UPSC CDS 2 2021 Math Paper

Concept:Total surface area of a solid formed by a cone placed on a hemisphere is the sum of the curved surface area of the cone and the curved surface area of the hemisphere.Explanation:Given: radius of cone and hemisphere = $x$; height of cone $= 2x$.Slant height of cone $l = \sqrt{r^2 + h^2} = \sqrt{x^2 + (2x)^2} = \sqrt{5x^2} = x\sqrt{5}$.Curved surface area of cone $= \pi r l = \pi \cdot x \cdot x\sqrt{5} = \pi x^2 \sqrt{5}$.Curved surface area of hemisphere $= 2\pi r^2 = 2\pi x^2$.Total surface area of solid $= \pi x^2 \sqrt{5} + 2\pi x^2 = \pi x^2 (\sqrt{5} + 2)$.Using $\pi \approx \frac{22}{7}$ and $\sqrt{5} \approx 2.236$:$(\sqrt{5} + 2) \approx 4.236$; then $\frac{22}{7} \times 4.236 \times x^2 \approx 13.31\, x^2$.Thus the approximate total surface area is $13.3\, x^2$.Answer:$13.3\, x^{2}$ (Option C)