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A TV tower is $120 \mathrm{~m}$ high. How much more height is to be added to it, if its coverage range is to become double?
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Verified Answer
The correct answer is:
$360 \mathrm{~m}$
Initial height of TV tower, $h_{1}=120 \mathrm{~m}$ Coverage range of TV tower is given as
$$
d=\sqrt{2 R h}
$$
where, $R$ is radius of the earth.
$$
\begin{aligned}
&\Rightarrow \quad \frac{d_{1}}{d_{2}}=\sqrt{\frac{h_{1}}{h_{2}}} \\
&\Rightarrow \quad \frac{d_{1}}{2 d_{1}}=\sqrt{\frac{120}{h_{2}}} \\
&\Rightarrow \quad \quad \frac{1}{4}=\frac{120}{h_{2}} \\
&\Rightarrow \quad h_{2}=480 \mathrm{~m} \\
&\left.\therefore \text { Heighi to be added }=h_{2}-h_{1}=2 d_{1}\right] \\
&\quad=480-120=360 \mathrm{~m}
\end{aligned}
$$
$$
d=\sqrt{2 R h}
$$
where, $R$ is radius of the earth.
$$
\begin{aligned}
&\Rightarrow \quad \frac{d_{1}}{d_{2}}=\sqrt{\frac{h_{1}}{h_{2}}} \\
&\Rightarrow \quad \frac{d_{1}}{2 d_{1}}=\sqrt{\frac{120}{h_{2}}} \\
&\Rightarrow \quad \quad \frac{1}{4}=\frac{120}{h_{2}} \\
&\Rightarrow \quad h_{2}=480 \mathrm{~m} \\
&\left.\therefore \text { Heighi to be added }=h_{2}-h_{1}=2 d_{1}\right] \\
&\quad=480-120=360 \mathrm{~m}
\end{aligned}
$$
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