The fundamental frequency of an air column in a pipe open at both ends is ' $\mathrm{f}_1$ '. Now $80 \%$ of…

The fundamental frequency of an air column in a pipe open at both ends is ' $\mathrm{f}_1$ '. Now $80 \%$ of its length is immersed in water, the fundamental frequency of the air column becomes $f_2$. The ratio of $f_1: f_2$ is
  1. $5: 2$
  2. $5: 4$
  3. $4: 5$
  4. $2: 5$

Solution

The fundamental frequency of the open pipe $=\mathrm{f}_1=\frac{\mathrm{V}}{2 \dot{\mathrm{~L}}}$
After dipping in water, the pipe behaves like a closed pipe. As the pipe is dipped $80 \%$ in water, $l=\frac{20}{100} \times \mathrm{L}=\frac{\mathrm{L}}{5}$ $\therefore \quad$ Fundamental frequency $\mathrm{f}_2=\frac{\mathrm{V}}{4 l}=\frac{5 \mathrm{~V}}{4 \mathrm{~L}}$ $\therefore \quad \frac{\mathrm{f}_1}{\mathrm{f}_2}=\frac{\mathrm{V}}{2 \mathrm{~L}} \times \frac{4 \mathrm{~L}}{5 \mathrm{~V}}=\frac{2}{5}$ ~

Asked in: MHT CET 2024 (03 May Shift 1)

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