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MHT CET 2024 (3 May, Shift 1) · Organ Pipes & Resonance Tube

Organ Pipes & Resonance Tube question from MHT CET 2024 (3 May, Shift 1)

Q52MHT CET 2024 · 3 MAY · SHIFT 1Easy

Organ Pipes & Resonance Tube

The fundamental frequency of an air column in a pipe open at both ends is ' f1f_1 '. Now 80%80\% of its length is immersed in water, the fundamental frequency of the air column becomes f2f_2. The ratio of f1:f2f_1 : f_2 is

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Answer: (d)

  1. Open pipe: f1=v2Lf_1 = \frac{v}{2L}.
  2. 80%80\% in water leaves a closed pipe of length 0.2L0.2L: f2=v4(0.2L)=v0.8Lf_2 = \frac{v}{4(0.2L)} = \frac{v}{0.8L}.
  3. f1f2=0.8L2L=25\frac{f_1}{f_2} = \frac{0.8L}{2L} = \frac{2}{5}
  4. So the answer is (d) 2:52 : 5.
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