#### Two solid cones A and B are placed in a cylindrical tube as shown in the Figure. The ratio of their capacities are 2:1. Find the heights and capacities of cones. Also, find the volume of the remaining portion of the cylinder.

Solution

Height of the tube = 21 cm

Base radius of the tube =3cm

Volume of tube $=\pi r ^2h$

$=\frac{22}{7}\times 3\times 3\times21=594cm^3$

Let the height of cone A is h cm

Height of cone $B=21-h cm$

Base radius of both A and B =3 cm

Volume of cone

$A=\frac{1}{3}\pi r^2h=\frac{1}{}3\pi (3)^2h$

$=3 \pi h$

Volume of cone

$B=\frac{1}{3}\pi r^2h=\frac{1}{}3\pi (3)^2(21-h)$

$=3 \pi (21-h)$

It is given that the ratio of the volume is 2: 1

$=\frac{3\pi h}{3 \pi (21-h)}=\frac{2}{1}$

$h=2(21-h)$

$h=42-2h$

$3h=42$

$h=14$

Height of cone A = 14 cm

Height of cone B =21-4=7 cm

Volume of cone $A=3 \pi h=3(3.14)(14)$

$=131.88cm^3$

Volume of cone $B = 3\pi (21-h)=2(3.14)(7)$

$=65.94cm^3$

The volume of remaining portion = Volume of the tube – the volume of cone A – the volume of cone B

$594-131.88 -65.94$

= 396.18 cm3

Volume of remaining portion  = 396.18 cm3