When two tuning forks A and B are sounded together, $4$ beats per second are heard. The frequency of fork B is $384Hz$. If one of the prongs of the fork A is filed and sounded with B, then the beat frequency increases. Therefore, the frequency of the fork A will be given as,
$\begin{align}
  & A.380Hz \\
 & B.388Hz \\
 & C.379Hz \\
 & D.389Hz \\
\end{align}$

Hint: The beat frequency can be found by taking the modulus of the difference between the frequency of the fork A and the frequency of the fork B. When there is a filing on the fork A, the beat frequency will get increased. This all will help you in answering this question.

Complete answer:
The beat frequency can be found by taking the modulus of the difference between the frequency of the fork A and the frequency of the fork B. that is we can write that,
$b=\left| {{\nu }_{A}}-{{\nu }_{B}} \right|$
Where ${{\nu }_{A}}$ be the frequency of the fork A and ${{\nu }_{B}}$ be the frequency of the fork B.
When we make filing on the fork A, then the beat frequency increases. Therefore the fork A will be having greater frequency than fork B as on filing, frequency of fork increases.
That is we can write that,
$b={{\nu }_{A}}-{{\nu }_{B}}$
It has been already mentioned in the question that the frequency of the fork B which can be written as,
${{\nu }_{B}}=384Hz$
And the beat frequency has been given as,
$b=4$
Substituting the values in it will give,
\[4={{\nu }_{A}}-384=388Hz\]

So, the correct answer is “Option B”.

Note:
Filing has been done in order to vary the frequency of vibration. If the prongs are loaded, their inertia will increase, which will cause a decrease of vibrational frequency. In the same sense, if the mass is removed by filing the prongs, their inertia gets decreased causing the increase in vibrational frequency.