JEE Main · 2019 · 10 Jan 2019 · Shift 1 · EM Waves

Line-of-sight range from an elevated transmitter to an elevated receiver adds the two horizon distances, $D=\sqrt{2h_{T}R}+\sqrt{2h_{R}R}$. From the $140$ m tower, $\sqrt{2\times 140\times 6.4\times 10^6}=\sqrt{1.792\times 10^9}\approx 4.23\times 10^4$ m, and from the $40$ m antenna, $\sqrt{2\times 40\times 6.4\times 10^6}=\sqrt{5.12\times 10^8}\approx 2.26\times 10^4$ m. Their sum is about $6.5\times 10^4$ m, roughly $65$ km. Both heights count because each station can see only as far as its own horizon, and the link bridges the gap where the two horizons meet.