91影视

1. Sound intensity, $I=0.0080\text{鈥塛}/{\text{m}}^{\text{2}}$.
2. Radius, $r=10\text{鈥尘}$.

Using the formula for power in terms of intensity we can get its value. Then, using the relation between intensity and distance, we can find the sound intensity at 5.0 m from the source. Inserting the value of intensity at 10 m in the formula for sound level, we can find its value at that distance.

Formulae:

$\mathit{I}\mathbf{=}\frac{\mathbf{P}}{\mathbf{4}\mathbf{蟺}{\mathbf{r}}^{\mathbf{2}}}$ 鈥�(1)

$\mathit{尾}\mathbf{=}\mathbf{\left(}\mathbf{10}\mathbf{\text{鈥塪叠}}\mathbf{\right)}\mathit{l}\mathit{o}\mathit{g}\frac{\mathbf{I}}{{\mathbf{I}}_{\mathbf{0}}}$ (2)

Intensity of waves is

$I=\frac{P}{4蟺r^2}$

Substitute all the value in the above equation.

$\begin{array}{c}0.0080\text{鈥塛}/{\text{m}}^{\text{2}}=\frac{P}{4脳3.14脳{\left(10\text{鈥尘}\right)}^2}\\ P=10.048\text{鈥塛}\\ P鈮�10\text{鈥塛}\end{array}$

Hence the power of source is, $10\text{鈥塛}$.

Using the above value of power, we can find the intensity for $r=5.0\text{鈥尘}$ as

$I=\frac{P}{4蟺r^2}$

Substitute all the value in the above equation.

$\begin{array}{c}I=\frac{10\text{鈥塛}}{4脳3.14脳{\left(5.0\text{鈥尘}\right)}^2}\\ I=0.03187\text{鈥塛}/{\text{m}}^{\text{2}}\\ I=0.032\text{鈥塛}/{\text{m}}^{\text{2}}\end{array}$

Hence the sound intensity is, $0.032\text{鈥塛}/{\text{m}}^{\text{2}}$.

The standard reference intensity is, $I_0=1.0脳{10}^{鈭�12}\text{鈥塛}/{\text{m}}^{\text{2}}$.

We know the equation for sound level

$尾=\left(10\text{鈥塪叠}\right)\log \frac{I}{I_0}$

Substitute all the value in the above equation.

With $I=0.0080\text{鈥塛}/{\text{m}}^{\text{2}}$ and $I_0=1.0脳{10}^{鈭�12}\text{鈥塛}/{\text{m}}^{\text{2}}$.

$\begin{array}{c}尾=\left(10\text{鈥塪叠}\right)\log \frac{0.0080\text{鈥塛}/{\text{m}}^{\text{2}}}{1.0脳{10}^{鈭�12}\text{鈥塛}/{\text{m}}^{\text{2}}}\\ 尾=99\text{鈥塪叠}\end{array}$

Hence the sound level is, $99\text{鈥塪叠}$.