MHT CET · Physics · Waves and Sound
A tuning fork of frequency ' \(n\) ' is held near the open end of tube which is closed at the other end and the length are adjusted until resonance occurs. The first resonance occurs at length \(\mathrm{L}_1\) and immediate next resonance occurs at length \(\mathrm{L}_2\). The speed of sound in air is
- A \(\mathrm{n}\left(\mathrm{L}_2-\mathrm{L}_1\right)\)
- B \(\frac{\mathrm{n}\left(\mathrm{L}_2-\mathrm{L}_1\right)}{2}\)
- C \(2 \mathrm{n}\left(\mathrm{L}_2-\mathrm{L}_1\right)\)
- D \(\frac{\mathrm{n}\left(\mathrm{L}_2+\mathrm{L}_1\right)}{2}\)
Answer & Solution
Correct Answer
(C) \(2 \mathrm{n}\left(\mathrm{L}_2-\mathrm{L}_1\right)\)
Step-by-step Solution
Detailed explanation
For first resonance \(\mathrm{L}_1=\frac{\lambda}{4}\)
For second resonance \(\mathrm{L}_2=\frac{3 \lambda}{4}\)
\(
\begin{aligned}
& \therefore \mathrm{L}_2-\mathrm{L}_1=\frac{\lambda}{2} \text { or } \lambda=2\left(\mathrm{~L}_2-\mathrm{L}_1\right) \\
& \mathrm{V}=\mathrm{n} \lambda=2 \mathrm{n}\left(\mathrm{L}_2-\mathrm{L}_1\right)
\end{aligned}
\)
For second resonance \(\mathrm{L}_2=\frac{3 \lambda}{4}\)
\(
\begin{aligned}
& \therefore \mathrm{L}_2-\mathrm{L}_1=\frac{\lambda}{2} \text { or } \lambda=2\left(\mathrm{~L}_2-\mathrm{L}_1\right) \\
& \mathrm{V}=\mathrm{n} \lambda=2 \mathrm{n}\left(\mathrm{L}_2-\mathrm{L}_1\right)
\end{aligned}
\)
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