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(C) The transmission range $d$ of a $TV$ tower of height $h$ is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth.Let the ini

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$10$

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(C) The transmission range $d$ of a $TV$ tower of height $h$ is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth.Let the initial height be $h_1$ and the initial range be $d_1 = \sqrt{2Rh_1}$.The new height $h_2$ is increased by $21 \%$,so $h_2 = h_1 + 0.21h_1 = 1.21h_1$.The new range $d_2$ is given by $d_2 = \sqrt{2Rh_2} = \sqrt{2R(1.21h_1)} = \sqrt{1.21} 	imes \sqrt{2Rh_1} = 1.1 	imes d_1$.The percentage increase in the range is given by $\frac{d_2 - d_1}{d_1} 	imes 100 \%$.Substituting the values: $\frac{1.1d_1 - d_1}{d_1} 	imes 100 \% = (1.1 - 1) 	imes 100 \% = 0.1 	imes 100 \% = 10 \%$.

By what percentage will the transmission range of a $TV$ tower be affected when the height of the tower is increased by $21 \%$ (in $\%$)?

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