Let the frequency of the first fork be \(f_1\) and that of second be \(f_2\).
We then have, \(f_1=\frac{v}{4 \times 24}\) and
\(f _2=\frac{ v }{4 \times 25}\)
We also see that \(f _1 > f _2\)
\(\therefore f _1- f _2=6\)
\(\text { and } \frac{ f _1}{ f _2}=\frac{24}{25}\)
Solving \((i)\) and \((ii)\), we get \(f _1=150\,Hz\) and \(f _2=144\,Hz\)
$\left(t_{0}\right.$ એ સમય દર્શાવે છે,કે જ્યારે ઉદગમ અને અવલોકન કાર વચ્ચેનું અંતર લઘુતમ થાય. $)$