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.
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n 1 = 1 n 1 + 2 n 1 + 3 n 1
n 1 = 1 n 1 + 2 n 1 + 3 n 1
n = 1 n + 2 n + 3 n
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.
4
5
7
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.
1332 Hz
1372 Hz
1412 Hz
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.
100 Hz
(2 ) 103 Hz
106 Hz
97 Hz
Solution
www.vedantu.com 28 t1=f0(vโv vโv s ) f1=100( vโ0 vโ(+9.7)) f1=100 v v(1โ 9.7 v) f1=100(1+ 3.7 330)=103Hz
155 Hz www.vedantu.com 34
205 Hz
10.5 Hz
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 โnv 2โ=420โ315 v 2โ=105 Hz Which is the min imum resonant frequency.
765 Hz
800 Hz
838 Hz
885 Hz
Solution
f0 = 800Hz Vsource= 15 m/s fa = 330 (330โ 15)800= 330 315ร 800 fa = 838 Hz
โ m1 m2
โ m1+m2 m2
โ m1 m2
โ m1+m2 m1
Solution
At bottom www.vedantu.com 8 ๐ฃ1 = โ๐2๐๐ฟ ๐1 โด ๐1 = โ๐2 ๐1 ๐๐ฟ 1 ๐ At top. โด ๐2 ๐1 = โ๐1 + ๐2 ๐2 ๐ฃ1 = โ(๐1 + ๐2)๐๐ฟ ๐1 โด ๐2 = โ(๐1 + ๐2)๐๐ฟ ๐1 1 ๐
66.7 cm
100 cm
150 cm
200 cm
Solution
First minimum resonating length for closed organ pipe = ๐ 4 = 50 ๐๐ โด Next larger length of air column = 3๐ 4 = 150 ๐๐
350 Hz
361 Hz
411 Hz
448 Hz
Solution
o A s vvff vv โกโค += โขโฅ -โฃโฆ 340 16.5400 340 22 +โกโค= โขโฅ -โฃโฆ fA = 448 Hz
10 Hz
20 Hz
30 Hz
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.
12ยท5 cm
8 cm
13ยท2 cm
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.
350 m/s
339 m/s
330 m/s
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.
523 Hz
524 Hz
536 Hz
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.
1 : 1
2:1
1: 2
1 : 2
Solution
We know, velocity of transverse wave = ยต Tv โด = ยตi Tv and 2= ยตf Tv โด 1 2 =i f v v
3 : 1
1 : 2
2 : 1
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.
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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