UPSC CDS 2 2022 Math Paper

Concept:Volume of water left equals volume of the cone minus volume of the submerged sphere.Explanation:The cone has radius $12$ cm and height $16$ cm.Its slant height $AO = \sqrt{12^2 + 16^2} = \sqrt{144+256} = \sqrt{400} = 20$ cm.Triangles $\triangle OBA$ and $\triangle OCD$ are similar.So $\frac{AB}{CD} = \frac{OA}{OD}$.Substitute $AB = 12$, $CD = r$ (sphere radius), $OA = 20$, $OD = OB - BD = 16 - r$.Thus $\frac{12}{r} = \frac{20}{16 - r}$.Cross-multiply: $12(16 - r) = 20r$ → $192 - 12r = 20r$ → $192 = 32r$ → $r = 6$ cm.Volume of cone = $\frac{1}{3}\pi (12)^2 (16) = \frac{1}{3}\pi \times 144 \times 16 = 768\pi$ cm³.Volume of sphere = $\frac{4}{3}\pi (6)^3 = \frac{4}{3}\pi \times 216 = 288\pi$ cm³.Remaining water = $768\pi - 288\pi = 480\pi$ cm³.Since $1$ cm³ = $1$ mL, water left = $480\pi$ mL.Answer:$480\pi$ mL (Option C).