Question

The ratio between the height of right circular cone of maximum volume inscribed in a given sphere and the diameter of the sphere is-

• A. $2:3$
• B. $3:4$
• C. $1:3$
• D. $1:4$

Answer 1

Answer: (A)

$2:3$

Let the height of the cone be $h$ and radius of the cone and sphere be $r$ and $R$ respectively.

In $\Delta OAB$

${\left(OB\right)}^2={\left(OA\right)}^2+{\left(AB\right)}^2$

$\Rightarrow R^2={\left(h-R\right)}^2+r^2$

$\Rightarrow R^2=h^2+R^2-2hR+r^2$

$\Rightarrow h^2+r^2=2hr-\left(1\right)$

Volume of cone$=\frac{1}{3}\pi r^{2}h$

$\Rightarrow V=\frac{1}{3}\pi h\left(2hR-h^2\right)$

$\Rightarrow V=\frac{\pi }{3}\left(2h^{2}R-h^3\right)$

$\Rightarrow \frac{dv}{dh}=\frac{\pi }{3}\left(4hR-3h^2\right)$

$\Rightarrow \frac{d^{2}v}{d{h}^2}=\frac{\pi }{3}\left(4R-6h\right)$

$\frac{dv}{dh}=0$

For critical points

$\frac{dv}{dh}=0$

$4hR-3h^2=0$

$\Rightarrow h(4R-3h)=0$

$\Rightarrow 3h=4R\left(h\ne 0\right)$

$\Rightarrow h=\frac{4}{3}R$

$\left(\frac{d^{2}v}{d{h}^2}\right)h=\frac{4}{3}R=\frac{\pi }{3}\left(4R-6\times \frac{4}{3}R\right)=\frac{\pi }{3}\left(-4R\right)<0$

so volume of cone is maximum

when $h=\frac{4}{3}R\Rightarrow \frac{h}{2R}=\frac{2}{3}$

$\Rightarrow \frac{heightofcone}{diameterofsphere}=\frac{2}{3}$