A solid cone of radius 20 cm is cut from its middle, parallel to its base, into two parts as shown. Find :

The ratio between the volumes of the two parts obtained.

Let the radius of the base of the cone be R = 20 cm and its height be H cm.

The cone is cut at the middle of its height by a plane parallel to the base. So the cut is at a height $\dfrac{H}{2}$ from the apex.

The upper part is a small cone similar to the whole cone. Since the height is halved, by similar triangles its radius is also halved.

∴ Radius of small cone (r) = $\dfrac{R}{2}$ = 10 cm and height = $\dfrac{H}{2}$.

Volume of small cone (upper part) :

$V_1 = \dfrac{1}{3}πr^2\left(\dfrac{H}{2}\right) = \dfrac{1}{3}π(10)^2 \times \dfrac{H}{2} = \dfrac{1}{3}π \times 50H.$

Volume of the whole cone :

$V = \dfrac{1}{3}πR^2H = \dfrac{1}{3}π(20)^2H = \dfrac{1}{3}π \times 400H.$

Volume of lower part (frustum) = Volume of whole conce - Volume of small cone

$V$2 = V - V1 = \dfrac{1}{3}π \times 400H - \dfrac{1}{3}π \times 50H = \dfrac{1}{3}π \times 350H.

∴ Ratio of the two parts :

$V$1 : V2 = \dfrac{1}{3}π \times 50H : \dfrac{1}{3}π \times 350H = 50 : 350 = 1 : 7.

Hence, the ratio between the volumes of the two parts (smaller cone : frustum) is 1 : 7.