If a line of sight radio communication, a distance of about 50 km is kept between the transmitting and receiving antennas. If the height of the receiving antenna is 70 m, then the minimum height of the transmitting antenna should be (Radius of the earth = $6.4 \times 10^{6}$ m)

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20 m

32 m

40 m

51 m

answer is B.

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Detailed Solution

$I n t h e l i n e o f s i g h t c o m m u n i c a t i o n, d i s \tan c e d b e t w e e n t r a n s m i t t i n g a n t e n n a a n d r e c e i v i n g a n t e n n a i s g i v e n b y d = \sqrt{2 R h_{T}} + \sqrt{2 R h_{R}} H e r e, h_{R} a n d h_{T} i s t h e h e i g h t o f r e c e i v i n g a n d t r a n s m i t t i n g a n t e n n a, r e s p e c t i v e l y . ∴ d = \sqrt{2 R h_{T}} + \sqrt{2 R h_{R}} \Rightarrow 50 \times 10^{3} = \sqrt{2 R} (\sqrt{h_{T}} + \sqrt{h_{R}}) \Rightarrow 50 \times 10^{3} = \sqrt{2 (6.4 \times 10^{6})} (\sqrt{h_{T}} + \sqrt{70}) \Rightarrow \sqrt{h_{T}} = \frac{50 \times 10^{3}}{3577.7} - 8.37 = 13.98 - 8.37 = 5.61 \Rightarrow h_{T} = 31.5 ≃ 32 m$