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(B) The range of a $TV$ transmission tower is given by $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth and $h$ is the height of the antenna.Case

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$50$

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(B) The range of a $TV$ transmission tower is given by $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth and $h$ is the height of the antenna.Case $(i)$: When the receiving antenna is at ground level,the range is $d_1 = \sqrt{2Rh}$,where $h = 20\, m$.Case $(ii)$: When the receiving antenna is at a height $h' = 5\, m$,the total range is $d_2 = \sqrt{2Rh} + \sqrt{2Rh'}$.The increase in range is $\Delta d = d_2 - d_1 = \sqrt{2Rh'}$.The percentage increase $n\%$ is given by $\frac{\Delta d}{d_1} 	imes 100\%$.$n = \frac{\sqrt{2Rh'}}{\sqrt{2Rh}} 	imes 100 = \sqrt{\frac{h'}{h}} 	imes 100$.Substituting the values $h = 20\, m$ and $h' = 5\, m$:$n = \sqrt{\frac{5}{20}} 	imes 100 = \sqrt{\frac{1}{4}} 	imes 100 = 0.5 	imes 100 = 50\%$.Therefore,the value of $n$ is $50$.

$A$ $TV$ transmission tower antenna is at a height of $20\, m$. Suppose that the receiving antenna is at:
$(i)$ ground level
$(ii)$ a height of $5\, m$.
The increase in antenna range in case $(ii)$ relative to case $(i)$ is $n\%$.
The value of $n$,to the nearest integer,is .....$\%$

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$A$ $TV$ transmission tower antenna is at a height of $20\, m$. Suppose that the receiving antenna is at:$(i)$ ground level$(ii)$ a height of $5\, m$.The increase in antenna range in case $(ii)$ relative to case $(i)$ is $n\%$.The value of $n$,to the nearest integer,is .....$\%$