A tuning fork produces 4 beats per second with another tuning fork of frequency 266Hz. The first one is now loaded with wax and the beat frequency is found to be 8 per second. What is the original frequency of the first tuning fork ?
A
270 Hz
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
266 Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
262 Hz
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
274 Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C262 Hz Given that f2=266Hz
Beat frequency =|f1−f2|=4 ⇒f1=262Hz or 270Hz
Case (i)
Let f1=262Hz
Now, as first tuning fork is loaded, f1 reduces.
If f1 reduces to 258Hz, we can hear beat frequency of 8 per second with f2.
Case (ii)
Let f1=270Hz
Now as first tuning fork is loaded, f1 reduces.
If f1 reduces to 258Hz, we can hear beat frequency of 8 per second with f2.