Two sitar strings A and B are slightly out of tune and produce beats of frequency 6 Hz. When the tension in string A is slightly decreased, the beat frequency is found to be reduced to 3 Hz. If the original frequency of A is 324 Hz, the frequency of B is

Correct option is C

The correct answer is: C) 327 Hz Beats = |f₁ – f₂|. Initially: |B – 324| = 6 => B = 330 or 318 On decreasing A (frequency decreases), beat becomes 3 => |B – new A| = 3 => B = 327 So, B = 327 Hz
Information Booster: • Beat frequency is the absolute difference between two frequencies.