Correct Answer: 384
Beat frequency is equal to the absolute difference of frequencies of the two tuning forks.
Initial beat frequency = 8 / 2 = 4 Hz
Let the original frequency of tuning fork A be f. Given frequency of tuning fork B is 380 Hz.
|f − 380| = 4
So possible values of f are 384 Hz or 376 Hz.
When wax is loaded on fork A, its frequency decreases. New beat frequency:
4 / 2 = 2 Hz
This means the difference between frequencies has reduced, so original frequency of A must be greater than 380 Hz.
Therefore, f = 384 Hz
Hence, the original frequency of tuning fork A is 384 Hz.
Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.