A transmitting antenna has a height of 320 m and that of receiving antenna is 2000 m. The maximum distance between them for satisfactory communication in line of sight mode is $$d$$. The value of $$d$$ is _________ km.

For line-of-sight (LOS) communication over the curved surface of the Earth, the maximum distance to the horizon from an antenna of height $$h$$ is obtained from the geometry of a tangent drawn to a sphere of radius $$R$$. The relation is stated as

$$d_{\text{horizon}} = \sqrt{2Rh}$$

where $$R$$ is the radius of the Earth and $$h$$ is in the same length unit as $$R$$. If two antennas are involved, the total LOS range $$d$$ is the sum of their individual horizon distances, so

$$d = \sqrt{2Rh_1} + \sqrt{2Rh_2}$$

We have

$$R = 6.4 \times 10^{6}\ \text{m}$$

Height of the transmitting antenna: $$h_1 = 320\ \text{m}$$

Height of the receiving antenna: $$h_2 = 2000\ \text{m}$$

First, we find the horizon distance for the transmitting antenna:

$$\sqrt{2Rh_1}=\sqrt{2 \times 6.4 \times 10^{6}\ \text{m} \times 320\ \text{m}}$$

$$= \sqrt{(2 \times 6.4) \times 320 \times 10^{6}}$$

$$= \sqrt{12.8 \times 320 \times 10^{6}}$$

$$= \sqrt{4096 \times 10^{6}}$$

$$= \sqrt{4.096 \times 10^{9}}$$

$$= \sqrt{4.096}\times\sqrt{10^{9}}$$

$$= 2.0249 \times 10^{4.5}$$

$$= 2.0249 \times 31\,622.776$$

$$\approx 64\,000\ \text{m}$$

$$= 64\ \text{km}$$

Next, we calculate the horizon distance for the receiving antenna:

$$\sqrt{2Rh_2}=\sqrt{2 \times 6.4 \times 10^{6}\ \text{m} \times 2000\ \text{m}}$$

$$= \sqrt{(2 \times 6.4) \times 2000 \times 10^{6}}$$

$$= \sqrt{12.8 \times 2000 \times 10^{6}}$$

$$= \sqrt{25\,600 \times 10^{6}}$$

$$= \sqrt{2.56 \times 10^{10}}$$

$$= \sqrt{2.56}\times\sqrt{10^{10}}$$

$$= 1.6 \times 10^{5}$$

$$= 160\,000\ \text{m}$$

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

Adding the two horizon distances gives the maximum LOS range:

$$d = 64\ \text{km} + 160\ \text{km}$$

$$d = 224\ \text{km}$$

Hence, the correct answer is Option C.