil-shaped wings.

[1 mark] (a)Compare the speed of the air flowing above and below the wing.

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The air above the wing flows faster than the air below it1

Tick each point you got.

[2 marks] (b)Compare the air pressure above and below the wing and state the direction of the resultant force.

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Pressure above is lower than the pressure below [1]1
The resultant force (lift force) acts upwards [1]1

Tick each point you got.

[1 mark] (c)State the principle that explains your answers in (b).

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Bernoulli's principle1

Tick each point you got.

[2 marks] (d)In a Venturi tube, which section of the tube has the lowest pressure? Explain.

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The narrowest section [1]1
Water flows fastest there, so by Bernoulli's principle the pressure is lowest [1]1

Tick each point you got.

For the teacher: Analysing, medium difficulty, standard 2.6

Textbook p. 82

Question 20

A driver finds that the hydraulic brake pedal of a car feels soft and has to be pushed much harder and further to stop the car. A mechanic finds air bubbles in the brake fluid.

[1 mark] (a)Explain why the air bubbles make the brakes less effective.

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Air is compressible, so part of the pressure and movement is used to compress the bubbles and less force reaches the brake pads1

Tick each point you got.

[2 marks] (b)Suggest two changes to the design of the braking system so that the car can be stopped with a greater braking force.

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Any 2 points

Use a larger cross-sectional area for the slave (wheel) cylinder compared to the master cylinder [1]1
Use a longer brake pedal bar (a bigger lever) to multiply the input force [1]1
Use a brake fluid that cannot be compressed, with a high boiling point [1]1

Tick each point you got.

For the teacher: Creating, high difficulty, standard 2.4

Textbook p. 89

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