91Ó°ÊÓ

1. Power,$P=1.0\text{W}$
2. Radius,$r_1=1.0\text{m}$
3. Radius,$r_2=2.5\text{″¾}$

If the source emits sound wavesisotropically, it means waves are emitted in an identical direction when energy is conserved. The intensity of the sound waves can be calculated by power per unit area.

Intensity of the sound wave is given as-

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

where P is power, A is area and I is intensity.

Since the intensity is the power emitted by the source to the area of the sphere. For radius 1.0 m²,

$\begin{array}{rcl}I& =& \frac{P}{A}\\ I& =& \frac{P}{4\mathrm{Ï€}{r}^2}\\ I& =& \frac{\text{1.0W}}{{\text{4×3.14×(1.0³¾)}}^{\text{2}}}\\ & & \begin{array}{c}I=0.0796\text{ W}/{\text{m}}^{\text{2}}\\ =0.080\text{ W}/{\text{m}}^{\text{2}}\end{array}\\ & & \\ & & \end{array}$

Hence,the intensity 1.0 m from the source is$0.080\text{ W}/{\text{m}}^{\text{2}}$ .

For radius 2.5 m,

$\begin{array}{c}I=\frac{P}{A}\\ =\frac{P}{4\mathrm{Ï€}{r}^2}\end{array}$

For the given values,

role="math" localid="1661492208414" $\begin{array}{l}I=\frac{\text{1.0W}}{{\text{4×3.14×(2.5³¾)}}^{\text{2}}}\\ =0.0127\text{ W}/{\text{m}}^{\text{2}}\\ =0.013\text{ W}/{\text{m}}^{\text{2}}\end{array}$

Hence, the intensity 2.5 m from the source is $0.013\text{ W}/{\text{m}}^{\text{2}}$.