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Question: Answered & Verified by Expert
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.
PhysicsWaves and SoundJEE MainJEE Main 2014 (06 Apr)
Options:
  • A 12
  • B 8
  • C 6
  • D 4
Solution:
1562 Upvotes Verified Answer
The correct answer is: 6

fundamental frequency

       


L=λ4v1=v4L


       


L = λ 2 + λ 4 = 3 λ 4

v2=3v4L

∴   v=oddv4L

Now L = 8 5 cm = 0.85 m

        v=340m/s

          = odd 3 4 0 4 × 0.85

         =odd×100 Hz

Hence possible frequencies below 1250 Hz are

100 Hz, 300 Hz, 500 Hz...........1100 Hz

Hence 6 

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