Question

The WCCO AM radio transmitter has a power of 48900 W. You measure the intensity of the signal to be 8×10−6 W/m^2. How far are you from the tower? (in km)

7.82×101 km

4.41×101 km

8.82×101 km

1.95×106 km

2.21×101 km

If you travel to a location that is 6 times further away from the tower than in the previous question, what intensity do you measure? (in W/m^2)

2.88×10-4 W/m2

1.33×10-6 W/m2

5.11×10-7 W/m2

3.64×10-17 W/m2

2.22×10-7 W/m2

Answer 1

1.

We know that relation between power and Intensity is given by:

Power = Intensity*Area

P = I*A

A = Cross-sectional area covered by tower = 4*pi*R^2

P = I*4*pi*R^2

R = sqrt (P/(4*pi*I))

P = Power transmitted = 48900 W

I = Intensity of signal = 8*10^-6 W/m^2

So,

R = sqrt (48900/(4*pi*8*10^-6))

R = 22054.87 m = 22100 m = 2.21*10^1*10^3 m

R = 2.21*10^1 km

Correct option is E.

2.

Now new intensity at a location that 6 times further away from above location will be:

I1 = P/A1

I1 = 48900/(4*pi*(6*22054.87)^2)

I1 = 2.22*10^-7 W/m^2

Correct option is E.

Let me know if you've any query.