Question

A pipe opened at both ends produces a note of frequency A. 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

Answer: (C)

$\frac{1}{2}$

Let the length of the open pipe be $L$.

Frequency produced in open pipe $f_1=\frac{v}{2L}$

Now $\frac{3}{4}th$ of the pipe is immersed in air.

Thus length of the pipe closed at one end $L^{\prime }=L-\frac{3}{4}L=\frac{L}{4}$

Frequency produced in closed pipe at one end $f_2=\frac{v}{4L^{\prime }}$

$\therefore$ $f_2=\frac{v}{4\times \frac{L}{4}}=\frac{v}{L}$

Ratio of frequency $\frac{f_1}{f_2}=\frac{v/2L}{v/L}$

$⟹$ $\frac{f_1}{f_2}=\frac{1}{2}$