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2026 Statistics
Key Facts: BGCSE Physics 1427 Exam
100
Practice Questions
OpenExamPrep Bank
1427
BEC Syllabus Code
Botswana Examinations Council
3 Papers
Official Exam Structure
Paper 1 (MCQ), Paper 2 (Theory), Paper 3 (Practical)
Grade C
Pass Benchmark
BEC Tertiary Admission Requirement
6 Domains
Core Content Areas
Outcome-Based Physics Curriculum
The BGCSE Physics 1427 practice bank offers 100 questions covering Mechanics, Energy, Thermal Physics, Waves, Electricity, and Nuclear Physics. It features step-by-step numerical calculations, unit conversions, formula applications, and detailed explanations for correct and incorrect choices to ensure top performance in BEC examinations.
Sample BGCSE Physics 1427 Practice Questions
Try these sample questions to test your BGCSE Physics 1427 exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1A vehicle travels along a straight road between Gaborone and Lobatse, covering a distance of 72 km in 48 minutes. What is the average speed of the vehicle in meters per second (m/s)?
A.25 m/s
B.1.5 m/s
C.90 m/s
D.15 m/s
Explanation: To calculate average speed in m/s, convert distance and time to SI units first: distance d = 72 km = 72,000 m, and time t = 48 minutes = 48 x 60 = 2,880 s. Using v = d / t, v = 72,000 / 2,880 = 25 m/s. Alternatively, 72 km / 0.8 h = 90 km/h, and 90 / 3.6 = 25 m/s.
2A sprinter starts from rest and accelerates uniformly at a rate of 2.5 m/s^2 along a straight track for 4.0 seconds. What is the final velocity of the sprinter?
A.10 m/s
B.6.25 m/s
C.20 m/s
D.1.6 m/s
Explanation: Using the first equation of motion v = u + at, where initial velocity u = 0 m/s, acceleration a = 2.5 m/s^2, and time t = 4.0 s. Substituting values gives v = 0 + (2.5 x 4.0) = 10 m/s. The sprinter reaches a speed of 10 m/s at the end of 4.0 seconds.
3A velocity-time graph shows an object accelerating uniformly from rest to 12 m/s in 6 seconds, traveling at a constant velocity of 12 m/s for 10 seconds, and then decelerating uniformly to rest in 4 seconds. What is the total displacement of the object?
A.180 m
B.240 m
C.150 m
D.120 m
Explanation: Total displacement is equal to the total area under the velocity-time graph (a trapezium). The shape has parallel sides of length b1 = 10 s (constant velocity phase) and b2 = 6 + 10 + 4 = 20 s (total time), with height h = 12 m/s. Area = 0.5 x (b1 + b2) x h = 0.5 x (10 + 20) x 12 = 0.5 x 30 x 12 = 180 m.
4An astronaut has a mass of 70 kg on Earth, where the gravitational field strength is g = 9.8 N/kg. If the gravitational field strength on the Moon is 1.6 N/kg, what are the mass and weight of the astronaut on the Moon?
A.Mass = 70 kg, Weight = 112 N
B.Mass = 11.4 kg, Weight = 112 N
C.Mass = 70 kg, Weight = 686 N
D.Mass = 43.75 kg, Weight = 70 N
Explanation: Mass is the amount of matter in an object and remains constant regardless of location (Mass = 70 kg). Weight is gravitational force W = m x g. On the Moon, W = 70 kg x 1.6 N/kg = 112 N. Therefore, mass is 70 kg and weight is 112 N.
5A solid metal rectangular block measures 5.0 cm by 4.0 cm by 2.0 cm and has a mass of 316 g. What is the density of the metal in grams per cubic centimeter (g/cm^3) and in SI units (kg/m^3)?
A.7.9 g/cm^3 (7,900 kg/m^3)
B.15.8 g/cm^3 (15,800 kg/m^3)
C.3.95 g/cm^3 (3,950 kg/m^3)
D.0.126 g/cm^3 (126 kg/m^3)
Explanation: Volume V = length x width x height = 5.0 x 4.0 x 2.0 = 40 cm^3. Density rho = mass / volume = 316 g / 40 cm^3 = 7.9 g/cm^3. To convert g/cm^3 to kg/m^3, multiply by 1,000: 7.9 x 1,000 = 7,900 kg/m^3 (which corresponds to iron).
6A wooden crate of mass 40 kg is pulled along a rough horizontal floor by a force of 140 N. If the frictional force opposing motion is 60 N, what is the acceleration of the crate?
A.2.0 m/s^2
B.3.5 m/s^2
C.5.0 m/s^2
D.1.5 m/s^2
Explanation: First find the resultant (net) horizontal force: F_net = Applied force - Friction = 140 N - 60 N = 80 N. Using Newton's second law F_net = m x a, acceleration a = F_net / m = 80 N / 40 kg = 2.0 m/s^2.
7Two forces of magnitude 6.0 N and 8.0 N act simultaneously at a point on a body at right angles (90 degrees) to each other. What is the magnitude of the resultant force?
A.10.0 N
B.14.0 N
C.2.0 N
D.48.0 N
Explanation: When two forces act at right angles, their resultant is calculated using Pythagoras' theorem: R = sqrt(F1^2 + F2^2) = sqrt(6.0^2 + 8.0^2) = sqrt(36 + 64) = sqrt(100) = 10.0 N.
8A uniform meter rule is pivoted at its 50 cm mark. A mass of 200 g is suspended at the 10 cm mark. At which mark must a 500 g mass be placed to balance the rule horizontally?
A.66 cm mark
B.16 cm mark
C.70 cm mark
D.80 cm mark
Explanation: Pivot is at 50 cm. Anti-clockwise moment = mass x distance from pivot = 200 g x (50 cm - 10 cm) = 200 x 40 = 8000 g cm. For rotational equilibrium, clockwise moment must equal anti-clockwise moment: 500 g x d = 8000 g cm -> d = 8000 / 500 = 16 cm to the right of the pivot. The position on the rule is 50 cm + 16 cm = 66 cm mark.
9Which modification will MOST effectively increase the stability of a bus or heavy transport vehicle?
A.Lowering the center of gravity and widening the wheel base
B.Raising the center of gravity and narrowing the wheel base
C.Placing all luggage on a high roof rack
D.Increasing the overall height of the passenger cabin
Explanation: Stability is maximized when the center of gravity is kept as low as possible and the base of support (wheel base) is as wide as possible. A low center of gravity means the line of action of weight remains within the base of support even at large tilt angles, preventing toppling.
10A helical spring has an unstretched length of 15.0 cm. When a load of 4.0 N is hung from it, its total length increases to 19.0 cm. Assuming the elastic limit is not exceeded, what will be the total length of the spring when loaded with a force of 10.0 N?
A.25.0 cm
B.10.0 cm
C.29.0 cm
D.21.0 cm
Explanation: Hooke's law states F = k x e. Extension for 4.0 N is e1 = 19.0 cm - 15.0 cm = 4.0 cm. Spring constant k = F / e = 4.0 N / 4.0 cm = 1.0 N/cm. For a load of 10.0 N, extension e2 = 10.0 N / 1.0 N/cm = 10.0 cm. Total length = original length + extension = 15.0 cm + 10.0 cm = 25.0 cm.
About the BGCSE Physics 1427 Practice Questions
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