A point source of sound has a power of 17 W - AQA - A-Level Physics - Question 3 - 2021 - Paper 5

To calculate the intensity level, we first need to determine the area over which the sound is distributed. The area (A) of a sphere is given by the formula:

$A = 4\pi r^2$

where r is the distance from the source. For our case:

$A = 4\pi (12^2) \approx 453.9 \text{ m}^2$

Next, we find the intensity (I) using the formula:

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

where P is the power of the sound. Therefore:

$I = \frac{17 \text{ W}}{453.9 \text{ m}^2} \approx 0.037 \text{ W/m}^2$

Finally, we calculate the intensity level (L) in decibels (dB) using:

$L = 10 \log_{10}\left( \frac{I}{I_0} \right)$

where the reference intensity, $I_0 = 10^{-12} \text{ W/m}^2$. Therefore:

$L = 10 \log_{10}\left( \frac{0.037}{10^{-12}} \right) \approx 10 \log_{10}(3.7 \times 10^{10}) \approx 10 \times 10.568 = 105.68 ext{ dB}$

Thus, the intensity level is approximately 105.7 dB.