A radar has a power of 1 kW and is operating | vedclass

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(B) Let $d$ be the maximum distance up to which the radar can detect an object on the Earth's surface. From the geometry of the right-angled triangle

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$80$

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(B) Let $d$ be the maximum distance up to which the radar can detect an object on the Earth's surface. From the geometry of the right-angled triangle $\Delta OAC$ (where $O$ is the center of the Earth,$A$ is the object on the surface,and $C$ is the radar on the mountain top):$OC^2 = AC^2 + OA^2$$(h + R)^2 = d^2 + R^2$$d^2 = (h + R)^2 - R^2$$d^2 = h^2 + 2hR + R^2 - R^2$$d = \sqrt{h^2 + 2hR}$Since $h = 500 \ m = 0.5 \ km$ and $R = 6400 \ km$,$h^2$ is negligible compared to $2hR$.$d \approx \sqrt{2hR} = \sqrt{2 	imes 0.5 	imes 6400} = \sqrt{6400} = 80 \ km$.

$A$ radar has a power of $1 \ kW$ and is operating at a frequency of $10 \ GHz$. It is located on a mountain top of height $500 \ m$. The maximum distance up to which it can detect an object located on the surface of the earth (Radius of earth $= 6.4 \times 10^6 \ m$) is ....... $km$.

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