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JEE Main Physics Class 11 Waves Part 1 Questions
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© examsnet.com
Question : 64
Total: 100
A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.
[2014]
12
8
6
4
Validate
Solution:
Length of pipe = 85 cm = 0.85m
Frequency of oscillations of air column in closed organ pipe is given by,
f
=
(
2
n
−
1
)
v
4
L
f
=
(
2
n
−
1
)
v
4
L
≤
1250
⇒
(
2
n
−
1
)
×
340
0.85
×
4
≤
1250
⇒
2
n
−
1
≤
12.5
≈
6
© examsnet.com
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