Q 12-08-206JEE MainJEE Main 2019 (8 Apr, Shift 2)Medium
In a line of sight radio communication, a distance of about $50\ \text{km}$ is kept between the transmitting and receiving antennas. If the height of the receiving antenna is $70\ \text{m}$, then the minimum height of the transmitting antenna should be (Radius of the Earth $= 6.4\times10^6\ \text{m}$)
Answer: (D) $32\ \text{m}$
Maximum line of sight distance: $d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$.
Receiver: $\sqrt{2\times6.4\times10^6\times70} = \sqrt{8.96\times10^8} \approx 29.9\ \text{km}$.
So $\sqrt{2Rh_T} = 50 - 29.9 = 20.1\ \text{km}$:
$$h_T = \frac{(2.01\times10^4)^2}{1.28\times10^7} \approx 32\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics