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100+ Free BGCSE Physics (0571) Practice Questions

Botswana General Certificate of Secondary Education Physics (Syllabus 0571) practice questions are available now; exam metadata is being verified.

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Key Facts: BGCSE Physics (0571) Exam

0571

BEC Syllabus Code

Botswana Examinations Council

40

Paper 1 MCQ Items

BEC Physics Syllabus 0571

45 min

Paper 1 Time Allowed

BEC Physics Syllabus 0571

Grade C

Credit Pass Benchmark

BEC Grading Guidelines

5

Core Content Domains

BEC Syllabus Framework

g = 10

Standard Gravity (N/kg)

BEC Physics Formula Standard

BGCSE Physics 0571 is the official Botswana national senior secondary physics qualification administered by the Botswana Examinations Council (BEC). Paper 1 consists of 40 compulsory multiple-choice questions testing core concepts and numerical calculations.

Sample BGCSE Physics (0571) Practice Questions

Try these sample questions to test your BGCSE Physics (0571) exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Which measuring instrument is best suited to measure the internal diameter of a test tube with high precision?
A.Metre rule
B.Vernier calipers (internal jaws)
C.Micrometer screw gauge
D.Measuring tape
Explanation: Vernier calipers have internal measuring jaws specifically designed to measure the inside diameters of hollow objects with a precision of 0.01 cm (0.1 mm). A metre rule lacks precision and internal jaws, while a micrometer screw gauge is designed for external dimensions such as wire thickness.
2A pendulum completes 20 full oscillations in 40 seconds. What is the period of the pendulum?
A.0.5 s
B.2.0 s
C.20 s
D.800 s
Explanation: The period T is the time taken for one complete oscillation. T = Total Time / Number of Oscillations = 40 s / 20 = 2.0 s.
3A car accelerates uniformly from rest to a speed of 25 m/s in 5.0 seconds. What is its acceleration?
A.0.2 m/s²
B.5.0 m/s²
C.20 m/s²
D.125 m/s²
Explanation: Acceleration is defined as a = (v - u) / t. Since the car starts from rest, u = 0 m/s, v = 25 m/s, and t = 5.0 s, giving a = (25 - 0) / 5.0 = 5.0 m/s².
4What quantity is represented by the area under a speed-time graph?
A.Acceleration
B.Total distance traveled
C.Average speed
D.Net force applied
Explanation: The area under a speed-time graph represents the total distance traveled by the object. In contrast, the gradient (slope) of a speed-time graph represents acceleration.
5A stone of mass 0.50 kg falls freely near the Earth's surface. Taking gravitational acceleration g = 10 N/kg, what is the weight of the stone?
A.0.05 N
B.2.0 N
C.5.0 N
D.50 N
Explanation: Weight W is calculated using W = mg. For mass m = 0.50 kg and g = 10 N/kg, W = 0.50 kg × 10 N/kg = 5.0 N.
6A rectangular block has dimensions 5 cm × 4 cm × 2 cm and a mass of 160 g. What is the density of the material?
A.0.25 g/cm³
B.4.0 g/cm³
C.40 g/cm³
D.6400 g/cm³
Explanation: Volume V = length × width × height = 5 cm × 4 cm × 2 cm = 40 cm³. Density ρ = mass / volume = 160 g / 40 cm³ = 4.0 g/cm³.
7Which statement correctly describes the mass and weight of an astronaut traveling from Earth to the Moon?
A.Both mass and weight decrease on the Moon.
B.Mass remains constant, but weight decreases on the Moon.
C.Weight remains constant, but mass decreases on the Moon.
D.Both mass and weight remain constant.
Explanation: Mass is the amount of matter in an object and is an intrinsic property that does not change with location. Weight is the gravitational force acting on an object (W = mg), which decreases on the Moon because lunar gravity is weaker than Earth's gravity.
8A net force of 12 N acts on a mass of 3.0 kg. What acceleration is produced?
A.0.25 m/s²
B.4.0 m/s²
C.9.0 m/s²
D.36 m/s²
Explanation: According to Newton's second law, F = ma. Rearranging for acceleration gives a = F / m = 12 N / 3.0 kg = 4.0 m/s².
9A spring extends by 3.0 cm when a load of 6.0 N is hung from it. Assuming Hooke's law is obeyed, what load causes an extension of 7.5 cm?
A.9.0 N
B.12 N
C.15 N
D.18 N
Explanation: Hooke's law states F = kx, where k is the spring constant. k = F / x = 6.0 N / 3.0 cm = 2.0 N/cm. For an extension of 7.5 cm, F = kx = 2.0 N/cm × 7.5 cm = 15 N.
10A force of 40 N acts at a perpendicular distance of 0.50 m from a pivot. What is the moment of the force about the pivot?
A.20 N·m
B.39.5 N·m
C.80 N·m
D.200 N·m
Explanation: Moment of a force = Force × perpendicular distance from pivot = 40 N × 0.50 m = 20 N·m.

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