A string is stretched between fixed points separated by $75.0 \mathrm{~cm}$. It is observed to have resonant frequencies of $420 \mathrm{~Hz}$ and $315 \mathrm{~Hz}$. There are no other resonant frequency for this string is: (1) $105 \mathrm{~Hz}$, (2) $155 \mathrm{~Hz}$ (3) $205 \mathrm{~Hz}$ (4) $105 \mathrm{~Hz}$

The correct Solution is (1)
Two consecutive resonant frequencies for a string fixed at both ends will be
$
\begin{array}{l}
\frac{n v}{2 \ell} \text { and } \frac{(n+1) v}{2 \ell} \\
\Rightarrow \frac{(n+1) v}{2 \ell}-\frac{n v}{2 \ell}=420-315 \\
\frac{v}{2 \ell}=105 \mathrm{~Hz}
\end{array}
$
that is the minimum resonant frequency.