Question

If a cone is cut into two parts by a horizontal plane passing through the mid-points of its axis, the ratio of the volumes of the upper part and the cone is:
(A) \[1:7\]
(B) \[1:4\]
(C) \[1:6\]
(D) \[1:8\]

Answer 1

Hint: In this question, we have to find out the ratio of the volumes on the basis of given information.
We first need to consider that by cutting the cone into two parts by a horizontal plane passing through the mid-points of its axis, we will get a smaller cone as an upper part. Then we can find the required specific by using the formula.

Formula used: The volume of a cone is \[V = \dfrac{1}{3}\pi {r^2}h\]
Where,
$r = $ Radius of the base of the cone
$h = $ Height of the cone

Complete step-by-step solution:
It is given that a cone is cut into two parts by a horizontal plane passing through the mid-points of its axis.
We need to find out the ratio of the volumes of the upper part and the cone.
Let the height and radius of the given cone be $h$ and $r$ respectively.

The cone is divided into two parts by drawing a plane through the mid-point of axis of cone and parallel to the axis of cone and parallel to the base.
One part is a smaller cone and the other part is a frustum of cone.
Then the upper part of the cone is also a cone with radius half of the original cone and with the height half of the original cone, since we got it by drawing a plane through the mid-point of axis of cone and parallel to the axis of cone and parallel to the base.
Thus volume of the smaller cone is \[\dfrac{1}{3}\pi D{E^2}AD\]
\[ \Rightarrow \dfrac{1}{3}\pi {\left( {\dfrac{r}{2}} \right)^2} \times \dfrac{h}{2}\]
The volume of the cone ABC is \[\dfrac{1}{3}\pi {r^2}h\]
The ratio of the volumes of the upper part and the cone is,
\[ \Rightarrow \dfrac{{\dfrac{1}{3}\pi {{\left( {\dfrac{r}{2}} \right)}^2} \times \dfrac{h}{2}}}{{\dfrac{1}{3}\pi {r^2} \times h}}\]
Squaring the terms,
\[ \Rightarrow \dfrac{{\dfrac{1}{3}\pi {{\dfrac{r}{4}}^2} \times \dfrac{h}{2}}}{{\dfrac{1}{3}\pi {r^2} \times h}}\]
Rearranging the terms,
\[ \Rightarrow \dfrac{{\dfrac{1}{8} \times \dfrac{1}{3}\pi {r^2} \times h}}{{\dfrac{1}{3}\pi {r^2} \times h}}\]
Dividing the similar terms we get,
\[ \Rightarrow \dfrac{{\dfrac{1}{8}}}{{\dfrac{1}{1}}}\]
Hence,
\[ \Rightarrow \dfrac{1}{8}\]
\[ \Rightarrow 1:8\]
Thus the ratio of the volumes of the upper part and the cone is \[1:8\].

$\therefore $ Option (D) is the correct answer.

Note: In geometry, the midpoint is the middle point of a line segment. It is equidistant from both endpoints, and it is the centroid both of the segment and of the endpoints. It bisects the segment.
A cone is a distinctive three-dimensional geometric figure that has a flat surface and a curved surface, pointed towards the top. The pointed end of the cone is called the apex, whereas the flat surface is called the base.