If the height of a cone of greatest volume that can be inscribed in a sphere of radius $\mathrm{R}$ is $\mathrm{kR}$, then ratio of the volume of the cone to the volume of the sphere is

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { In } 	riangle O D C 	ext {, } \ & r^2=R^2-(h-R)^2 \ & =2 h R-h^2\end{aligned}$ Hence volume of cone $A B C=\frac{1}{3} \pi r^2 h$ $\Rightarrow \mathrm{V}=\frac{1}{3} \pi\left(2 h R-h^2ight) h=\frac{1}{3} \pi\left(2 R h^2-h^3ight)$ Now for volume to be maximum, $ \begin{aligned} & \frac{d V}{d h}=0 \ & \Rightarrow \frac{1}{3 \pi}\left[4 h R-3 h^2ight]=0 \ & \Rightarrow h 
eq 0 	ext { Hence } h=\frac{4 R}{3} \ & 	ext { at } h=\frac{4 R}{3}, \frac{d^2 V}{d h^2} < 0 \end{aligned} $ Hence at $h=\frac{4 R}{3}$, volume of cone is maximum. $ \begin{aligned} & \frac{	ext { Volume of cone }}{	ext { Volume of sphere }}=\frac{\frac{1}{3} \pi r^2 h}{\frac{4}{3} \pi R^3}=\frac{\left(2 h R-h^2ight) h}{4 R^3} \ & =\frac{\left(\frac{8 R^2}{3}-\frac{16 R^2}{9}ight) \frac{4 R}{3}}{4 R^3}=\frac{8}{27} \end{aligned} $

If the height of a cone of greatest volume that can be inscribed in a sphere of radius $\mathrm{R}$ is…

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