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JEE Mains · Physics · STD 11 - Waves and sound
  • JEE Mains 2023
  • Medium
  • Numerical

A wire of density \(8 \times 10^3\,kg / m ^3\) is stretched between two clamps \(0.5\,m\) apart. The extension developed in the wire is \(3.2 \times 10^{-4}\,m\). If \(Y =8 \times 10^{10}\,N / m ^2\), the fundamental frequency of vibration in the wire will be \(......\,Hz\).

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Answer & Solution

Correct Answer

(A) \(80\)

Step-by-step Solution

Detailed explanation

\(f =\frac{1}{2 L } \sqrt{\frac{ T }{\mu}}=\frac{1}{2 L } \sqrt{\frac{ YA \Delta L }{\rho A L}}\)

\(f =80\,Hz\)

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