The height of a TV tower is 180 m. To what distance the transmission can be made from this tower, if the radius of earth is \( 6.4\times10^{6}\,\mathrm{m} \)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 48 km
TV signals travel in line of sight, so the range is the distance d from the top of the tower to the point where the ray just grazes the earth's surface. By Pythagoras for the tangent drawn from the tower top:
\( d^{2} + R^{2} = (R + h)^{2} \Rightarrow d^{2} = 2Rh + h^{2} \approx 2Rh \)
The height h is very small compared with R, so \( h^{2} \) is neglected.
\( d = \sqrt{2Rh} = \sqrt{2\times6.4\times10^{6}\,\mathrm{m}\times180\,\mathrm{m}} = \sqrt{2304\times10^{6}}\,\mathrm{m} \)
\( d = 48\times10^{3}\,\mathrm{m} = 48\,\mathrm{km} \)
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 – “… Space waves are used for line-of-sight (LOS) communication as well as satellite communication. … If the transmitting antenna is at a height hT, then you can show that the distance to the horizon dT is given as
\( \displaystyle d_T = \sqrt{2Rh_T} \)
, where R is the radius of the earth (approximately 6400 km).”