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Organ Pipes & Resonance Tube question from JEE Main 2014

Q115JEE Main 2014Difficult

Organ Pipes & Resonance Tube

A pipe of length 85 cm85\text{ cm} is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz1250\text{ Hz}. The velocity of sound in air is 340 m/s340\text{ m/s}.

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Answer: (c)

  1. Closed pipe: f=(2n−1)v4Lf = (2n-1)\frac{v}{4L} (odd harmonics only)
  2. v4L=3404×0.85=100 Hz\frac{v}{4L} = \frac{340}{4 \times 0.85} = 100\text{ Hz}
  3. Allowed: 100,300,500,700,900,1100 Hz100, 300, 500, 700, 900, 1100\text{ Hz} (next is 1300>12501300 > 1250)
  4. So the answer is (c) 66.
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