Laws of Motion – NDA Physics PYQs

NDA 2026-I

Q. 1. Two forces of equal magnitude simultaneously act at a point. It is observed that the resultant force is equal in magnitude to the individual forces. Which of the following statements is/are correct?

I.   The angle between one of the individual forces and the resultant force is π/3.

II.  The angle between the individual forces is 2π/3.

Select the answer using the code given below.

(a)  I only

(b)  II only

(c)  Both I and II

(d)  Neither I nor II

Answer: (c) Both I and II

Explanation: Let each force have magnitude F and the angle between them be θ. Resultant R = √(F² + F² + 2F²cosθ) = F√(2 + 2cosθ). For R = F: F² = 2F² + 2F²cosθ, so 1 = 2 + 2cosθ, giving cosθ = −1/2, so θ = 120° = 2π/3. Statement II is correct. The resultant bisects the angle between equal forces, so the angle between each force and the resultant = θ/2 = 60° = π/3. Statement I is correct. Both are correct. Concept Tested: Vector addition of equal forces: angle between forces and resultant when R = F

Q. 2. What is the upward force acting on a skydiver of mass 60 kg who is falling at a uniform speed of 2 m/s?

(a)  0

(b)  60 N

(c)  588 N

(d)  708 N

Answer: (c) 588 N

Explanation: At uniform (constant) speed, net force = 0. Therefore upward drag force = downward weight = mg = 60 × 9.8 = 588 N. The skydiver is in equilibrium: drag exactly equals weight. The speed (2 m/s) is terminal velocity: not relevant to the force calculation. Concept Tested: Terminal velocity: drag force equals weight at constant falling speed

Q. 3. The force-time (F-t) graph of a 10 kg moving object is shown below (trapezoidal shape, rises to 20 N at t = 5s, stays at 20 N till t = 20s, drops to 0 at t = 25s). What is the final speed of the object?

(a)  0

(b)  15 m/s

(c)  30 m/s

(d)  40 m/s

Answer: (d) 40 m/s

Explanation: The area under an F–t graph equals impulse = change in momentum. Trapezoidal area: Area = ½ × (5) × 20 (rising triangle) + 15 × 20 (rectangle) + ½ × 5 × 20 (falling triangle) = 50 + 300 + 50 = 400 N·s. Impulse = Δp = m × Δv. If the object starts from rest: Δv = 400/10 = 40 m/s. Final speed = 40 m/s. Concept Tested: Area under F-t graph equals impulse: finding final speed from trapezoidal F-t graph

NDA 2025-II

Q. 4. Which one of the following statements for second law of motion is NOT correct?

(a)  The net force on a body is proportional to that body’s acceleration

(b)  The net force on a body is in the same direction as the acceleration

(c)  The time rate of change of momentum is equal to force

(d)  The net force on a body is proportional to that body’s momentum

Answer: (d) The net force on a body is proportional to that body’s momentum

Explanation: Option (a) is correct: F ∝ a (from F = ma). Option (b) is correct: force and acceleration are in the same direction. Option (c) is correct: F = dp/dt. Option (d) is wrong: force is proportional to acceleration, not to momentum. A body can have large momentum with zero net force (e.g., a moving object with no friction). F ∝ p would mean a stationary body (p = 0) has no force acting on it: but a body can be pushed from rest. Concept Tested: Newton’s Second Law: force ∝ acceleration (NOT ∝ momentum)

★ JOVIK Exam Insight Force is proportional to acceleration, NOT to momentum. This is the most precisely tested statement of the Second Law in recent NDA papers (2025-II). Large momentum with zero force is possible (uniform motion). Force = rate of change of momentum: not momentum itself.

NDA 2024-II

Q. 5. An astronaut whose weight on the Earth is 600 N experiences weightlessness on International Space Station orbiting around the Earth. It means that

(a)  acceleration of the astronaut is zero

(b)  normal reaction of the space-station floor on the astronaut is zero

(c)  gravitational pull of earth on the astronaut is zero

(d)  space station applies a centrifugal force on the astronaut

Answer: (b) normal reaction of the space-station floor on the astronaut is zero

Explanation: In orbit, the space station and astronaut are both in free fall around Earth. Gravity acts on the astronaut: providing the centripetal force for orbital motion. The astronaut is not resting on the floor in the usual sense; the floor exerts no normal reaction. Weightlessness = zero normal reaction, not zero gravity. Concept Tested: Weightlessness in orbit: normal reaction = 0, not absence of gravity

★ JOVIK Exam Insight Weightlessness does NOT mean gravity is absent. Gravity provides centripetal force for the orbit. The ISS and astronaut are in the same free-fall trajectory: so the floor exerts no reaction. This is the same as the lift-cable-breaking scenario.

Q. 6. Which of the following statements give characteristics of contact forces?

1.  It appears between an object when it is in contact with some other object

2.  It satisfies the third law of motion

3.  It may appear between a pair of solid and fluid

Select the answer using the code given below:

(a)  1 and 3 only

(b)  2 and 3 only

(c)  1 and 2 only

(d)  1, 2 and 3

Answer: (d) 1, 2 and 3

Explanation: All three statements are correct. Statement 1: contact forces require physical contact. Statement 2: contact forces satisfy Newton’s Third Law: action and reaction. Statement 3: contact forces can act between solids and fluids (e.g., drag, buoyancy, viscous friction). All characteristics apply. Concept Tested: Characteristics of contact forces: physical contact, Third Law, solid-fluid interface

Q. 7. A car weighs 1000 kg. It is moving with a uniform velocity of 72 km/h towards a straight road. The driver suddenly presses the brakes. The car stops in 0·2 s. The retarding force applied on the car to stop it is

(a)  100 N

(b)  1000 N

(c)  10 kN

(d)  100 kN

Answer: (d) 100 kN

Explanation: Convert velocity: 72 km/h = 20 m/s. Retarding force = Δp/Δt = (m × Δv)/Δt = (1000 × 20)/0.2 = 20,000/0.2 = 100,000 N = 100 kN. Concept Tested: Impulse-momentum: retarding force from rate of change of momentum

Q. 8. A block of mass 2·0 kg slides on a rough horizontal plane. Speed at a particular instant is 10 m/s. It comes to rest after travelling 20 m. Magnitude of frictional force could be:

(a)  10 N

(b)  20 N

(c)  40 N

(d)  50 N

Answer: (a) 10 N

Explanation: Using v² = u² + 2as: 0 = 100 + 2a(20), so a = −100/40 = −2.5 m/s². Friction force = ma = 2 × 2.5 = 5 N… Checking: The question says ‘could be’ and 5 N is not among the options. Rechecking using energy: work by friction = KE lost = ½mv² = ½ × 2 × 100 = 100 J. Friction force = 100/20 = 5 N. The closest available option is (a) 10 N. This may reflect a source discrepancy: verify against the original paper. Concept Tested: Friction: calculating frictional force from kinematics (energy method)

Note: The calculated frictional force (5 N using both kinematic and energy methods) does not match any listed option exactly. Option (a) 10 N is the nearest and is taken from the source. Verify against the original NDA 2024-II paper.

Q. 9. The figure given below shows the direction of the two forces P and Q acting on a skydiver: Which one among the following statements is correct about the two forces?

(a)  Force P is caused by the gravity and force Q is caused by the friction

(b)  When the force P is bigger than the force Q, the speed of the skydiver remains the same

(c)  After the parachute opens, force P remains the same while force Q increases

(d)  After the parachute opens, force P decreases while force Q increases

Answer: (c) After the parachute opens, force P remains the same while force Q increases

Explanation: In the standard skydiver figure, P is the downward gravitational force (weight = mg) and Q is the upward drag force (air resistance). Weight P = mg remains constant: mass does not change when the parachute opens. When the parachute opens, the much larger surface area dramatically increases drag Q. The skydiver decelerates until a new lower terminal velocity is reached where P = Q again. Concept Tested: Forces on a skydiver: weight (P) constant, drag (Q) increases when parachute opens

NDA 2024-I

Q. 10. A uniform meter scale of mass 0·24 kg is made of steel. It is kept on two wedges, W₁ and W₂, in a horizontal position. W₁ is at a distance of 0·2 m from one of its ends, while W₂ is at distance of 0·4 m from the other end. If the force on the scale is N₁ due to W₁ and N₂ due to W₂, then: (take g = 10·0 m s⁻²)

(a)  N₁ = 1·6 N and N₂ = 0·8 N

(b)  N₁ = 0·8 N and N₂ = 1·6 N

(c)  N₁ = 0·6 N and N₂ = 1·8 N

(d)  N₁ = 1·8 N and N₂ = 0·6 N

Answer: (c) N₁ = 0·6 N and N₂ = 1·8 N

Explanation: The scale has length 1 m. W₁ is at 0.2 m from one end (so 0.2 m from left). W₂ is at 0.4 m from the other end (so 0.6 m from left). Centre of scale = 0.5 m from left. Total weight = 0.24 × 10 = 2.4 N. Taking moments about W₁ (at 0.2 m from left): N₂ × (0.6 − 0.2) = 2.4 × (0.5 − 0.2). N₂ × 0.4 = 2.4 × 0.3 = 0.72. N₂ = 1.8 N. N₁ = 2.4 − 1.8 = 0.6 N. Concept Tested: Principle of moments: torque balance on a uniform metre scale with two supports

Q. 11. A ball of 0·1 kg mass is dropped on a hard floor from a height of 0·45 m and rises to a height of 0·20 m. If it was in touch with the floor for 0·1 s, the net force it applied on the floor while bouncing is: (take the gravitational acceleration g = 10 m s⁻²)

(a)  1·0 N

(b)  6·0 N

(c)  3·0 N

(d)  5·0 N

Answer: (d) 5·0 N

Explanation: Velocity just before impact: v₁ = √(2 × 10 × 0.45) = 3 m/s (downward). Velocity just after: v₂ = √(2 × 10 × 0.20) = 2 m/s (upward). Change in momentum = m(v₂ − (−v₁)) = 0.1 × (2 + 3) = 0.5 kg m/s. Force = Δp/Δt = 0.5/0.1 = 5 N. This is the net impulse force. By Newton’s Third Law, the ball applies the same force on the floor. Concept Tested: Impulse-momentum theorem: force from change in momentum during bounce

Q. 12. Which one of the following about different frictional forces is correct?

(a)  Kinetic friction > Static friction > Rolling friction

(b)  Static friction > Rolling friction > Kinetic friction

(c)  Static friction > Kinetic friction > Rolling friction

(d)  Static friction > Kinetic friction = Rolling friction

Answer: (c) Static friction > Kinetic friction > Rolling friction

Explanation: Static friction is largest: it must be overcome before motion begins. Once motion starts, kinetic (sliding) friction is lower than static friction. Rolling friction is the smallest because the deformation at the contact point is minimal for a rolling object. The correct order is: static > kinetic > rolling. Concept Tested: Types of friction: magnitude order: static > kinetic > rolling

Q. 13. A metallic bob X of mass m is released from position A and collides elastically with identical bob Y at rest on a frictionless table. Angle AOB = 30°. How high does bob X rise immediately after collision?

(a)  To the same height as position A on the other side in same trajectory

(b)  To half the height of position A along same trajectory

(c)  The same height at position A

(d)  It stops at position B

Answer: (d) It stops at position B

Explanation: In a perfectly elastic collision between two identical masses, the moving mass comes to a complete stop and the stationary mass moves off with the original velocity. Bob X transfers all its kinetic energy and momentum to bob Y: it stops at the collision point B. Bob Y swings up. This is the principle behind Newton’s cradle. Concept Tested: Elastic collision between identical masses: moving mass stops; stationary mass takes full velocity

Q. 14. If block P as shown in figure is to be at rest, what should magnitude of force F be? (Forces: 10N left, 15N left, 2N right, 15N right, F right)

(a)  5 N

(b)  6 N

(c)  8 N

(d)  10 N

Answer: (c) 8 N

Explanation: Net force without F: Leftward = 10 + 15 = 25 N. Rightward = 2 + 15 = 17 N. Net = 25 − 17 = 8 N leftward. For equilibrium, F must act rightward with magnitude 8 N to cancel the net leftward force. Concept Tested: Static equilibrium: force balance (ΣF = 0)

Q. 15. A block of mass 2 kg, moving with initial speed 3 m/s, comes to rest on rough horizontal surface after travelling 3 m. Magnitude of frictional force is:

(a)  9 N

(b)  3 N

(c)  18 N

(d)  1 N

Answer: (b) 3 N

Explanation: Using v² = u² + 2as: 0 = 9 + 2a(3), so a = −9/6 = −1.5 m/s². Frictional force = ma = 2 × 1.5 = 3 N. The frictional force decelerates the block to rest. Concept Tested: Friction from kinematics: F = ma using v² = u² + 2as

NDA 2023-II

Q. 16. The motion of a particle of mass m is described by the relation, y = ut − ½gt², where u is the initial velocity of the particle. The force acting on the particle is

(a)  F = m(du/dt)

(b)  F = mg

(c)  F = m(dy/dt)

(d)  F = −mg

Answer: (d) F = −mg

Explanation: Differentiating y = ut − ½gt² twice: dy/dt = u − gt (velocity), d²y/dt² = −g (acceleration). By Newton’s Second Law, F = m × d²y/dt² = m × (−g) = −mg. The negative sign indicates force acts downward (opposite to the positive y-direction, which is upward). This is the gravitational force: written correctly in vector form as −mg. Concept Tested: Newton’s Second Law: force from second derivative of position equation

NDA 2023-I

Q. 17. Two identical spring balances S₁ and S₂ are connected one after the other and are held vertically. A mass of 10 kg is hanging from S₂. If the readings on S₁ and S₂ are W₁ and W₂ respectively, then:

(a)  W₁ = 5 kg and W₂ = 10 kg

(b)  W₁ = 10 kg and W₂ = 5 kg

(c)  W₁ = 5 kg and W₂ = 5 kg

(d)  W₁ = 10 kg and W₂ = 10 kg

Answer: (d) W₁ = 10 kg and W₂ = 10 kg

Explanation: The two spring balances are in series. S₂ directly supports the 10 kg mass: it reads 10 kg. S₁ supports S₂ and the 10 kg mass together, but spring balances show the tension in their own spring, which equals the load hanging below them. S₁ is in the transmission path of the same 10 kg pull. Both balances read 10 kg. Concept Tested: Spring balances in series: both read the full suspended load

★ JOVIK Exam Insight A common error is to assume the load splits between two balances (5 kg each). This is wrong: in a series arrangement, each balance transmits the full load. The reading does not split.

Q. 18. Two forces of 5.0 N each are acting on a point mass. If the angle between the forces is 60°, then the net force acting on the point mass has magnitude close to:

(a)  8.6 N

(b)  4.3 N

(c)  5.0 N

(d)  6.7 N

Answer: (a) 8.6 N

Explanation: Using the parallelogram law: R = √(F₁² + F₂² + 2F₁F₂ cos θ) = √(25 + 25 + 2×25×cos 60°) = √(25 + 25 + 25) = √75 = 5√3 ≈ 8.66 N ≈ 8.6 N. Note: cos 60° = 0.5. Concept Tested: Vector addition of equal forces at 60°: resultant = √3 × F ≈ 8.66 N

Q. 19. Sand falls vertically on a conveyor belt at a rate of 0.1 kg/s. In order to keep the belt moving at a uniform speed of 2 m/s, the force required to be applied on the belt is:

(a)  0 N

(b)  0.2 N

(c)  1.0 N

(d)  2.0 N

Answer: (b) 0.2 N

Explanation: Sand lands at rest and must be accelerated to belt speed v = 2 m/s. This is a variable-mass problem. The force required = (dm/dt) × v = 0.1 × 2 = 0.2 N. If no force were applied (option a), the belt would slow down due to momentum transfer from the falling sand. Concept Tested: Variable-mass force: conveyor belt: F = (dm/dt) × v

Q. 20. One block of 2.0 kg mass is placed on top of another block of 3.0 kg mass. The coefficient of static friction between the two blocks is 0.2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s², the maximum value of the frictional force is:

(a)  50 N

(b)  30 N

(c)  4 N

(d)  10 N

Answer: (c) 4 N

Explanation: The frictional force on the top block (2 kg) from the bottom block is what causes the top block to accelerate. Maximum static friction = μₛ × N = μₛ × m_top × g = 0.2 × 2 × 10 = 4 N. This is the maximum friction available to keep the blocks moving together. Beyond this, the top block slips. Concept Tested: Maximum static friction between stacked blocks: μₛ × m_top × g

Q. 21. Which one of the following statements is true?

(a)  The force of gravity of the Earth on the Moon is greater than the force of gravity of the Moon on the Earth.

(b)  The force of gravity of the Moon on the Earth is greater than the force of gravity of the Earth on the Moon.

(c)  The force of gravity of the Earth on the Moon and of the Moon on the Earth are equal in magnitude and are in the same direction.

(d)  The force of gravity of the Earth on the Moon and of the Moon on the Earth are equal in magnitude but are in opposite directions.

Answer: (d) The force of gravity of the Earth on the Moon and of the Moon on the Earth are equal in magnitude but are in opposite directions.

Explanation: This is Newton’s Third Law applied to gravity. The Earth pulls the Moon with a gravitational force, and the Moon pulls the Earth with an equal and opposite gravitational force. Neither is larger. The two forces act on different bodies and are directed toward each other: equal magnitude, opposite directions. Concept Tested: Newton’s Third Law and gravity: Earth-Moon gravitational forces are equal and opposite

NDA 2022-II

Q. 22. A bullet of mass 10 g is horizontally fired with velocity 300 m s⁻¹ from a pistol of mass 1 kg. What is the recoil velocity of the pistol?

(a)  0.3 m s⁻¹

(b)  3 m s⁻¹

(c)  −3 m s⁻¹

(d)  −0.3 m s⁻¹

Answer: (c) −3 m s⁻¹

Explanation: Before firing, total momentum = 0 (system at rest). After firing: m_bullet × v_bullet + m_pistol × v_pistol = 0. (0.01)(300) + (1)(v) = 0. 3 + v = 0. v = −3 m s⁻¹. The negative sign means the pistol recoils in the direction opposite to the bullet. Option (c) is correct. Concept Tested: Conservation of momentum: gun recoil (pistol-bullet system)

★ JOVIK Exam Insight Recoil velocity is a standard conservation-of-momentum calculation. Initial momentum = 0. Bullet forward → pistol backward. Always take care of units: mass in kg, velocity in m/s.

NDA 2022-I

Q. 23. A 5 N force is defined when a mass of 10 kg is accelerated with

(a)  5.0 cm/s²

(b)  0.5 m/s²

(c)  0.5 cm/s²

(d)  5.0 m/s²

Answer: (b) 0.5 m/s²

Explanation: From F = ma: a = F/m = 5 N / 10 kg = 0.5 m/s². The SI result is always in m/s², not cm/s². Option (a) 5.0 cm/s² = 0.05 m/s² is wrong. Option (c) 0.5 cm/s² = 0.005 m/s² is wrong. Concept Tested: Newton’s Second Law: F = ma calculation; acceleration in SI (m/s²)

Q. 24. A boy of mass 52 kg jumps with a horizontal velocity of 2 m/s onto a stationary cart of mass 3 kg. The cart is fixed with frictionless wheels. Which one of the following would be the speed of the cart?

(a)  2.15 m/s

(b)  1.89 m/s

(c)  1.51 m/s

(d)  2.51 m/s

Answer: (b) 1.89 m/s

Explanation: Conservation of momentum: initial momentum = 52 × 2 = 104 kg m/s. After the boy lands on the cart, both move together. Total mass = 52 + 3 = 55 kg. Final speed v = 104/55 ≈ 1.89 m/s. The speed is less than 2 m/s because the total mass has increased. Concept Tested: Conservation of momentum: perfectly inelastic collision (boy jumps onto cart)

NDA 2021-II

Q. 25. Weight and mass of an object are defined with Newton’s laws of motion. Which among the following is true?

(a)  Weight is a constant of proportionality

(b)  Mass is a constant of proportionality

(c)  Mass is not a constant of proportionality

(d)  Weight is a universal constant

Answer: (b) Mass is a constant of proportionality

Explanation: In Newton’s Second Law, F = ma: mass m is the constant of proportionality between force and acceleration. Weight W = mg is a force (the gravitational pull on a body), not a constant: it varies from place to place. Mass is fixed for a given body; weight depends on g. Concept Tested: Newton’s Second Law: mass is the constant of proportionality in F = ma

Q. 26. A uniform motion of a car along a circular path experiences

(a)  a change in speed due to a change in its direction of motion

(b)  a change in velocity due to a change in its direction of motion

(c)  a change in momentum due to no change in its direction of motion

(d)  a constant momentum due to a change in its direction of motion

Answer: (b) a change in velocity due to a change in its direction of motion

Explanation: In uniform circular motion, speed is constant but the direction of motion continuously changes. Since velocity is a vector (magnitude + direction), changing direction means changing velocity: even though speed is unchanged. Changing velocity means changing momentum and non-zero acceleration. Speed does not change; velocity does. Concept Tested: Uniform circular motion: velocity changes direction; speed and magnitude are constant

★ JOVIK Exam Insight Circular motion is the most persistently tested conceptual trap in this chapter. Speed = constant. Velocity ≠ constant (direction changes). Momentum ≠ constant. Acceleration ≠ zero (centripetal). NDA tested this in 2013-I and 2021-II.

NDA 2021-I

Q. 27. The statement “friction force is a contact force while magnetic force is a non-contact force” is

(a)  always true

(b)  true only at 0°C

(c)  a false statement

(d)  either true or false depending upon the temperature of the surroundings

Answer: (a) always true

Explanation: Friction is a contact force: it acts only when two surfaces are in direct physical contact. Magnetic force is a non-contact force: it acts at a distance through the magnetic field without requiring physical contact. This classification is independent of temperature. The statement is always true. Concept Tested: Contact vs non-contact forces: friction (contact) and magnetic force (non-contact)

NDA 2019-II

Q. 28. Let there be an object having some chemicals in it. It starts moving with a uniform velocity v and a chemical reaction starts happening. In this case, which of the following statement/s is/are correct?

1.  Chemical reactions happening in the system cannot change the velocity v of the center of mass of the object.

2.  Chemical reactions happening in the system cannot change kinetic energy of the particles inside with respect to the center of mass of object.

Select the correct answer using the code given below:

(a)  1 only

(b)  2 only

(c)  Both 1 and 2

(d)  Neither 1 nor 2

Answer: (a) 1 only

Explanation: Statement 1 is correct: internal forces (including chemical reactions) cannot change the velocity of the centre of mass of a system, because they produce equal and opposite impulses within the system. No external net force acts, so the centre of mass continues at velocity v. Statement 2 is wrong: internal reactions can convert chemical energy into kinetic energy of individual parts relative to the centre of mass: they definitely change internal kinetic energies. Concept Tested: Conservation of momentum: internal forces cannot change centre-of-mass velocity

Q. 29. When a ball bounces off the ground, which of the following changes suddenly? (Assume no loss of energy to the floor)

(a)  Its speed

(b)  Its momentum

(c)  Its kinetic energy

(d)  Its potential energy

Answer: (b) Its momentum

Explanation: In an elastic bounce (no energy loss), speed is unchanged: the ball leaves the ground at the same speed it arrived. Therefore kinetic energy (½mv²) is also unchanged. Potential energy (at ground level) is zero before and after. But momentum changes direction: from downward to upward: so the change in momentum is Δp = 2mv (a vector reversal). Momentum changes suddenly. Concept Tested: Elastic bounce: momentum reverses direction; speed and kinetic energy unchanged

★ JOVIK Exam Insight A classic vector-nature-of-momentum question. In elastic bounce: speed = constant, KE = constant, but momentum direction reverses → momentum changes. This tests whether students understand momentum as a vector.

NDA 2019-I

Q. 30. Which one of the following forces is non-central and non-conservative?

(a)  Frictional force

(b)  Electric force

(c)  Gravitational force

(d)  Mechanical force

Answer: (a) Frictional force

Explanation: A central force acts along the line joining two interacting bodies. A conservative force allows energy recovery: work done is path-independent. Friction acts opposite to motion (not along the line joining bodies, so non-central) and dissipates energy as heat (non-conservative). Gravitational and electric forces are both central and conservative. Concept Tested: Classification of forces: friction is non-central and non-conservative

NDA 2018-I

Q. 31. Which one of the following has maximum inertia?

(a)  An atom

(b)  A molecule

(c)  A one-rupee coin

(d)  A cricket ball

Answer: (d) A cricket ball

Explanation: Inertia is directly proportional to mass. Among an atom, a molecule, a one-rupee coin, and a cricket ball, the cricket ball has the greatest mass: and therefore the greatest inertia. A cricket ball is about 0.16 kg; a one-rupee coin is about 4.85 g; atoms and molecules are far smaller. Concept Tested: Inertia: directly proportional to mass; greatest mass has greatest inertia

Q. 32. A ball balanced on a vertical rod is an example of

(a)  stable equilibrium

(b)  unstable equilibrium

(c)  neutral equilibrium

(d)  perfect equilibrium

Answer: (b) unstable equilibrium

Explanation: In unstable equilibrium, any slight displacement moves the centre of gravity away from the equilibrium position, and forces push the body further away rather than restoring it. A ball on top of a rod has its centre of gravity high above a tiny base: any nudge topples it. In stable equilibrium (like a ball in a bowl), the centre of gravity rises with displacement and the body returns. Concept Tested: Types of equilibrium: unstable: slight displacement causes further departure

NDA 2016-II

Q. 33. Which one of the following is not a contact force?

(a)  Push force

(b)  Gravitational force

(c)  Frictional force

(d)  Strain force

Answer: (b) Gravitational force

Explanation: Contact forces act through direct physical contact between bodies: push, friction, tension, normal reaction, and strain forces are all contact forces. Gravitational force acts at a distance: no contact is required between the two masses. Electric and magnetic forces are also non-contact forces. Concept Tested: Contact vs non-contact forces: gravitational force acts at a distance

★ JOVIK Exam Insight Contact vs non-contact forces have been tested in NDA 2016-II, 2021-I, and 2024-II. Contact forces: push, friction, tension, normal. Non-contact forces: gravity, electric, magnetic.

Q. 34. When a force of 1 newton acts on a mass of 1 kg which is able to move freely, the object moves in the direction of force with a/an

(a)  speed of 1 km/s

(b)  acceleration of 1 m/s²

(c)  speed of 1 m/s

(d)  acceleration of 1 km/s²

Answer: (b) acceleration of 1 m/s²

Explanation: F = ma. With F = 1 N and m = 1 kg: a = F/m = 1 m/s². This is the dimensional definition of the Newton: 1 N produces 1 m/s² acceleration in 1 kg. The force produces acceleration, not a constant speed. Speed of 1 m/s would only apply at one specific instant. Concept Tested: Newton’s Second Law: 1 N on 1 kg produces 1 m/s² acceleration (definition of Newton)

NDA 2016-I

Q. 35. Mass of a particular amount of substance

1.  is the amount of matter present in it.

2.  does not vary from place to place.

3.  changes with change in gravitational force.

Select the correct answer using the code given below:

(a)  1, 2 and 3

(b)  1 and 2 only

(c)  2 and 3 only

(d)  1 only

Answer: (b) 1 and 2 only

Explanation: Statement 1 is correct: mass is the measure of amount of matter. Statement 2 is correct: mass does not vary with location: it is the same at the poles, equator, or on the Moon. Statement 3 is wrong: mass does not change with gravitational force. Weight (= mg) changes with g; mass does not. Concept Tested: Mass vs weight: mass is location-independent; weight changes with g

★ JOVIK Exam Insight Mass is constant everywhere. Weight = mg varies with g. NDA has tested this distinction across 2016-I and 2021-II. Mass is the constant of proportionality in F = ma: not weight.

Q. 36. Two bodies A and B are moving with equal velocities. The mass of B is double that of A. In this context, which one of the following statements is correct?

(a)  Momentum of B will be double that of A.

(b)  Momentum of A will be double that of B.

(c)  Momentum of B will be four times that of A.

(d)  Momenta of both A and B will be equal.

Answer: (a) Momentum of B will be double that of A.

Explanation: Momentum p = mv. Both A and B have equal velocity v. If m_B = 2m_A, then p_B = 2m_A × v = 2(m_A × v) = 2p_A. Momentum scales linearly with mass at constant velocity: not as the square of mass. Concept Tested: Linear momentum: p = mv; doubling mass doubles momentum at constant velocity

Q. 37. A man weighing 70 kg is coming down in a lift. If the cable of the lift breaks suddenly, the weight of the man would become

(a)  70 kg

(b)  35 kg

(c)  140 kg

(d)  Zero

Answer: (d) Zero

Explanation: If the cable breaks, the lift falls freely with acceleration g. In free fall, apparent weight = m(g − a). When a = g (free fall), apparent weight = m(g − g) = 0. The man experiences complete weightlessness. The gravitational pull still acts: but the normal reaction from the floor of the lift is zero. Concept Tested: Apparent weight in free fall: weightlessness when lift cable breaks

Q. 38. The impulse on a particle due to a force acting on it during a given time interval is equal to the change in its

(a)  force

(b)  momentum

(c)  work done

(d)  energy

Answer: (b) momentum

Explanation: Impulse = F × Δt. By the impulse-momentum theorem, impulse equals the change in linear momentum of the body: J = Δp = mv_f − mv_i. Impulse does not equal change in force, work done, or energy. Concept Tested: Impulse-momentum theorem: impulse equals change in linear momentum

NDA 2015-II

Q. 39. Conservation of momentum in a collision between particles can be understood on the basis of:

(a)  Newton’s first law of motion

(b)  Newton’s second law of motion only

(c)  Both Newton’s second law of motion and Newton’s third law of motion

(d)  Conservation of energy

Answer: (c) Both Newton’s second law of motion and Newton’s third law of motion

Explanation: During a collision, Newton’s Third Law tells us that the forces two bodies exert on each other are equal and opposite. Newton’s Second Law tells us that impulse equals change in momentum. Equal and opposite forces acting for the same time produce equal and opposite changes in momentum: so total momentum is conserved. Both laws together are required. Concept Tested: Conservation of momentum: derived from Newton’s Second and Third Laws combined

Q. 40. Two forces, one of 3 newton and another of 4 newton are applied on a standard 1 kg body, placed on a horizontal and frictionless surface, simultaneously along the x-axis and the y-axis, respectively, as shown below: The magnitude of the resultant acceleration is:

(a)  7 m/s²

(b)  1 m/s²

(c)  5 m/s²

(d)  √7 m/s²

Answer: (c) 5 m/s²

Explanation: The two forces are perpendicular (along x and y axes). Resultant force = √(3² + 4²) = √(9 + 16) = √25 = 5 N. Since mass = 1 kg, acceleration = F/m = 5/1 = 5 m/s². Perpendicular forces are combined by Pythagoras, not by simple addition (which would give 7 N: a common error). Concept Tested: Resultant force from perpendicular components: Pythagorean theorem (3-4-5 triangle)

NDA 2013-I

Q. 41. A car is moving with a uniform speed. However its momentum is changing. Then the car

(a)  may be on an elliptical path

(b)  is moving on a straight path without acceleration

(c)  is moving on a straight path with acceleration

(d)  is moving without any acceleration

Answer: (a) may be on an elliptical path

Explanation: Momentum is a vector (p = mv). Uniform speed means constant |v|, but if direction changes, velocity changes and hence momentum changes. This happens in any curved path: circular, elliptical, or any non-straight path. On a straight path with constant speed, momentum is constant. So changing momentum with constant speed implies a curved path. Concept Tested: Vector nature of momentum: constant speed but changing momentum implies curved path

★ JOVIK Exam Insight Circular motion causes continuous momentum change even at constant speed. The centripetal force changes direction of velocity at every instant. This concept links Newton’s First and Second Laws.

Q. 42. A motor vehicle is moving on a circle with a uniform speed. The net acceleration of the vehicle is

(a)  zero

(b)  towards the centre of the circle

(c)  away from the centre along the radius of the circle

(d)  perpendicular to the radius and along the velocity

Answer: (b) towards the centre of the circle

Explanation: Uniform circular motion has constant speed but continuously changing direction. Changing direction of velocity means there is acceleration. This acceleration (centripetal acceleration) is directed toward the centre of the circle at every instant. It is NOT zero, not outward, and not along the velocity. Concept Tested: Uniform circular motion: centripetal acceleration directed toward the centre

Q. 43. An object is undergoing a non-accelerated motion. Its rate of change in momentum is

(a)  a non-zero constant

(b)  zero

(c)  not a constant

(d)  None of the above

Answer: (b) zero

Explanation: Non-accelerated motion means zero acceleration. By Newton’s Second Law, F = ma = 0 when a = 0. Force also equals dp/dt (rate of change of momentum). If F = 0, then dp/dt = 0: rate of change of momentum is zero. Concept Tested: Newton’s Second Law: zero acceleration implies zero rate of change of momentum

Q. 44. A force F is applied on a body (which moves on a straight line) for a duration of 3 s. The momentum of the body changes from 10 g cm/s to 40 g cm/s. The magnitude of the force F is

(a)  10 dynes

(b)  10 newtons

(c)  120 dynes

(d)  12 dynes

Answer: (a) 10 dynes

Explanation: Change in momentum = 40 − 10 = 30 g cm/s. Force = Δp/Δt = 30 g cm/s ÷ 3 s = 10 g cm/s². In CGS units, 1 dyne = 1 g cm/s². So F = 10 dynes. Option (c) 120 dynes would be the answer if Δp were not divided by time: a common error. Concept Tested: Impulse-momentum theorem: F = Δp/Δt in CGS units (dynes)

NDA 2011-II

Q. 45. Momentum of a body is

1.  a vector quantity.

2.  a conserved quantity in an isolated system.

3.  same as force in linear motion.

Select the correct answer using the code given below:

(a)  1 and 3 only

(b)  2 and 3 only

(c)  1 and 2 only

(d)  1, 2 and 3

Answer: (c) 1 and 2 only

Explanation: Statement 1 is correct: momentum (p = mv) is a vector. Statement 2 is correct: in an isolated system with no external force, total momentum is conserved. Statement 3 is wrong: force is the rate of change of momentum (F = dp/dt), not momentum itself. Momentum and force are distinct quantities with different dimensions. Concept Tested: Nature of linear momentum: vector quantity and conservation in isolated systems

Q. 46. If the ratio of the weight of a man in a stationary lift and when it is moving downwards with uniform acceleration ‘a’ is 3 : 2, then the value of ‘a’ is

(a)  3g/2

(b)  g/3

(c)  g

(d)  2g/3

Answer: (b) g/3

Explanation: Actual weight = mg. Apparent weight in a downward-accelerating lift = m(g − a). Ratio: mg / m(g − a) = 3/2. So 2g = 3(g − a), giving 2g = 3g − 3a, hence 3a = g, so a = g/3. Concept Tested: Apparent weight in a lift: downward acceleration reduces apparent weight

★ JOVIK Exam Insight Apparent weight in lifts is a recurring NDA calculation. Remember: going down with acceleration a → apparent weight = m(g − a). Cable breaks → a = g → apparent weight = 0.

Q. 47. Which one among the following is correct for resultant of balanced forces?

(a)  It is zero

(b)  It is non-zero

(c)  It varies continuously

(d)  None of the above

Answer: (a) It is zero

Explanation: Balanced forces are forces that are in equilibrium: they are equal and opposite in all directions. Their vector sum (resultant) is zero. This is the condition for a body to remain in rest or uniform motion (Newton’s First Law). Concept Tested: Balanced forces: vector resultant is zero

NDA 2011-I

Q. 48. A man is at rest in the middle of a horizontal plane of perfectly smooth surface of ice. He can move himself to the shore by making use of Newton’s:

(a)  First law of motion

(b)  Second law of motion

(c)  Third law of motion

(d)  First, second and third laws of motion

Answer: (c) Third law of motion

Explanation: The man can throw an object (clothing, keys, etc.) away from the shore. By Newton’s Third Law, the reaction force propels him toward the shore. With no friction on the ice, this reaction is the only horizontal force available. This is the same principle used by rockets and jet engines. Concept Tested: Newton’s Third Law: action-reaction on a frictionless surface

Q. 49. When a moving bus suddenly applies brakes, the passengers sitting in it fall in the forward direction. This can be explained by:

(a)  the theory of relativity

(b)  Newton’s first law

(c)  Newton’s second law

(d)  Newton’s third law

Answer: (b) Newton’s first law

Explanation: Newton’s First Law states that a body in motion continues in the same state of motion unless acted upon by an external force. When the bus brakes, the passengers’ bodies continue moving forward due to inertia: no net forward force has acted on them yet. The bus decelerates; the passengers do not. Concept Tested: Newton’s First Law: inertia of motion (passengers lurching forward when bus brakes)

★ JOVIK Exam Insight Bus-braking and coin-tossing-in-train are both Newton’s First Law examples. Both test inertia of motion. NDA has tested this concept across 2011-I and 2014-I (kinematics chapter).

Q. 50. A Jet engine works on the principle of conservation of:

(a)  linear momentum

(b)  angular momentum

(c)  energy

(d)  mass

Answer: (a) linear momentum

Explanation: A jet engine expels hot gases backward at high velocity. By conservation of linear momentum (and equivalently Newton’s Third Law), the equal and opposite reaction propels the aircraft forward. The total momentum of the system (aircraft + expelled gas) is conserved. Concept Tested: Conservation of linear momentum: principle behind jet and rocket propulsion

NDA 2010-II

Q. 51. A body is at rest on the surface of the earth. Which one among the following statements is correct regarding this?

(a)  No force is acting on the body

(b)  Only weight of the body acts on it

(c)  Net downward force is equal to the net upward force

(d)  None of the above statements is correct

Answer: (c) Net downward force is equal to the net upward force

Explanation: A stationary body on the Earth’s surface is acted upon by two forces: its weight (mg, downward) and the normal reaction from the surface (N, upward). For equilibrium, N = mg. The net downward force equals the net upward force. The body is NOT force-free: it is in force balance. Concept Tested: Newton’s First Law: equilibrium of a body at rest (force balance, not force absence)

★ JOVIK Exam Insight A stationary body is never force-free. Two forces act: weight downward and normal reaction upward. They are equal and opposite: net force is zero. NDA has tested this misconception directly.

Q. 52. A circus performer of mass M is walking along a wire as shown in the figure given above. The tension T in the wire is: (g = acceleration due to gravity)

(a)  approximately Mg

(b)  less than Mg

(c)  more than Mg

(d)  depends on whether the performer stands on one or two feet

Answer: (c) more than Mg

Explanation: The performer’s weight Mg acts downward. The wire makes a very small angle θ with the horizontal at each side. The vertical component of tension in each segment (T sin θ) must support half the performer’s weight. Since sin θ is very small, T must be very large to provide the required vertical support. T = Mg/(2 sin θ) >> Mg. The tension in a nearly horizontal wire always exceeds the suspended load. Concept Tested: Tension in a nearly horizontal wire supporting a vertical load: tension > load

Q. 53. A heavy ball is suspended as shown in the figure given above. A quick jerk on the lower string will break that string but a slow pull on the lower string will break the upper string. The first result occurs because:

(a)  the force is too small to move the ball

(b)  air friction holds the ball back

(c)  of action and reaction

(d)  the ball has inertia

Answer: (d) the ball has inertia

Explanation: A quick jerk creates a large impulsive force confined to the lower string. The ball’s inertia resists the sudden change: so the impulse does not have time to be transmitted through the ball to the upper string. The lower string breaks before the ball moves. A slow pull gradually builds tension through the ball’s weight, loading the upper string until it breaks. Concept Tested: Inertia: role of inertia in string-breaking under impulsive vs sustained force

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