If the sum of the heights of transmitting and receiving antennas in the line of sight of communication is fixed at 160 m, then the maximum range of LOS communication is _________ km (Take radius of Earth = 6400 km)

For line-of-sight (LOS) communication over the curved surface of the Earth, the farthest distance to the visible horizon from the top of an antenna of height $$h$$ is obtained from the well-known geometry relation

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

where

$$R = 6400 \text{ km}$$ (radius of the Earth) and $$h$$ is measured in kilometres, while $$d$$ comes out in kilometres.

When two antennas of heights $$h_1$$ and $$h_2$$ communicate, the total maximum LOS range is the sum of their individual horizon distances:

$$D \;=\; \sqrt{2 R h_1}\;+\;\sqrt{2 R h_2}$$

According to the question we have a fixed total height

$$h_1 + h_2 = 160 \text{ m}$$

To keep units consistent with the radius in kilometres we convert metres to kilometres:

$$160 \text{ m} = 0.160 \text{ km}$$

Thus

$$h_1 + h_2 = 0.160 \text{ km} \quad\text{(1)}$$

Our objective is to maximise

$$D(h_1,h_2) \;=\; \sqrt{2R\,h_1}\;+\;\sqrt{2R\,h_2}$$

Substituting $$R = 6400 \text{ km}$$ first:

$$D(h_1,h_2) = \sqrt{2 \times 6400 \times h_1}\;+\;\sqrt{2 \times 6400 \times h_2}$$

Simplifying the constant factor inside each square root, we factor out the common number:

$$2 \times 6400 = 12800$$ so

$$D(h_1,h_2) = \sqrt{12800\,h_1}\;+\;\sqrt{12800\,h_2}$$

Using the property $$\sqrt{ab} = \sqrt{a}\,\sqrt{b}$$ we write

$$D(h_1,h_2) = \sqrt{12800}\,\left(\sqrt{h_1} + \sqrt{h_2}\right)$$

The factor $$\sqrt{12800}$$ is a positive constant, hence maximising $$D$$ is equivalent to maximising

$$f(h_1,h_2) = \sqrt{h_1} + \sqrt{h_2}$$

subject to the constraint (1) $$h_1 + h_2 = 0.160.$$ Because the square-root function is concave, the sum $$\sqrt{h_1} + \sqrt{h_2}$$ is largest when $$h_1$$ and $$h_2$$ are equal (this can be shown either by calculus with a Lagrange multiplier or by the inequality of arithmetic and geometric means).

Hence, for maximum range, we set

$$h_1 = h_2 = \frac{0.160}{2} = 0.080 \text{ km} = 80 \text{ m}$$

Now we compute the horizon distance for one antenna of 80 m height:

$$d_1 = \sqrt{2 R h_1} = \sqrt{2 \times 6400 \times 0.080}$$

Multiplying inside the root:

$$2 \times 6400 = 12800$$

$$12800 \times 0.080 = 1024$$

Thus

$$d_1 = \sqrt{1024} \text{ km}$$

Since $$1024 = 32^2$$, we get

$$d_1 = 32 \text{ km}$$

Because both antennas have the same height, the second antenna gives the same distance $$d_2 = 32 \text{ km}$$. Hence the total maximum LOS range is

$$D_{\max} = d_1 + d_2 = 32 \text{ km} + 32 \text{ km} = 64 \text{ km}$$

Hence, the correct answer is Option 64.