MHT CET · Physics · Waves and Sound
A resonance tube completely filled with water has a small hole at the bottom.
Length of the tube is \(0.8 \mathrm{~m}\). A vibrating tuning fork of frequency \(500 \mathrm{~Hz}\) is held
near the open end of tube. Water is slowly removed from the bottom. The
maximum number of resonances heard will be (Neglect end correction. Speed of
sound in air \(=340 \mathrm{~m} / \mathrm{s}\) )
- A \(5\)
- B \(4\)
- C \(2\)
- D \(3\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
\(
\begin{array}{l}
\ell=0.8 \mathrm{~m} \quad \mathrm{f}=500 \mathrm{~Hz} \quad \mathrm{v}=340 \mathrm{~m} / \mathrm{s} \\
\mathrm{v}=\mathrm{f} \lambda \quad \lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{340}{500}=\frac{34}{50}=\frac{34}{50}=\frac{17}{25} \mathrm{~m} \\
\frac{\lambda}{4}=\frac{17}{100}=0.17 \mathrm{~m} \\
\frac{3 \lambda}{4}=0.51 \mathrm{~m} \\
\frac{5 \lambda}{4}=0.85 \mathrm{~m}
\end{array}
\)
So, only 2 resonances will be heard.
\begin{array}{l}
\ell=0.8 \mathrm{~m} \quad \mathrm{f}=500 \mathrm{~Hz} \quad \mathrm{v}=340 \mathrm{~m} / \mathrm{s} \\
\mathrm{v}=\mathrm{f} \lambda \quad \lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{340}{500}=\frac{34}{50}=\frac{34}{50}=\frac{17}{25} \mathrm{~m} \\
\frac{\lambda}{4}=\frac{17}{100}=0.17 \mathrm{~m} \\
\frac{3 \lambda}{4}=0.51 \mathrm{~m} \\
\frac{5 \lambda}{4}=0.85 \mathrm{~m}
\end{array}
\)
So, only 2 resonances will be heard.
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