The volume of the largest possible right circular cylinder that can be inscribed in a sphere of radius $=\sqrt{3}$ is:

Question

Aicharya · Accepted Answer

Given, radius of sphere $=\sqrt{3}$ Now, In $\triangle \mathrm{OAB}$, by Pythagoras theorem $ (\mathrm{OA})^2=(\mathrm{OB})^2+(\mathrm{AB})^2 $

$ \begin{aligned} &(\sqrt{3})^2=\left(\frac{h}{2}\right)^2+r^2 \\ &3=\frac{h^2}{4}+r^2 \Rightarrow r^2=3-\frac{h^2}{4} \end{aligned} $ Now, volume of cylinder $=\pi r^2 h$ $ \begin{aligned} &V=\pi\left(3-\frac{h^2}{4}\right) h \\ &V=3 \pi h-\frac{\pi h^3}{4} \end{aligned} $ Now, for largest possible right circular cylinder the volume must be maximum $\therefore$ For maximum volume, $\frac{d V}{d h}=0$ Now, Differentiating eq. (2) w.r.t. $h$ $V^1=\frac{d V}{d h}=3 \pi-\frac{3}{4} \pi h^2$ or $3 \pi-\frac{3}{4} \pi h^2=0 \Rightarrow 3 \pi=\frac{3}{4} \pi h^2$ $\Rightarrow h^2=4 \Rightarrow h=2$ Now, volume $(V)$ of the cylinder $ =\pi\left(3-\frac{h^2}{4}\right) h=\pi(6-2)=4 \pi $

The volume of the largest possible right circular cylinder that can be inscribed in a sphere of radius…

Solution