question_answer

The height of a cone is 30 cm. A small cone is cut-off at the top by a plane parallel to the base. If its volume is\[\frac{1}{27}\text{th}\] of the volume of the given cone, at what height above the base is the section made? [SSC (CGL) Mains 2014]

A) 19 cm              

B) 20 cm

C) 12 cm  

D) 15 cm

Correct Answer: B

Solution :

Let the height of small cone be h cm.

Given, volume of small cone

\[=\frac{1}{27}\times \] volume of large cone

\[\Rightarrow \]\[{{V}_{s}}=\frac{1}{27}{{V}_{B}}\]\[\Rightarrow \]\[\frac{{{V}_{s}}}{{{V}_{B}}}=\frac{1}{27}\]                        (i)

Then, \[\frac{\frac{1}{3}\pi {{r}^{2}}h}{\frac{1}{3}\pi {{R}^{2}}\times 30}=\frac{1}{27}\]      \[[\because h=30]\]

\[\Rightarrow \]   \[\frac{h}{30}\times {{\left( \frac{r}{R} \right)}^{2}}=\frac{1}{27}\]                   (ii)

In above figure, \[\Delta ABC\sim \Delta ADE\]

\[\Rightarrow \]\[\frac{AB}{BC}=\frac{AD}{DE}\]\[\Rightarrow \]\[\frac{h}{r}=\frac{30}{R}\]\[\Rightarrow \]\[\frac{r}{R}=\frac{h}{30}\]

Now, from Eq. (ii)

\[\frac{1}{27}={{\left( \frac{h}{30} \right)}^{2}}\times \frac{h}{30}\]\[\Rightarrow \]\[{{(30)}^{3}}=27{{h}^{3}}\]

\[\Rightarrow \]\[{{(30)}^{3}}={{(3h)}^{3}}\]\[\Rightarrow \]\[h=10\]

Then,    \[BD=30-h=30-10=20\,\,cm\]