Q 11-14-134JEE MainJEE Main 2018 (16 Apr, Shift 1)Easy
Two sitar strings, $A$ and $B$ playing the note 'Dha' are slightly out of tune and produce beats of frequency $5\ \text{Hz}$. The tension of the string $B$ is slightly increased and the beat frequency is found to decrease by $3\ \text{Hz}$. If the frequency of $A$ is $425\ \text{Hz}$, the original frequency of $B$ is
Answer: (D) $420\ \text{Hz}$
With 5 beats, $f_B = 425 \pm 5 = 420$ or $430\ \text{Hz}$.
Increasing the tension raises $f_B$ (since $f \propto \sqrt T$). The beat frequency falls (from 5 to 2 Hz) only if $f_B$ moves closer to 425 Hz, i.e. if $f_B$ was below 425 Hz. So
$$f_B = 420\ \text{Hz}$$
Solution by Sreeraj P, M.Sc Physics