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(D) Let $d$ be the coverage range of the tower of height $h$.The range $d$ is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the Ear

Vedclass · Accepted Answer

$900$

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(D) Let $d$ be the coverage range of the tower of height $h$.The range $d$ is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth.Thus,$d \propto \sqrt{h}$.Let the initial height be $h_1 = 100\,m$ and the initial range be $d_1$.Let the new height be $h_2$ and the new range be $d_2 = 3d_1$.Since $d \propto \sqrt{h}$,we have:$\frac{d_2}{d_1} = \sqrt{\frac{h_2}{h_1}}$Substituting the values:$3 = \sqrt{\frac{h_2}{100}}$Squaring both sides:$9 = \frac{h_2}{100}$$h_2 = 900\,m$.Therefore,the height of the tower should be increased to $900\,m$.

At a particular station,the $TV$ transmission tower has a height of $100\,m$. To triple its coverage range,the height of the tower should be increased to ......$m$.

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