In 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)

For line of sight radio links the curving surface of the Earth restricts how far one can “see”. For a single antenna of height $$h$$ above the Earth’s surface, simple geometry of a tangent from the antenna top to the Earth gives the range (distance to the horizon)

$$d \;=\;\sqrt{2Rh},$$

where $$R$$ is the radius of the Earth and $$h \ll R$$. When two antennas are used, their individual ranges add, so for a link of length $$D$$ we must have

$$D \;=\; d_1 + d_2 \;=\; \sqrt{2Rh_r}\;+\;\sqrt{2Rh_t},$$

with $$h_r$$ the height of the receiving antenna and $$h_t$$ that of the transmitting antenna. The given data are

$$D = 50\ \text{km} = 50\,000\ \text{m},\qquad h_r = 70\ \text{m},\qquad R = 6.4\times10^{6}\ \text{m}.$$

We first find the range of the receiving antenna:

$$d_1 = \sqrt{2Rh_r} = \sqrt{2 \times 6.4\times10^{6}\times 70}.$$

Multiplying inside the root,

$$2 \times 6.4\times10^{6}\times 70 = 12.8 \times 70 \times 10^{6} = 896 \times 10^{6} = 8.96 \times 10^{8}.$$

Hence

$$d_1 = \sqrt{8.96 \times 10^{8}} = \sqrt{8.96}\times10^{4} \approx 2.993 \times 10^{4}\ \text{m} \approx 29.93\ \text{km}.$$

Now the remaining distance that must be covered by the transmitting antenna is

$$d_2 = D - d_1 = 50\ \text{km} - 29.93\ \text{km} \approx 20.07\ \text{km} = 20\,070\ \text{m}.$$

Again using the horizon formula for the transmitting antenna,

$$d_2 = \sqrt{2Rh_t}\quad\Longrightarrow\quad (\,d_2\,)^2 = 2Rh_t.$$

So

$$h_t = \frac{d_2^{\,2}}{2R} = \frac{(20\,070)^2}{2 \times 6.4\times10^{6}}.$$

Calculating the numerator,

$$(20\,070)^2 = 20\,070 \times 20\,070 = 402\,804\,900 = 4.028049 \times 10^{8}.$$

And since the denominator is

$$2R = 1.28 \times 10^{7},$$

we get

$$h_t = \frac{4.028049 \times 10^{8}}{1.28 \times 10^{7}} = 3.148 \times 10^{1}\ \text{m} \approx 31.5\ \text{m}.$$

Taking the minimum integral value that safely meets the requirement,

$$h_t \approx 32\ \text{m}.$$

Hence, the correct answer is Option D.