AIEEE 2002 Physics Solved Paper

A tuning fork produces 4 beats\/sec with another tuning fork of frequency 288‌cps. From this information we can conclude that the frequency of unknown fork is 288+4‌cps or 288−4‌cps i.e. 292‌cps or 284‌cps.
Here when a little wax is placed on the unknown fork, it decreases the frequency of unknown fork. Here also beats per second decreases to 2 from 4 . So the difference between frequency decreases.
This is possible only when before placing the wax, the frequency of unknown fork is greater than the frequency of the given tuning fork.
So the frequency of the unknown tuninh fork is =292‌Hz