Question

A siren on a tall pole radiates sound waves uniformly in all directions. At a distance of 15 m from the siren, the sound intensity is 0.250 W/m\(^2\). The intensity of sound at distance 75 m from siren is:

• A. 0.250 W/m\(^2\)
• B. 0.010 W/m\(^2\)
• C. 0.100 W/m\(^2\)
• D. 6.250 W/m\(^2\)

Hint:

The inverse square law is fundamental for any quantity that spreads out uniformly from a point source, including sound intensity, light intensity, and gravitational and electrostatic fields. If the distance increases by a factor of \(n\), the intensity decreases by a factor of \(n^2\).

Answer 1

The Correct Option is B

Step 1: Understand the inverse square law for isotropic sources. The intensity \(I\) of radiation from an isotropic source is inversely proportional to the square of the distance \(r\) from the source, expressed as \( I \propto \frac{1}{r^2} \). This implies that \(I \cdot r^2\) is a constant, so \(I_1 r_1^2 = I_2 r_2^2\).
Step 2: List the provided data. - Initial intensity, \(I_1 = 0.250\) W/m\(^2\). - Initial distance, \(r_1 = 15\) m. - Final distance, \(r_2 = 75\) m. - The objective is to determine the final intensity, \(I_2\).
Step 3: Calculate \(I_2\). Using the formula \(I_2 = I_1 \left( \frac{r_1}{r_2} \right)^2\): \[ I_2 = 0.250 \, \text{W/m}^2 \times \left( \frac{15 \, \text{m}}{75 \, \text{m}} \right)^2 \] \[ I_2 = 0.250 \times \left( \frac{1}{5} \right)^2 = 0.250 \times \frac{1}{25} \] \[ I_2 = \frac{0.250}{25} = 0.010 \, \text{W/m}^2 \]