Question

A cylindrical container of radius 21 units and height 20 units is filled with chocolate. The chocolate is to be filled into cones of height 3 units and radius $\frac{21}{2}$ units having a hemispherical shape on the top. Find the number of cones that can be filled with chocolate. $\text{[Use}\pi =\frac{2}{7}]$

Answer 1

Cylinder:
$\text{Volume of cylinder}=\pi \times r^2\times h$

Given values:
Radius = 21 units
Height = 20 units

So, volume of the cylinder
$=\pi r^{2}h=\frac{22}{7}\times (21{)}^2\times 20=27720\text{cubic units}$

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

Given value:
$\text{Radius}=\frac{21}{2}\text{units}$
Height = 3 units

So, volume of the cone
$=\frac{1}{3}\pi r^{\prime 2}h=\frac{1}{3}\times \frac{22}{7}\times (\frac{21}{2}{)}^2\times 3$
$=346.5\text{cubic units}$

Hemisphere:
$\text{Volume of hemisphere}=\frac{2}{3}\times \pi r^3$

$\text{Radius}=\frac{21}{2}\text{units}$

Volume of a hemisphere = $\frac{2}{3}\pi r^3=\frac{2}{3}\times \frac{22}{7}\times (\frac{21}{2}{)}^3$
$=2425.5\text{cubic units}$

Volume of the new shape so formed is
= Volume of cone + Volume of hemisphere
= 2425.5 + 346.5 = 2772 cubic units

Let the number of cones that can be filled be $x$.

$x=\frac{\text{Volume of cylinder}}{\text{Total volume of the solid}}$

(Here, solid = Cone + Hemisphere)

$=\frac{27720}{2772}$

$=10$

Therefore, 10 cones can be filled with chocolate.