i) Relation between r and h if the volume of cone is maximum is given by
a)4r=3h b)3r=4h c)r=h/3 d)2r/3
ii) Ratio of volume of cone to volume of sphere, when volume of cone is maximum is
a)4/27 b)8/27 c)1/3 d)4/9

Dear Student,
Here I have considered radius of cone as 'r' and radius of sphere as 'R' .Please consider these variables at required places
Please find below the solution to the asked query:


$$\begin{gathered} \mathrm{Let} \mathrm{radius} \mathrm{and} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{inscribed} \mathrm{right} \mathrm{circular} \mathrm{cone} \mathrm{be} \mathrm{r} \mathrm{and} \mathrm{h} \mathrm{respectively}. \\ \mathrm{In} ∆\mathrm{ABC}, \mathrm{by} \mathrm{Pythagoras} \mathrm{theorem}, \\ {\left(\mathrm{h}-\mathrm{R}\right)}^2+{\mathrm{r}}^2={\mathrm{R}}^2 \\ \Rightarrow {\mathrm{h}}^2+{\mathrm{R}}^2-2\mathrm{hR}+{\mathrm{r}}^2={\mathrm{R}}^2 \\ \Rightarrow {\mathrm{r}}^2=2\mathrm{hR}-{\mathrm{h}}^2 \\ \mathrm{Let} \mathrm{curved} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{cone} \mathrm{be} \mathrm{S}. \\ \Rightarrow \mathrm{S}=\mathrm{\pi rl} \\ \Rightarrow \mathrm{S}=\mathrm{\pi r}\sqrt{{\mathrm{h}}^2+{\mathrm{r}}^2} \\ \Rightarrow {\mathrm{S}}^2={\mathrm{\pi }}^2{\mathrm{r}}^2\left({\mathrm{h}}^2+{\mathrm{r}}^2\right)={\mathrm{\pi }}^2\left(2\mathrm{hR}-{\mathrm{h}}^2\right)\left({\mathrm{h}}^2+2\mathrm{hR}-{\mathrm{h}}^2\right) \\ \Rightarrow {\mathrm{S}}^2={\mathrm{\pi }}^2\left(2\mathrm{hR}-{\mathrm{h}}^2\right)\left(2\mathrm{hR}\right) \\ \Rightarrow {\mathrm{S}}^2={\mathrm{\pi }}^2\left(4{\mathrm{h}}^2{\mathrm{R}}^2-2{\mathrm{h}}^3\mathrm{R}\right) \\ \Rightarrow \mathrm{Z} = {\mathrm{\pi }}^2\left(4{\mathrm{h}}^2{\mathrm{R}}^2-2{\mathrm{h}}^3\mathrm{R}\right) \\ \mathrm{Differentiating} \mathrm{both} \mathrm{sides} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{h}, \mathrm{we} \mathrm{get}, \\ \frac{dZ}{\mathrm{dh}}= {\mathrm{\pi }}^2\left(8{\mathrm{hR}}^2-6{\mathrm{h}}^2\mathrm{R}\right) \\ \mathrm{For} \mathrm{maxima} \mathrm{or} \mathrm{minima} \frac{dZ}{\mathrm{dh}}=0 \\ \Rightarrow {\mathrm{\pi }}^2\left(8{\mathrm{hR}}^2-6{\mathrm{h}}^2\mathrm{R}\right)=0 \\ \Rightarrow 8{\mathrm{hR}}^2-6{\mathrm{h}}^2\mathrm{R}=0 \\ \mathrm{Dividing} \mathrm{each} \mathrm{term} \mathrm{by} \mathrm{hR} \mathrm{we} \mathrm{get}, \\ 8\mathrm{R}-6\mathrm{h}=0 \\ \Rightarrow \mathrm{h}=\frac{4}{3}\mathrm{R} \\ \frac{d^{2}Z}{d{h}^2} = {\mathrm{\pi }}^2\left[16\mathrm{hR} - 6{\mathrm{h}}^2\right] \\ {\left[\frac{{\mathrm{d}}^2\mathrm{Z}}{{\mathrm{dh}}^2}\right]}_{\mathrm{h}=\frac{4\mathrm{R}}{3}} {\mathrm{\pi }}^2\left[\frac{64{\mathrm{R}}^2}{9} - \frac{96{\mathrm{R}}^2}{9}\right] = -\frac{32{\mathrm{\pi R}}^2}{9} <\hspace{0.17em}0 \\ \mathrm{hence}, \mathrm{Z} \mathrm{is} \mathrm{maximum} \mathrm{at} \mathrm{h}=\frac{4}{3}\mathrm{R} \\ \mathrm{Or} {\mathrm{S}}^2 \mathrm{is} \mathrm{maximum} \mathrm{at} \mathrm{h}=\frac{4}{3}\mathrm{R} \\ \mathrm{or} \mathrm{S} \mathrm{is} \mathrm{maximum} \mathrm{at} \mathrm{h}=\frac{4}{3}\mathrm{R}. \\ \mathbf{Altitude}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{a}\mathbf{ }\mathbf{right}\mathbf{ }\mathbf{circular}\mathbf{ }\mathbf{cone}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{maximum}\mathbf{ }\mathbf{curved}\mathbf{ }\mathbf{surface}\mathbf{ }\mathbf{area}\mathbf{ }\mathbf{which}\mathbf{ }\mathbf{can}\mathbf{ }\mathbf{be}\mathbf{ }\mathbf{inscribed}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{a}\mathbf{ }\mathbf{sphere}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{radius}\mathbf{ }\mathbf{R}\mathbf{ }\mathbf{is}\mathbf{ }\frac{\mathbf{4}}{\mathbf{3}}\mathbf{R}\mathbf{.} \end{gathered}$$

Hope this information will clear your doubts about this topic.
Regards