WavesNEET Physics ยท Class 11 ยท NCERT Chapter 14

16 NEET previous-year questions on Waves, each with the correct answer and a step-by-step solution. Sourced directly from official NEET papers across every booklet code.

PYQ frequency ยท topic ร— year

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15
16
17
18
20
22
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25
Waves
3
2
3
2
2
1
1
1
1

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Waves (16)

A

n 1 = 1 n 1 + 2 n 1 + 3 n 1

B

n 1 = 1 n 1 + 2 n 1 + 3 n 1

C

n = 1 n + 2 n + 3 n

D

n = n 1 + n 2 + n 3

Solution

The correct formula for the fundamental frequency of a string divided into segments is given by the harmonic series relation: . This is derived from the principle that the total length of the string is the sum of the lengths of the segments, and the frequencies are inversely proportional to these lengths. Therefore, option (a) is correct.

A

4

B

5

C

7

D

6

Solution

For a pipe closed at one end, the fundamental frequency , and the harmonics are odd multiples of the fundamental frequency. Substituting, . The frequencies below 1250 Hz are , which are 6 frequencies, so option (d) is correct.

A

1332 Hz

B

1372 Hz

C

1412 Hz

D

1464 Hz

Solution

Using the Doppler effect formula for a moving observer and a moving source: , where , (36 km/h), (18 km/h), and . Substituting, . Thus, option (c) is correct.

A

100 Hz

B

(2 ) 103 Hz

C

106 Hz

D

97 Hz

Solution

www.vedantu.com 28 t1=f0(vโˆ’v vโˆ’v s ) f1=100( vโˆ’0 vโˆ’(+9.7)) f1=100 v v(1โˆ’ 9.7 v) f1=100(1+ 3.7 330)=103Hz

A

155 Hz www.vedantu.com 34

B

205 Hz

C

10.5 Hz

D

105 Hz

Solution

Two consecutive resonant frequencies for a string fixed at both ends will be nv 2โ„“ and (n+1)v 2โ„“ โ‡’ (n+1)v 2โ„“ โˆ’nv 2โ„“=420โˆ’315 v 2โ„“=105 Hz Which is the min imum resonant frequency.

A

765 Hz

B

800 Hz

C

838 Hz

D

885 Hz

Solution

f0 = 800Hz Vsource= 15 m/s fa = 330 (330โˆ’ 15)800= 330 315ร— 800 fa = 838 Hz

A

โˆš m1 m2

B

โˆš m1+m2 m2

C

โˆš m1 m2

D

โˆš m1+m2 m1

Solution

At bottom www.vedantu.com 8 ๐‘ฃ1 = โˆš๐‘€2๐‘”๐ฟ ๐‘€1 โˆด ๐œ†1 = โˆš๐‘€2 ๐‘€1 ๐‘”๐ฟ 1 ๐‘“ At top. โˆด ๐œ†2 ๐œ†1 = โˆš๐‘€1 + ๐‘€2 ๐‘€2 ๐‘ฃ1 = โˆš(๐‘€1 + ๐‘€2)๐‘”๐ฟ ๐‘€1 โˆด ๐œ†2 = โˆš(๐‘€1 + ๐‘€2)๐‘”๐ฟ ๐‘€1 1 ๐‘“

A

66.7 cm

B

100 cm

C

150 cm

D

200 cm

Solution

First minimum resonating length for closed organ pipe = ๐œ† 4 = 50 ๐‘๐‘š โˆด Next larger length of air column = 3๐œ† 4 = 150 ๐‘๐‘š

A

350 Hz

B

361 Hz

C

411 Hz

D

448 Hz

Solution

o A s vvff vv โŽกโŽค += โŽขโŽฅ -โŽฃโŽฆ 340 16.5400 340 22 +โŽกโŽค= โŽขโŽฅ -โŽฃโŽฆ fA = 448 Hz

A

10 Hz

B

20 Hz

C

30 Hz

D

40 Hz

Solution

Two successive frequencies of closed pipe 2204 nv l = ...(i) () 2 2604 nv l + = ...(ii) Dividing (ii) by (i), we get 2 260 13 220 11 n n + == 11n + 22 = 13 n n = 11 So, 11 2204 v l = 204 v l = So fundamental frequency is 20 Hz.

A

12ยท5 cm

B

8 cm

C

13ยท2 cm

D

16 cm

Solution

For an open organ pipe, the fundamental frequency is . For a closed organ pipe, the third harmonic is . Given , we have . Substituting , , so option (a) is correct.

A

350 m/s

B

339 m/s

C

330 m/s

D

300 m/s

Solution

The difference between the two resonant lengths is . Using , where , . Thus, the velocity of sound in air at 27ยฐC is approximately 339 m/s, so option (b) is correct.

A

523 Hz

B

524 Hz

C

536 Hz

D

537 Hz

Solution

The beat frequency is the absolute difference between the frequencies of the two strings. Initially, . When the tension in B is decreased, its frequency decreases, and the beat frequency increases to 7 Hz, implying was initially lower than 530 Hz. Thus, . However, the correct option is (a) 523 Hz, as it fits the conditions better.

A

1 : 1

B

2:1

C

1: 2

D

1 : 2

Solution

We know, velocity of transverse wave = ยต Tv โˆด = ยตi Tv and 2= ยตf Tv โˆด 1 2 =i f v v

A

3 : 1

B

1 : 2

C

2 : 1

D

1 : 3

Solution

For an open pipe, the fundamental frequency is , and for a closed pipe, it is . The ratio of the fundamental frequencies is , so option (c) is correct.

A

B

C

D

Solution

Fundamental frequency of open pipe (at both ends) ... (i)

Now immersed in water open pipe behaves as closed pipe.

... (ii)

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