Question

Question 8
Two solid cones A and B are placed in a cylindrical tube. 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.

Answer 1

Let volume of cone A be 2V and volume of cone B be V. Again, let height of the cone A = $h_{1}cm$ the height of cone B = $(21–h_1)cm.$

Given , diameter of the cone = 6 cm
$\therefore Radiusofthecone=\frac{6}{2}=3cmNow,volumeofthecone,A\Rightarrow 2V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi (3{)}^2{h}_1\Rightarrow V=\frac{1}{6}\pi 9h_1=\frac{3}{2}{h}_1\pi ...(i)$

And volume of the cone B
$\Rightarrow V=\frac{1}{3}\pi \left(3{)}^2\right(21-h_1)=3\pi (21-h_1)...(ii)$

From Eqs. $\left(i\right)$ and $\left(ii\right)$
$\Rightarrow V=V$
$\Rightarrow \frac{3}{2}{h}_1\pi =3\pi (21-h_1)\Rightarrow h_1=2(21-h_1)\Rightarrow 3h_1=42\Rightarrow h_1=\frac{42}{3}=14cm$

$\therefore Heightofcone,A=h_1=14cmHeightofcone,B=(21–h_1)=21–14=7cm$

Now, Volume of the cone A $=3\times 14\times \frac{22}{7}=132c{m}^3[UsingEq.(i\left)\right]$
And Volume of the cone , $B=V=\frac{1}{3}\times \frac{22}{7}\times 9\times 7=66c{m}^3[UsingEq.(ii\left)\right]$

Now, volume of the cylinder
$=\pi r^{2}h=\frac{22}{7}(3{)}^2\times 21=594c{m}^3$

$\therefore$ Required volume of the remaining portion
= (Vol. of the cylinder) – (Vol. of cone A + Vol. of cone B)
$=594c{m}^3–(132c{m}^3+66c{m}^3)$
$=396c{m}^3$