Question

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 in water it produces a note of frequency f${}_2$. The ratio $\frac{f_1}{f_2}$ is

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{1}{2}$
• D. $2$

Answer 1

The correct option is C $\frac{1}{2}$

Open pipe frequency, $f_1=\frac{v}{2L}$

Frequency, $f_2=\frac{v}{4\left(L/4\right)}$

( $\because$ when $3/4th$ of length is dipped, it acts as closed pipe of length $\frac{1}{4}$ of length )

So, $\frac{f_1}{f_2}=\frac{{}^{\nu }/2L}{{}^{\nu }/4\left(L/4\right)}=\frac{1}{2}$