JEE Main · 2012 · 29 Apr 2012 · Offline · Semiconductors

From a height $h$ the farthest point on the ground the radar can see lies along the line tangent to the Earth, where $(R+h{)}^2=R^2+d^2$. Expanding and dropping the tiny $h^2$ term leaves $d^2=2Rh$, so $d=\sqrt{2Rh}=\sqrt{2\times 6.4\times 10^6\times 500}=\sqrt{6.4\times 10^9}=8\times 10^4$ m, that is $80$ km. The quoted power and frequency play no part in this line-of-sight geometry.