A transmitting antenna at the top of the tower has a height 18 m and the height of the receiving antenna is 32m. The maximum distance between them for satisfactory communciation in line of sight mode is (Radius of earth`= 6.4xx10^(6)m`)

Options

A. 15.15km
B. 21.25km
C. 30.45km
D. 35.42km.

Explanation

To find the maximum distance for satisfactory communication in line of sight mode between a transmitting antenna and a receiving antenna, we can use the formula for the maximum line of sight distance \( D_m \):

\[ D_m = \sqrt{2h_T R} + \sqrt{2h_R R} \]

where:
- \( h_T \) is the height of the transmitting antenna,
- \( h_R \) is the height of the receiving antenna, and
- \( R \) is the radius of the Earth.

### Step 1: Identify the given values
- Height of the transmitting antenna, \( h_T = 18 \, \text{m} \)
- Height of the receiving antenna, \( h_R = 32 \, \text{m} \)
- Radius of the Earth, \( R = 6.4 \times 10^6 \, \text{m} \)

### Step 2: Calculate the first term \( \sqrt{2h_T R} \)
\[ \sqrt{2h_T R} = \sqrt{2 \times 18 \, \text{m} \times 6.4 \times 10^6 \, \text{m}} \]
Calculating the values:
\[ = \sqrt{36 \times 6.4 \times 10^6} \]
\[ = \sqrt{230.4 \times 10^6} = \sqrt{2.304 \times 10^8} = 15.2 \times 10^4 \, \text{m} \]

### Step 3: Calculate the second term \( \sqrt{2h_R R} \)
\[ \sqrt{2h_R R} = \sqrt{2 \times 32 \, \text{m} \times 6.4 \times 10^6 \, \text{m}} \]
Calculating the values:
\[ = \sqrt{64 \times 6.4 \times 10^6} \]
\[ = \sqrt{409.6 \times 10^6} = \sqrt{4.096 \times 10^8} = 20.2 \times 10^4 \, \text{m} \]

### Step 4: Add the two terms to find \( D_m \)
\[ D_m = \sqrt{2h_T R} + \sqrt{2h_R R} \]
\[ D_m = (15.2 \times 10^4) + (20.2 \times 10^4) = 35.4 \times 10^4 \, \text{m} \]

### Step 5: Convert to kilometers
\[ D_m = 35.4 \times 10^4 \, \text{m} = 354000 \, \text{m} = 354 \, \text{km} \]

### Final Answer
The maximum distance between the transmitting and receiving antennas for satisfactory communication in line of sight mode is approximately **354 km**.

---