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JEE Main Physics Class 11 Waves Part 1 Questions
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© examsnet.com
Question : 38
Total: 100
A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be: (Assume that the highest frequency a person can hear is 20,000 Hz)
[10 Jan. 2019 (I)]
6
4
7
5
Validate
Solution:
If a closed pipe vibration in
N
th
mode then frequency of vibration
n
=
(
2
N
−
1
)
v
4
l
=
(
2
N
−
1
)
n
1
(where
n
1
=
fundamental frequency of vibration)
Hence
20
,
000
=
(
2
N
−
1
)
×
1500
⇒
N
=
7.1
≈
7
∴
Number of over tones
=
(
No. of mode of vibration
)
−
1
=
7
−
1
=
6
© examsnet.com
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