NEETAceToday's NEET Daily

Force, inertia and Newton’s laws of motion: NEET MCQs with answers

59 questions on this topic of Laws of Motion (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

Practise Laws of Motion freeAll Laws of Motion questions

Free with a Google sign-in; NEET Daily needs no sign-in at all.

easy

1. A constant force acts on an object of mass 5 kg initially at rest. If the object acquires a velocity of 20 m/s in 4 seconds, the magnitude of the force applied is:

  1. 50 N
  2. 20 N
  3. 100 N
  4. 25 N
Show answer and explanation

Answer: (D) 25 N

Acceleration a = (v - u)/t = (20 - 0)/4 = 5 m/s². From Newton's second law, Force F = m × a = 5 kg × 5 m/s² = 25 N.

medium

2. Masses of 7 kg and 1 kg hang from the two ends of a light string over a frictionless pulley. Taking g = 10 m/s², the acceleration of the system is:

  1. 7.5 m/s²
  2. 13.3 m/s²
  3. 10 m/s²
  4. 8.57 m/s²
Show answer and explanation

Answer: (A) 7.5 m/s²

a = (m₁ − m₂)g/(m₁ + m₂) = 6 × 10/8 = 7.5 m/s².

hard

3. A horizontal force F acts on a body of mass 4 kg resting on a smooth horizontal table. The force varies with time as F(t) = (12 - 3t) N, from t = 0 to the instant when F becomes zero. The velocity acquired by the body during this time is:

  1. 6 m/s
  2. 12 m/s
  3. 8 m/s
  4. 16 m/s
Show answer and explanation

Answer: (A) 6 m/s

The force becomes zero when 12 - 3t = 0, so t = 4 s. Impulse J = ∫ F dt from 0 to 4 = [12t - (3t²/2)] from 0 to 4 = 48 - 24 = 24 N·s. Since J = m × Δv, we have 24 = 4 × v, which yields v = 6 m/s.

medium

4. Masses of 6 kg and 2 kg hang from the two ends of a light string over a frictionless pulley. Taking g = 10 m/s², the tension in the string is:

  1. 30 N
  2. 60 N
  3. 20 N
  4. 40 N
Show answer and explanation

Answer: (A) 30 N

a = (m₁ − m₂)g/(m₁ + m₂) = 5 m/s². T = 2m₁m₂g/(m₁ + m₂) = 2 × 6 × 2 × 10/8 = 30 N.

easy

5. Which of the following statements is fundamentally correct regarding Newton's third law of motion?

  1. Action and reaction act on the same body to produce equilibrium.
  2. Action precedes reaction by a very small interval of time.
  3. Action and reaction forces act simultaneously on two different bodies.
  4. Reaction exists only when the bodies are in physical contact.
Show answer and explanation

Answer: (C) Action and reaction forces act simultaneously on two different bodies.

According to Newton's third law, forces always occur in matched pairs. The action and reaction forces act simultaneously on two distinct interacting bodies, so they never cancel each other out.

medium

6. A 5 kg block on a frictionless surface experiences a net force of 2 N. Its acceleration is:

  1. 2.5 m/s²
  2. 0.4 m/s²
  3. 0.2 m/s²
  4. 10 m/s²
Show answer and explanation

Answer: (B) 0.4 m/s²

Unit check: Convert to SI, then a = F/m = 2/5 = 0.4 m/s².

medium

7. Masses of 9 kg and 4 kg hang from the two ends of a light string over a frictionless pulley. Taking g = 10 m/s², the tension in the string is:

  1. 50 N
  2. 90 N
  3. 55.4 N
  4. 40 N
Show answer and explanation

Answer: (C) 55.4 N

a = (m₁ − m₂)g/(m₁ + m₂) = 3.85 m/s². T = 2m₁m₂g/(m₁ + m₂) = 2 × 9 × 4 × 10/13 = 55.4 N.

medium

8. A 2 kg block on a frictionless surface experiences a net force of 0.02 kN. Its acceleration is:

  1. 10 m/s²
  2. 40 m/s²
  3. 0.1 m/s²
  4. 5 m/s²
Show answer and explanation

Answer: (A) 10 m/s²

Unit check: Convert to SI, then a = F/m = 20/2 = 10 m/s².

medium

9. Masses of 10 kg and 7 kg hang from the two ends of a light string over a frictionless pulley. Taking g = 10 m/s², the tension in the string is:

  1. 30 N
  2. 82.4 N
  3. 100 N
  4. 70 N
Show answer and explanation

Answer: (B) 82.4 N

a = (m₁ − m₂)g/(m₁ + m₂) = 1.76 m/s². T = 2m₁m₂g/(m₁ + m₂) = 2 × 10 × 7 × 10/17 = 82.4 N.

medium

10. Masses of 9 kg and 8 kg hang from the two ends of a light string over a frictionless pulley. Taking g = 10 m/s², the acceleration of the system is:

  1. 0.588 m/s²
  2. 1.11 m/s²
  3. 10 m/s²
  4. 170 m/s²
Show answer and explanation

Answer: (A) 0.588 m/s²

a = (m₁ − m₂)g/(m₁ + m₂) = 1 × 10/17 = 0.588 m/s².

medium

11. Masses of 6 kg and 1 kg hang from the two ends of a light string over a frictionless pulley. Taking g = 10 m/s², the tension in the string is:

  1. 60 N
  2. 17.1 N
  3. 50 N
  4. 10 N
Show answer and explanation

Answer: (B) 17.1 N

a = (m₁ − m₂)g/(m₁ + m₂) = 7.14 m/s². T = 2m₁m₂g/(m₁ + m₂) = 2 × 6 × 1 × 10/7 = 17.1 N.

medium

12. A 2 kg block on a frictionless surface experiences a net force of 5 N. Its acceleration is:

  1. 0.4 m/s²
  2. 10 m/s²
  3. 2.5 m/s²
  4. 1.25 m/s²
Show answer and explanation

Answer: (C) 2.5 m/s²

Unit check: Convert to SI, then a = F/m = 5/2 = 2.5 m/s².

Keep going with Laws of Motion. All 290 questions with instant answers, your mistakes saved for revision and your accuracy by topic.
Start practising

Other topics in Laws of Motion