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(A) Let the volume of the smaller cone (top part) be $V_1$ and the volume of the larger cone be $V_2$. Since the plane divides the cone into two parts

Vedclass · Accepted Answer

$1: \sqrt[3]{2}-1$

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(A) Let the volume of the smaller cone (top part) be $V_1$ and the volume of the larger cone be $V_2$. Since the plane divides the cone into two parts of equal volume,the volume of the smaller cone is half the volume of the larger cone. Thus,$\frac{V_1}{V_2} = \frac{1}{2}$. For a cone cut by a plane parallel to its base,the ratio of the volumes of the smaller cone and the larger cone is equal to the cube of the ratio of their corresponding heights: $\frac{V_1}{V_2} = \left(\frac{h_1}{h_2}\right)^3 = \frac{1}{2}$. Taking the cube root on both sides,we get $\frac{h_1}{h_2} = \frac{1}{\sqrt[3]{2}}$. Here,$h_1$ is the height of the smaller cone and $h_2$ is the height of the larger cone. The height of the frustum (the bottom part) is $h_2 - h_1$. The ratio in which the height is divided is $h_1 : (h_2 - h_1)$. Substituting the values,we get $1 : (\sqrt[3]{2} - 1)$.

$A$ plane divides a right circular cone into two parts of equal volume. If the plane is parallel to the base,then the ratio,in which the height of the cone is divided,is

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