A tuning fork produces 6 beats per second with both, 60\ cm and 40\...

Let the frequency of the tuning fork be $n$.

The frequency of a stretched sonometer wire is given by $f = \dfrac{v}{2L}$, which implies $f \propto \dfrac{1}{L}$.

Let $f_1$ and $f_2$ be the frequencies of the wire of lengths $L_1 = 60\ \text{cm}$ and $L_2 = 40\ \text{cm}$ respectively.

Since $L_1 > L_2$, we have $f_1 < f_2$.

The tuning fork produces $6$ beats per second with both lengths. Thus, $n$ must lie between $f_1$ and $f_2$.

$n - f_1 = 6 \Rightarrow f_1 = n - 6$

$f_2 - n = 6 \Rightarrow f_2 = n + 6$

Since $f \propto \dfrac{1}{L}$, we have $f_1 L_1 = f_2 L_2$.

Substituting the values:

$(n - 6) \times 60 = (n + 6) \times 40$

$3(n - 6) = 2(n + 6)$

$3n - 18 = 2n + 12$

$n = 30\ \text{Hz}$

Answer: $30\ \text{Hz}$