Chapter Exam

Chapter-4 Forces — Online MCQ Test

Q: The spring constant of a spring is 200 N/m. What is the extension when a force of 50 N is applied?

Answer: A. 0.25 m — Using F = kx, extension x = F/k = 50/200 = 0.25 m.

Q: A student applies a force of 30 N to a door handle at a perpendicular distance of 0.8 m from the hinge. What is the moment produced?

Answer: A. 24 N·m — Moment = Force × perpendicular distance = 30 × 0.8 = 24 N·m.

Q: A car of mass 1000 kg moves in a horizontal circular path of radius 50 m at constant speed of 10 m/s. What is the centripetal force required?

Answer: A. 2000 N — Centripetal force F = mv²/r = (1000 × 10²)/50 = 100,000/50 = 2000 N.

PHYSICS · Grade 10 · IGCSE cambridge

Practice Chapter-4 Forces with a free chapter-wise online MCQ test for IGCSE cambridge Grade 10 PHYSICS. This chapter covers: The effects of forces Hooke's law friction circular motion and turning effects including moments.. AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter-4 Forces — Important Questions & Answers (FAQ)

Frequently asked questions from IGCSE cambridge Grade 10 PHYSICS — Chapter-4 Forces, with answers and explanations. These are sample questions; the exam has its own separate question set.

What does Hooke's law state about the relationship between force and extension in a spring?

A. Force is inversely proportional to extension
B. Force is directly proportional to extension, up to the limit of proportionality ✓
C. Force is equal to the extension in all cases
D. Force causes permanent deformation immediately

Answer: B. Force is directly proportional to extension, up to the limit of proportionality
Hooke's law states F = kx, where force is directly proportional to extension within the elastic limit.

The spring constant of a spring is 200 N/m. What is the extension when a force of 50 N is applied?

A. 0.25 m ✓
B. 0.5 m
C. 2 m
D. 4 m

Answer: A. 0.25 m
Using F = kx, extension x = F/k = 50/200 = 0.25 m.

A student applies a force of 30 N to a door handle at a perpendicular distance of 0.8 m from the hinge. What is the moment produced?

A. 24 N·m ✓
B. 37.5 N·m
C. 3.75 N·m
D. 26.8 N·m

Answer: A. 24 N·m
Moment = Force × perpendicular distance = 30 × 0.8 = 24 N·m.

A car of mass 1000 kg moves in a horizontal circular path of radius 50 m at constant speed of 10 m/s. What is the centripetal force required?

A. 2000 N ✓
B. 1000 N
C. 5000 N
D. 500 N

Answer: A. 2000 N
Centripetal force F = mv²/r = (1000 × 10²)/50 = 100,000/50 = 2000 N.

A material is stretched beyond its elastic limit and does not return to its original length when the force is removed. Which of the following best explains this phenomenon?

A. The spring constant has increased permanently
B. Hooke's law no longer applies because permanent deformation has occurred ✓
C. The material has gained elastic potential energy
D. The force applied was insufficient to cause any change

Answer: B. Hooke's law no longer applies because permanent deformation has occurred
Once the elastic limit is exceeded, the material undergoes plastic deformation and does not obey Hooke's law; permanent deformation occurs as atomic bonds are broken or rearranged.