A transmitting antenna at the top of a tower has a height 320 m, the height of the receiving antenna is 150 m. What is the maximum distance between them for satisfactory communication in LOS mode? (taking radius of earth \( 6.4\times10^{6}\,\mathrm{m} \))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 107.8 km
In line-of-sight (LOS) communication the signal travels straight from one antenna to the other and is cut off by the curvature of the earth. The maximum distance is the sum of the two radio horizons.
\( d_M = \sqrt{2Rh_T} + \sqrt{2Rh_R} \)
\( \sqrt{2Rh_T} = \sqrt{2\times6.4\times10^{6}\times320} = \sqrt{4.096\times10^{9}} = 6.4\times10^{4}\,\mathrm{m} = 64\,\mathrm{km} \)
\( \sqrt{2Rh_R} = \sqrt{2\times6.4\times10^{6}\times150} = \sqrt{1.92\times10^{9}} \approx 4.38\times10^{4}\,\mathrm{m} = 43.8\,\mathrm{km} \)
\( d_M = 64\,\mathrm{km} + 43.8\,\mathrm{km} \approx 107.8\,\mathrm{km} \)
Shortcut: With \( R = 6.4\times10^{6}\,\mathrm{m} \), \( \sqrt{2R} \approx 3.58\times10^{3} \), so the radio horizon is about 3.58 km for every \( \sqrt{h} \) (h in metres). Then \( d_M \approx 3.58\,(\sqrt{320} + \sqrt{150}) = 3.58\times(17.9 + 12.2) \approx 107.8\,\mathrm{km} \).
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Space Wave – …
