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Tube $A$ has both ends open while tube $B$ has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube $A$ and $B$ is
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Verified Answer
The correct answer is:
$2: 1$
Fundamental freq. for,
Tube $\mathrm{A} \Rightarrow \mathrm{f}_{\mathrm{A}}=\frac{\mathrm{v}}{2 \mathrm{~L}}$
Tube $\mathrm{B} \Rightarrow \mathrm{f}_{\mathrm{B}}=\frac{\mathrm{v}}{4 \mathrm{~L}}$
Now,
$$
\begin{array}{l}
\frac{\mathrm{f}_{\mathrm{A}}}{\mathrm{f}_{\mathrm{B}}}=\frac{\mathrm{v}}{2 \mathrm{~L}} \times \frac{4 \mathrm{~L}}{\mathrm{v}}=\frac{2}{1} \\
\mathrm{f}_{\mathrm{A}}: \mathrm{f}_{\mathrm{B}}=2: 1
\end{array}
$$
Tube $\mathrm{A} \Rightarrow \mathrm{f}_{\mathrm{A}}=\frac{\mathrm{v}}{2 \mathrm{~L}}$
Tube $\mathrm{B} \Rightarrow \mathrm{f}_{\mathrm{B}}=\frac{\mathrm{v}}{4 \mathrm{~L}}$
Now,
$$
\begin{array}{l}
\frac{\mathrm{f}_{\mathrm{A}}}{\mathrm{f}_{\mathrm{B}}}=\frac{\mathrm{v}}{2 \mathrm{~L}} \times \frac{4 \mathrm{~L}}{\mathrm{v}}=\frac{2}{1} \\
\mathrm{f}_{\mathrm{A}}: \mathrm{f}_{\mathrm{B}}=2: 1
\end{array}
$$
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