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A glass tube of $1 \mathrm{~m}$ length is filled with water. The water can be drained out slowly from the bottom of the tube. If vibrating tuning fork of frequency $500 \mathrm{~Hz}$ is brought at the upper end of the tube then total number of resonances obtained are [Velocity of sound in air is $320 \mathrm{~ms}^{-1}$ ]
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Verified Answer
The correct answer is:
3
$$
\begin{aligned}
& \mathrm{f}=500 \mathrm{~Hz}, \mathrm{v}=320 \mathrm{~m} / \mathrm{s} \\
& \lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{320}{500}=0.64 \mathrm{~m}=64 \mathrm{~cm}
\end{aligned}
$$
Resonances will be obtained at air columns of lengths
$$
\begin{aligned}
& \frac{\lambda}{4}, \frac{3 \lambda}{4}, \frac{5 \lambda}{4}, \frac{7 \lambda}{4}, \ldots \ldots \\
& \frac{\lambda}{4}=\frac{64}{4}=16 \mathrm{~cm}
\end{aligned}
$$
Resonance can be obtained at $16 \mathrm{~cm}, 48 \mathrm{~cm}, 80 \mathrm{~cm}, 112 \mathrm{~cm}, \ldots$ Since the length of the tube is $100 \mathrm{~cm}$, only first three resonances can be obtained.
\begin{aligned}
& \mathrm{f}=500 \mathrm{~Hz}, \mathrm{v}=320 \mathrm{~m} / \mathrm{s} \\
& \lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{320}{500}=0.64 \mathrm{~m}=64 \mathrm{~cm}
\end{aligned}
$$
Resonances will be obtained at air columns of lengths
$$
\begin{aligned}
& \frac{\lambda}{4}, \frac{3 \lambda}{4}, \frac{5 \lambda}{4}, \frac{7 \lambda}{4}, \ldots \ldots \\
& \frac{\lambda}{4}=\frac{64}{4}=16 \mathrm{~cm}
\end{aligned}
$$
Resonance can be obtained at $16 \mathrm{~cm}, 48 \mathrm{~cm}, 80 \mathrm{~cm}, 112 \mathrm{~cm}, \ldots$ Since the length of the tube is $100 \mathrm{~cm}$, only first three resonances can be obtained.
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