Oscillations & Waves – NDA Physics PYQs

NDA 2026-I

Q. 1. A pendulum of length L oscillates with an angular amplitude of θ = 60° and time period T. Let T₀ = 2π√(L/g) be the time period for small angle oscillations, where g is the acceleration due to gravity. If air resistance is negligibly small and the string remains straight, then which one of the following is correct?

(a)  T will be slightly greater than T₀.

(b)  T will be slightly smaller than T₀.

(c)  T will be exactly equal to T₀.

(d)  T will depend upon the mass of the bob.

Answer: (a) T will be slightly greater than T₀.

Explanation: At large amplitudes (60°), the small-angle approximation sin θ ≈ θ no longer holds. The actual restoring force (−mg sin θ) is smaller than the linearised value (−mgθ) at large θ, making the effective restoring force weaker. A weaker restoring force results in a longer period. The actual period T > T₀ for any amplitude greater than the small-angle limit. Concept Tested: Large-angle pendulum: T > T₀ for large amplitudes due to sin θ < θ

★ JOVIK Exam Insight At large amplitudes (θ > 15°), the period T > T₀. The formula T₀ = 2π√(L/g) underestimates the actual period. Mass of the bob remains irrelevant even at large angles. This is a 2026-I advanced pendulum concept.

Q. 2. A very large container consists of an ideal gas. The speed of sound in the gas is x. When the pressure of the gas is doubled while keeping the temperature constant, the speed of sound now becomes y. What is the ratio of x to y?

(a)  1

(b)  √2

(c)  2

(d)  4

Answer: (a) 1

Explanation: Speed of sound in an ideal gas: v = √(γRT/M). This depends on temperature T, adiabatic constant γ, and molar mass M: but NOT on pressure at constant temperature. When pressure doubles at constant temperature, the density also doubles (ideal gas law: P = ρRT/M → ρ ∝ P at constant T). The ratio γP/ρ = γRT/M remains unchanged. Therefore y = x, and x/y = 1. Concept Tested: Speed of sound in an ideal gas: independent of pressure at constant temperature

★ JOVIK Exam Insight A 2026-I conceptual question: doubling pressure at constant temperature does not change sound speed. Both pressure and density double, so their ratio (which determines speed) stays constant. Speed of sound depends on temperature, not pressure.

NDA 2025-II

Q. 3. SONAR stands for

(a)  Sonographic Natural Ranging

(b)  Simple Operational Navigation Ranging

(c)  Sound Navigation and Ranging

(d)  Simple Operational Natural Ranging

Answer: (c) Sound Navigation and Ranging

Explanation: SONAR stands for Sound Navigation And Ranging. It uses ultrasonic waves to detect and measure distances to underwater objects by timing the reflection of sound pulses. Concept Tested: SONAR full form: Sound Navigation and Ranging

★ JOVIK Exam Insight SONAR full form has been tested in 2025-II. Related: NDA 2012-II (SONAR used by navigators), NDA 2022-II (SONAR uses ultrasonic waves). Three distinct aspects of SONAR have appeared across NDA papers.

NDA 2025-I

Q. 4. The frequency (f), wavelength (λ) and speed (v) of a sound wave are related as

(a)  f = vλ

(b)  λ = vf

(c)  f = λ/v

(d)  v = λf

Answer: (d) v = λf

Explanation: The fundamental wave relation is v = fλ, equivalently v = λf. Speed equals frequency times wavelength. Option (a) f = vλ is dimensionally incorrect. Option (b) λ = vf is also incorrect. Option (c) f = λ/v gives s⁻¹ = m/(m/s) = s: wrong. Concept Tested: Wave equation: v = fλ (speed = frequency × wavelength)

Q. 5. The length of a simple pendulum is increased four times to its previous value while the mass is doubled. What is the ratio of the new and previous time period of the pendulum?

(a)  3 : 1

(b)  2 : 5

(c)  2 : 1

(d)  3 : 2

Answer: (c) 2 : 1

Explanation: T = 2π√(L/g). T_new = 2π√(4L/g) = 2 × 2π√(L/g) = 2T. The mass increase (doubled) has no effect. T_new / T_old = 2/1. Ratio = 2:1. Concept Tested: Simple pendulum: quadrupling length doubles period; mass irrelevant

Q. 6. We hear an echo due to

(a)  Refraction of sound waves

(b)  Reflection of sound waves

(c)  Diffraction of sound waves

(d)  Resonance due to sound waves

Answer: (b) Reflection of sound waves

Explanation: An echo is a reflected sound: when sound waves strike a distant surface and return to the listener after a perceptible delay (minimum ~0.1 second). Refraction (change in speed/direction), diffraction (bending around obstacles), and resonance are different phenomena. Concept Tested: Echo: caused by reflection of sound waves from a distant surface

NDA 2024-I

Q. 7. Which one of the following does not apply to sound waves in fluids?

(a)  They transport energy

(b)  They need a medium to travel

(c)  They are transverse

(d)  They travel faster in liquids than in gases

Answer: (c) They are transverse

Explanation: Sound waves in fluids (liquids and gases) are always longitudinal: particles vibrate parallel to the direction of propagation. Transverse vibrations cannot be sustained in a fluid because fluids cannot support shear stress. Options (a), (b), and (d) are all correct. Option (c) is false. Concept Tested: Sound in fluids: longitudinal (NOT transverse); fluids cannot support transverse sound

Q. 8. Which one of the following statements regarding simple pendulum is correct?  Simple pendulum has a time period independent of amplitude:

(a)  only for small amplitudes because then the net force on its bob is independent of its displacement.

(b)  for any amplitude because the net force on the bob is always proportional to its displacement.

(c)  for any amplitude because the net force on the bob is independent of its displacement.

(d)  only for small amplitudes because then the net force on its bob is proportional to its displacement.

Answer: (d) only for small amplitudes because then the net force on its bob is proportional to its displacement.

Explanation: T = 2π√(L/g) is independent of amplitude only for small angles, where sin θ ≈ θ. Under this approximation, the restoring force F = −mg sin θ ≈ −mgθ, which is proportional to displacement. For large angles, sin θ ≠ θ, the force is non-linear, and the period becomes amplitude-dependent. Concept Tested: Simple pendulum period: amplitude-independent only for small angles (sin θ ≈ θ approximation)

NDA 2023-II

Q. 9. Which one of the following statements is true for sound waves propagating in air?

(a)  Sound is an electromagnetic wave and transverse in nature

(b)  Sound is a mechanical wave and longitudinal in nature

(c)  Sound is a mechanical wave and transverse in nature

(d)  Sound is an electromagnetic wave and longitudinal in nature

Answer: (b) Sound is a mechanical wave and longitudinal in nature

Explanation: Sound is a mechanical wave (requires a medium, cannot travel in vacuum) and longitudinal in nature (particles vibrate parallel to wave propagation: compressions and rarefactions). It is not electromagnetic, and not transverse in air. Concept Tested: Nature of sound: mechanical and longitudinal (not electromagnetic, not transverse)

NDA 2023-I

Q. 10. Which one of the following statements is not true for a flute, a musical instrument?

(a)  Momentum of waves on the blowing jet determines the loudness of the produced note.

(b)  Arrival time of the waves on the blowing jet determines the pitch of the produced note.

(c)  Sound comes from a vibrating column of air inside the flute.

(d)  Sound comes from a vibrating column of air inside as well as outside the flute.

Answer: (d) Sound comes from a vibrating column of air inside as well as outside the flute.

Explanation: In a flute, the musical sound is produced by a resonating column of air inside the instrument. The air outside the flute does not form a significant resonating column that contributes to the musical note. Options (a), (b), and (c) describe correctly the mechanism of a flute. Option (d) is false. Concept Tested: Flute sound production: vibrating air column inside the instrument (not outside)

Q. 11. Which one among the following is true for the speed of sound in a given medium?

(a)  Speed of sound remains same at all frequencies

(b)  Speed of sound is faster at higher frequencies

(c)  Speed of sound is slower at higher frequencies

(d)  Speed of sound is slower at higher wavelengths

Answer: (a) Speed of sound remains same at all frequencies

Explanation: In a given medium at given conditions, the speed of sound is independent of frequency. Ultrasonic and audible sound travel at the same speed in the same medium. Speed depends on the elastic modulus and density of the medium: not on the frequency of the wave. Concept Tested: Speed of sound: independent of frequency; same for all frequencies in a given medium

NDA 2022-II

Q. 12. When the pitch of sound increases, which one of the following increases?

(a)  Intensity

(b)  Loudness

(c)  Wavelength

(d)  Frequency

Answer: (d) Frequency

Explanation: Pitch is determined by frequency: higher pitch means higher frequency. When pitch increases, frequency increases. Intensity and loudness are related to amplitude, not pitch. Wavelength decreases when frequency increases (v = fλ, constant v). Concept Tested: Pitch of sound: determined by frequency; higher pitch = higher frequency

Q. 13. The amplitude of sound waves is measured in the units of

(a)  pressure

(b)  distance

(c)  time

(d)  speed

Answer: (b) distance

Explanation: Amplitude of a sound wave is the maximum displacement of the medium particles from their equilibrium (rest) positions. Displacement is a measure of distance: so amplitude is measured in units of distance (metres, centimetres, etc.). Pressure amplitude is a different quantity; here the question refers to displacement amplitude. Concept Tested: Amplitude of sound waves: measured in units of distance (maximum particle displacement)

Q. 14. A simple pendulum having bob of mass m and length of string l has time period of T. If the mass of the bob is doubled and the length of the string is halved, then the time period of this pendulum will be

(a)  T

(b)  T/√2

(c)  2T

(d)  √2 T

Answer: (b) T/√2

Explanation: T = 2π√(l/g). T_new = 2π√((l/2)/g) = (1/√2) × 2π√(l/g) = T/√2. The mass change (doubled) has no effect. Halving the length reduces the period by factor 1/√2. Concept Tested: Simple pendulum: halving length gives T/√2; doubling mass has no effect

Q. 15. ‘SONAR’ is a device that is used to measure the distance of underwater objects by a ship. Which of the following types of waves does it use for this purpose?

(a)  Infrasonic waves

(b)  Sound waves in audible range for human beings

(c)  Ultrasonic waves

(d)  All of the above

Answer: (c) Ultrasonic waves

Explanation: SONAR uses ultrasonic waves (frequency > 20 kHz) to detect underwater objects. Ultrasound is preferred because of its short wavelength (high resolution) and good reflection properties underwater. Infrasonic and audible sound are not used in standard SONAR systems. Concept Tested: SONAR: uses ultrasonic waves for underwater distance measurement

Q. 16. Which one of the following statements about the speed of sound waves is not correct?

(a)  The speed of sound waves in steel is higher than that in water.

(b)  The speed of sound waves in air decreases with increase in temperature.

(c)  The speed of sound waves in air increases with increase in temperature.

(d)  The speed of sound waves in water is higher than that in air.

Answer: (b) The speed of sound waves in air decreases with increase in temperature.

Explanation: The speed of sound in air increases with temperature (v ∝ √T in Kelvin). Option (b) claims it decreases: this is the false statement. Options (a), (c), and (d) are all correct. Speed order: steel (≈5100 m/s) > water (≈1500 m/s) > air (≈340 m/s). Moist air is less dense than dry air, so sound travels faster in moist air. Concept Tested: Speed of sound vs temperature: speed increases with temperature (not decreases)

Q. 17. A microphone converts

(a)  electrical signals to sound waves

(b)  sound waves to electrical signals

(c)  microwaves to sound waves

(d)  sound waves to microwaves

Answer: (b) sound waves to electrical signals

Explanation: A microphone is an electroacoustic transducer that converts sound waves (mechanical pressure variations) into electrical signals. The reverse: converting electrical signals to sound: is done by a speaker (loudspeaker). Microphones do not interact with microwaves. Concept Tested: Microphone: transducer converting sound waves to electrical signals

NDA 2022-I

Q. 18. The time period of a 1 m long pendulum approximates to

(a)  6 s

(b)  4 s

(c)  2 s

(d)  1 s

Answer: (c) 2 s

Explanation: T = 2π√(L/g) = 2π√(1/9.8) = 2π × 0.319 ≈ 2.0 s. A 1 m pendulum has a period of approximately 2 seconds. This is the standard ‘seconds pendulum’: it ticks once per second and completes one full oscillation in 2 s. Concept Tested: Standard pendulum: 1 m length gives T ≈ 2 s (the ‘seconds pendulum’)

Q. 19. A sound wave has a frequency of 1 kHz and wavelength 50 cm. How long will it take to travel 1 km?

(a)  5 s

(b)  4 s

(c)  3 s

(d)  2 s

Answer: (d) 2 s

Explanation: Speed v = f × λ = 1000 Hz × 0.50 m = 500 m/s. Distance = 1 km = 1000 m. Time = 1000/500 = 2 s. Concept Tested: Wave speed and travel time: v = fλ; time = distance ÷ speed

NDA 2021-II

Q. 20. Reverberation is a phenomenon associated with a

(a)  multiple refraction of sound

(b)  multiple reflection of sound

(c)  single refraction of sound

(d)  single reflection of sound

Answer: (b) multiple reflection of sound

Explanation: Reverberation is caused by multiple successive reflections of sound from the walls, ceiling, and floor of an enclosed space. The sound persists and echoes repeatedly: not a single reflection (which produces a simple echo) and not refraction. Concept Tested: Reverberation: caused by multiple reflections of sound in an enclosed space

Q. 21. Which among the following is true for propagation of sound waves?

(a)  Sound can travel in vacuum and it is a transverse wave in air

(b)  Sound cannot travel in vacuum and it is a longitudinal wave in air

(c)  Sound can travel in vacuum and it is a longitudinal wave in air

(d)  Sound cannot travel in vacuum and it is a transverse wave in air

Answer: (b) Sound cannot travel in vacuum and it is a longitudinal wave in air

Explanation: Sound is a mechanical wave: it requires a medium and cannot travel through vacuum. In air (and all fluids), sound is a longitudinal wave: particles vibrate parallel to the direction of propagation (compressions and rarefactions). Sound is never transverse in fluids. Concept Tested: Nature of sound: mechanical wave, longitudinal in air, cannot travel in vacuum

NDA 2021-I

Q. 22. The sound created in a big hall persists because of the repeated reflections. The phenomenon is called

(a)  Reverberation

(b)  Dispersion

(c)  Refraction

(d)  Diffraction

Answer: (a) Reverberation

Explanation: Reverberation is the persistence of sound in an enclosed space due to multiple successive reflections from walls, ceiling, and floor. The sound continues after the source stops. This is what gives large halls and auditoriums their characteristic ‘echo-like’ quality. Dispersion, refraction, and diffraction are different wave phenomena. Concept Tested: Reverberation: persistence of sound due to multiple reflections in an enclosed space

Q. 23. Which one of the following cannot be the unit of frequency of a sound wave?

(a)  dB

(b)  s⁻¹

(c)  Hz

(d)  min⁻¹

Answer: (a) dB

Explanation: Hz (hertz) = s⁻¹: this is the standard SI unit of frequency. min⁻¹ (cycles per minute) is also a valid unit of frequency (convertible to Hz by dividing by 60). dB (decibels) is a unit of sound intensity level: a logarithmic scale of loudness: not of frequency. dB cannot express frequency. Concept Tested: Units of frequency: Hz and s⁻¹ are valid; dB measures intensity level (not frequency)

Q. 24. ‘Beats’ is a phenomenon that occurs when frequencies of two harmonic waves are

(a)  equal

(b)  far apart

(c)  multiples of each other

(d)  nearly same

Answer: (d) nearly same

Explanation: Beats occur when two sound waves of nearly equal (but not identical) frequencies superpose. The resulting amplitude fluctuates periodically at the beat frequency = |f₁ − f₂|. When frequencies are exactly equal, no beats occur (continuous steady sound). When frequencies are far apart, no coherent periodic amplitude variation is perceived. Concept Tested: Beats phenomenon: occurs when two frequencies are nearly equal (small difference)

Q. 25. Which one of the following statements is true for a simple harmonic oscillator?

(a)  Force acting is directly proportional to the displacement from the mean position and is in same direction

(b)  Force acting is directly proportional to the displacement from the mean position and is in opposite direction

(c)  Acceleration of the oscillator is constant

(d)  The velocity of the oscillator is not periodic

Answer: (b) Force acting is directly proportional to the displacement from the mean position and is in opposite direction

Explanation: The defining condition for SHM is F = −kx: the restoring force is proportional to displacement from mean position and directed opposite to displacement (toward mean position). Option (a) has the wrong direction (same direction would be unstable). Option (c) is wrong: acceleration changes continuously. Option (d) is wrong: velocity is periodic. Concept Tested: SHM condition: F ∝ −x (proportional to displacement, directed toward mean position)

NDA 2020-I & II

Q. 26. Which of the following statements is NOT correct regarding the travel of sound waves?

(a)  Sound waves can travel through water

(b)  Sound waves can travel through air

(c)  Sound waves can travel through steel

(d)  Sound waves can travel through vacuum

Answer: (d) Sound waves can travel through vacuum

Explanation: Sound is a mechanical wave: it requires a material medium (solid, liquid, or gas) to propagate. It can travel through water, air, and steel. It cannot travel through vacuum. This is a fundamental property that distinguishes mechanical waves from electromagnetic waves. Concept Tested: Sound requires a medium: cannot travel through vacuum

★ JOVIK Exam Insight Sound cannot travel through vacuum. This has been tested in at least 4 NDA papers (2020-I & II, 2021-II, 2023-II, 2024-I). The most tested fact about the nature of sound in NDA Physics.

Q. 27. Which one of the following statements about sound is NOT correct?

(a)  Sound travels at a speed slower than the speed of light

(b)  Sound waves are transverse waves

(c)  Sound waves are longitudinal waves

(d)  Sound travels faster in moist air than in dry air

Answer: (b) Sound waves are transverse waves

Explanation: Sound in air is a longitudinal wave: particles vibrate parallel to the direction of propagation. Option (b) saying sound is transverse is false. Option (a) is correct (light travels at ~3×10⁸ m/s; sound at ~340 m/s). Option (c) is correct. Option (d) is correct: moist air is less dense than dry air (water vapour is lighter), so sound travels slightly faster. Concept Tested: Sound waves are longitudinal: NOT transverse (this is the false statement)

NDA 2019-II

Q. 28. A pendulum clock is lifted to a height where the gravitational acceleration has a certain value g. Another pendulum clock of same length but of double the mass of the bob is lifted to another height where the gravitational acceleration is g/2. The time period of the second pendulum would be: (in terms of period T of the first pendulum)

(a)  √2 T

(b)  (1/√2) T

(c)  2√2 T

(d)  T

Answer: (a) √2 T

Explanation: T₁ = 2π√(L/g). T₂ = 2π√(L/(g/2)) = 2π√(2L/g) = √2 × 2π√(L/g) = √2 × T₁. The doubled mass has no effect. Halving g multiplies the period by √2. Concept Tested: Simple pendulum at different gravity: T ∝ 1/√g; halving g multiplies T by √2

Q. 29. The loudness of sound depends upon the

(a)  velocity of sound waves in the medium.

(b)  amplitude of the sound waves.

(c)  frequency of the sound waves.

(d)  frequency and velocity of the sound waves.

Answer: (b) amplitude of the sound waves.

Explanation: Loudness is the subjective perception of sound intensity, which is determined by the amplitude of the sound waves. Greater amplitude means greater energy transferred per unit area per unit time: louder sound. Velocity and frequency determine other properties of sound but not its loudness. Concept Tested: Loudness: depends on amplitude of sound waves

Q. 30. Compared to audible sound waves, ultrasound waves have

(a)  higher speed.

(b)  higher frequency.

(c)  longer wavelength.

(d)  both higher speed and frequency.

Answer: (b) higher frequency.

Explanation: Ultrasound has frequency > 20 kHz: higher than audible sound. In the same medium at the same temperature, ultrasound and audible sound travel at the same speed (speed depends on the medium, not frequency). Since speed is the same and frequency is higher, ultrasound has shorter (not longer) wavelength (v = fλ → λ = v/f). Concept Tested: Ultrasound vs audible sound: higher frequency only; same speed in same medium; shorter wavelength

★ JOVIK Exam Insight A very common NDA trap: ultrasound does NOT travel faster than audible sound. Speed depends on medium properties, not frequency. Ultrasound = higher f, shorter λ, same v as audible sound in the same medium.

NDA 2019-I

Q. 31. At 20°C, the speed of sound in water is approximately

(a)  330 m/s

(b)  800 m/s

(c)  1500 m/s

(d)  5000 m/s

Answer: (c) 1500 m/s

Explanation: Speed of sound at 20°C: in air ≈ 340 m/s; in water ≈ 1500 m/s; in steel ≈ 5100 m/s. Water is much denser than air but far more elastic (incompressible), giving a speed about 4–5 times greater than in air. Option (a) 330 m/s is the air value. Option (d) 5000 m/s is approximately the steel value. Concept Tested: Speed of sound in water: approximately 1500 m/s at 20°C

★ JOVIK Exam Insight Speed of sound in media: air ≈ 340 m/s; water ≈ 1500 m/s; steel ≈ 5100 m/s. Order: solids > liquids > gases. NDA tested this in 2019-I and the comparison indirectly in 2022-II.

NDA 2018-II

Q. 32. The frequency of ultrasound waves is

(a)  less than 20 Hz

(b)  between 20 Hz and 2 kHz

(c)  between 2 kHz and 20 kHz

(d)  greater than 20 kHz

Answer: (d) greater than 20 kHz

Explanation: Ultrasound has frequency above the upper limit of human hearing: > 20 kHz (20,000 Hz). Option (a) describes infrasound. Options (b) and (c) are within the audible range. This is the same definition as in NDA 2011-II: a repeated examination point. Concept Tested: Ultrasound frequency: greater than 20 kHz

Q. 33. The time period of oscillation of a simple pendulum having length L and mass of the bob m is given as T. If the length of the pendulum is increased to 4L and the mass of the bob is increased to 2m, then which one of the following is the new time period of oscillation?

(a)  T

(b)  2T

(c)  4T

(d)  T/2

Answer: (b) 2T

Explanation: T = 2π√(L/g). New T’ = 2π√(4L/g) = 2 × 2π√(L/g) = 2T. The mass increase (2m → 2m) has no effect on the period. Quadrupling the length doubles the period. Concept Tested: Simple pendulum period: quadrupling length doubles T; mass has no effect

★ JOVIK Exam Insight Simple pendulum period comparisons appear across 6 different NDA papers: 2018-II, 2019-II, 2022-I, 2022-II, 2025-I, 2026-I. The standard method: T_new/T_old = √(L_new/L_old) × √(g_old/g_new). Mass is always irrelevant.

NDA 2018-I

Q. 34. If T is the time period of an oscillating pendulum, which one of the following statements is NOT correct?

(a)  The motion repeats after time T only once

(b)  T is the least time after which motion repeats itself

(c)  The motion repeats itself after nT, where n is a positive integer

(d)  T remains the same only for small angular displacements

Answer: (a) The motion repeats after time T only once

Explanation: Option (a) is false: the motion repeats after T, and also after 2T, 3T, 4T, and every integer multiple nT. T is the minimum repeat interval, not the only one. Option (b) is correct: T is the least (minimum) time after which motion repeats. Option (c) is correct. Option (d) is correct: T = 2π√(L/g) applies strictly only for small angular displacements. Concept Tested: Time period T: motion repeats every nT (not ‘only once’); T is the minimum repeat interval

Q. 35. Which one of the following frequency ranges is sensitive to human ears?

(a)  0 – 200 Hz

(b)  20 – 20,000 Hz

(c)  200 – 20,000 Hz only

(d)  2,000 – 20,000 Hz only

Answer: (b) 20 – 20,000 Hz

Explanation: The human audible range is 20 Hz to 20,000 Hz (20 kHz). Below 20 Hz is infrasonic (felt but not heard). Above 20 kHz is ultrasonic. Option (a) starts at 0 Hz (wrong lower boundary). Options (c) and (d) miss the lower part of the audible range. Concept Tested: Human audible frequency range: 20 Hz to 20,000 Hz (20 kHz)

Q. 36. A vibrating pendulum bob, acting under the force of gravity of the Earth, is an example of which one of the following?

(a)  Applied force

(b)  Frictional force

(c)  Restoring force

(d)  Virtual force

Answer: (c) Restoring force

Explanation: In a simple pendulum, gravity provides the restoring force: the component of gravitational force that acts along the arc toward the equilibrium position. This force brings the pendulum back to the mean position whenever it is displaced. It is not an applied (external) force, not friction, and not a virtual (pseudo) force. Concept Tested: Simple pendulum: gravity acts as restoring force directed toward equilibrium

Q. 37. Which of the following statements about electromagnetic waves, sound waves and water waves is/are correct?

1.  They exhibit reflection

2.  They carry energy

3.  They exert pressure

4.  They can travel in vacuum

Select the correct answer using the code given below:

(a)  1, 2 and 3

(b)  2 and 4

(c)  1 and 3 only

(d)  1 only

Answer: (a) 1, 2 and 3

Explanation: Statement 1: All waves (EM, sound, water) exhibit reflection. Correct. Statement 2: All waves carry energy. Correct. Statement 3: All waves exert pressure (radiation pressure for EM; mechanical pressure for sound and water). Correct. Statement 4: Only electromagnetic waves travel in vacuum: sound and water waves require a medium. Statement 4 is false. So 1, 2, and 3 are correct. Concept Tested: Common properties of waves: all exhibit reflection and carry energy; only EM waves travel in vacuum

NDA 2017-II

Q. 38. Bats detect obstacles in their path by receiving the reflected?

(a)  Infrasonic waves

(b)  Ultrasonic waves

(c)  Radio waves

(d)  Microwaves

Answer: (b) Ultrasonic waves

Explanation: Bats use echolocation: they emit ultrasonic waves (> 20 kHz) from their bodies, and detect the reflected ultrasonic waves that return from nearby objects. Radio waves and microwaves are electromagnetic and are not produced by bats. Infrasonic waves are below the frequency bats use for navigation. Concept Tested: Bat echolocation: uses reflected ultrasonic waves to detect obstacles

Q. 39. Which one of the following waves does not belong to the category of the other three?

(a)  X-rays

(b)  Microwaves

(c)  Radio waves

(d)  Sound waves

Answer: (d) Sound waves

Explanation: X-rays, microwaves, and radio waves are all electromagnetic waves: transverse, travel at the speed of light, can propagate through vacuum. Sound waves are mechanical, longitudinal, require a medium, and travel at ~340 m/s in air. Sound does not belong to the electromagnetic wave category. Concept Tested: Wave classification: sound waves are mechanical; X-rays, microwaves, radio waves are electromagnetic

Q. 40. Which one of the following statements is not correct?

(a)  Ultrasonic waves cannot get reflected, refracted or absorbed.

(b)  Ultrasonic waves are used to detect the presence of defects like cracks, porosity, etc. in the internal structure of common structure materials.

(c)  Ultrasonic waves can be used for making holes in very hard materials like diamond.

(d)  Ultrasonic waves cannot travel through vacuum.

Answer: (a) Ultrasonic waves cannot get reflected, refracted or absorbed.

Explanation: Option (a) is false: ultrasonic waves can be reflected, refracted, and absorbed, just like any other mechanical wave. Their ability to be reflected is the basis of SONAR and non-destructive testing. Options (b) and (c) are correct applications of ultrasound. Option (d) is correct: ultrasound is mechanical and cannot travel through vacuum. Concept Tested: Ultrasonic waves: can be reflected, refracted, and absorbed (option (a) is false)

NDA 2017-I

Q. 41. The following figure shows displacement versus time curve for a particle executing simple harmonic motion. Which one of the following statements is correct?

(a)  Phase of the oscillating particle is same at t = 1 s and t = 3 s

(b)  Phase of the oscillating particle is same at t = 2 s and t = 8 s

(c)  Phase of the oscillating particle is same at t = 3 s and t = 7 s

(d)  Phase of the oscillating particle is same at t = 4 s and t = 10 s

Answer: (b) Phase of the oscillating particle is same at t = 2 s and t = 8 s

Explanation: Two points are in phase when separated by exactly one complete period T (or integer multiples). For a standard NDA 2017-I graph with T = 6 s: t = 2 s and t = 8 s are separated by 6 s = T: they are in phase. Points separated by T/2 or non-integer multiples of T are not in phase. Concept Tested: Phase in SHM: same phase when time separation equals integer multiple of period T

Q. 42. Which one of the following is the correct relation between frequency f and angular frequency ω?

(a)  f = πω

(b)  ω = 2πf

(c)  f = 2ω/π

(d)  f = 2πω

Answer: (b) ω = 2πf

Explanation: Angular frequency ω = 2πf, where f is the frequency in Hz. Equivalently, f = ω/(2π). One complete oscillation subtends an angle of 2π radians, so angular frequency is 2π times the ordinary frequency. Options (a), (c), and (d) are all incorrect rearrangements. Concept Tested: Relation between ω and f: ω = 2πf

NDA 2016-II

Q. 43. When sound waves are propagated through a medium, the physical quantity/quantities transmitted is/are

(a)  matter only

(b)  energy only

(c)  energy and matter only

(d)  energy, momentum and matter

Answer: (b) energy only

Explanation: Sound waves are mechanical waves: medium particles oscillate about their equilibrium positions but do not travel with the wave. The wave transmits energy (and momentum) through the medium, but does not transport matter from place to place. The correct answer in the NDA source context is energy only: the momentum transmission of sound is negligible for practical purposes compared to the dominant energy transport. Concept Tested: Sound wave propagation: transmits energy through the medium; matter does not travel

Q. 44. A particle is executing simple harmonic motion. Which one of the following statements about the acceleration of the oscillating particle is true?

(a)  It is always in the opposite direction to velocity

(b)  It is proportional to the frequency of oscillation

(c)  It is minimum when the speed is maximum

(d)  It decreases as the potential energy increases

Answer: (c) It is minimum when the speed is maximum

Explanation: In SHM, speed is maximum at the mean position (where displacement = 0). Acceleration a = −ω²x is zero when x = 0: so acceleration is minimum (zero) when speed is maximum. Option (a) is wrong: acceleration is opposite to displacement, not necessarily opposite to velocity. Option (b) is wrong: a ∝ displacement, not frequency. Option (d) is wrong: as PE increases, |x| increases, so acceleration increases. Concept Tested: SHM acceleration: minimum (zero) at mean position where speed is maximum

Q. 45. Which one of the following four particles, whose displacement x and acceleration aₓ are related as follows, is executing simple harmonic motion?

(a)  aₓ = + 3x

(b)  aₓ = + 3x²

(c)  aₓ = − 3x²

(d)  aₓ = − 3x

Answer: (d) aₓ = − 3x

Explanation: SHM requires acceleration proportional to displacement and directed opposite to it: a = −ω²x. Option (d) gives a = −3x: negative sign (directed toward mean position) and linear in x. This satisfies the SHM condition with ω² = 3. Option (a) has positive sign (away from mean: unstable). Options (b) and (c) are non-linear (x²: not proportional to x). Concept Tested: Condition for SHM: a ∝ −x (negative sign, linear in displacement)

Q. 46. Which one of the following statements is correct?

(a)  The speed of sound waves in a medium depends upon the elastic property of the medium but not on inertia property

(b)  The speed of sound waves in a medium depends upon the inertia property of the medium but not on elastic property

(c)  The speed of sound waves in a medium depends neither on its elastic property nor on its inertia property

(d)  The speed of sound waves in a medium depends both on elastic and inertia properties of the medium

Answer: (d) The speed of sound waves in a medium depends both on elastic and inertia properties of the medium

Explanation: Speed of sound: v = √(B/ρ), where B is the bulk modulus (elastic property) and ρ is the density (inertia property). Both properties together determine the speed. A stiffer medium (higher B) gives faster sound; a denser medium (higher ρ) gives slower sound. Concept Tested: Speed of sound: depends on both elastic property (bulk modulus) and inertia property (density)

Q. 47. Which one of the following statements is not correct?

(a)  Pitch of a sound is its characteristic by which we can generally differentiate between a male voice and a female voice

(b)  The loudness of sound is related to its frequency

(c)  A musical sound has certain well defined frequencies which are generally harmonics of a fundamental frequency

(d)  The timbre of a particular musical sound is related to the waveform of the sound wave

Answer: (b) The loudness of sound is related to its frequency

Explanation: Pitch (high/low tone) corresponds to frequency: males have lower pitch, females higher. Musical sounds consist of harmonics of a fundamental. Timbre (tone quality) depends on the waveform (mixture of harmonics). Loudness depends on amplitude, NOT frequency. Option (b) is the false statement. Concept Tested: Sound characteristics: loudness depends on amplitude (not frequency); pitch depends on frequency

Q. 48. A particle executes linear simple harmonic motion with amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitudes of the velocity and the acceleration are equal. Then its time period (in seconds) is

(a)  2π/√3

(b)  √3/2π

(c)  √3/π

(d)  1/2π√3

Answer: (a) 2π/√3

Explanation: In SHM: velocity v = ω√(A²−x²); acceleration |a| = ω²x. At x = 1 cm, A = 2 cm. Setting |v| = |a|: ω√(4−1) = ω² × 1 → ω√3 = ω² → ω = √3. Time period T = 2π/ω = 2π/√3. Concept Tested: SHM numerical: finding period from equal velocity and acceleration at given displacement

NDA 2015-II

Q. 49. The loudness of sound is related to:

(a)  its frequency

(b)  its amplitude

(c)  its speed

(d)  its pitch

Answer: (b) its amplitude

Explanation: Loudness is determined by the amplitude (or intensity) of the sound wave: a larger amplitude means greater energy and thus louder sound. Pitch is determined by frequency. Speed is a property of the medium, not of perceived loudness. Pitch and loudness are independent characteristics. Concept Tested: Loudness of sound: determined by amplitude (not frequency or speed)

★ JOVIK Exam Insight Loudness (amplitude) vs pitch (frequency) is a recurring NDA distinction. Three papers test it directly: 2015-II, 2019-II, 2022-II. Loudness = amplitude. Pitch = frequency. Always separate.

NDA 2015-I

Q. 50. Ultrasonic waves of frequency 3 × 10⁵ Hz are passed through a medium where speed of sound is 10 times that in air (Speed of sound in air is 300 m/s). The wavelength of this wave in that medium will be of the order of

(a)  1 cm

(b)  10 cm

(c)  100 cm

(d)  0.1 cm

Answer: (a) 1 cm

Explanation: Speed in medium = 10 × 300 = 3000 m/s. Wavelength λ = v/f = 3000/(3 × 10⁵) = 10⁻² m = 1 cm. Concept Tested: Wavelength from speed and frequency: λ = v/f in a different medium

NDA 2014-I

Q. 51. The displacement (x)-time (t) graph given above approximately represents the motion of a

(a)  simple pendulum placed in vacuum

(b)  simple pendulum immersed in water

(c)  simple pendulum placed in outer space

(d)  point mass moving in air

Answer: (b) simple pendulum immersed in water

Explanation: A displacement–time graph showing decreasing amplitude over successive oscillations represents damped oscillation. A simple pendulum in vacuum has no damping: amplitude stays constant. A pendulum immersed in water experiences strong viscous damping: the amplitude decays rapidly. This matches a graph of exponentially decreasing amplitude. Concept Tested: Damped oscillation: decreasing amplitude over time indicates viscous damping (pendulum in water)

Q. 52. Sound waves are similar to the waves

(a)  of laser light passing through air

(b)  generated in a stretched wire by hitting or plucking the wire

(c)  generated in a pipe filled with air by moving the piston attached to the pipe up and down

(d)  generated by the mobile phone towers

Answer: (c) generated in a pipe filled with air by moving the piston attached to the pipe up and down

Explanation: Sound waves in air are longitudinal pressure waves: particles vibrate parallel to the direction of wave travel (compressions and rarefactions). Waves in an air-filled pipe created by piston movement are also longitudinal pressure waves. Waves in a stretched wire are transverse. Laser light and mobile phone waves are transverse electromagnetic waves. Concept Tested: Sound waves are longitudinal: most similar to pressure waves in an air-filled pipe

Q. 53. A sound wave has frequency of 2 kHz and wavelength of 35 cm. If an observer is 1.4 km away from the source, after what time interval could the observer hear the sound?

(a)  2 s

(b)  20 s

(c)  0.5 s

(d)  4 s

Answer: (a) 2 s

Explanation: Speed v = f × λ = 2000 Hz × 0.35 m = 700 m/s. Distance = 1.4 km = 1400 m. Time = distance / speed = 1400/700 = 2 s. Concept Tested: Wave speed calculation: v = fλ; travel time = distance ÷ speed

NDA 2013-I

Q. 54. The displacement of a particle is given by x = cos²ωt. The motion is

(a)  simple harmonic

(b)  periodic but not simple harmonic

(c)  non-periodic

(d)  None of the above

Answer: (b) periodic but not simple harmonic

Explanation: cos²ωt = (1 + cos 2ωt)/2. This is a constant (½) plus a sinusoidal term (½ cos 2ωt). The motion is periodic: it repeats with period π/ω. However, it is not SHM because it contains a constant offset, and the equivalent restoring force does not satisfy F ∝ −x about the stated equilibrium. SHM requires a pure sinusoidal displacement with zero mean offset. Concept Tested: x = cos²ωt: periodic but not SHM (contains constant offset after trigonometric identity)

NDA 2012-II

Q. 55. A swinging pendulum has its maximum acceleration at

(a)  the bottom of the swing

(b)  the two extremities of the swing

(c)  every point on the swing

(d)  no particular portion of the pendulum

Answer: (b) the two extremities of the swing

Explanation: In SHM, acceleration a = −ω²x. Acceleration is proportional to displacement and is maximum when displacement is maximum: at the two extreme positions of the swing. At the bottom (mean position), displacement is zero so acceleration is zero. This is consistent with the restoring force being zero at the mean position. Concept Tested: Acceleration in SHM: maximum at extremes (maximum displacement), zero at mean position

Q. 56. SONAR is mostly used by

(a)  Doctors

(b)  Engineers

(c)  Astronauts

(d)  Navigators

Answer: (d) Navigators

Explanation: SONAR (Sound Navigation And Ranging) uses ultrasonic waves to detect and measure distances to underwater objects. It is used primarily by navigators: on ships, submarines, and naval vessels: to map the seabed and detect underwater obstacles. Doctors use ultrasound (medical imaging), not SONAR specifically. Concept Tested: SONAR: used by navigators for underwater distance measurement using ultrasound

Q. 57. For a simple pendulum, the graph between T² and L (where T is the time period & L is the length) is

(a)  straight line passing through origin

(b)  parabolic

(c)  circle

(d)  none of the above

Answer: (a) straight line passing through origin

Explanation: T = 2π√(L/g). Squaring: T² = (4π²/g) × L. This is a linear equation: T² ∝ L, with constant of proportionality 4π²/g. The T²–L graph is a straight line through the origin: not a parabola or circle. Concept Tested: T²–L graph for simple pendulum: straight line through origin confirming T² ∝ L

NDA 2012-I

Q. 58. Which one among the following is not produced by sound waves in air?

(a)  Polarization

(b)  Diffraction

(c)  Reflection

(d)  Refraction

Answer: (a) Polarization

Explanation: Sound waves in air are longitudinal: particles vibrate parallel to the direction of propagation. Polarization is a property of transverse waves only (oscillations perpendicular to propagation can be restricted to one plane). Sound can undergo diffraction (bend around obstacles), reflection (echo), and refraction (change speed in different media): but not polarization. Concept Tested: Polarization not possible for sound: sound is longitudinal; only transverse waves can be polarized

NDA 2011-II

Q. 59. Ultrasonic waves are those sound waves having frequency

(a)  between 20 hertz and 1000 hertz

(b)  between 1000 hertz and 20000 hertz

(c)  more than 20 kilohertz

(d)  less than 20 hertz

Answer: (c) more than 20 kilohertz

Explanation: Sound frequency ranges: below 20 Hz = infrasonic; 20 Hz to 20,000 Hz (20 kHz) = audible (human hearing range); above 20 kHz = ultrasonic. Option (c) correctly defines ultrasonic as more than 20 kHz. Option (d) describes infrasonic. Concept Tested: Ultrasonic frequency range: above 20 kHz (20,000 Hz)

★ JOVIK Exam Insight Sound frequency classification: infrasonic < 20 Hz; audible 20 Hz–20 kHz; ultrasonic > 20 kHz. NDA has tested this definition in 2011-II and 2018-II. Always: ultrasonic = above human hearing = above 20,000 Hz.

Q. 60. It is impossible for two oscillators, each executing simple harmonic motion, to remain in phase with each other if they have different

(a)  time periods

(b)  amplitudes

(c)  spring constants

(d)  kinetic energy

Answer: (a) time periods

Explanation: Phase in SHM is defined relative to the oscillation cycle. Two oscillators with different time periods (frequencies) will continuously drift in their relative phase: sometimes aligned, sometimes opposed. Differences in amplitude, spring constant, or kinetic energy do not prevent the two oscillators from maintaining a fixed phase relationship. Concept Tested: Phase in SHM: different time periods prevent permanent phase-locking

NDA 2011-I

Q. 61. Which one among the following statements is not correct?

(a)  In progressive waves, the amplitude may be constant and neighbouring points are out of phase with each other

(b)  In air or other gases, a progressive antinode occurs at a displacement node and a progressive node occurs at a displacement antinode

(c)  Transverse wave can be polarized while longitudinal wave can not be polarized

(d)  Longitudinal wave can be polarized while transverse wave can not be polarized

Answer: (d) Longitudinal wave can be polarized while transverse wave can not be polarized

Explanation: Option (a) is correct: in progressive waves, all particles have the same amplitude but different phases. Option (b) is correct for standing waves (nodes and antinodes). Option (c) is correct: transverse waves (like light) can be polarized: longitudinal waves (like sound) cannot. Option (d) reverses the correct statement: it is false. Concept Tested: Polarization: transverse waves can be polarized; longitudinal waves cannot

Q. 62. Bats can ascertain distances, directions, nature and size of the obstacles at night. This is possible by reflection of the emitted:

(a)  ultrasonic waves from the bat

(b)  ultrasonic waves from the distant objects

(c)  supersonic waves from the bat

(d)  supersonic waves from the distant objects

Answer: (a) ultrasonic waves from the bat

Explanation: Bats use echolocation: they emit ultrasonic waves (frequency > 20 kHz) from their own bodies, and detect the reflected waves that return from nearby objects. The waves originate from the bat: not from the objects. They are ultrasonic (high-frequency sound), not supersonic (which refers to speed exceeding the speed of sound). Concept Tested: Bat echolocation: ultrasonic waves emitted by the bat, reflected from objects

★ JOVIK Exam Insight Bats emit the ultrasonic waves: the waves do not come from the objects. Supersonic refers to speed (faster than sound), not frequency. Echolocation is frequency-based (ultrasound), not speed-based (supersonic). NDA tested bats in 2011-I and 2017-II.

NDA 2010-II

Q. 63. A pendulum beats faster than a standard pendulum. In order to bring it to the standard beat, the length of the pendulum is to be?

(a)  reduced

(b)  increased

(c)  reduced and the mass of the bob increased

(d)  reduced and also the mass of the bob reduced

Answer: (b) increased

Explanation: T = 2π√(L/g). A faster-beating pendulum has a shorter period: meaning it completes more oscillations per unit time. To increase the period to the standard value, length L must be increased (T ∝ √L). The mass of the bob has no effect on the period. Concept Tested: Simple pendulum period: increasing length increases period; mass has no effect

NDA 2010-I

Q. 64. A particle oscillates in one dimension about the equilibrium position subject to a force Fₓ(x) that has an associated potential energy U(x). If k is the force constant, which one of the following relations is true?

(a)  Fₓ(x) = −kx²

(b)  Fₓ(x) = −kx

(c)  U(x) = ½kx

(d)  U(x) = ½k²x

Answer: (b) Fₓ(x) = −kx

Explanation: The defining force law for SHM is F = −kx: force proportional to displacement and directed opposite to it. The associated potential energy is U = ½kx² (not ½kx or ½k²x). Option (a) F = −kx² is non-linear and does not give SHM. Only option (b) is correct. Concept Tested: SHM force law: F = −kx and potential energy U = ½kx²

Q. 65. When a body moves with simple harmonic motion, then the phase difference between the velocity and the acceleration is

(a)  0°

(b)  90°

(c)  180°

(d)  270°

Answer: (b) 90°

Explanation: In SHM with displacement x = A sin(ωt), velocity v = Aω cos(ωt) and acceleration a = −Aω² sin(ωt). Velocity is a cosine function while acceleration is a negative sine: the phase difference between them is 90° (π/2 radians). Velocity leads acceleration by 90°. Concept Tested: Phase difference between velocity and acceleration in SHM: 90°

Q. 66. For a simple pendulum in simple harmonic motion, which of the following statements is/are correct?

1.  The kinetic energy is maximum at the mean position.

2.  The potential energy is maximum at the mean position.

3.  Acceleration is maximum at the mean position.

Select the correct answer using the code given below:

(a)  1 only

(b)  2 only

(c)  1 and 3

(d)  2 and 3

Answer: (a) 1 only

Explanation: At the mean (equilibrium) position: velocity is maximum → KE is maximum (Statement 1 is correct). Displacement is zero → PE is zero (not maximum; Statement 2 is wrong). Restoring force F = −kx = 0 → acceleration is zero (not maximum; Statement 3 is wrong). PE and acceleration are maximum at the extreme positions. Concept Tested: Energy and acceleration in SHM: KE max at mean position; PE and acceleration max at extremes

★ JOVIK Exam Insight A classic multi-statement NDA trap. Mean position: KE = max, PE = 0, acceleration = 0. Extreme position: KE = 0, PE = max, acceleration = max. These three pairs are always reversed from each other.

---