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MHT CET 2024 (3 May, Shift 2) · Organ Pipes & Resonance Tube

Organ Pipes & Resonance Tube question from MHT CET 2024 (3 May, Shift 2)

Q60MHT CET 2024 · 3 MAY · SHIFT 2Medium

Organ Pipes & Resonance Tube

The end correction of resonance tube is 1 cm. If the shortest length resonating with a tuning fork is 15 cm, the next resonating length will be

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Answer: (c)

  1. l1+e=λ4⇒λ4=15+1=16l_1 + e = \frac{\lambda}{4} \Rightarrow \frac{\lambda}{4} = 15 + 1 = 16 cm.
  2. l2+e=3λ4=48l_2 + e = \frac{3\lambda}{4} = 48 cm ⇒l2=48−1=47\Rightarrow l_2 = 48 - 1 = 47 cm
  3. So the answer is (c) 47 cm.
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