A cylindrical container of radius 6 cm and height 15 cm is filled with ice-cream.
Question:

A cylindrical container of radius 6 cm and height 15 cm is filled with ice-cream. The whole ice-cream has to be distributed to 10 children in equal cones with hemispherical tops. If the height of the conical portion is 4 times the radius of its base, then find the radius of the ice-cream cone.

Solution:

We have,

the base radius of the cylindrical container, $R=6 \mathrm{~cm}$,

the height of the container, $H=15 \mathrm{~cm}$,

Let the base radius and the height of the ice – cream cone be $r$ and $h$, respectively.

Also, $h=4 r$

Now, the volume of the cylindrical container $=\pi R^{2} H$

$=\frac{22}{7} \times 6 \times 6 \times 15$

$=\frac{11880}{7} \mathrm{~cm}^{3}$

$\Rightarrow$ the volume of the ice $-$ cream distributed to 10 children $=\frac{11880}{7} \mathrm{~cm}^{3}$

$\Rightarrow 10 \times$ Volume of a ice $-$ cream cone $=\frac{11880}{7}$

$\Rightarrow 10 \times($ Volume of the cone $+$ Volume of the hemisphere $)=\frac{11880}{7}$

$\Rightarrow 10 \times\left(\frac{1}{3} \pi r^{2} h+\frac{2}{3} \pi r^{3}\right)=\frac{11880}{7}$

$\Rightarrow 10 \times\left(\frac{1}{3} \pi r^{2} \times 4 r+\frac{2}{3} \pi r^{3}\right)=\frac{11880}{7}$

$\Rightarrow 10 \times\left(\frac{4}{3} \pi r^{3}+\frac{2}{3} \pi r^{3}\right)=\frac{11880}{7}$

$\Rightarrow 10 \times\left(\frac{6}{3} \pi r^{3}\right)=\frac{11880}{7}$

$\Rightarrow 10 \times 2 \times \frac{22}{7} \times r^{3}=\frac{11880}{7}$

$\Rightarrow r^{3}=\frac{11880 \times 7}{7 \times 10 \times 2 \times 22}$

$\Rightarrow r^{3}=27$

$\Rightarrow r=\sqrt[3]{27}$

$\therefore r=3 \mathrm{~cm}$

So, the radius of the ice-cream cone is 3 cm.