Question

Two solid cones A and B are placed in a cylindrical tube as shown in fig .16.76. The ratio of their capacities are 2: 1 . Find the heights and capacities of the cones . Also, find the volume of the remaining portion of the cylinder.

figure

Answer 1

V1 : V2 = 2 : 1
Diameter of the cylinder = 6 cm
Radius, r = 3 cm
Height of the cylinder = 21 cm
Let the height of one cone be H.
So, the height of the other cone will be 21 − H.
$$\begin{gathered} \frac{V_1}{V_2}=\frac{\mathrm{\pi }{\left(3\right)}^2\mathrm{H}}{\mathrm{\pi }{\left(3\right)}^2\left(21-\mathrm{H}\right)} \\ \Rightarrow \frac{2}{1}=\frac{H}{21-H} \\ \Rightarrow 42-2H=H \\ \Rightarrow H=14cm \end{gathered}$$
Height of one of the cones will be 14 cm and of the other will be 21 − H = 21 − 14 = 7 cm
Volume of cone with height 14 cm = $V_1=\mathrm{\pi }{\left(3\right)}^2\times 14=396{\mathrm{cm}}^3$
Volume of cone with height 7 cm = $V_2=\frac{1}{3}\mathrm{\pi }{\left(3\right)}^2\times 7=66{\mathrm{cm}}^3$
Volume of the remaining portion of the cylinder = $Volumeofthecylinder-volumeofcone1-volumeofcone2$
$$\begin{gathered} \Rightarrow V=\mathrm{\pi }{\left(3\right)}^2\times 21-396-66 \\ =594-396-66 \\ =132{\mathrm{cm}}^3 \end{gathered}$$