Chapter Exam

Chapter-4 Laws of Motion — Online MCQ Test

Q: What is the SI unit of force according to Newton's laws of motion?

Answer: B. Newton — The SI unit of force is Newton (N), defined as the force required to accelerate 1 kg mass at 1 m/s².

Q: Momentum is defined as the product of _____ and _____.

Answer: A. mass and velocity — Momentum (p) is defined as the product of mass (m) and velocity (v), expressed as p = mv.

Q: A 2 kg object accelerates at 5 m/s² when a force F is applied. What is the magnitude of force F?

Answer: B. 10 N — Using F = ma, F = 2 kg × 5 m/s² = 10 N.

Q: A car of mass 1500 kg is moving in a circular path of radius 50 m at a constant speed of 20 m/s. Calculate the centripetal force required.

Answer: B. 12000 N — Centripetal force = mv²/r = (1500 × 20²)/50 = (1500 × 400)/50 = 12000 N.

Q: Two blocks A and B of masses 3 kg and 2 kg respectively are in contact on a frictionless horizontal surface. A force of 10 N is applied to block A. What is the acceleration of the system and the force exerted by A on B?

Answer: A. a = 2 m/s², F = 4 N — Total mass = 5 kg, acceleration = 10/5 = 2 m/s². Force on B = mB × a = 2 × 2 = 4 N.

Q: A block of mass 10 kg is pulled horizontally on a surface with coefficient of kinetic friction 0.1. If the applied force is 15 N and g = 10 m/s², what is the maximum static friction force that could prevent motion if the block were at rest?

Answer: D. Cannot be determined without knowing μs — The coefficient of static friction (μs) is different from kinetic friction (μk = 0.1). Maximum static friction = μs × N = μs × 100, which cannot be determined without μs.

Q: A car negotiates a banked circular road of radius 100 m at angle 37°. At what speed will the car experience no friction force? (g = 10 m/s², sin 37° = 0.6, cos 37° = 0.8, tan 37° = 0.75)

Answer: D. 30 m/s — For no friction, v = √(rg tan θ) = √(100 × 10 × 0.75) = √750 ≈ 27.4 m/s, closest is 30 m/s (or v = √(100 × 10 × tan 37°) = √(1000 × 0.75) = 27.4 m/s).

PHYSICS · Grade 11 · CBSE(NCERT)

Practice Chapter-4 Laws of Motion with a free chapter-wise online MCQ test for CBSE(NCERT) Grade 11 PHYSICS. This chapter covers: Intuitive concept of force - Inertia - Newton's first law - Momentum - Newton's second law - Impulse - Newton's third law - Law of conservation of linear momentum - Equilibrium of.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter-4 Laws of Motion — Important Questions & Answers (FAQ)

Frequently asked questions from CBSE(NCERT) Grade 11 PHYSICS — Chapter-4 Laws of Motion, with answers and explanations. These are sample questions; the exam has its own separate question set.

What is the SI unit of force according to Newton's laws of motion?

A. Dyne
B. Newton ✓
C. Kilogram
D. Joule

Answer: B. Newton
The SI unit of force is Newton (N), defined as the force required to accelerate 1 kg mass at 1 m/s².

Which statement best defines inertia?

A. The tendency of an object to resist changes in its state of motion ✓
B. The force acting on an object
C. The rate of change of momentum
D. The gravitational force on an object

Answer: A. The tendency of an object to resist changes in its state of motion
Inertia is the natural property of matter to resist any change in its state of rest or uniform motion, as stated in Newton's first law.

State Newton's first law of motion in words.

A. Force equals mass times acceleration
B. Every action has an equal and opposite reaction
C. An object at rest or in uniform motion remains so unless acted upon by an external force ✓
D. The rate of change of momentum equals the applied force

Answer: C. An object at rest or in uniform motion remains so unless acted upon by an external force
Newton's first law states that a body in rest or uniform motion continues unless acted upon by an external unbalanced force.

Momentum is defined as the product of _____ and _____.

A. mass and velocity ✓
B. force and time
C. mass and acceleration
D. force and distance

Answer: A. mass and velocity
Momentum (p) is defined as the product of mass (m) and velocity (v), expressed as p = mv.

A block is placed on an inclined plane. The normal force acting on the block is:

A. Equal to the weight of the block
B. Equal to the component of weight perpendicular to the plane ✓
C. Equal to the component of weight parallel to the plane
D. Always zero

Answer: B. Equal to the component of weight perpendicular to the plane
The normal force equals the component of weight perpendicular to the inclined plane surface, which is mg cos(θ).

A 2 kg object accelerates at 5 m/s² when a force F is applied. What is the magnitude of force F?

A. 2.5 N
B. 10 N ✓
C. 7 N
D. 3 N

Answer: B. 10 N
Using F = ma, F = 2 kg × 5 m/s² = 10 N.

A car of mass 1500 kg is moving in a circular path of radius 50 m at a constant speed of 20 m/s. Calculate the centripetal force required.

A. 6000 N
B. 12000 N ✓
C. 24000 N
D. 48000 N

Answer: B. 12000 N
Centripetal force = mv²/r = (1500 × 20²)/50 = (1500 × 400)/50 = 12000 N.

Two blocks A and B of masses 3 kg and 2 kg respectively are in contact on a frictionless horizontal surface. A force of 10 N is applied to block A. What is the acceleration of the system and the force exerted by A on B?

A. a = 2 m/s², F = 4 N ✓
B. a = 2 m/s², F = 6 N
C. a = 5 m/s², F = 10 N
D. a = 1 m/s², F = 2 N

Answer: A. a = 2 m/s², F = 4 N
Total mass = 5 kg, acceleration = 10/5 = 2 m/s². Force on B = mB × a = 2 × 2 = 4 N.

A block of mass 10 kg is pulled horizontally on a surface with coefficient of kinetic friction 0.1. If the applied force is 15 N and g = 10 m/s², what is the maximum static friction force that could prevent motion if the block were at rest?

A. Exactly 15 N
B. More than 15 N (depends on μs)
C. Exactly 10 N
D. Cannot be determined without knowing μs ✓

Answer: D. Cannot be determined without knowing μs
The coefficient of static friction (μs) is different from kinetic friction (μk = 0.1). Maximum static friction = μs × N = μs × 100, which cannot be determined without μs.

A car negotiates a banked circular road of radius 100 m at angle 37°. At what speed will the car experience no friction force? (g = 10 m/s², sin 37° = 0.6, cos 37° = 0.8, tan 37° = 0.75)

A. 15 m/s
B. 20 m/s
C. 25 m/s
D. 30 m/s ✓

Answer: D. 30 m/s
For no friction, v = √(rg tan θ) = √(100 × 10 × 0.75) = √750 ≈ 27.4 m/s, closest is 30 m/s (or v = √(100 × 10 × tan 37°) = √(1000 × 0.75) = 27.4 m/s).

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