Q111 / 150 · JEE Main 2006 · Physics

A string is stretched between fixed points separated by $75 \mathrm{~cm}$. It is observed to have resonant frequencies of $420 \mathrm{~Hz}$ and $315 \mathrm{~Hz}$. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

  • A
    $10.5 \mathrm{~Hz}$
  • Correct answer
    B
    $105 \mathrm{~Hz}$
  • C
    $1.05 \mathrm{~Hz}$
  • D
    $1050 \mathrm{~Hz}$

Solution

$\frac{\mathrm{n}}{2 \ell}(\mathrm{v})=315, \frac{(\mathrm{n}+1)}{2 \ell} \mathrm{v}=420$ Solving $\frac{\mathrm{v}}{2 \ell}=105$

Hence, the correct answer is option: 'B'.

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