Right choice is (c) 11.19dB

Explanation: Given Pt=5W, Pr=150μW, f=500MHz, R=100m and Gt in dB=10dB

Gt in dB=10log10 Gt=10dB

Gt=10

⇨ \(\lambda = \frac{c}{f} = \frac{3×10^8}{500Mhz} = 0.6m\)

From Friss transmission equation, \(\frac{P_r}{P_t} = \frac{G_t G_r \lambda^2}{(4\pi R)^2} \)

⇨ \(G_r=\frac{P_r (4\pi R)^2}{P_t G_t \lambda^2} = \frac{150\mu(4\pi ×100)^2}{5×10×0.6^2}=13.16\)

⇨ Gr in dB=10log10 Gr=10log10 13.16=11.19dB