Question

The pressure amplitude in a sound waveform radio receiver is $2.0\times {10}^{-3}{\text{N/m}}^2$ and the intensity at a point is ${10}^{-8}{\text{W/m}}^2$. If by turning the "volume" knob the pressure amplitude is increased to $3\times {10}^{-3}{\text{N/m}}^2$. The intensity is equal to $x\times {10}^{-10}{\text{W/m}}^2$, where $x$ is:

• A. 225
• B. 250
• C. 275
• D. 300

Answer 1

The correct option is A 225

The intensity of sound waves is directly proportional to the square of the pressure amplitude,
$I\propto P_0^2$
by this we have, $\frac{I_1}{I_2}=\frac{P_{01}^2}{P_{02}^2}$,
given $P_{01}=2.0\times {10}^{-3}{\text{N/m}}^2$, $P_{02}=3\times {10}^{-3}{\text{N/m}}^2$, $I_1={10}^{-8}{\text{W/m}}^2$,
therefore, $I_2=I_1\left(\frac{P_{02}^2}{P_{01}^2}\right)$,
$I_2={10}^{-8}{\left(\frac{3\times {10}^{-3}}{2.0\times {10}^{-3}}\right)}^2=2.25\times {10}^{-8}{\text{W/m}}^2$