Question

A man observed the pressure amplitude in a sound wave from a wireless receiver as $4\times {10}^{-2}{\text{N/m}}^2$ and the intensity at a point as ${10}^{-6}{\text{W/m}}^2$. If by turning the volume knob, the pressure amplitude is increased to $5\times {10}^{-2}{\text{N/m}}^2$, evaluate the intensity.

• A. ${10}^{-6}{\text{W/m}}^2$
• B. $1.25\times {10}^{-6}{\text{W/m}}^2$
• C. $1.56\times {10}^{-6}{\text{W/m}}^2$
• D. $2\times {10}^{-6}{\text{W/m}}^2$

Answer 1

The correct option is C $1.56\times {10}^{-6}{\text{W/m}}^2$

We know that ,
$I=\frac{P_o^2}{2fv}$
For same medium at same condition
$I\propto P_o^2$
So, $\frac{I_1}{I_2}=\frac{P_{o1}^2}{P_{o2}^2}\Rightarrow I_2={\left(\frac{P_{o2}}{P_{o1}}\right)}^2\times I_1={\left(\frac{5\times {10}^{-2}}{4\times {10}^{-2}}\right)}^2\times {10}^{-6}$
$=1.56\times {10}^{-6}{\text{W/m}}^2$