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If the end correction of an open pipe is 0.8 \mathrm{~cm} then the inner radius of that pipe will be
A) \frac{1}{3} \mathrm{~cm}
B) \frac{2}{3} \mathrm{~cm}
C) \frac{3}{2} \mathrm{~cm}
D) 0.2 \mathrm{~cm}
Physics
If the end correction of an open pipe is 0.8 \mathrm{~cm} then the inner radius of that pipe will be
A) \frac{1}{3} \mathrm{~cm}
B) \frac{2}{3} \mathrm{~cm}
C) \frac{3}{2} \mathrm{~cm}
D) 0.2 \mathrm{~cm}

Answer is (B)
For open organ pipe
\begin{array}{l} \Delta \mathrm{L}=1.2 \times \mathrm{r} \\ \therefore \mathrm{r}=\Delta \mathrm{L} \times \frac{1}{1.2}=\frac{0.8}{1.2}=\frac{2}{3} \mathrm{~cm} . \end{array}

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