Question:
A TV transmitting antenna is 70 m tall. If the receiving antenna is at ground level, then the area it can cover is: (Radius of Earth = 6400 km)
A TV transmitting antenna is 70 m tall. If the receiving antenna is at ground level, then the area it can cover is: (Radius of Earth = 6400 km)
Show Hint
For horizon problems, use \(d = \sqrt{2Rh}\) and then compute area as \(\pi d^2\).
Updated On: Jun 19, 2026
- \(2562\, km^2\)
- \(3256\, km^2\)
- \(710\pi\, km^2\)
- \(2816\, km^2\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Convert height into km.
Height of antenna: \[ h = 70\,m = 0.07\,km \]
Step 2: Use line-of-sight distance formula.
For small height compared to Earth radius: \[ d = \sqrt{2Rh} \] where \(R = 6400\,km\).
Step 3: Substitute values.
\[ d = \sqrt{2 \times 6400 \times 0.07} \]
Step 4: Simplify inside root.
\[ d = \sqrt{896} \approx 29.93\,km \]
Step 5: Calculate covered area.
Coverage area is circular: \[ A = \pi d^2 = \pi \times 896 \]
Step 6: Final simplification.
\[ A = 896\pi \approx 2816\,km^2 \]
Final Answer: \[ \boxed{2816\,km^2} \]
Height of antenna: \[ h = 70\,m = 0.07\,km \]
Step 2: Use line-of-sight distance formula.
For small height compared to Earth radius: \[ d = \sqrt{2Rh} \] where \(R = 6400\,km\).
Step 3: Substitute values.
\[ d = \sqrt{2 \times 6400 \times 0.07} \]
Step 4: Simplify inside root.
\[ d = \sqrt{896} \approx 29.93\,km \]
Step 5: Calculate covered area.
Coverage area is circular: \[ A = \pi d^2 = \pi \times 896 \]
Step 6: Final simplification.
\[ A = 896\pi \approx 2816\,km^2 \]
Final Answer: \[ \boxed{2816\,km^2} \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET communication systems Questions
- The process of retrieval of information from the carrier wave
- The maximum number of TV signals that can be transmitted along a co-axial cable is:
- An oscillator is an amplifier with
- A message signal of frequency \(20~\text{kHz}\) and peak voltage \(20~\text{V}\) is used to modulate a carrier of peak voltage \(100~\text{V}\). The modulation index is
- The frequency suitable beyond the horizon communication using sky wave is:
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions