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(B) Let $h$ be the height of the given cone. On dividing the cone through the midpoint of its axis and parallel to its base into two parts,we obtain a

Vedclass · Accepted Answer

$1:7$

Vedclass · Answer

(B) Let $h$ be the height of the given cone. On dividing the cone through the midpoint of its axis and parallel to its base into two parts,we obtain a smaller cone at the top and a frustum at the bottom.In two similar triangles,the ratio of corresponding sides is equal. Let $r$ be the radius of the smaller cone. By similarity of triangles,we have $\frac{r}{4} = \frac{h/2}{h} = \frac{1}{2}$.Therefore,$r = 2 \, cm$.Volume of the smaller cone $(V_1)$ = $\frac{1}{3} \pi r^2 (h/2) = \frac{1}{3} \pi (2)^2 (h/2) = \frac{2}{3} \pi h$.Volume of the original cone $(V)$ = $\frac{1}{3} \pi (4)^2 h = \frac{16}{3} \pi h$.Volume of the frustum $(V_2)$ = $V - V_1 = \frac{16}{3} \pi h - \frac{2}{3} \pi h = \frac{14}{3} \pi h$.Ratio of the volume of the smaller cone to the volume of the frustum = $\frac{V_1}{V_2} = \frac{\frac{2}{3} \pi h}{\frac{14}{3} \pi h} = \frac{2}{14} = \frac{1}{7}$.Thus,the ratio is $1:7$.

$A$ cone of radius $4 \, cm$ is divided into two parts by drawing a plane through the midpoint of its axis and parallel to its base. Compare the volumes of the two parts.

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