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: Recall the relationship between intensity and distance for an isotropic source. For a source radiating uniformly in all directions, the intensity \(I\) decreases with the square of the distance \(r\) from the source. This is the inverse square law: \[ I \propto \frac{1}{r^2} \] This means that the product \(I \cdot r^2\) is constant. Therefore, we can write \(I_1 r_1^2 = I_2 r_2^2\).
Step 2: Identify the given values. - Initial intensity \(I_1 = 0.250\) W/m\(^2\). - Initial distance \(r_1 = 15\) m. - Final distance \(r_2 = 75\) m. - We need to find the final intensity \(I_2\).
Step 3: Solve for \(I_2\). \[ I_2 = I_1 \left( \frac{r_1}{r_2} \right)^2 \] \[ I_2 = 0.250 \, \text{W/m}^2 \times \left( \frac{15 \, \text{m}}{75 \, \text{m}} \right)^2 \] \[ I_2 = 0.250 \times \left( \frac{1}{5} \right)^2 = 0.250 \times \frac{1}{25} \] \[ I_2 = \frac{0.250}{25} = 0.010 \, \text{W/m}^2 \]