Suppose that the earth is a sphere of radius 6400 kilometers. The height from the earth's surface from where exactly a fourth of the earth's surface is visible, is

• A

  $3200 \mathrm{~km}$
• B

  $3200 \sqrt{2} \mathrm{~km}$
• C

  $3200 \sqrt{3} \mathrm{~km}$
• Correct answer
  D

  $6400 \mathrm{~km}$

Solution

$$ \mathrm{AP}=\mathrm{R} \sin \theta $$ $$ \text { area of ring }=(2 \pi \operatorname{Rsin} \theta) \cdot \operatorname{Rd} \theta $$ Total area required, $\pi \mathrm{R}^{2}=\int_{0}^{\theta} 2 \pi R^{2} \sin \theta d \theta \mid$ $$ \begin{array}{l} \frac{1}{2}=1-\cos \theta \\ \cos \theta=\frac{1}{2} \end{array} $$ Now,
$$ \begin{array}{l} \cos \theta=\frac{R}{R+H} \\ \frac{1}{2}=\frac{R}{R+H}=R=H=6400 \end{array} $$

Hence, the correct answer is option: 'D'.