Q 11-14-097JEE MainJEE Main 2020 (9 Jan, Shift 2)Easy
A wire of length $L$ and mass per unit length $6.0\times10^{-3}$ kg m$^{-1}$ is put under tension of $540$ N. Two consecutive frequencies that it resonates at are $420$ Hz and $490$ Hz. Then $L$ in meters is:
Answer: (A) $2.1$ m
Wave speed: $v = \sqrt{\dfrac{540}{6\times10^{-3}}} = 300$ m/s.
For a string fixed at both ends, consecutive resonant frequencies differ by the fundamental $\dfrac{v}{2L}$:
$$\frac{v}{2L} = 490 - 420 = 70 \Rightarrow L = \frac{300}{140} \approx 2.1\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics