91Ó°ÊÓ

1. The density of the air, $\left(\mathrm{ÒÏ}\right)=1.10kg/m^3$
2. Resonant frequency$\left(f_r\right)=7.00Hz$
3. Bulk modulus$\left(\mathrm{β}\right)=1.33×{10}^{5}Pa.$

We can find the sound velocity from bulk modulus and air density using the corresponding relation. By inserting it in the formula for the resonant frequency of pipe closed at one end for the lowest frequency, we can get how far down in the well the water surface is.

Formula:

The velocity of the sound wave in air,$\mathit{v}\mathbf{=}\sqrt{\frac{\mathbf{β}}{\mathbf{ÒÏ}}}$ …(1)

The resonant frequency of a wave, ${\mathit{f}}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{nv}}{\mathbf{4}\mathbf{L}}$ …(2)

Sound velocity is given by the formula from equation (1) and the given values as:

$\begin{array}{rcl}v& =& \sqrt{\frac{1.33×{10}^{5}Pa}{1.10s}}\\ & =& 347.71m/s\end{array}$

The length of the wall using equation (2) is given as:

$L=\frac{1}{4×f_r}v$

Substitute all the value in the above equation.

$\begin{array}{rcl}L& =& \frac{1}{4×f_r}v\\ & =& \frac{1}{4×7Hz}×347.71m/s\\ & =& 12.418m\\ & â‰\dots & 12.4m.\end{array}$

Hence, the level of the water in the well is 12.4m down.