An open pipe is in second harmonic with frequency $f_1$. One end of the tube is closed and frequency is increased to $f_2$ and it resonates again with nth harmonic. For what value of n , the ratio of $\dfrac{f_1}{f_2}$ is $\dfrac{4}{5}$ ?

For an open pipe of length $L$, the frequency of the second harmonic is given by $f_1 = 2 \times \dfrac{v}{2L} = \dfrac{v}{L}$, where $v$ is the speed of sound.

When one end is closed, the pipe becomes a closed pipe of length $L$. The frequencies of a closed pipe are given by $f = m \dfrac{v}{4L}$, where $m$ is an odd integer ($1, 3, 5, \dots$). The $n$-th harmonic of a closed pipe corresponds to the $n$-th odd harmonic, so $f_2 = n \dfrac{v}{4L}$, where $n$ is an odd integer.

The ratio of the frequencies is given as $\dfrac{f_1}{f_2} = \dfrac{4}{5}$.

Substituting the expressions for $f_1$ and $f_2$:

$\dfrac{v/L}{n v / 4L} = \dfrac{4}{5}$

$\dfrac{4}{n} = \dfrac{4}{5}$

Therefore, $n = 5$.

Answer: 5