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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 = $330 \mathrm{~m} / \mathrm{s}$ )
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$375 \mathrm{~Hz}$ 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 = $330 \mathrm{~m} / \mathrm{s}$ )
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Verified Answer
The correct answer is:
$0.58 \mathrm{~m}$
$\ell=0.8 \mathrm{~m} \quad \mathrm{f}=375 \mathrm{~Hz} \quad \mathrm{v}=330 \mathrm{~m}$
$\mathrm{n}_{0}=\frac{\mathrm{v}}{4 \ell}=375$
$\therefore \ell=\frac{330}{4 \times 375}=0.22 \mathrm{~m}$
$\therefore \mathrm{L}-\ell=0.8-0.22 \mathrm{~m}=0.58 \mathrm{~m}$
$\mathrm{n}_{0}=\frac{\mathrm{v}}{4 \ell}=375$
$\therefore \ell=\frac{330}{4 \times 375}=0.22 \mathrm{~m}$
$\therefore \mathrm{L}-\ell=0.8-0.22 \mathrm{~m}=0.58 \mathrm{~m}$
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