MCQ
A pipe open at both ends produces a note of frequency $f_1$. When the pipe is kept with $\frac{3}{4}$th of its length it water, it produced a note of frequency $f_2$. The ratio $\frac{{{f_1}}}{{{f_2}}}$ is
- A$0.75$
- B$1.33$
- ✓$0.5$
- D$2$
${f_2} = \frac{v}{{4 \times \left( {\frac{l}{4}} \right)}} = \frac{v}{l} = 2{f_1}$ ==> $\frac{{{f_1}}}{{{f_2}}} = \frac{1}{2}$
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