In resonance tube of length $0.8 \mathrm{~m}$, air column vibrates with a source of frequency $375…

In resonance tube of length $0.8 \mathrm{~m}$, air column vibrates with a source of frequency $375 \mathrm{~Hz}$ for a certain height of water from bottom of the tube. Water level corresponding to fundamental frequency is (Neglect end correction, speed of sound in air = $330 \mathrm{~m} / \mathrm{s}$ )
  1. $0.45 \mathrm{~m}$
  2. $0.58 \mathrm{~m}$
  3. $0.8 \mathrm{~m}$
  4. $0.65 \mathrm{~m}$

Solution

$\ell=0.8 \mathrm{~m} \quad \mathrm{f}=375 \mathrm{~Hz} \quad \mathrm{v}=330 \mathrm{~m}$ $\mathrm{n}_{0}=\frac{\mathrm{v}}{4 \ell}=375$ $\therefore \ell=\frac{330}{4 \times 375}=0.22 \mathrm{~m}$ $\therefore \mathrm{L}-\ell=0.8-0.22 \mathrm{~m}=0.58 \mathrm{~m}$

Asked in: MHT CET 2020 (20 Oct Shift 1)

Practice more Waves and Sound questions on Aicharya