Waves NEET PYQReal NEET & AIPMT 2005–2020 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Waves question from the NEET & AIPMT papers on this site (2005–2020), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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32 questions. Tap an option to check your answer.
Q1NEET 2020 · Q147
In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency 6 Hz. When tension in B is slightly decreased, the beat frequency increases to 7 Hz. If the frequency of A is 530 Hz, the original frequency of B will be :
Not quite — the answer is B.
With 6 Hz beats, B is 524 Hz or 536 Hz. Lowering tension lowers B's frequency (f ∝ √T); the beat frequency rises only if B moves further from 530 Hz, so B must start below A at 524 Hz. Had B been 536 Hz, lowering it would bring it closer to 530 Hz and reduce the beats.
Q2NEET 2018 · Q24
A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of 27°C two successive resonances are produced at 20 cm and 73 cm of column length. If the frequency of the tuning fork is 320 Hz, the velocity of sound in air at 27°C is
Not quite — the answer is B.
Successive resonance lengths differ by λ/2: 73 − 20 = 53 cm, so λ = 1.06 m. v = fλ = 320 × 1.06 ≈ 339 m/s. Using 4 × 20 cm (ignoring end correction) would give a wrong value.
Q3NEET 2017 · Q146
Two cars moving in opposite directions approach each other with speed of 22 m/s and 16.5 m/s respectively. The driver of the first car blows a horn having a frequency 400 Hz. The frequency heard by the driver of the second car is [velocity of sound 340 m/s]:
Not quite — the answer is D.
Source and observer approach each other, so f′ = f(v + v₀)/(v − vₛ) = 400 × (340 + 16.5)/(340 − 22) = 400 × 356.5/318 ≈ 448 Hz. The heard frequency must be higher than 400 Hz, which rules out 350 Hz and 361 Hz.
Q4NEET 2017 (Gujarati paper) · Q159
Two open organ pipes of fundamental frequencies n₁ and n₂ are joined in series. The fundamental frequency of the new pipe so obtained will be:
Not quite — the answer is C.
For an open pipe n = v/2L, so L₁ = v/2n₁ and L₂ = v/2n₂. The joined pipe has length L₁ + L₂, so n = v/2(L₁ + L₂) and 1/n = 1/n₁ + 1/n₂, giving n = n₁n₂/(n₁ + n₂). A longer pipe must have a lower frequency, so n₁ + n₂ cannot be right.
Q5NEET 2016 Phase 2 · Q21
The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe L metre long. The length of the open pipe will be
Not quite — the answer is B.
The first overtone of a closed pipe is 3v/(4L). The second overtone of an open pipe of length L′ is the third harmonic 3v/(2L′). Equating, 2L′ = 4L, so L′ = 2L. Treating the second overtone as the second harmonic would give a different length.
Q6NEET 2016 Phase 1 · Q10
A uniform rope of length L and mass m₁ hangs vertically from a rigid support. A block of mass m₂ is attached to the free end of the rope. A transverse pulse of wavelength λ₁ is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is λ₂. The ratio λ₂/λ₁ is:
Not quite — the answer is B.
Frequency stays the same, so λ ∝ v ∝ √T. Tension at the bottom is m₂g and at the top is (m₁ + m₂)g, so λ₂/λ₁ = √((m₁ + m₂)/m₂). Putting m₁ in the denominator would wrongly use the rope's own weight as the bottom tension.
Q7NEET 2016 Phase 1 · Q12
An air column, closed at one end and open at the other, resonates with a tuning fork when the smallest length of the column is 50 cm. The next larger length of the column resonating with the same tuning fork is:
Not quite — the answer is C.
For a closed pipe resonating with a fixed fork, resonant lengths are λ/4, 3λ/4, 5λ/4, … The first is 50 cm, so the next is 3 × 50 = 150 cm. 100 cm would be the next length for a pipe open at both ends.
Q8AIPMT 2015
The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is:
Not quite — the answer is D.
Closed pipe fundamental = v/(4 × 20 cm) = v/80. Second overtone of an open pipe is the third harmonic, 3v/(2L). Equating, 3v/(2L) = v/80 gives L = 120 cm. Taking second overtone as the second harmonic gives 80 cm.
Q9AIPMT 2014
If n₁, n₂ and n₃ are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency n of the string is given by:
Not quite — the answer is D.
At fixed tension and linear density the fundamental frequency is n = (1/2l)√(T/μ), so l ∝ 1/n. Since the segment lengths add up to the whole length, 1/n = 1/n₁ + 1/n₂ + 1/n₃; simply adding the frequencies would wrongly treat n ∝ l.
Q10AIPMT 2014
The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are: (velocity of sound = 340 ms⁻¹)
Not quite — the answer is C.
For a closed pipe the fundamental is v/4L = 340/(4 × 0.85) = 100 Hz and only odd harmonics exist: 100, 300, 500, 700, 900, 1100 Hz. That makes 6 below 1250 Hz; counting all harmonics would wrongly give more.
Q11AIPMT 2014
A speeding motorcyclist sees traffic jam ahead of him. He slows down to 36 km/hour. He finds that traffic has eased and a car moving ahead of him at 18 km/hour is honking at a frequency of 1392 Hz. If the speed of sound is 343 m/s, the frequency of the honk as heard by him will be:
Not quite — the answer is B.
Motorcyclist (10 m/s) moves towards the car, and the car (5 m/s, the source) moves away from him. f′ = 1392 × (343 + 10)/(343 + 5) = 1392 × 353/348 = 1412 Hz.
Q12NEET 2013
A wave travelling in the +ve x-direction having displacement along y-direction as 1 m, wavelength 2π m and frequency of 1/π Hz is represented by
Not quite — the answer is A.
k = 2π/λ = 2π/2π = 1 m⁻¹ and ω = 2πf = 2π × (1/π) = 2 rad s⁻¹, amplitude 1 m. A wave along +x is y = A sin(kx − ωt) = sin(x − 2t). A plus sign between the terms would describe a wave moving along −x.
Q13NEET 2013
A source of unknown frequency gives 4 beats/s when sounded with a source of known frequency 250 Hz. The second harmonic of the source of unknown frequency gives five beats per second when sounded with a source of frequency 513 Hz. The unknown frequency is
Not quite — the answer is A.
4 beats with 250 Hz means f = 254 Hz or 246 Hz. Second harmonics are 508 Hz or 492 Hz; only 508 Hz gives 513 − 508 = 5 beats per second. So f = 254 Hz; 246 Hz would give 21 beats.
Q14AIPMT 2012 Mains
The equation of a simple harmonic wave is given by y = 3 sin (π/2)(50t − x), where x and y are in metres and t is in seconds. The ratio of maximum particle velocity to the wave velocity is
Not quite — the answer is B.
y = 3 sin(25πt − πx/2), so A = 3 m, ω = 25π rad/s, k = π/2 m⁻¹. Wave speed = ω/k = 50 m/s and maximum particle speed = Aω = 75π m/s, giving a ratio of 75π/50 = 3π/2.
Q15AIPMT 2012 Mains
A train moving at a speed of 220 m s⁻¹ towards a stationary object emits a sound of frequency 1000 Hz. Some of the sound reaching the object gets reflected back to the train as echo. The frequency of the echo as detected by the driver of the train is (speed of sound in air is 330 m s⁻¹)
Not quite — the answer is C.
The object receives f₁ = f·v/(v − u) and reflects it; the moving driver then hears f₂ = f₁(v + u)/v = f(v + u)/(v − u) = 1000 × 550/110 = 5000 Hz. Treating only one Doppler shift gives 3000 Hz, which is wrong.
Q16AIPMT 2012 Prelims
When a string is divided into three segments of length l₁, l₂ and l₃, the fundamental frequencies of these three segments are ν₁, ν₂ and ν₃ respectively. The original fundamental frequency (ν) of the string is
Not quite — the answer is D.
For the same string (same tension and linear density) the fundamental frequency ν = v/(2l) is inversely proportional to length. Since l = l₁ + l₂ + l₃, 1/ν = 1/ν₁ + 1/ν₂ + 1/ν₃. Adding frequencies directly would wrongly treat ν as proportional to length.
Q17AIPMT 2012 Prelims
Two sources of sound placed close to each other are emitting progressive waves given by y₁ = 4 sin 600πt and y₂ = 5 sin 608πt. An observer located near these two sources of sound will hear
Not quite — the answer is A.
ω₁ = 600π gives f₁ = 300 Hz and ω₂ = 608π gives f₂ = 304 Hz, so beat frequency = 4 per second. Intensity ratio I_max/I_min = (A₁ + A₂)²/(A₁ − A₂)² = 9²/1² = 81 : 1. 25 : 16 is just the ratio of individual intensities, and 8 beats comes from using ω/2π wrongly.
Q18AIPMT 2011 Mains
Two identical piano wires, kept under the same tension T, have a fundamental frequency of 600 Hz. The fractional increase in the tension of one of the wires which will lead to occurrence of 6 beats/s when both the wires oscillate together would be
Not quite — the answer is C.
f ∝ √T, so Δf/f = ½ ΔT/T. Six beats means Δf = 6 Hz, so 6/600 = ½ ΔT/T and ΔT/T = 0.02. Forgetting the square-root dependence gives 0.01.
Q19AIPMT 2011 Prelims
Two waves are represented by the equations y₁ = a sin(ωt + kx + 0.57) m and y₂ = a cos(ωt + kx) m, where x is in metre and t in s. The phase difference between them is
Not quite — the answer is B.
Write y₂ = a cos(ωt + kx) = a sin(ωt + kx + π/2). The phase difference is π/2 − 0.57 = 1.57 − 0.57 = 1.0 radian. 1.57 rad would be the phase of the cosine wave alone, ignoring the 0.57 in the first wave.
Q20AIPMT 2011 Prelims
Sound waves travel at 350 m/s through warm air and at 3500 m/s through brass. The wavelength of a 700 Hz acoustic wave as it enters brass from warm air
Not quite — the answer is D.
Frequency is unchanged when a wave enters a new medium, so λ = v/f is proportional to speed. λ goes from 350/700 = 0.5 m to 3500/700 = 5 m, an increase by a factor of 10. It increases, not decreases, because sound is faster in brass.
Q21AIPMT 2010 Prelims · Q20
A transverse wave is represented by y = A sin(ωt − kx). For what value of the wavelength is the wave velocity equal to the maximum particle velocity?
Not quite — the answer is D.
Wave speed ω/k equals the maximum particle speed Aω when 1/k = A, i.e. λ/(2π) = A, so λ = 2πA. Choosing πA drops the factor 2 in k = 2π/λ.
Q22AIPMT 2010 Prelims · Q21
A tuning fork of frequency 512 Hz makes 4 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per sec when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was
Not quite — the answer is A.
The string frequency is 512 ± 4, i.e. 508 or 516 Hz. Increasing tension raises the string's frequency; the beats fall to 2, so the frequency moved towards 512, meaning it was below: 508 Hz. If it were 516 Hz, raising it would increase the beats to more than 4.
Q23AIPMT 2009 Prelims · Q16
Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to 1 g/m. When both the strings vibrate simultaneously the number of beats is:
Not quite — the answer is A.
Wave speed v = √(T/μ) = √(20/10⁻³) ≈ 141.4 m/s. Fundamentals: f = v/2l gives 141.4/1.032 ≈ 137.0 Hz and 141.4/0.982 ≈ 144.0 Hz. Beat frequency = 144 − 137 = 7 per second.
Q24AIPMT 2009 Prelims · Q21
The driver of a car travelling with speed 30 m/sec towards a hill sounds a horn of frequency 600 Hz. If the velocity of sound in air is 330 m/s, the frequency of reflected sound as heard by driver is:
Not quite — the answer is B.
The hill receives f₁ = 600 × 330/(330 − 30) from the approaching car and reflects it; the driver, moving towards the hill, hears f = f₁ × (330 + 30)/330 = 600 × 360/300 = 720 Hz. The reflected sound is heard at a higher, not lower, frequency.
Q25AIPMT 2009 Prelims · Q31
A wave in a string has an amplitude of 2 cm. The wave travels in the + ve direction of x axis with a speed of 128 m/sec. and it is noted that 5 complete waves fit in 4 m length of the string. The equation describing the wave is:
Not quite — the answer is C.
λ = 4/5 = 0.8 m, so k = 2π/0.8 ≈ 7.85 m⁻¹ and ω = kv = 7.85 × 128 ≈ 1005 s⁻¹. A wave moving towards +x has the form sin(kx − ωt), so y = 0.02 sin(7.85x − 1005t); the '+' form travels towards −x and 15.7 uses λ = 0.4 m.
Q26AIPMT 2008 Prelims · Q18
The wave described by y = 0.25 sin(10πx − 2πt), where x and y are in metres and t in seconds, is a wave travelling along the
Not quite — the answer is D.
Compare with y = A sin(kx − ωt): k = 10π gives λ = 2π/k = 0.2 m, and ω = 2π gives f = 1 Hz. The minus sign between the kx and ωt terms means the wave moves along +x. Reading ω = 2π as 'frequency π Hz' confuses ω with 2πf.
Q27AIPMT 2008 Prelims · Q20
Two points are located at a distance of 10 m and 15 m from the source of oscillation. The period of oscillation is 0.05 sec and the velocity of the wave is 300 m/sec. What is the phase difference between the oscillations of two points?
Not quite — the answer is C.
Wavelength λ = vT = 300 × 0.05 = 15 m. Path difference = 15 − 10 = 5 m, so phase difference = (2π/λ) × 5 = 2π/3. Using λ = 5 m by mistake would give 2π, not one of the options.
Q28AIPMT 2006 Prelims · Q35
Two sound waves with wavelengths 5.0 m and 5.5 m respectively, each propagate in a gas with velocity 330 m/s. We expect the following number of beats per second :
Not quite — the answer is D.
The frequencies are 330/5.0 = 66 Hz and 330/5.5 = 60 Hz. The beat frequency is the difference, 66 − 60 = 6 per second. Adding or halving these does not give the beat count.
Q29AIPMT 2006 Prelims · Q38
A transverse wave propagating along x-axis is represented by : y(x, t) = 8.0 sin (0.5πx − 4πt − π/4) where x is in metres and t is in seconds. The speed of the wave is :
Not quite — the answer is D.
Comparing with y = A sin(kx − ωt + φ), k = 0.5π rad/m and ω = 4π rad/s. The wave speed is v = ω/k = 4π/0.5π = 8 m/s. The value 8π m/s comes from mixing the amplitude with ω and is wrong.
Q30AIPMT 2006 Prelims · Q40
Which one of the following statements is true ?
Not quite — the answer is B.
Sound in air travels as compressions and rarefactions, so it is longitudinal; light is an electromagnetic wave with transverse fields. Light can travel through vacuum but sound needs a material medium, so the statement that both travel in vacuum is false.
Q31AIPMT 2005 Prelims · Q3
A point source emits sound equally in all directions in a non-absorbing medium. Two points P and Q are at distance of 2m and 3m respectively from the source. The ratio of the intensities of the waves at P and Q is :
Not quite — the answer is A.
A point source spreads energy over a sphere, so I ∝ 1/r². I_P/I_Q = (3/2)² = 9/4, i.e. 9 : 4. Writing 4 : 9 inverts the ratio by taking I ∝ r².
Q32AIPMT 2005 Prelims · Q36
Two vibrating tuning forks produce progressive waves given by y₁ = 4 sin 500πt and y₂ = 2 sin 506πt. Number of beats produced per minute is :
Not quite — the answer is B.
ω₁ = 500π gives f₁ = 250 Hz and ω₂ = 506π gives f₂ = 253 Hz. Beat frequency = 3 per second = 180 per minute. 360 per minute would double the true beat frequency of 3 Hz.
How often Waves came up
Questions from this chapter in each paper on this site.
- NEET 2020: 1
- NEET 2018: 1
- NEET 2017: 1
- NEET 2017 (Gujarati paper): 1
- NEET 2016 Phase 2: 1
- NEET 2016 Phase 1: 2
- AIPMT 2015: 1
- AIPMT 2014: 3
- NEET 2013: 2
- AIPMT 2012 Mains: 2
- AIPMT 2012 Prelims: 2
- AIPMT 2011 Mains: 1
- AIPMT 2011 Prelims: 2
- AIPMT 2010 Prelims: 2
- AIPMT 2009 Prelims: 3
- AIPMT 2008 Prelims: 2
- AIPMT 2006 Prelims: 3
- AIPMT 2005 Prelims: 2
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