91Ó°ÊÓ

$\begin{array}{c}\mathrm{P}=1.00\text{μ°Â}\\ =1×{10}^{−6}\text{W}\end{array}$

• Radius,$\mathrm{r}=3.00\text{m}$
• ${\mathrm{I}}_0=1.00×{10}^{−12}{\text{W/m}}^2$.

The source of sound, with power, can emit sound with some intensity. This intensity can be calculated using the formula, $\mathbf{I}\mathbf{=}\mathbf{P}\mathbf{/}\mathbf{A}$. Let that source be centered at the sphere having area A. Also, this emitted sound has its sound level$\mathbf{β}\mathbf{.}\mathbf{}$Sound level is the ratio of$\mathbf{\text{10log}}$role="math" localid="1661403876712" $\mathbf{(}{\mathbf{I}}_1\mathbf{/}{\mathbf{I}}_0\mathbf{)}\mathbf{,}\mathbf{}$(where${\mathbf{I}}_0$is the reference intensity having value,${\mathbf{I}}_0\mathbf{=}{\mathbf{\text{1.00×10}}}^{\mathbf{\text{-12}}}{\mathbf{\text{W/m}}}^{\mathbf{\text{2}}}$.

The expression for the power is given by,

$\mathrm{P}=\mathrm{I}×\mathrm{A}$

Here$\mathrm{I}$ is the intensity and$A$ is the area.

To find the intensity of the source,

$\mathrm{I}=\frac{\mathrm{P}}{\mathrm{A}}$

Substitute $4{\mathrm{Ï€°ù}}^2$for $\mathrm{A}$into the above equation,

$\mathrm{I}=\frac{\mathrm{P}}{4{\mathrm{Ï€°ù}}^2}$

Substitute$1×{10}^{−6}\text{W}$for$\mathrm{P}$and$3\text{ }\mathrm{m}$for$\mathrm{r}$into the above equation,

$\begin{array}{c}\mathrm{I}=\frac{1×{10}^{−6}}{4\left(3.14\right){\left(3\right)}^2}\\ =8.84×{10}^{−9}{\text{W/m}}^{\text{2}}\end{array}$

Hence, the intensity 3.00m away is $8.84×{10}^{−9}{\text{W/m}}^{\text{2}}$.

To find the sound level,

$\mathrm{β}=10\text{ }\log \left(\frac{\mathrm{I}}{{\mathrm{I}}_0}\right)$

Substitute $8.84×{10}^{−9}{\text{W/m}}^{\text{2}}$for role="math" localid="1661404350240" $\mathit{l}$and $1.00×{10}^{−12}{\text{W/m}}^2$for $I_0$into the above equation,

$\begin{array}{c}\mathrm{β}=10\log \frac{8.84×{10}^{−9}}{1.00×{10}^{−12}}\\ =39.5\text{ }\mathrm{dB}\end{array}$

Hence, the sound level in decibels at that distance is $39.5\text{ }\mathrm{dB}$.