JEE Mains · Physics · STD 11 - 14. waves and sound
A wire of length \(L\) and mass per unit length \(6.0\times 10^{-3}\; \mathrm{kgm}^{-1}\) is put under tension of \(540\; \mathrm{N}\). Two consecutive frequencies that it resonates at are : \(420\; \mathrm{Hz}\) and \(490 \;\mathrm{Hz}\). Then \(\mathrm{L}\) in meters is
- A \(8.1 \;m\)
- B \(5.1 \;m\)
- C \(1.1 \;m\)
- D \(2.1 \;m\)
Answer & Solution
Correct Answer
(D) \(2.1 \;m\)
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{nv}}{2 \ell}=420\) \(\frac{(\mathrm{n}+1) \mathrm{v}}{2 \ell}=490\) \(\frac{v}{2 \ell}=70\) \(\ell=\frac{\mathrm{v}}{140}=\frac{1}{140} \sqrt{\frac{540}{6 \times 10^{-3}}}=\frac{1}{140} \sqrt{90 \times 10^{3}}\) \(\ell=\frac{300}{140}=2.142\)
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