Physics - Sound Wave Question with Solution | TestHub

Let the length of the pipe be $L$ and the speed of sound in air be $v$.

For a pipe open at both ends, the fundamental frequency is given by:
$f_1=\frac{v}{2L}$

When $80\%$ of the pipe is immersed in water, the length of the air column remaining above the water surface is:
$L^{\prime }=L-0.8L=0.2L=\frac{L}{5}$

The pipe now acts as a closed pipe at one end (the water surface acts as a closed end). The fundamental frequency of this closed pipe is:
$f_2=\frac{v}{4L^{\prime }}$

Substituting $L^{\prime }=\frac{L}{5}$:
$f_2=\frac{v}{4\left(\frac{L}{5}\right)}=\frac{5v}{4L}$

The ratio of the frequencies is:
$\frac{f_1}{f_2}=\frac{\frac{v}{2L}}{\frac{5v}{4L}}=\frac{v}{2L}\times \frac{4L}{5v}=\frac{4}{10}=\frac{2}{5}$

Thus, the ratio $f_1:f_2$ is $2:5$.

Answer: $2:5$