Answer:
The reasonable power will be "0.234 μW".
Explanation:
The given values are:
P = 15 μW
d = 175 m
As we know,
Propagation follows inverse cube power law, then
∴ [tex]Power \ \alpha \ \frac{1}{d^3}[/tex]
⇒ [tex]Power=\frac{K}{d^3}[/tex]
On substituting the estimated values, we get
⇒ [tex]15\times 10^{-6}=\frac{K}{(173)^3}[/tex]
⇒ [tex]K=15\times (175)^3\times 10^{-6}[/tex]
Now,
"P" at 0.7 km or 700 m from BS will be:
⇒ [tex]P=\frac{K}{d^3}[/tex]
⇒ [tex]P=\frac{15\times (175)^3\times 10^{-6}}{(700)^3}[/tex]
⇒ [tex]P=0.234 \ \mu W[/tex]