JEE Mains · Physics · STD 11 - 14. waves and sound
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 \(...... {m}\)
- A \(50\)
- B \(100\)
- C \(10\)
- D \(2\)
Answer & Solution
Correct Answer
(C) \(10\)
Step-by-step Solution
Detailed explanation
\(\mu=9.0 \times 10^{-4}\, \frac{{kg}}{{m}}\) \({T}=900\, {N}\) \({V}=\sqrt{\frac{{T}}{\mu}}=\sqrt{\frac{900}{9 \times 10^{-4}}}=1000 \,{m} / {s}\) \({f}_{1}=500 \,{Hz}\) \({f}=550\) \(\frac{{nV}}{2 \ell}=500 \ldots . \text { (i) }\) \(\frac{({n}+1) {V}}{2 \ell}=500 \ldots(ii)\)…
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