To determine the new height required for the TV transmission tower to triple its coverage range, we need to understand how the coverage range of a tower is related to its height.
The coverage range (\(R\)) of a transmission tower is given by the following formula:
\(R = \sqrt{2 \times h \times R_E}\)
Where:
To triple the coverage range, the new range \(R_{\text{new}}\) should be:
\(R_{\text{new}} = 3 \times R\)
This implies:
\(\sqrt{2 \times h_{\text{new}} \times R_E} = 3 \times \sqrt{2 \times h \times R_E}\)
By squaring both sides, we get:
\(2 \times h_{\text{new}} \times R_E = 9 \times (2 \times h \times R_E)\)
This simplifies to:
\(h_{\text{new}} = 9 \times h\)
Given the original height \(h = 100 \, \text{m}\), the new height \(h_{\text{new}}\\) should be:
\(h_{\text{new}} = 9 \times 100 \, \text{m} = 900 \, \text{m}\)
Therefore, the height of the tower should be increased to 900 m to triple its coverage range.
Thus, the correct answer is: 900 m.
Match List-I with List-II:
| List-I (Modulation Schemes) | List-II (Wave Expressions) |
|---|---|
| (A) Amplitude Modulation | (I) \( x(t) = A\cos(\omega_c t + k m(t)) \) |
| (B) Phase Modulation | (II) \( x(t) = A\cos(\omega_c t + k \int m(t)dt) \) |
| (C) Frequency Modulation | (III) \( x(t) = A + m(t)\cos(\omega_c t) \) |
| (D) DSB-SC Modulation | (IV) \( x(t) = m(t)\cos(\omega_c t) \) |
Choose the correct answer: