91Ó°ÊÓ

1. Sound intensity, $I=0.0080\text{ W}/{\text{m}}^{\text{2}}$.
2. Radius, $r=10\text{″¾}$.

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:

$\mathit{I}\mathbf{=}\frac{\mathbf{P}}{\mathbf{4}\mathbf{π}{\mathbf{r}}^{\mathbf{2}}}$ …(1)

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

Intensity of waves is

$I=\frac{P}{4\mathrm{Ï€}{r}^2}$

Substitute all the value in the above equation.

$\begin{array}{c}0.0080\text{ W}/{\text{m}}^{\text{2}}=\frac{P}{4×3.14×{\left(10\text{″¾}\right)}^2}\\ P=10.048\text{ W}\\ P≃10\text{ W}\end{array}$

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

Using the above value of power, we can find the intensity for $r=5.0\text{″¾}$ as

$I=\frac{P}{4\mathrm{Ï€}{r}^2}$

Substitute all the value in the above equation.

$\begin{array}{c}I=\frac{10\text{ W}}{4×3.14×{\left(5.0\text{″¾}\right)}^2}\\ I=0.03187\text{ W}/{\text{m}}^{\text{2}}\\ I=0.032\text{ W}/{\text{m}}^{\text{2}}\end{array}$

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

The standard reference intensity is, $I_0=1.0×{10}^{−12}\text{ W}/{\text{m}}^{\text{2}}$.

We know the equation for sound level

$\mathrm{β}=\left(10\text{ dµþ}\right)\log \frac{I}{I_0}$

Substitute all the value in the above equation.

With $I=0.0080\text{ W}/{\text{m}}^{\text{2}}$ and $I_0=1.0×{10}^{−12}\text{ W}/{\text{m}}^{\text{2}}$.

$\begin{array}{c}\mathrm{β}=\left(10\text{ dµþ}\right)\log \frac{0.0080\text{ W}/{\text{m}}^{\text{2}}}{1.0×{10}^{−12}\text{ W}/{\text{m}}^{\text{2}}}\\ \mathrm{β}=99\text{ dµþ}\end{array}$

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