In a resonance tube experiment when the tube is filled with water up to a height of $17.0 \mathrm{~cm}$ from bottom, it resonates with a given tuning fork. When the water level is raised the next resonance with the same tuning fork occurs at a height of $24.5 \mathrm{~cm}$. If the velocity of sound in air is $330 \mathrm{~m} / \mathrm{s}$, the tuning fork frequency is :
Correct Option: 1
(1) Here, $l_{1}=17 \mathrm{~cm}$ and $l_{2}=24.5 \mathrm{~cm}, V=330 \mathrm{~m} / \mathrm{s}, f=?$
$\lambda=2\left(l_{2}-l_{1}\right)=2 \times(24.5-17)=15 \mathrm{~cm}$
Now, from $v=f \lambda \Rightarrow 330=\lambda \times 15 \times 10^{-2}$
$\therefore \lambda=\frac{330}{15} \times 100=\frac{1100 \times 100}{5}=2200 \mathrm{~Hz}$
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