Question 1
Post-UTME 2006A 70kg man ascends a flight of stairs of height 4m in 7s. The power expended by the man is;
- a.40W
- b.100W
- c.280W
- d.400W
Post-UTME Physics Past Questions
Real Post-UTME Physics past questions, each with the correct answer and a worked solution. Below are 8 to work through free — no account needed.
Sample drawn from papers of 2012, 2011, 2006.
8 Post-UTME Physics questions with answers
Question 2
Post-UTME 2006A 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?
- a.45cm
- b.30cm
- c.15cm
- d.20cm
Question 3
Post-UTME 2011An airplane increases its speed 36 km/h to 360 km/h in 20.0 s. How far does it travel while accelerating.
- a.4.4 km
- b.1.1 km
- c.2.3 km
- d.1.0 km
Question 4
Post-UTME 2012In order to remove the error of parallax when taking measurements with a metre rule, the eye should be focused
- a.slantingly towards the right on the markings
- b.slantingly towards the left on the markings
- c.vertically downwards on the markings
- 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 5
Post-UTME 2012A 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?
- a.1524N
- b.1350N
- c.1260N
- 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 6
Post-UTME 2012Calculate 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
- a.5500m
- b.4500m
- c.4200m
- 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 7
Post-UTME 2012A 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
- a.6km
- b.12km
- c.18km
- 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
Question 8
Post-UTME 2012A rocket is fired from the earth's surface to a distant planet. By Newton's law of universal gravitation, the force F will
- a.increase as r reduces
- b.increase as G varies
- c.remains constant
- d.increases as r increases
Solution
By Newton's law of universal gravitation, the gravitational force increases as the distance (r) reduces. This is because gravitational force is inversely proportional to the square of the distance between the objects (F ∝ 1/r²). Practical example: As the rocket approaches the distant planet, it experiences increasingly stronger gravitational pull from that planet. Conversely, as it moves away from Earth, Earth's gravitational pull weakens.
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