c
(c)
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 $f _2=144\,Hz$