Q74 / 75 · JEE Main 2026 (28 January Shift 2) · Physics
Two tuning forks $A$ and $B$ are sounded together giving rise to 8 beats in 2 s. When fork $A$ is loaded with wax, the beat frequency is reduced to 4 beats in 2 s. If the original frequency of tuning fork $B$ is 380 Hz then original frequency of tuning fork $A$ is $\_\_\_\_$ Hz.
Correct answer: 384
Solution
$|f_A - f_B| = 4$
$|f_A - 380| = 4$
So, $f_A = 384$ Hz or $376$ Hz
on loading with wax $f_A$ decreases
So, $f_A = 384$ Hz
$|f_A - 380| = 4$
So, $f_A = 384$ Hz or $376$ Hz
on loading with wax $f_A$ decreases
So, $f_A = 384$ Hz
Hence, the correct answer is 384.