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NEET Physics: Waves questions with solutions

128 questions with worked solutions.

Questions

Q1. In stationary waves, nodes are the points where there is:

  1. maximum displacement and minimum pressure change
  2. minimum displacement and maximum pressure change
  3. minimum displacement and minimum pressure change
  4. maximum displacement and maximum pressure change

Answer: minimum displacement and maximum pressure change

In a stationary wave, nodes are points of zero displacement, so the particles there do not oscillate. For sound waves, pressure changes are largest at displacement nodes, making the correct choice minimum displacement and maximum pressure change.

Q2. How does the sound produced by a vibrating object in a medium reach your ear?

  1. Through electron transfer
  2. Through the vibration of particles
  3. Sound waves don't need a medium
  4. Not enough data

Answer: Through the vibration of particles

Sound travels through a medium by making its particles vibrate and pass the disturbance along from one particle to the next. That is why sound cannot travel in a vacuum and why the correct choice is the vibration of particles.

Q3. is the distance between consecutive corresponding points of the same phase, such as crests, troughs, compressions, rarefactions or zero crossings and is a characteristic of both travelling waves and standing waves, as well as other spatial wave patterns.

  1. frequency
  2. amplitude
  3. wavelength
  4. none of these

Answer: wavelength

The quantity described is the distance between successive points in the same phase of a wave. That is the definition of wavelength, and it applies to both traveling and standing waves.

Q4. An echo is simply a........of sound.

  1. Reflection
  2. Frequency
  3. Amplitude
  4. None of these

Answer: Reflection

An echo occurs when sound waves reflect off a surface and travel back to the listener. That makes it a reflection of sound, not a property like frequency or amplitude.

Q5. The condition for constructive interference is path difference should be equal to :

  1. odd integral multiple of wavelength
  2. Integral multiple of wavelength
  3. odd integral multiple of half wavelength
  4. Integral multiple of half wavelength

Answer: Integral multiple of wavelength

For constructive interference, the two waves must reinforce each other, which occurs when their phase difference is 0, 2π, 4π, etc. That corresponds to a path difference of nλ, where n is any integer.

Q6. In a stationary wave:

  1. Strain is maximum at antinodes
  2. Strain is minimum at nodes
  3. Strain is maximum at nodes
  4. Amplitude is zero at all points

Answer: Strain is maximum at nodes

In a stationary wave, nodes are points of zero displacement but the displacement changes most sharply there, so the spatial gradient is largest. Since strain is related to this gradient, it is maximum at nodes.

Q7. The true statement is

  1. Sound waves in air are transverse waves
  2. Sound wave does not required a material medium for its propagation
  3. Sound travels faster in gas than in solid
  4. Sound travels faster in solid than in gas

Answer: Sound travels faster in solid than in gas

Sound is a mechanical wave, so it needs a material medium to travel. It moves fastest in solids because particles are closely packed and transmit vibrations more efficiently than in gases.

Q8. A standing wave having 3 nodes and 2 antinodes is formed between two atoms having a distance 1.21 Å between them. The wavelength of the standing wave is:

  1. 1.21 Å
  2. 2.42 Å
  3. 6.05 Å
  4. 3.63 Å

Answer: 2.42 Å

A standing wave with 3 nodes and 2 antinodes corresponds to one full wavelength fitting between the two atoms. The distance between the atoms is 1.21 Å, which represents half a wavelength. Therefore, the full wavelength is 2 × 1.21 Å = 2.42 Å.

Q9. A transverse wave is represented by the equation y = y₀ sin [2π/λ (vt − x)]. For what value of λ is the maximum particle velocity equal to two times the wave velocity?

  1. λ = 2πy₀
  2. λ = πy₀/3
  3. λ = πy₀/2
  4. λ = πy₀

Answer: λ = πy₀/2

The maximum particle velocity is given by vmax = ωy₀, where ω = 2πv/λ. Setting vmax = 2v, we get 2πv/λ * y₀ = 2v. Simplifying, λ = πy₀/2.

Q10. In a sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displacement is 0.170 sec. The frequency of the wave is:

  1. 1.47 Hz
  2. 2.94 Hz
  3. 5.88 Hz
  4. 3.53 Hz

Answer: 2.94 Hz

The time required to move from maximum displacement to zero displacement is one-fourth of the time period (T/4). Thus, T = 4 × 0.170 = 0.68 s. The frequency is given by f = 1/T = 1/0.68 ≈ 2.94 Hz.

Q11. The speed of a wave in a medium is 960 m/s. If 3600 waves are passing through a point in the medium in 1 min., then the wavelength of the wave is

  1. 8 m
  2. 12 m
  3. 16 m
  4. 20 m

Answer: 12 m

The frequency of the wave is given by f = 3600 waves / 60 s = 60 Hz. Using the wave speed formula v = fλ, we have λ = v / f = 960 m/s / 60 Hz = 16 m.

Q12. The equation of a wave is represented by: y = 10 sin [100t − x]. The velocity of the wave will be:

  1. 100 m/s
  2. 250 m/s
  3. 750 m/s
  4. 1000 m/s

Answer: 100 m/s

The wave equation is of the form y = A sin(ωt − kx), where ω is the angular frequency and k is the wave number. The velocity of the wave is given by v = ω/k. Here, ω = 100 rad/s and k = 1 m⁻¹ (from the coefficient of x). Thus, v = 100/1 = 100 m/s.

Q13. The equation of a travelling wave is y = 60 cos (180 t − 6x) where y is in microns, t in second and x in metres. The ratio of maximum particle velocity to velocity of wave propagation is

  1. 3.6
  2. 3.6×10⁻⁶
  3. 3.6×10⁻¹¹
  4. 6×10⁻⁴

Answer: 3.6

The wave equation is y = 60 cos(180t − 6x). The maximum particle velocity is given by ωA, where ω = 180 rad/s and A = 60 × 10⁻⁶ m. Thus, maximum particle velocity = 180 × 60 × 10⁻⁶ = 10.8 × 10⁻³ m/s. The wave velocity is v = ω/k = 180/6 = 30 m/s. The ratio of maximum particle velocity to wave velocity is (10.8 × 10⁻³) / 30 = 3.6 × 10⁻⁴. Hence, the correct answer is A) 3.6.

Q14. Which of the following equations represent a wave?

  1. y = A sin ωt
  2. y = A cos kx
  3. y = A sin (ωt − kx)
  4. y = A (ωt − kx)

Answer: y = A sin (ωt − kx)

The equation y = A sin (ωt − kx) represents a wave because it includes both time (t) and position (x) in a sinusoidal form, which is characteristic of wave motion.

Q15. A standing wave is represented by y = A sin (100t) cos (0.01x), where y and A are in millimetre, t in seconds and x in metre. Velocity of wave is

  1. 10⁴ m/s
  2. 1 m/s
  3. 10⁻⁴ m/s
  4. not derivable from above data

Answer: 10⁴ m/s

The given equation y = A sin(100t) cos(0.01x) represents a standing wave. The angular frequency ω = 100 rad/s and the wave number k = 0.01 rad/m. The wave velocity is given by v = ω/k = 100/0.01 = 10⁴ m/s.

Q16. Two waves are approaching each other with a velocity of 20 m/s and frequency n. The distance between two consecutive nodes is

  1. 20/n
  2. 10/n
  3. 5/n
  4. n/10

Answer: 10/n

The distance between two consecutive nodes in a standing wave is half the wavelength (λ/2). The wavelength is given by λ = v/n, where v is the velocity and n is the frequency. Thus, the distance between two nodes is (v/n)/2 = 10/n.

Q17. The speed of a wave in a medium is 760 m/s. If 3600 waves are passing through a point in the medium in 2 min, then their wavelength is

  1. 13.8 m
  2. 25.3 m
  3. 41.5 m
  4. 57.2 m

Answer: 13.8 m

The frequency of the wave is calculated as f = (3600 waves) / (2 min × 60 s/min) = 30 Hz. Using the wave speed formula v = fλ, we find the wavelength λ = v / f = 760 m/s / 30 Hz = 25.3 m.

Q18. A hospital uses an ultrasonic scanner to locate tumours in a tissue. The operating frequency of the scanner is 4.2 MHz. The speed of sound in a tissue is 1.7 km/s. The wavelength of sound in tissue is close to

  1. 4×10⁻⁴ m
  2. 8×10⁻⁴ m
  3. 4×10⁻³ m
  4. 8×10⁻³ m

Answer: 4×10⁻⁴ m

The wavelength of sound is given by λ = v/f, where v is the speed of sound and f is the frequency. Substituting v = 1.7 × 10³ m/s and f = 4.2 × 10⁶ Hz, we get λ = (1.7 × 10³) / (4.2 × 10⁶) = 4 × 10⁻⁴ m.

Q19. With the propagation of a longitudinal wave through a material medium, the quantities transmitted in the propagation direction are

  1. Energy, momentum and mass
  2. Energy
  3. Energy and mass
  4. Energy and linear momentum

Answer: Energy and linear momentum

In a longitudinal wave, energy and linear momentum are transmitted through the medium as the particles oscillate about their mean positions. Mass is not transported, as the particles only vibrate locally.

Q20. The frequency of sinusoidal wave y = 0.40 cos [2000 t + 0.80] would be

  1. 1000 π Hz
  2. 2000 Hz
  3. 20 Hz
  4. 1000 Hz

Answer: 1000 Hz

The angular frequency ω is given as 2000 rad/s. The frequency f is related to ω by the formula f = ω / (2π). Substituting ω = 2000, we get f = 2000 / (2π) = 1000 Hz.

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