An unknown tuning fork produces 5 beats per second with a second...

The fundamental frequency of a closed organ pipe is given by $n_2 = \dfrac{v}{4L}$.

Substituting the given values, $n_2 = \dfrac{320}{4 \times 0.4} = 200 \text{ Hz}$.

Since the unknown tuning fork produces $5$ beats per second with the second tuning fork, its frequency is $n_1 = n_2 \pm 5 = 200 \pm 5$.

Thus, $n_1$ is either $205 \text{ Hz}$ or $195 \text{ Hz}$.

When a tuning fork is loaded with wax, its frequency decreases.

If $n_1 = 205 \text{ Hz}$, a decrease in frequency will bring it closer to $200 \text{ Hz}$, resulting in a beat frequency less than $5 \text{ Hz}$. This matches the given condition.

If $n_1 = 195 \text{ Hz}$, a decrease in frequency will take it further away from $200 \text{ Hz}$, resulting in a beat frequency greater than $5 \text{ Hz}$. This contradicts the given condition.

Therefore, the original frequency of the first tuning fork is $205 \text{ Hz}$.