Question
A T.V tower has a height of $80 \mathrm{~m}$. The maximum distance up to which T.V transmission can be received is equal to (radius of earth $=6.4 \times 10^{6} \mathrm{~m}$ )(A) $16 \mathrm{~km}$(B) $32 \mathrm{~km}$(C) $80 \mathrm{~km}$(D) $160 \mathrm{~km}$
Step 1
Step 1: The maximum distance up to which TV transmission can be received is given by the formula $d = \sqrt{2hR}$, where $h$ is the height of the tower and $R$ is the radius of the earth. Show more…
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Round 2
A transmitting antenna at the top of a tower has a height of $50 \mathrm{~m}$ and the height of the receiving antenna is $32 \mathrm{~m}$. The maximum distance between them for satisfactory communication in line of sight mode is Given radius of earth $\mathrm{R}=6400 \mathrm{~km}$ (A) $25.29 \times 10^{3} \mathrm{~km}$ (B) $20.23 \times 10^{3} \mathrm{~km}$ (C) $45.5 \mathrm{~km}$ (D) None of these.
The maximum distance a up to which T.V transmission from a T.V. tower of height $h$ can be received is Proportional to (A) $\mathrm{h}^{(1 / 2)}$ (B) $\mathrm{h}$ (C) $\mathrm{h}^{(3 / 2)}$ (D) $\mathrm{h}^{2}$
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