JEE Main 2022 — Waves and Sound Question with Solution

We know that, $v=\lambda f$, then, $\lambda =\frac{v}{f}=\frac{340 \mathrm{m} {\mathrm{s}}^{-1}}{340 \mathrm{Hz}}=\frac{340 \mathrm{m} {\mathrm{s}}^{-1}}{340 {\mathrm{s}}^{-1}}=1 \mathrm{m}$

First resonating length, $l_1=\frac{\lambda }{4}=\frac{100}{4}=25 \mathrm{cm}$

Second resonating length, $l_2=\frac{3\lambda }{4}=\frac{3\times 100}{4}=75 \mathrm{cm}$

Third resonating length, $l_3=\frac{5\lambda }{4}=\frac{5\times 100}{4}=125 \mathrm{cm}$

So third resonance is not possible since the length of the tube is $125 \mathrm{cm}$

Thus, minimum height of water necessary for resonance $=125-75=50 \mathrm{cm}$