Tube $A$ has both ends open while tube $B$ has one end closed, otherwise they are identical. The ratio of…
- $1: 2$
- $1: 4$
- $2: 1$
- $4: 1$
Solution
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,
$\begin{array}{l}
\frac{\mathrm{f}_{\mathrm{A}}}{\mathrm{f}_{\mathrm{B}}}=\frac{\mathrm{v}}{2 \mathrm{~L}} \times \frac{4 \mathrm{~L}}{\mathrm{v}}=\frac{2}{1} \\
\mathrm{f}_{\mathrm{A}}: \mathrm{f}_{\mathrm{B}}=2: 1
\end{array}$
Asked in: MHT CET Full Test 12