MHT CET · Physics · Waves and Sound
In a resonance pipe the first and second resonances are obtained at depths \(22.7 \mathrm{~cm}\) and \(70.2 \mathrm{~cm}\) respectively. What will be the end correction ?
- A \(1.05 \mathrm{~cm}\)
- B \(115.5 \mathrm{~cm}\)
- C \(92.5 \mathrm{~cm}\)
- D \(113.5 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(A) \(1.05 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
For end correction \(x\),
\(
\frac{l_{2}+x}{l_{1}+x}=\frac{3 \lambda / 4}{\lambda / 4}=3
\)
\(
\begin{aligned}
x &=\frac{l_{2}-3 l_{1}}{2} \\
&=\frac{70.2-3 \times 22.7}{2}=1.05 \mathrm{~cm}
\end{aligned}
\)
\(
\frac{l_{2}+x}{l_{1}+x}=\frac{3 \lambda / 4}{\lambda / 4}=3
\)
\(
\begin{aligned}
x &=\frac{l_{2}-3 l_{1}}{2} \\
&=\frac{70.2-3 \times 22.7}{2}=1.05 \mathrm{~cm}
\end{aligned}
\)
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