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Chapter 20: Problem 2683

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) \(\mathrm{h}^{2}\)

Short Answer

Expert verified
The maximum distance up to which TV transmission from a TV tower of height \(h\) can be received is proportional to \(h^{(1/2)}\).

Step by step solution

01

Understand the problem situation

A transmitter sends signals from the top of a tower of height h. Signals can be received in a direct line of sight before being obstructed by the Earth's curve. Let d be the maximum distance of reception, we'll find the proportionality between d and h.
02

Set up the formula for line of sight distance

The formula for the distance between the line of sight of two objects with given heights is: \(d = \sqrt{2Rh_1} +\sqrt{2Rh_2}\) In our case, the receiver is on the ground (height \(h_2\) = 0) and the transmitter is on the tower (height \(h_1\) = h), thus the formula becomes \(d = \sqrt{2Rh}\), where R is the radius of Earth.
03

Find the proportionality between d and h

Now we have the formula \(d = \sqrt{2Rh}\). To find the proportionality between d and h, we can rearrange the formula: \(d = \sqrt{2R} * \sqrt{h}\) We can see that d is proportional to \(\sqrt{h}\)
04

Choose the correct answer

The correct answer is the one that represents the proportionality between d and h: \(d \propto h^{(1/2)}\) Therefore, the correct answer is option (A): \(h^{(1/2)}\)

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