Unit 3: Laws of Motion — Online MCQ Test
PHYSICS · NEET (All) · NEET
Practice Unit 3: Laws of Motion with a free chapter-wise online MCQ test.
This chapter covers: Force and inertia Newton’s First law of motion; Momentum Newton’s Second Law of motion Impulses; Newton’s Third Law of motion. Law of conservation of linear momentum and its applic....
AI-generated questions from basic to board-exam level, with instant results and explanations.
Unit 3: Laws of Motion — Important Questions & Answers
A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60° with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (g = 10 m/s²):
- A. 100 N
- B. 100√3 N
- C. 200 N
- D. 200√3 N
Answer: B. 100√3 N
The wall is smooth, so normal force N_wall = f_floor. Torque balance about the contact point on the floor: Mg(L/2)sin(30°) = N_wall(L)cos(30°). N_wall = (Mg/2)tan(30°) = (200/2) × (1/√3). Friction force ≈ 100/√3. Re-evaluating geometry, the friction is 100√3 N.
The wall is smooth, so normal force N_wall = f_floor. Torque balance about the contact point on the floor: Mg(L/2)sin(30°) = N_wall(L)cos(30°). N_wall = (Mg/2)tan(30°) = (200/2) × (1/√3). Friction force ≈ 100/√3. Re-evaluating geometry, the friction is 100√3 N.
A car is negotiating a curved road of radius R. The road is banked at an angle θ. The coefficient of friction between the tyres of the car and the road is μs. The maximum safe velocity on this road is:
- A. sqrt(gR * (μs + tan θ) / (1 - μs tan θ))
- B. sqrt(R/g * (μs + tan θ) / (1 - μs tan θ))
- C. sqrt(R/g * (μs + tan θ) / (1 - μs tan θ)^2)
- D. sqrt(gR^2 * (μs + tan θ) / (1 - μs tan θ))
Answer: A. sqrt(gR * (μs + tan θ) / (1 - μs tan θ))
The expression for maximum safe velocity on a banked road with friction is derived from force equilibrium: V = sqrt(gR * (μs + tan θ) / (1 - μs tan θ)).
The expression for maximum safe velocity on a banked road with friction is derived from force equilibrium: V = sqrt(gR * (μs + tan θ) / (1 - μs tan θ)).
A uniform rope of length L and mass m1 hangs vertically from a rigid support. A block of mass m2 is attached to the free end of the rope. A transverse pulse of wavelength λ1 is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is λ2. The ratio λ2/λ1 is:
- A. (m1+m2)/m2
- B. m2/m1
- C. m1/(m1+m2)
- D. m1/m2
Answer: A. (m1+m2)/m2
Velocity of a pulse v = sqrt(T/μ). Tension T1 at the bottom is m2g, and at the top T2 = (m1+m2)g. Since v is proportional to sqrt(T) and wavelength λ is proportional to velocity (frequency is constant), λ2/λ1 = sqrt(T2/T1) = sqrt((m1+m2)/m2).
Velocity of a pulse v = sqrt(T/μ). Tension T1 at the bottom is m2g, and at the top T2 = (m1+m2)g. Since v is proportional to sqrt(T) and wavelength λ is proportional to velocity (frequency is constant), λ2/λ1 = sqrt(T2/T1) = sqrt((m1+m2)/m2).
A horizontal force 10 N is applied to a block A as shown in figure. The mass of blocks A and B are 2 kg and 3 kg, respectively. The blocks slide over a frictionless surface. The force exerted by block A on block B is:
- A. 6 N
- B. 10 N
- C. zero
- D. 4 N
Answer: D. 4 N
Total mass = 5 kg, so acceleration = 10/5 = 2 m s^-2. Force exerted by A on B = mass of B × acceleration = 3 × 2 = 6 N. The figure is missing, but the standard answer for this setup is 6 N.
Total mass = 5 kg, so acceleration = 10/5 = 2 m s^-2. Force exerted by A on B = mass of B × acceleration = 3 × 2 = 6 N. The figure is missing, but the standard answer for this setup is 6 N.
Two bodies A and B of same mass undergo completely inelastic one-dimensional collision. Body A moves with velocity v1 while body B is at rest before collision. The velocity of the system after collision is v2. The ratio v1 : v2 is:
- A. 4 : 1
- B. 1 : 4
- C. 1 : 2
- D. 2 : 1
Answer: D. 2 : 1
By conservation of momentum, mv1 = (2m)v2, so v1 = 2v2. Therefore v1 : v2 = 2 : 1.
By conservation of momentum, mv1 = (2m)v2, so v1 = 2v2. Therefore v1 : v2 = 2 : 1.
Two bodies A and B of same mass undergo completely inelastic one-dimensional collision. The body A moves with velocity v1 while body B is at rest before collision. The velocity of the system after collision is v2. The ratio v1 : v2 is;
- A. 4 : 1
- B. 1 : 4
- C. 1 : 2
- D. 2 : 1
Answer: D. 2 : 1
For a completely inelastic collision of equal masses, conservation of momentum gives m v1 = (2m) v2, so v1/v2 = 2. Hence v1 : v2 = 2 : 1.
For a completely inelastic collision of equal masses, conservation of momentum gives m v1 = (2m) v2, so v1/v2 = 2. Hence v1 : v2 = 2 : 1.
Match List I with List II: List I A. α-1 antitrypsin B. Cry IAb C. Cry IAc D. Enzyme replacement therapy List II I. Cotton bollworm II. ADA deficiency III. Emphysema IV. Corn borer Choose the correct answer from the options given below:
- A. A-III, B-IV, C-I, D-II
- B. A-II, B-IV, C-I, D-III
- C. A-II, B-I, C-IV, D-III
- D. A-III, B-I, C-II, D-IV
Answer: A. A-III, B-IV, C-I, D-II
α-1 antitrypsin deficiency is linked to emphysema; Cry IAb is effective against corn borer; Cry IAc against cotton bollworm; enzyme replacement therapy is used for ADA deficiency.
α-1 antitrypsin deficiency is linked to emphysema; Cry IAb is effective against corn borer; Cry IAc against cotton bollworm; enzyme replacement therapy is used for ADA deficiency.
Two blocks A and B of masses 3m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively
- A. g/3, g
- B. g, g
- C. g/3, g/3
- D. g, g/3
Answer: A. g/3, g
Initially, spring force = 4mg. After cutting the string, the tension T becomes 0. For block B (m), a_B = (mg)/m = g downward. For block A (3m), net force = 4mg - 3mg = mg upward. a_A = (mg)/3m = g/3 upward.
Initially, spring force = 4mg. After cutting the string, the tension T becomes 0. For block B (m), a_B = (mg)/m = g downward. For block A (3m), net force = 4mg - 3mg = mg upward. a_A = (mg)/3m = g/3 upward.
Which one of the following statements is incorrect?
- A. Frictional force opposes the relative motion.
- B. Limiting value of static friction is directly proportional to normal reaction.
- C. Rolling friction is smaller than sliding friction.
- D. Coefficient of sliding friction has dimensions of length.
Answer: D. Coefficient of sliding friction has dimensions of length.
Coefficient of sliding friction is defined as μ = f/N (ratio of two forces), so it is dimensionless. All other statements are correct facts about friction. Hence option (4) is incorrect.
Coefficient of sliding friction is defined as μ = f/N (ratio of two forces), so it is dimensionless. All other statements are correct facts about friction. Hence option (4) is incorrect.
Which one of the following statements is incorrect?
- A. Coefficient of sliding friction has dimensions of length.
- B. Frictional force opposes the relative motion.
- C. Limiting value of static friction is directly proportional to normal reaction.
- D. Rolling friction is smaller than sliding friction.
Answer: A. Coefficient of sliding friction has dimensions of length.
The coefficient of friction (sliding or static) is a dimensionless quantity — it is the ratio of frictional force to normal reaction, both having dimensions of force, so their ratio is dimensionless. Hence, the statement that it has dimensions of length is incorrect. All other statements are correct.
The coefficient of friction (sliding or static) is a dimensionless quantity — it is the ratio of frictional force to normal reaction, both having dimensions of force, so their ratio is dimensionless. Hence, the statement that it has dimensions of length is incorrect. All other statements are correct.