Question

Question 8
Two solid cones A and B are placed in a cylindrical tube as shown in the Fig.12.9. 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 2 V 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
$\therefore Radiusofthecone=\frac{6}{2}=3cmNow,volumeofthecone,A=2v=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi (3{)}^2{h}_1\Rightarrow V=\frac{1}{6}\pi 9h_1=\frac{3}{2}{h}_1\pi$ ...................(i)
And volume of the cone, $B=V=\frac{1}{3}\pi \left(3{)}^2\right(21-h_1)=3\pi (21-h_1)........(ii)$
From Eqs. (i) and (ii)
$\frac{3}{2}{h}_1\pi =3\pi (21-h_1)\Rightarrow h_1=2(21-h_1)\Rightarrow 3h_1=42\Rightarrow h_1=\frac{42}{3}=14cm\therefore Heightofcone,B=21–h_1=21–14=7cm$
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 = Volume of the cylinder – ( Volume of cone A + Volume of cone B)
= 594 – ( 132 + 66)
$=396c{m}^3$