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

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$(\sqrt{3/2}-1) d$

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(B) The range $d$ of a $TV$ transmission tower of height $h$ is given by the formula $d = \sqrt{2hR}$,where $R$ is the radius of the Earth.Initially,$d = \sqrt{2hR}$.When the height is increased to $h' = 3/2 h$,the new range $d'$ becomes:$d' = \sqrt{2h'R} = \sqrt{2(3/2 h)R} = \sqrt{3/2} \sqrt{2hR} = \sqrt{3/2} d$.The change in range is $\Delta d = d' - d$.Substituting the values,we get $\Delta d = \sqrt{3/2} d - d = (\sqrt{3/2} - 1) d$.

$A$ $TV$ transmission tower of height $h$ covers a range of distance $d$. By how much will the range change if the height is increased to $3/2 h$?

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