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WAEC Physics Past Questions

Real WAEC Physics past questions, each with the correct answer and a worked solution. NECO candidates sit the same paper, so these apply there too. Below are 8 to work through free — no account needed.

Sample drawn from papers of 2012, 2011, 2006.

8 WAEC Physics questions with answers

Question 1

Two identical small spheres X and Y carry charges of +4 μC and -6 μC respectively. They are brought into contact and then separated. What will be the final charge on sphere X?

  1. a.+1 μC
  2. b.-1 μC
  3. c.+5 μC
  4. d.-5 μC

Solution

When the spheres touch, the total charge (+4 μC and -6 μC = -2 μC) is shared equally since the spheres are identical, so each gets -1 μC. The most tempting wrong answer is '+1 μC', which results from adding the charges and forgetting to divide equally between the spheres.

Question 2

WAEC 2006

A 70kg man ascends a flight of stairs of height 4m in 7s. The power expended by the man is;

  1. a.40W
  2. b.100W
  3. c.280W
  4. d.400W

Question 3

WAEC 2006

A convex lens of focal length 10.0cm is used to form a real image which is half the size of the object. How far from the object is the image?

  1. a.45cm
  2. b.30cm
  3. c.15cm
  4. d.20cm

Question 4

WAEC 2011

An airplane increases its speed 36 km/h to 360 km/h in 20.0 s. How far does it travel while accelerating.

  1. a.4.4 km
  2. b.1.1 km
  3. c.2.3 km
  4. d.1.0 km

Question 5

WAEC 2012

In order to remove the error of parallax when taking measurements with a metre rule, the eye should be focused

  1. a.slantingly towards the right on the markings
  2. b.slantingly towards the left on the markings
  3. c.vertically downwards on the markings
  4. d.vertically upwards on the markings

Solution

To remove parallax error, the eye should be focused vertically downwards on the markings, ensuring the line of sight is perpendicular to the scale. This eliminates angular displacement and ensures accurate readings. Practical tip: When using a metre rule, always position your eye directly above (perpendicular to) the point you're measuring to avoid parallax error. Some instruments have mirrors behind the scale to help verify correct eye position - your eye and its reflection should align.

Question 6

WAEC 2012

A load is pulled at a uniform speed along a horizontal floor by a rope at 45° to the floor. If the force in the rope is 1500N, what is the frictional force on the load?

  1. a.1524N
  2. b.1350N
  3. c.1260N
  4. d.1061N

Solution

Given information: Force in rope (F) = 1500 N Angle to floor (θ) = 45° Load moves at uniform speed (constant velocity) Frictional force = ? Key principle: Since the load moves at uniform speed, the net force is zero (equilibrium). This means: Horizontal component of applied force = Frictional force Step 1: Resolve the rope force into components Horizontal component: F_horizontal = F cos θ F_horizontal = 1500 × cos 45° F_horizontal = 1500 × (1/√2) F_horizontal = 1500 × 0.7071 F_horizontal = 1060.7 N F_horizontal ≈ 1061 N Vertical component: F_vertical = F sin θ F_vertical = 1500 × sin 45° F_vertical = 1500 × (1/√2) F_vertical = 1060.7 N Step 2: Apply equilibrium condition For uniform speed (no acceleration): Horizontal forces must balance: Frictional force = Horizontal component of applied force f = F_horizontal = 1061 N

Question 7

WAEC 2012

Calculate the total distance covered by a train before coming to rest if its initial speed is 30ms⁻¹ with a constant retardation of 0.1ms⁻²

  1. a.5500m
  2. b.4500m
  3. c.4200m
  4. d.3000m

Solution

Given information: Initial velocity (u) = 30 m/s Final velocity (v) = 0 m/s (comes to rest) Retardation (deceleration, a) = -0.1 m/s² (negative because it's slowing down) Distance (s) = ? Using the equation of motion: v² = u² + 2as Substituting values: 0² = 30² + 2(-0.1)s 0 = 900 - 0.2s 0.2s = 900 s = 900/0.2 s = 4500 m

Question 8

WAEC 2012

A car starts from rest and moves with a uniform acceleration of 30ms⁻² for 20s. Calculate the distance covered at the end of the motion

  1. a.6km
  2. b.12km
  3. c.18km
  4. d.24km

Solution

Given information: Initial velocity (u) = 0 m/s (starts from rest) Acceleration (a) = 30 m/s² Time (t) = 20 s Distance (s) = ? Using equation of motion: s = ut + ½at² Substituting values: s = 0(20) + ½(30)(20)² s = 0 + ½(30)(400) s = ½ × 12,000 s = 6,000 m s = 6 km

Practise the full set under exam conditions

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Other WAEC subjects

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