Subject: General Aptitude · Chapter: Volume and Surface Area · Exam: · Marks: · Difficulty:
A right circular cone is divided into two portions by a plane parallel to the base and passing through a point which is $\frac{1}{3} \mathrm{rd}$ of the height from the top. The ratio of the volume of the smaller cone to that of the remaining frustum of the cone is
(a)1 : 3
(b)1 : 9
(c)1 : 26
(d)1 : 27
Answer
Answer (as printed): C
Explanation
Let the radius and height of the cone be $r$ and $h$ respectively. Then, $A B=r, O A=h, O C=\frac{h}{3}$. Now, $\triangle A O B \sim \triangle C O D$. $$\therefore \frac{A B}{C D}=\frac{O A}{O C} \Rightarrow \frac{r}{C D}=\frac{h}{h / 3} \Rightarrow C D=\frac{r}{3} .$$ Volume of bigger cone $=\frac{1}{3} \pi r^{2} h$. Volume of smaller cone $=\frac{1}{3} \pi\left(\frac{r}{3}\right)^{2}\left(\frac{h}{3}\right)=\frac{1}{27}\left(\frac{1}{3} \pi r^{2} h\right)$.
Explanation as extracted from the printed page; notation may be imperfect.