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MHT CET 2025 (20 Apr, Shift 2) · Organ Pipes & Resonance Tube

Organ Pipes & Resonance Tube question from MHT CET 2025 (20 Apr, Shift 2)

Q43MHT CET 2025 · 20 APR · SHIFT 2Medium

Organ Pipes & Resonance Tube

The frequency of a stretched uniform wire of length $L$ under tension is in resonance with the fundamental frequency of a closed pipe of same length. If the tension in the wire is increased by 8 N, it is in resonance with the first overtone of the same closed pipe. The initial tension in the wire is

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Answer: (d)

String frequency $f \propto \sqrt{T}$ ; closed pipe first overtone $= 3 \times$ fundamental.
$\sqrt{\frac{T + 8}{T}} = 3 \Rightarrow T + 8 = 9T$
$T = 1$ N
So the answer is (d) 1 N.

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