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JEE Mains · Physics · STD 11 - Waves and sound
  • JEE Mains 2021
  • Hard
  • Numerical

A wire having a linear mass density \(9.0 \times 10^{-4} \;{kg} / {m}\) is stretched between two rigid supports with a tension of \(900\; {N}\). The wire resonates at a frequency of \(500\;{Hz}\). The next higher frequency at which the same wire resonates is \(550\; {Hz}\). The length of the wire is \(...... {m}\)

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Answer & Solution

Correct Answer

(C) \(10\)

Step-by-step Solution

Detailed explanation

\(\mu=9.0 \times 10^{-4}\, \frac{{kg}}{{m}}\)

\({T}=900\, {N}\)

\({V}=\sqrt{\frac{{T}}{\mu}}=\sqrt{\frac{900}{9 \times 10^{-4}}}=1000 \,{m} / {s}\)

\({f}_{1}=500 \,{Hz}\)

\({f}=550\)

\(\frac{{nV}}{2 \ell}=500 \ldots . \text { (i) }\)

\(\frac{({n}+1) {V}}{2 \ell}=500 \ldots(ii)\)

\((ii)\, (i)\) \(\frac{{V}}{2 \ell}=50\)

\(\ell=\frac{1000}{2 \times 50}=10\)

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