Question:medium

A wire of length \(1.2\,\text{m}\) is subjected to a tension of \(240\,\text{N}\). If the frequencies of two successive modes of vibration of the wire are \(200\,\text{Hz}\) and \(250\,\text{Hz}\), then the mass of the wire is

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For a stretched string, \[ \boxed{ f_n=\frac{n}{2L}\sqrt{\frac{T}{\mu}}. } \] The difference between two successive frequencies is \[ \boxed{ f_{n+1}-f_n=\frac{1}{2L}\sqrt{\frac{T}{\mu}}. } \]
Updated On: Jul 18, 2026
  • \(10\,\text{g}\)
  • \(20\,\text{g}\)
  • \(30\,\text{g}\)
  • \(40\,\text{g}\)
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The Correct Option is B

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