Two tuning forks when sounded together produced $4beats/\sec $. the frequency of one fork is $256$. The number of beats heard increases when the fork of frequency $256$ is loaded with wax. The frequency of the other fork is
A. $504$
B. $520$
C. $260$
D. $252$

Hint: When two waves of nearly equal frequency produce sound. We cannot distinguish between them. We hear the sound of the frequency of their average frequencies. This is the condition of base formation.

Complete step by step answer:
Let us first write the information given in the question.
When two tuning forks sound together producing $4beats/\sec $, the frequency of one fork is $256$. Now when this fork of frequency $256$ is loaded with wax the number of beats heard is increased.
We have to calculate the frequency of the other fork.
Now, we can have two possibilities for the frequency of the second fork.
${f_{\sec ond - fork}} = {f_{first - fork}} \pm beat - frequency$………….(1)
Now, according to the question, the first fork, fork with a given frequency, is loaded with wax; this means its frequency will decrease but the beat frequency is increasing. This also means that the frequency difference between the two forks will increase. This is only possible when the following condition is followed.
${f_{\sec ond - fork}} = {f_{first - fork}} + beat - frequency$
Now, on substituting the values we get the following.
${f_{\sec ond - fork}} = 256 + 4 = 260$
Therefore, the frequency of the other or second fork is $260Hz$.
Hence, the correct option is (C) $260Hz$.

Note: The beats are when two nearly equal frequency sounds reach our ears, we hear an alternating increasing and decreasing sound.
The frequency of beat is the difference between the frequencies of the two sound waves, due to which the beat is formed.