Question

A radio transmitter sends out 60 W of radiation. Assuming that the radiation is uniform on a sphere with the transmitter at its centre, the intensity (in W/m\(^2\)) of the wave at a distance 12 km is

• A. \( 5.33 \times 10^{-8} \)
• B. \( 3.33 \times 10^{-6} \)
• C. \( 2.12 \times 10^{-8} \)
• D. \( 6.66 \times 10^{-8} \)
• E. \( 3.33 \times 10^{-8} \)

Hint:

Remember the Inverse Square Law: Intensity is inversely proportional to the square of the distance. If you double the distance, the intensity drops to one-fourth of its original value.

Answer 1

Answer: (E)

Concept: The intensity \(I\) of a wave is defined as the power \(P\) distributed over an area \(A\). For a point source radiating uniformly in all directions, the wavefronts are spherical. The surface area of a sphere of radius \(r\) is given by \(4\pi r^2\). \[ I = \frac{P}{4\pi r^2} \]

Step 1: Identify the given values and convert units.
Power \(P = 60 \, \text{W}\) Distance (radius) \(r = 12 \, \text{km} = 12 \times 10^3 \, \text{m}\)

Step 2: Apply the intensity formula.
Substitute the values into the equation: \[ I = \frac{60}{4 \times \pi \times (12 \times 10^3)^2} \] \[ I = \frac{60}{4 \times 3.14159 \times 144 \times 10^6} \]

Step 3: Calculate the final value.
\[ I = \frac{15}{3.14159 \times 144 \times 10^6} \] \[ I \approx \frac{15}{452.39 \times 10^6} \approx 0.03315 \times 10^{-6} \] Adjusting the scientific notation: \[ I \approx 3.315 \times 10^{-8} \, \text{W/m}^2 \] The closest value among the options is \(3.33 \times 10^{-8}\).