Question

At a particular station, the TV transmission tower has a height of 100 m. To triple its coverage range, height of the tower should be increased to

• A. 200 m
• B. 300 m
• C. 600 m
• D. 900 m

Answer 1

The Correct Option is D

To solve this question, we need to understand the relationship between the height of a TV transmission tower and its coverage range. The range of coverage of a transmission tower is directly proportional to the square root of its height. Mathematically, this can be expressed as:

\(R \propto \sqrt{h}\) 

Where \(R\) is the range and \(h\) is the height of the tower.

Given that initially, the tower has a height of 100 m, and we want to triple its coverage range, the calculation can be done as follows:

1. \(R_1 = k\sqrt{100} = 10k\) (for initial height where \(k\) is a proportionality constant)
2. We want the new range \(R_2\) to be triple the initial range, so \(R_2 = 3R_1 = 3 \times 10k = 30k\)
3. For the new range, the formula becomes \(R_2 = k\sqrt{h_2}\)
4. Equating, \(k\sqrt{h_2} = 30k\)
5. Divide both sides by \(k\) (assuming \(k\neq0\)), \(\sqrt{h_2} = 30\)
6. Square both sides to solve for \(h_2\), we get \(h_2 = 30^2 = 900\)

Thus, to triple the coverage range, the height of the tower should be increased to 900 m.

Correct Answer: 900 m

The options provided were 200 m, 300 m, 600 m, and 900 m, and based on our calculations, the correct option is indeed 900 m.

 

Tip: Remember that the relationship between tower height and coverage range follows the formula \(R \propto \sqrt{h}\). This can help quickly solve similar problems.

Answer 2

The correct answer is (D) : 900 m
\(∴r_m=\sqrt{2Rh}\)
\(⇒\frac{r_1}{r_2}=\sqrt{\frac{h_1}{h_2}}\)
\(⇒\frac{1}{3}=\sqrt{\frac{100}{h_2}}\)
\(⇒h_2=900\ m\)
\(\therefore\) height of the tower should be increased to 900 m