The transmission range (\( d \)) of a TV tower is given by:
\( d = \sqrt{2Rh} \)
where \( R \) is the radius of the Earth and \( h \) is the height of the tower.
If the height is increased by 21%, the new height (\( h' \)) is:
\( h' = h + 0.21h = 1.21h \)
The new transmission range (\( d' \)) is:
\( d' = \sqrt{2Rh'} = \sqrt{2R(1.21h)} = \sqrt{1.21} \sqrt{2Rh} = 1.1\sqrt{2Rh} \)
Since \( d = \sqrt{2Rh} \), the new range is:
\( d' = 1.1d \)
The percentage increase in the transmission range is:
\( \frac{d' - d}{d} \times 100 = \frac{1.1d - d}{d} \times 100 = 0.1 \times 100 = 10\% \)
The transmission range increases by 10% (Option 4).
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:


Let \( a \in \mathbb{R} \) and \( A \) be a matrix of order \( 3 \times 3 \) such that \( \det(A) = -4 \) and \[ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} \] where \( I \) is the identity matrix of order \( 3 \times 3 \).
If \( \det\left( (a + 1) \cdot \text{adj}\left( (a - 1) A \right) \right) \) is \( 2^m 3^n \), \( m, n \in \{ 0, 1, 2, \dots, 20 \} \), then \( m + n \) is equal to: