IGCSE Physics Physical Quantities and Measurement Techniques Questions and Worked Solutions (1.1) | 0625

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IGCSE Physics 0625 · Topic 1.1

IGCSE Physics Physical Quantities and Measurement Techniques Questions and Worked Solutions

50 topic-wise practice questions on Physical Quantities and Measurement Techniques (Unit 1: Motion, Forces and Energy), written to the Cambridge IGCSE Physics 0625 syllabus for 2026–2028, each with a step-by-step worked solution.

50 questions22 Core18 Extended10 Challenge129 marksTextbook §1.1 · pp. 2–11
Questions tagged Core are for all candidates (Papers 1 & 3). Questions tagged Extended are Supplement content (Papers 2 & 4). Questions 41–50 are harder Challenge questions. Try each question first, then tap Show solution to check your working.

Topic guide: Physical Quantities and Measurement Techniques

Physics is built on measurement. This topic covers how to measure length, volume and time accurately, how to reduce the effect of errors by measuring multiples, and (Extended) the difference between scalar and vector quantities and how to combine perpendicular vectors.

Key formulas

average = sum of readings ÷ number of readingsused to reduce random errorCore
value of one = total measurement ÷ number of itemse.g. thickness of one sheet, period of one oscillationCore
resultant R = √(A² + B²)two vectors at right angles (N, m/s, m)Extended
tan θ = B ÷ Adirection of the resultant of two perpendicular vectorsExtended

Prerequisite knowledge

  • SI units and prefixes (milli, centi, kilo, mega)
  • Reading scales and stopwatches
  • Basic Pythagoras and trigonometry (Extended)

Exam tips & common mistakes

  • Always subtract the starting reading if a stopwatch or cylinder was not at zero
  • State the instrument's smallest division when asked about precision
  • For vectors, give both size and direction (e.g. 50 N at 37° north of east)

Key facts from the syllabus (1.1)

  • Rulers measure lengths of a few mm to 1 m; measuring cylinders measure liquid volumes; clocks and digital timers measure time
  • Read a measuring cylinder at eye level at the bottom of the meniscus to avoid parallax error
  • Measure several oscillations or a stack of objects and divide, to reduce the effect of reaction time and instrument precision
  • Scalars have magnitude only: distance, speed, time, mass, energy, temperature
  • Vectors have magnitude and direction: force, weight, velocity, acceleration, momentum, electric field strength, gravitational field strength
  • A resultant of two perpendicular vectors can be found by calculation or by a scale drawing

Textbook: §1.1 Physical quantities and measurement techniques, pp. 2–11

1
Core[1 mark]

State the most suitable instrument for measuring the length of a laboratory bench of about 2 m.

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Method: A long length with millimetre precision needs a long scale.

A metre rule (or measuring tape)
2
Core[2 marks]

A measuring cylinder contains water. Describe how a student should read the volume accurately.

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Method: The water surface curves (meniscus). Read at eye level to avoid parallax error.

Place the cylinder on a flat surface, keep the eye level with the meniscus and read the bottom of the meniscus
3
Core[2 marks]

A measuring cylinder reads 46 cm³. A stone is lowered in and the reading rises to 71 cm³. Calculate the volume of the stone.

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Method: Volume of stone = final reading − initial reading.

V = 71 − 46 = 25 cm³
4
Core[2 marks]

A stack of 50 identical sheets of paper is 4.5 mm thick. Calculate the thickness of one sheet.

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Method: Measuring a multiple and dividing reduces the effect of measurement uncertainty.

Thickness = 4.5 ÷ 50 = 0.090 mm
5
Core[2 marks]

A student times 20 oscillations of a pendulum as 32.6 s. Calculate the period of the pendulum.

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Method: Period = total time ÷ number of oscillations.

T = 32.6 ÷ 20 = 1.63 s
6
Core[2 marks]

Explain why timing 20 oscillations gives a more accurate period than timing one oscillation.

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Method: Human reaction time (about 0.2–0.3 s) is a fixed error. Spread over 20 oscillations, the error in each period is about 20 times smaller.

The reaction-time error becomes a much smaller fraction of the measured time
7
Core[3 marks]

The initial stopwatch reading is 2.40 s. The student starts the stopwatch and stops it after 25 oscillations; the final reading is 44.90 s. Calculate the period.

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Method: The stopwatch was not reset, so subtract the initial reading first.

Time for 25 = 44.90 − 2.40 = 42.50 s; T = 42.50 ÷ 25 = 1.70 s
8
Core[1 mark]

State one advantage of using a digital timer instead of a hand-operated stopwatch for timing a falling ball.

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Method: A digital timer triggered by light gates removes human reaction time.

It removes reaction-time error (more accurate/precise)
9
Core[2 marks]

40 identical coins in a stack have a height of 6.8 cm. Calculate the thickness of one coin in mm.

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Method: 6.8 cm = 68 mm. Divide by the number of coins.

Thickness = 68 ÷ 40 = 1.70 mm
10
Core[2 marks]

Describe how to find the diameter of a thin wire using a pencil and a ruler.

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Method: Wind the wire closely around a pencil for N turns, measure the length L of the coil with the ruler, then divide.

Diameter = length of coil ÷ number of turns (e.g. 20 turns)
11
Core[1 mark]

Name an instrument used to measure time intervals in the laboratory.

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Method: Timing devices.

A stopwatch (or digital timer)
12
Core[1 mark]

Name an instrument used to measure the volume of a liquid.

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Method: Liquid volume.

A measuring cylinder
13
Core[2 marks]

A ruler is used to measure a pencil. One end is at 2.0 cm and the other at 15.4 cm. Calculate the length of the pencil.

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Method: Subtract the start reading.

L = 15.4 − 2.0 = 13.4 cm
14
Core[2 marks]

Explain why a ruler should be read with the eye directly above the mark.

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Method: Parallax.

To avoid parallax error, which gives an incorrect reading
15
Core[2 marks]

A pile of 25 identical tiles is 37.5 cm high. Calculate the thickness of one tile.

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Method: Divide by number of tiles.

t = 37.5 ÷ 25 = 1.50 cm
16
Core[2 marks]

A measuring cylinder contains 60 cm³ of water. Ten identical marbles are added and the level rises to 85 cm³. Calculate the volume of one marble.

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Method: Total volume ÷ 10.

V = (85 − 60) ÷ 10 = 2.50 cm³
17
Core[2 marks]

A runner's time for 100 m is shown on a stopwatch as 00:12.48. State the time in seconds and the precision of the stopwatch.

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Method: Read display.

12.48 s; precision 0.01 s
18
Core[3 marks]

A student measures the time for one swing of a pendulum five times: 1.4 s, 1.6 s, 1.5 s, 1.5 s, 1.5 s. Calculate the average period.

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Method: Add and divide by 5.

Average = 7.5 ÷ 5 = 1.50 s
19
Core[2 marks]

A metre rule has 1 mm divisions. State the smallest length it can measure precisely and give one length that would be better measured with a different instrument.

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Method: Precision of rule.

1 mm; e.g. the thickness of a wire (use a micrometer) or a long corridor (use a tape)
20
Core[2 marks]

Describe how to measure the period of a pendulum as accurately as possible with a stopwatch.

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Method: Many oscillations, fiducial mark.

Time 20 oscillations, starting and stopping as the bob passes the centre; repeat and average; divide by 20
21
Core[3 marks]

A ball of plasticine is dropped into a measuring cylinder containing 32.0 cm³ of water. The reading becomes 45.5 cm³. Calculate the volume of the plasticine and state one precaution.

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Method: Displacement.

V = 45.5 − 32.0 = 13.5 cm³; read at eye level at the bottom of the meniscus (or lower the object gently to avoid splashing)
22
Core[1 mark]

State the SI unit of length.

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Method: Base unit.

metre (m)
23
Extended[1 mark]

State the difference between a scalar quantity and a vector quantity.

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Method: Scalars have size only; vectors have size and direction.

Scalar: magnitude only. Vector: magnitude and direction
24
Extended[2 marks]

From the list, identify the vectors: mass, weight, speed, velocity, energy, momentum.

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Method: Vectors have a direction.

Weight, velocity, momentum
25
Extended[2 marks]

State two scalar quantities from the 0625 syllabus list.

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Method: Scalars: distance, speed, time, mass, energy, temperature.

Any two of: distance, speed, time, mass, energy, temperature
26
Extended[3 marks]

Two forces of 6.0 N east and 8.0 N north act on an object. Calculate the size of the resultant force.

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Method: Forces at right angles: use Pythagoras, R = √(6.0² + 8.0²).

R = √(36 + 64) = 10.0 N
27
Extended[3 marks]

For the forces in the previous question (6.0 N east, 8.0 N north), calculate the angle between the resultant and the east direction.

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Method: tan θ = 8.0 ÷ 6.0.

θ = tan⁻¹(8.0/6.0) = 53.1° north of east
28
Extended[3 marks]

A boat moves at 3.0 m/s across a river that flows at 4.0 m/s at right angles to the boat's direction. Calculate the boat's resultant speed.

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Method: Velocities at right angles: v = √(3.0² + 4.0²).

v = 5.00 m/s
29
Extended[3 marks]

A plane flies north at 120 m/s. A wind blows east at 35 m/s. Calculate the resultant velocity of the plane (magnitude).

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Method: v = √(120² + 35²).

v = √(14400 + 1225) = 125 m/s
30
Extended[2 marks]

Explain why weight is a vector quantity but mass is a scalar quantity.

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Method: Weight is a force, acting in a direction (towards the centre of the Earth). Mass is an amount of matter with no direction.

Weight is a force with a direction (downwards); mass has magnitude only
31
Extended[3 marks]

Two forces of 5.0 N and 12.0 N act at right angles on a point. Calculate the resultant force.

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Method: R = √(5.0² + 12.0²).

R = √(25 + 144) = 13.0 N
32
Extended[2 marks]

A resultant of 10 N is made of two perpendicular forces. One force is 6.0 N. Calculate the other force.

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Method: 10² = 6.0² + F².

F = √(100 − 36) = 8.00 N
33
Extended[2 marks]

Explain why distance is a scalar and displacement is a vector.

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Method: Direction.

Distance has magnitude only; displacement has magnitude and a direction
34
Extended[2 marks]

Classify each as scalar or vector: acceleration, temperature, force, time.

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Method: Direction needed?

Vectors: acceleration, force. Scalars: temperature, time
35
Extended[3 marks]

A walker goes 400 m east then 300 m north. Calculate the magnitude of the displacement.

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Method: Pythagoras.

s = √(400² + 300²) = 500 m
36
Extended[3 marks]

For the walker in the previous question, calculate the distance walked and compare it with the displacement.

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Method: Add distances.

Distance = 700 m; displacement = 500 m (smaller because it is the straight-line separation)
37
Extended[3 marks]

A force of 9.0 N acts north and 12 N acts west. Calculate the resultant and its angle west of north.

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Method: Pythagoras and tan.

R = 15.0 N; θ = tan⁻¹(12/9.0) = 53.1° west of north
38
Extended[3 marks]

A swimmer swims at 1.5 m/s across a river flowing at 0.80 m/s. Calculate the swimmer's resultant speed.

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Method: Perpendicular velocities.

v = √(1.5² + 0.80²) = 1.70 m/s
39
Extended[3 marks]

Two tugs pull a ship with forces of 20 kN and 15 kN at right angles. Calculate the resultant force on the ship.

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Method: Pythagoras.

R = √(20² + 15²) = 25.0 kN
40
Extended[4 marks]

Describe how to find the resultant of two perpendicular forces of 30 N and 40 N by scale drawing.

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Method: Choose a scale, draw tip to tail.

Choose a scale (e.g. 1 cm = 10 N); draw 3.0 cm and 4.0 cm arrows at right angles, tip to tail; measure the line from start to finish (5.0 cm = 50 N) and its angle with a protractor (about 53° from the 30 N force)

Challenge questions

Harder, multi-step questions at the top end of the exam. Attempt these once you are confident with the questions above.

41
ChallengeCore[4 marks]

A student times 25 oscillations of a pendulum. The stopwatch reads 3.6 s at the start (it was not reset) and 41.2 s at the end. The student's reaction time is about 0.2 s. (a) Calculate the period. (b) Estimate the uncertainty in the period.

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Method: (a) Subtract the initial reading, then divide by 25. (b) Reaction time affects start and stop: about ±0.4 s over 25 oscillations.

(a) T = (37.6) ÷ 25 = 1.50 s (b) ±0.4 ÷ 25 ≈ ±0.016 s
42
ChallengeCore[4 marks]

100 sheets of paper make a stack 1.05 cm thick, measured with a ruler to ±0.05 cm. (a) Calculate the thickness of one sheet in mm. (b) Explain why measuring one sheet directly with the ruler would be useless.

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Method: (a) 10.5 mm ÷ 100. (b) One sheet is far smaller than the ruler's smallest division.

(a) 0.105 mm (b) 0.1 mm is much smaller than the 1 mm divisions, so the uncertainty would be larger than the measurement
43
ChallengeCore[4 marks]

A measuring cylinder holds 50.0 cm³ of water. 20 identical ball bearings are added, raising the level to 57.6 cm³. The total mass of the bearings is 59.6 g. Calculate (a) the volume of one bearing, (b) the density of the steel.

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Method: (a) (57.6 − 50.0) ÷ 20. (b) Total mass ÷ total volume.

(a) 0.380 cm³ (b) ρ = 59.6 ÷ 7.6 = 7.84 g/cm³
44
ChallengeCore[3 marks]

A thin copper wire is wound tightly around a pencil for 40 turns. The coil measures 18.0 mm. Calculate the diameter of the wire and suggest why the result may be slightly too large.

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Method: Coil length ÷ turns.

d = 18.0 ÷ 40 = 0.450 mm; small gaps between turns make the coil longer than the true total
45
ChallengeCore[4 marks]

A stopwatch measures the time for a ball to roll 2.00 m as 1.62 s, 1.55 s, 1.70 s, 1.59 s. (a) Calculate the mean time. (b) Calculate the average speed. (c) Suggest why the readings differ.

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Method: Mean of four; v = d/t.

(a) 1.61 s (b) v = 2.00 ÷ 1.61 = 1.24 m/s (c) Human reaction time varies
46
ChallengeExtended[4 marks]

A boat heads due north at 4.0 m/s across a river flowing east at 3.0 m/s. The river is 60 m wide. Calculate (a) the resultant speed, (b) the time to cross, (c) how far downstream the boat lands.

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Method: (a) Pythagoras. (b) Only the northward velocity crosses the river. (c) East speed × time.

(a) 5.00 m/s (b) t = 60 ÷ 4.0 = 15.0 s (c) 3.0 × 15 = 45.0 m
47
ChallengeExtended[4 marks]

A pilot wants to fly due north. The wind blows from west to east at 30 m/s. The plane's airspeed is 150 m/s. Calculate the angle west of north the pilot must head, and the resulting ground speed.

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Method: sin θ = 30/150; ground speed = √(150² − 30²).

θ = sin⁻¹(0.20) = 11.5° west of north; v = 147 m/s
48
ChallengeExtended[4 marks]

Two forces act at right angles on an object: 24 N east and an unknown force north. The resultant is 26 N. Calculate the unknown force and the angle of the resultant from east.

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Method: Pythagoras then tan.

F = √(26² − 24²) = 10.0 N; θ = tan⁻¹(10/24) = 22.6° north of east
49
ChallengeExtended[5 marks]

A hiker walks 3.0 km north, then 4.0 km east, then 3.0 km south. (a) Calculate the total distance. (b) Calculate the displacement. (c) The trip took 2.5 h; calculate the average speed and average velocity.

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Method: Distance adds; displacement: north and south cancel.

(a) 10 km (b) 4.0 km east (c) speed = 4.0 km/h; velocity = 1.6 km/h east
50
ChallengeExtended[4 marks]

A sign of weight 200 N hangs from two ropes at right angles to each other, each at 45° to the vertical. By drawing a vector triangle (or calculation), find the tension in each rope.

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Method: Two equal perpendicular tensions whose resultant is 200 N upward: T√2 = 200.

T = 200 ÷ √2 = 141 N each
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