UPSC CDS 2 2022 Math Paper

Concept:The sphere is inscribed in a cone, and the similarity of triangles is used to relate the dimensions of the cone and the sphere.Explanation:Given the cone: radius $r_{\text{cone}} = 12$ cm, height $h = 16$ cm.Let the radius of the sphere be $r$.First, find the slant height ($l$) of the cone:$l = \sqrt{r_{\text{cone}}^2 + h^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20$ cm.In the vertical cross-section, consider triangles $\text{triangle OBA}$ (from the cone's vertex to the base edge) and $\text{triangle OCD}$ (from the sphere's center to the point of tangency on the slant side).These triangles are similar because the line from the cone's vertex to the sphere's center is parallel to the cone's axis.From similarity:$\frac{AB}{CD} = \frac{OA}{OD}$Here $AB = r_{\text{cone}} = 12$, $CD = r$, $OA = l = 20$, and $OD = \text{height} - r = 16 - r$.So $\frac{12}{r} = \frac{20}{16 - r}$.Cross-multiply:$12(16 - r) = 20r$$192 - 12r = 20r$$192 = 32r$$r = \frac{192}{32} = 6$ cm.Answer:$6$ cm.