MHT CET · Physics · Waves and Sound
If the end correction of an open pipe is \(0.8 \mathrm{~cm}\), then the inner radius of that pipe is
- A \(\frac{1}{3} \mathrm{~cm}\)
- B \(\frac{2}{3} \mathrm{~cm}\)
- C \(\frac{3}{2} \mathrm{~cm}\)
- D \(0.2 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(B) \(\frac{2}{3} \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
For an open pipe, \(\mathrm{e}=0.6 \mathrm{~d}\)
\(\begin{aligned}
& \therefore \quad \mathrm{d}=\frac{\mathrm{e}}{0.6} \\
& \therefore \quad 2 \mathrm{r}=\frac{\mathrm{e}}{0.6} \\
& \therefore \quad \mathrm{r}=\frac{0.8}{1.2}=\frac{2}{3} \mathrm{~cm}
\end{aligned}\)
\(\begin{aligned}
& \therefore \quad \mathrm{d}=\frac{\mathrm{e}}{0.6} \\
& \therefore \quad 2 \mathrm{r}=\frac{\mathrm{e}}{0.6} \\
& \therefore \quad \mathrm{r}=\frac{0.8}{1.2}=\frac{2}{3} \mathrm{~cm}
\end{aligned}\)
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