ExamBro
ExamBro
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

  1. A \(8.1 \;m\)
  2. B \(5.1 \;m\)
  3. C \(1.1 \;m\)
  4. D \(2.1 \;m\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(2.1 \;m\)

Step-by-step Solution

Detailed explanation

\(\frac{\mathrm{nv}}{2 \ell}=420\) \(\frac{(\mathrm{n}+1) \mathrm{v}}{2 \ell}=490\) \(\frac{v}{2 \ell}=70\) \(\ell=\frac{\mathrm{v}}{140}=\frac{1}{140} \sqrt{\frac{540}{6 \times 10^{-3}}}=\frac{1}{140} \sqrt{90 \times 10^{3}}\) \(\ell=\frac{300}{140}=2.142\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app