QJEEM23Ph1a7b: JEE Main 2023 Physics - Class 12 JEEM23Ph1a7b Solution: A transmitting antenna is kept on the surface of the earth. The …

To determine the minimum height of the receiving antenna required to receive the signal in line of sight at a distance of 4 km from the transmitting antenna, we need to consider the curvature of the Earth. Let's denote the height of the receiving antenna as $h$ and the distance between the antennas as $d = 4 \mathrm{~km}$ .

We can use the Pythagorean theorem to relate the Earth's radius $R$ , the height of the receiving antenna $h$ , and the distance between the antennas $d$ :

$(R + h)^2 = R^2 + d^2$

Now, we can plug in the given values for $R$ and $d$ :

$(6400 + h)^2 = 6400^2 + 4^2$

Expanding the equation and subtracting $6400^2$ from both sides gives:

$2 \times 6400 \times h + h^2 = 16$

Since $h$ is much smaller than the Earth's radius, we can ignore the $h^2$ term:

$2 \times 6400 \times h \approx 16$

Now, we can solve for $h$ :

$h \approx \frac{16}{2 \times 6400} = \frac{1}{800} \mathrm{~km}$

Converting to meters and multiplying by $10^2$ to match the desired format:

$x = h \times 10^3 \times 10^2 = \frac{1}{800} \times 10^5 = 125$

So, the value of $x$ is 125