Q153 / 180 · COMEDK 2023 (Morning Shift) · Physics
A TV tower has a height $72 \mathrm{~m}$. The maximum distance upto which TV transmission can be received is (Take, radius of the earth as $\left.6.4 \times 10^6 \mathrm{~m}\right)$
- Correct answerA30.36 km
- B25.12 km
- C27.52 km
- D34.25 km
Solution
The maximum distance $d$ up to which the TV transmission can be received from a tower of height $h$ is given by the formula $d = \sqrt{2Rh}$, where $R$ is the radius of the Earth.
Given values are $h = 72 \text{ m}$ and $R = 6.4 \times 10^6 \text{ m}$.
Substituting these values into the formula:
$d = \sqrt{2 \times (6.4 \times 10^6 \text{ m}) \times 72 \text{ m}}$
$d = \sqrt{12.8 \times 10^6 \times 72}$
$d = \sqrt{921.6 \times 10^6}$
$d = \sqrt{921.6} \times 10^3 \text{ m}$
$d = 30.3578 \times 10^3 \text{ m}$
$d \approx 30.36 \text{ km}$.
Answer: 30.36 km
Given values are $h = 72 \text{ m}$ and $R = 6.4 \times 10^6 \text{ m}$.
Substituting these values into the formula:
$d = \sqrt{2 \times (6.4 \times 10^6 \text{ m}) \times 72 \text{ m}}$
$d = \sqrt{12.8 \times 10^6 \times 72}$
$d = \sqrt{921.6 \times 10^6}$
$d = \sqrt{921.6} \times 10^3 \text{ m}$
$d = 30.3578 \times 10^3 \text{ m}$
$d \approx 30.36 \text{ km}$.
Answer: 30.36 km
Hence, the correct answer is option: 'A'.