KVPY SA Exam 2018 Question Paper

Leh Height of cone = h
Radius of base = r
And slant height $=\mathit{l};=\mathit{l}=\sqrt{{\mathit{r}}^2+{\mathit{h}}^2}$
Given volume = surface area.
⇒ $\frac{1}{3}\mathit{\pi }{\mathit{r}}^2\mathit{h}=\mathit{\pi }\mathit{r}\mathit{l}+\mathit{\pi }{\mathit{r}}^2$
⇒ $\mathit{r}\mathit{h}=3\mathit{l}+3\mathit{r}\Rightarrow \mathit{l}=\frac{1}{3}\left(\mathit{r}\mathit{h}-3\mathit{r}\right)$
⇒ $\sqrt{{\mathit{r}}^2+{\mathit{h}}^2}=\frac{\mathit{r}}{3}\left(\mathit{h}-3\right)$
⇒ ${\mathit{r}}^2+{\mathit{h}}^2=\frac{{\mathit{r}}^2}{9}\left({\mathit{h}}^2-6\mathit{h}+9\right)$
⇒ ${\mathit{h}}^2=\frac{{\mathit{r}}^2\mathrm{}{\mathit{h}}^2}{9}-\frac{2\mathit{h}{\mathit{r}}^2}{3}$
⇒ $\mathit{h}=\frac{6{\mathit{r}}^2}{{\mathit{r}}^2-9}=6+\frac{54}{{\mathit{r}}^2-9}$
Since h and r must be integers, and ${\mathit{r}}^2-\mathit{g}$ must be a factor of 54
${\mathit{r}}^2-9$ must be divisible by 3
⇒ r = 3 k.
$\mathit{h}=6+\frac{54}{9{\mathit{k}}^2-9}=6+\frac{6}{{\mathit{k}}^2-1}$
Since
⇒ k = 2 is only such value; for which h is integer.
So, $\mathit{r}=2\times 3=6\mathit{i}$
$\mathit{h}=6+\frac{6}{3}=8$
$\mathit{l}=10$
is the only possible values for r and h.