Work, Energy & Power – NDA Physics PYQs

NDA 2026-I

Q. 1. A ball is dropped from rest from height h above the ground in frame S (the Earth’s frame). Another frame S’ is moving upwards with constant speed u relative to the Earth. In frame S’, the ball has initial downward speed v. Pertaining to the change in kinetic energy (ΔK) of the ball from release to just before hitting the ground as measured in frames S and S’ separately, which one of the following is correct?

(a)  ΔK is the same in both the frames.

(b)  ΔK is larger in frame S.

(c)  ΔK is larger in frame S’.

(d)  ΔK cannot be compared without the information of mass of the ball.

Answer: (c) ΔK is larger in frame S’.

Explanation: In frame S (Earth): ball drops from rest, final speed = √(2gh). ΔKE_S = mgh. In frame S’ (moving upward at speed u): initial speed of ball = u (downward), final speed = √(2gh) + u (downward). ΔKE_S’ = ½m[(√(2gh)+u)² − u²] = ½m[2gh + 2u√(2gh)] = mgh + mu√(2gh) > mgh. ΔKE in S’ is larger because the ball has a larger effective displacement in the upward-moving frame. Concept Tested: Frame dependence of kinetic energy change: ΔKE is larger in the upward-moving frame S’

★ JOVIK Exam Insight A high-difficulty NDA 2026-I question: kinetic energy changes ARE frame-dependent. In the frame moving upward, the ball effectively falls further relative to that frame, so it gains more KE as measured in that frame. This is a university-level concept now appearing in NDA.

Q. 2. A constant power machine pulls a block on a smooth horizontal surface. Which one of the following correctly describes the relation between speed of the block (v) and time (t)?

(a)  v ∝ t

(b)  v ∝ √t

(c)  v ∝ t²

(d)  v ∝ t³/²

Answer: (b) v ∝ √t

Explanation: At constant power P: P = Fv = mav. So ma = P/v. Then m(dv/dt) = P/v. Rearranging: mv dv = P dt. Integrating: ½mv² = Pt. Therefore v² = 2Pt/m ∝ t. So v ∝ √t. Speed increases as the square root of time when power is constant. Concept Tested: Constant power kinematics: v ∝ √t (from P = Fv and F = ma)

★ JOVIK Exam Insight A post-2023 NDA trend: constant-power kinematics. P = Fv = mav → integrate → v ∝ √t. This is now at NDA-level after appearing in 2026-I. Contrast with constant force: v ∝ t.

Q. 3. A particle is moved from point X to point Y under a force field. The work done depends only on the initial and final speeds. Which one of the following is correct?

(a)  The force is conservative.

(b)  The net work equals the change in kinetic energy.

(c)  The force is constant.

(d)  The velocity must always increase.

Answer: (b) The net work equals the change in kinetic energy.

Explanation: The work-energy theorem (W_net = ΔKE = ½mv_f² − ½mv_i²) holds universally: for all forces, conservative or not. When work depends only on initial and final speeds, it equals the change in kinetic energy. Option (a) is wrong: conservative force means work is path-independent (depends on positions, not speeds): the conditions are related but distinct. Option (c) is wrong: the force need not be constant. Option (d) is wrong: velocity need not always increase. Concept Tested: Work-energy theorem: W_net = ΔKE applies universally; does not prove conservative force

★ JOVIK Exam Insight A subtle NDA 2026-I trap: work depending on initial and final speeds proves W = ΔKE (work-energy theorem): but does NOT prove the force is conservative. Conservative force means work depends on initial and final POSITIONS (path-independent). These are related but different conditions.

NDA 2025-II

Q. 4. Which one of the following statements for the work done by gravity on a body is NOT correct?

(a)  Work done is independent of the path followed by the body

(b)  Work done depends only on the vertical distance separating the initial and the final position of the body

(c)  Work done depends on the path followed by the body

(d)  Work done is zero if the body is displaced horizontally to the force of gravity

Answer: (c) Work done depends on the path followed by the body

Explanation: Gravity is a conservative force. Work done by gravity depends only on vertical displacement, not on the path. Option (a) is correct: path-independent. Option (b) is correct: only vertical displacement matters. Option (d) is correct: horizontal displacement perpendicular to gravity gives zero work. Option (c) is the false statement: work done by gravity does NOT depend on path. Concept Tested: Conservative force (gravity): work is path-independent; depends only on vertical displacement

★ JOVIK Exam Insight Path-independence of gravitational work has been directly tested in NDA 2025-II. Gravity is the standard example of a conservative force. Work = mgh regardless of the path taken to change height by h.

Q. 5. The work done by the force acting on an object is zero if the displacement of the object

(a)  is in the opposite direction of the direction of force

(b)  is in the same direction of the direction of force

(c)  is in perpendicular direction of the direction of force

(d)  none of above

Answer: (c) is in perpendicular direction of the direction of force

Explanation: Work = F × s × cos θ. When force and displacement are perpendicular, θ = 90°, cos 90° = 0, so W = 0. Opposite direction (θ = 180°) gives negative work. Same direction (θ = 0°) gives maximum positive work. Zero work requires a 90° angle between force and displacement. Concept Tested: Zero work condition: force and displacement are perpendicular (θ = 90°)

Q. 6. The energy is always conserved for a system which is

(a)  isolated only

(b)  non-isolated only

(c)  both isolated and non-isolated

(d)  none of above

Answer: (c) both isolated and non-isolated

Explanation: The law of conservation of energy is universal: energy is always conserved in any system. In an isolated system, total internal energy remains constant. In a non-isolated system, any energy lost by the system equals energy gained by the surroundings: the total universe energy is still conserved. Option (a) is a common misconception. Concept Tested: Law of conservation of energy: applies to all systems, not isolated systems only

★ JOVIK Exam Insight A precise NDA conceptual trap: students often say conservation of energy applies only to isolated systems. Wrong: it applies universally. Energy is always conserved. In a non-isolated system, energy flows to/from the surroundings, but total energy (system + surroundings) is constant.

Q. 7. Suppose, a ball of mass M is thrown upwards from a point A and it reaches up to the highest point B and returns back to point A, which one among the following is correct?

(a)  Kinetic Energy at A = Potential Energy at B

(b)  Kinetic Energy at A = Potential Energy at A

(c)  Kinetic Energy at B = Potential Energy at B

(d)  Kinetic Energy at B = Kinetic Energy at A

Answer: (a) Kinetic Energy at A = Potential Energy at B

Explanation: At point A (launch point), the ball has KE (velocity = u) and zero PE (reference level). At point B (maximum height), the ball has zero KE (velocity = 0) and maximum PE. By conservation of energy: KE at A = PE at B. At B, KE = 0, so KE at B ≠ PE at B (unless both are zero, which contradicts the scenario). KE at A ≠ KE at B since the ball decelerates. Concept Tested: Conservation of energy: KE at launch equals PE at maximum height

NDA 2024-II

Q. 8. An incandescent electric bulb converts 20% of power into light. Filament resistance 200Ω, current 2A, ON for 10h, rate ₹5/unit. Money spent on producing light:

(a)  ₹5

(b)  ₹6

(c)  ₹7

(d)  ₹8

Answer: (d) ₹8

Explanation: Total power = I²R = 2² × 200 = 800 W = 0.8 kW. Light power = 20% of 0.8 kW = 0.16 kW. Energy for light in 10 h = 0.16 × 10 = 1.6 kWh. Cost = 1.6 × ₹5 = ₹8. Concept Tested: Electrical power efficiency: P = I²R, then 20% efficiency, then energy cost

Q. 9. There is a ball of mass 320 g. It has 625 J potential energy when released freely from a height. The speed with which it will hit the ground is

(a)  62·5 m/s

(b)  2·0 m/s

(c)  50 m/s

(d)  40 m/s

Answer: (a) 62·5 m/s

Explanation: At the ground, all PE converts to KE: ½mv² = 625. Mass = 320 g = 0.32 kg. v² = 2 × 625/0.32 = 3906.25. v = √3906.25 = 62.5 m/s. Do not forget to convert grams to kilograms. Using g = 10 m/s²: mgh = 625 → h = 625/(0.32 × 10) = 195.3 m, which is consistent. Concept Tested: Conservation of energy: PE to KE conversion; speed = √(2 × PE/m)

NDA 2023-I

Q. 10. Two identical containers X and Y are connected at the bottom by a thin tube of negligible volume. The tube has a valve in it. Initially container X has a liquid filled up to height h in it and container Y is empty. When the valve is opened, both containers have equal amount of liquid in equilibrium. If the initial (before the valve is opened) potential energy of the liquid is P₁ and the final potential energy is P₂, then:

(a)  P₁ = P₂

(b)  P₁ = 4P₂

(c)  P₁ = 2P₂

(d)  P₁ = 8P₂

Answer: (c) P₁ = 2P₂

Explanation: Let liquid have total mass m, total height in X = h. Initial centre of mass of liquid is at h/2, so P₁ = mg(h/2). After equalising, each container has mass m/2 at height h/4 (centre of mass of each column of height h/2). Final PE: P₂ = 2 × (m/2)g(h/4) = mg(h/4). Ratio: P₁/P₂ = (h/2)/(h/4) = 2. So P₁ = 2P₂. The lost energy is dissipated as heat through the connecting tube. Concept Tested: Liquid potential energy redistribution: P₁ = 2P₂ when liquid equalises between two containers

Q. 11. The power required to lift a mass of 8.0 kg up a vertical distance of 4 m in 2 s is (taking acceleration due to gravity as 10 m/s²):

(a)  80 W

(b)  160 W

(c)  320 W

(d)  640 W

Answer: (b) 160 W

Explanation: Work done against gravity = mgh = 8 × 10 × 4 = 320 J. Power = Work/Time = 320/2 = 160 W. Option (a) 80 W would result from using g = 5 or from omitting the factor of 2 in mass conversion. Option (c) 320 W would result from forgetting to divide by time. Concept Tested: Power: P = mgh/t; lifting mass against gravity

NDA 2022-I

Q. 12. A mass M is dragged by a pulley on a horizontal plane by a force anti-parallel to its displacement. The work done in pulling the mass M is

(a)  zero

(b)  positive

(c)  infinite

(d)  negative

Answer: (d) negative

Explanation: Work = F · s · cos θ. The force is anti-parallel to displacement: the angle between them is 180°, so cos 180° = −1. Therefore W = −Fs (negative). The physical setup of pulling through a pulley can create a force directed opposite to the displacement. The sign depends entirely on the angle, not on the act of pulling. Concept Tested: Sign of work: anti-parallel force gives negative work in pulley arrangement

★ JOVIK Exam Insight Anti-parallel force and displacement always gives negative work. NDA has tested this directly in 2022-I (pulley arrangement) and asked about it conceptually in 2021-II. Force opposing motion = negative work.

Q. 13. The energy possessed by a body due to its change in position or shape is called

(a)  thermal energy

(b)  potential energy

(c)  kinetic energy

(d)  electric energy

Answer: (b) potential energy

Explanation: Potential energy is stored energy by virtue of position (gravitational PE: height above a reference) or configuration/shape (elastic PE: stretched spring or compressed rubber). Kinetic energy is energy of motion. Thermal energy is energy of random molecular motion. Concept Tested: Definition of potential energy: energy due to position or shape/configuration

NDA 2021-II

Q. 14. Work is said to be one Joule when a force of

(a)  4 N moves an object by 25 cm

(b)  2 N moves an object by 1 m

(c)  1 N moves an object by 1 cm

(d)  1 N moves an object by 50 cm

Answer: (a) 4 N moves an object by 25 cm

Explanation: Work = Force × Displacement (when force and displacement are parallel). Option (a): W = 4 × 0.25 = 1 J ✓. Option (b): W = 2 × 1 = 2 J ✗. Option (c): W = 1 × 0.01 = 0.01 J ✗. Option (d): W = 1 × 0.50 = 0.5 J ✗. Only option (a) gives exactly 1 J. Concept Tested: Definition of 1 Joule: F × d = 1 J; arithmetic check of all four options

★ JOVIK Exam Insight A precision-arithmetic trap: the obvious answer (1 N through 1 m) is not among the options. Convert cm to m before calculating: 25 cm = 0.25 m; 1 cm = 0.01 m; 50 cm = 0.50 m. Only 4 N × 0.25 m = 1 J.

Q. 15. An object of mass 2000 g possesses 100 J kinetic energy. The object must be moving with a speed of

(a)  10.0 m/s

(b)  11.1 m/s

(c)  11.2 m/s

(d)  12.1 m/s

Answer: (a) 10.0 m/s

Explanation: Mass = 2000 g = 2 kg. KE = ½mv² → 100 = ½ × 2 × v² → v² = 100 → v = 10 m/s exactly. The distractors (11.1, 11.2, 12.1) result from forgetting to convert grams to kilograms or from incorrect formula rearrangement. Concept Tested: Kinetic energy formula: KE = ½mv²; unit conversion: g to kg

Q. 16. A negative work is done when an applied force F and the corresponding displacement S are

(a)  perpendicular to each other

(b)  parallel to each other

(c)  anti-parallel to each other

(d)  equal in magnitude

Answer: (c) anti-parallel to each other

Explanation: Work W = F · s · cos θ. When force and displacement are anti-parallel, θ = 180°, cos 180° = −1, so W = −Fs (negative). When perpendicular, θ = 90°, W = 0. When parallel, θ = 0°, W = +Fs (positive). Equal magnitude has no effect on the sign of work. Concept Tested: Sign of work: anti-parallel force and displacement gives negative work (W = F·s·cos 180° = −Fs)

NDA 2020-I & II

Q. 17. The cost of energy to operate an industrial refrigerator that consumes 5 kW power working 10 hours per day for 30 days will be (Given that the charge per kW.h of energy = ₹4)

(a)  ₹600

(b)  ₹6,000

(c)  ₹1,200

(d)  ₹1,500

Answer: (b) ₹6,000

Explanation: Energy consumed = Power × Time = 5 kW × 10 h/day × 30 days = 1500 kWh. Cost = 1500 kWh × ₹4/kWh = ₹6,000. Option (a) ₹600 is the result if only 1 day is calculated instead of 30. Always multiply by all three factors: power, hours per day, and number of days. Concept Tested: Electrical energy cost: Power (kW) × Time (hours) × Rate (₹/kWh)

★ JOVIK Exam Insight Electrical energy cost is a three-factor calculation: power × daily hours × days. NDA has tested this type in 2020-I & II and 2024-II. Always express power in kW and time in hours before multiplying by the unit rate.

NDA 2019-II

Q. 18. A rigid body of mass 2 kg is dropped from a stationary balloon kept at a height of 50 m from the ground. The speed of the body when it just touches the ground and the total energy when it is dropped from the balloon are respectively (acceleration due to gravity = 9·8 m/s²)

(a)  980 m.s⁻¹ and 980 J

(b)  √980 m.s⁻¹ and √980 J

(c)  980 m.s⁻¹ and √980 J

(d)  √980 m.s⁻¹ and 980 J

Answer: (d) √980 m.s⁻¹ and 980 J

Explanation: Total energy at release = PE = mgh = 2 × 9.8 × 50 = 980 J. On reaching the ground, all PE converts to KE: ½mv² = 980. v² = 980. v = √980 m/s ≈ 31.3 m/s. Total energy is 980 J (conserved throughout). Speed is √980 m/s: not 980 m/s, which would be impossible. Concept Tested: Conservation of energy in free fall: speed = √(2gh), total energy = mgh = constant

★ JOVIK Exam Insight A common trap: the speed is √980 m/s (about 31 m/s), NOT 980 m/s. 980 is the numerical value of total energy in joules. Options (a) and (c) confuse the numerical value of energy with the speed. Always take the square root when using KE = ½mv².

NDA 2019-I

Q. 19. Which one of the following energy is stored in the links between the atoms?

(a)  Nuclear energy

(b)  Chemical energy

(c)  Potential energy

(d)  Thermal energy

Answer: (b) Chemical energy

Explanation: Chemical energy is stored in the chemical bonds (links) between atoms. When these bonds are broken and reformed (during chemical reactions), energy is released or absorbed. Nuclear energy is stored in the nucleus: between nucleons. Thermal energy is the kinetic energy of random molecular motion. Concept Tested: Chemical energy: stored in interatomic bonds (links between atoms)

Q. 20. The correct sequence of energy transfer that occurs when an apple falls to the ground is

(a)  Gravitational potential energy → heat energy to air → kinetic energy → heat energy to ground and apple → sound energy

(b)  Gravitational potential energy → sound energy → kinetic energy → heat energy to air → heat energy to ground and apple

(c)  Gravitational potential energy → kinetic energy → heat energy to air → heat energy to ground and apple → sound energy

(d)  Gravitational potential energy → kinetic energy → sound energy → heat energy to air → heat energy to ground and apple

Answer: (c) Gravitational potential energy → kinetic energy → heat energy to air → heat energy to ground and apple → sound energy

Explanation: As the apple falls, PE converts to KE. Air resistance continuously converts some KE to heat during the fall. On impact with the ground, the remaining KE converts to heat in the ground and apple, plus sound energy at the moment of impact. Sound is produced at impact: not during the fall. Option (a) places heat before KE, which reverses the primary conversion sequence. Concept Tested: Energy transformation sequence: falling apple: PE → KE → heat (air resistance) → heat + sound (impact)

NDA 2017-II

Q. 21. Which one of the following statements about energy is correct?

(a)  Energy can be created as well as destroyed.

(b)  Energy can be created but not destroyed.

(c)  Energy can neither be created nor destroyed.

(d)  Energy cannot be created but can be destroyed.

Answer: (c) Energy can neither be created nor destroyed.

Explanation: This is the law of conservation of energy. Energy cannot be created from nothing, nor can it be destroyed into nothing. It can only be transformed from one form to another: mechanical to heat, chemical to electrical, etc. The total energy of the universe remains constant. Concept Tested: Law of conservation of energy: energy can only be transformed, not created or destroyed

NDA 2016-II

Q. 22. How is the kinetic energy of a moving object affected if the net work done on it is positive?

(a)  Decreases

(b)  Increases

(c)  Remains constant

(d)  Becomes zero

Answer: (b) Increases

Explanation: The work-energy theorem states: W_net = ΔKE = KE_final − KE_initial. If net work done is positive, ΔKE > 0, meaning kinetic energy increases. Positive work adds energy to the object, accelerating it. Concept Tested: Work-energy theorem: positive net work increases kinetic energy

NDA 2016-I

Q. 23. A body has a free fall from a height of 20 m. After falling through a distance of 5 m, the body would

(a)  lose one-fourth of its total energy

(b)  lose one-fourth of its potential energy

(c)  gain one-fourth of its potential energy

(d)  gain three-fourth of its total energy

Answer: (b) lose one-fourth of its potential energy

Explanation: Initial PE = mgh = mg × 20. After falling 5 m (one-quarter of 20 m), height remaining = 15 m. PE remaining = mg × 15 = ¾ of initial PE. PE lost = ¼ of initial PE. Total mechanical energy remains constant: the lost PE converts to kinetic energy. The body has not lost any total energy. Concept Tested: Free fall: PE lost after falling ¼ of total height equals ¼ of initial PE; total energy conserved

★ JOVIK Exam Insight A common NDA trap: after falling 5 m out of 20 m, students think the body has lost ¼ of total energy. Wrong: total energy is conserved. Only PE changes (converted to KE). The loss is ¼ of potential energy, not total energy.

Q. 24. Which one of the following is NOT a form of stored energy?

(a)  Nuclear energy

(b)  Potential energy

(c)  Electrical energy

(d)  Chemical energy

Answer: (c) Electrical energy

Explanation: Nuclear energy (stored in nuclear bonds), potential energy (stored by virtue of position or configuration), and chemical energy (stored in chemical bonds) are all forms of stored energy. Electrical energy in a circuit is energy in transit or flow: not stored energy in the conventional sense. This question contrasts energy storage with energy transfer. Concept Tested: Stored energy vs energy in transit: electrical energy flows, not stored like nuclear or chemical

★ JOVIK Exam Insight Stored energy forms: nuclear, potential, chemical, elastic. Electrical energy is energy in transit (current flowing). NDA has tested this distinction directly in 2016-I. Related: chemical energy is stored in bonds between atoms: tested in 2019-I.

NDA 2015-I

Q. 25. Which one among the following happens when a swing rises to a certain height from its rest position?

(a)  Its potential energy decreases while kinetic energy increases

(b)  Its kinetic energy decreases while potential energy increases

(c)  Both potential and kinetic energy decrease

(d)  Both potential and kinetic energy increase

Answer: (b) Its kinetic energy decreases while potential energy increases

Explanation: At the lowest (rest) position, the swing has maximum kinetic energy and minimum potential energy. As it rises, it slows down (KE decreases) and gains height (PE increases). The total mechanical energy is conserved: the decrease in KE exactly equals the increase in PE. Concept Tested: KE ↔ PE conversion in a swinging pendulum: rising reduces KE and increases PE

NDA 2011-I

Q. 26. A body initially at rest is acted upon by a constant force. The rate of change of its kinetic energy varies:

(a)  linearly with square root of time

(b)  linearly with time

(c)  linearly with square of time

(d)  inversely with time

Answer: (b) linearly with time

Explanation: With constant force F on mass m (starting from rest): acceleration a = F/m. Velocity v = at. KE = ½mv² = ½m(at)² = ½ma²t². Rate of change of KE = dKE/dt = ma²t. This is linear in time: KE rate increases proportionally with t, not with t² or √t. Concept Tested: Rate of change of kinetic energy under constant force: linear with time

Q. 27. Mass of B is four times that of A. B moves with a velocity half that of A. Then B has:

(a)  kinetic energy equal to that of A

(b)  half the kinetic energy of A

(c)  twice the kinetic energy of A

(d)  kinetic energy one-fourth of A

Answer: (a) kinetic energy equal to that of A

Explanation: KE_A = ½m_A v_A². KE_B = ½(4m_A)(v_A/2)² = ½ × 4m_A × v_A²/4 = ½m_A v_A² = KE_A. The four-times mass and one-half velocity together cancel out: the squared velocity penalty (÷4) exactly offsets the mass advantage (×4). Both bodies have equal kinetic energy. Concept Tested: Kinetic energy ratio: mass ×4 and speed ÷2 give equal KE

★ JOVIK Exam Insight A classic NDA KE ratio trap: four times mass at half speed gives equal KE (not double or quadruple). Always square the velocity before comparing. This is among the most precisely tested KE comparison problems in NDA Physics.

NDA 2010-I

Q. 28. A body is thrown vertically upwards and then falls back on the ground. Its potential energy is maximum

(a)  on the ground

(b)  at the maximum height

(c)  during the return journey

(d)  both on the ground and at the maximum height

Answer: (b) at the maximum height

Explanation: As the body rises, kinetic energy converts to potential energy. At the maximum height, velocity = 0, so kinetic energy = 0. All the initial kinetic energy has become potential energy: this is the maximum. On the ground, height = 0 so potential energy = 0 (reference level). During the return journey, PE decreases as the body falls. Concept Tested: Gravitational PE: maximum when velocity is zero at the highest point

★ JOVIK Exam Insight KE ↔ PE conversion under gravity is the most tested concept in this chapter. At max height: KE = 0, PE = maximum. At ground level: PE = 0, KE = maximum. NDA tests this from multiple angles every few years.

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