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).
Let $ f(x) = \begin{cases} (1+ax)^{1/x} & , x<0 \\1+b & , x = 0 \\\frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & , x>0 \end{cases} $ be continuous at x = 0. Then $ e^a bc $ is equal to
Total number of nucleophiles from the following is: \(\text{NH}_3, PhSH, (H_3C_2S)_2, H_2C = CH_2, OH−, H_3O+, (CH_3)_2CO, NCH_3\)