UPSC CDS 1 2020 Math Paper

Concept:The centers of the four balls form a regular tetrahedron of edge $2r$. The distance from the top ball's centre to the plane is the tetrahedron's height plus the radius of a bottom ball.Explanation:Step 1: The three bottom balls lie on a plane and touch each other. Their centres form an equilateral triangle with side $2r$.Step 2: The fourth ball touches all three bottom balls, so its centre is at distance $2r$ from each bottom centre. Hence the four centres form a regular tetrahedron of edge $2r$.Step 3: Height of a regular tetrahedron with edge $a$ is $\sqrt{\frac{2}{3}}a$. Here $a=2r$, so height from the top centre to the base plane (plane of the bottom centres) is $\sqrt{\frac{2}{3}} \times 2r = \frac{2\sqrt{2}r}{\sqrt{3}}$.Step 4: Bottom centres are at height $r$ above the given plane (each bottom ball touches the plane). Therefore, distance from the top centre to the given plane is $r$ plus the tetrahedron height.Step 5: Required distance = $\frac{2\sqrt{2}r}{\sqrt{3}} + r = \frac{2\sqrt{2} + \sqrt{3}}{\sqrt{3}} r = \frac{\sqrt{3} + 2\sqrt{2}}{\sqrt{3}} r$.Answer:Option D: $\frac{\sqrt{3} + 2\sqrt{2}}{\sqrt{3}} r$ unit.