question_answer

The height of the cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume is \[\frac{1}{27}\]of the volume of the cone, at what height above the base is the section made?                                           [SSC (10+2) 2008]

A) 6 cm                            

B) 8 cm 

C) 10 cm              

D) 20 cm

Correct Answer: D

Solution :

Let H and R be the height and radius of bigger cone respectively and h and r that of smaller cone.

Since, \[\Delta AOB\] and \[\Delta AMN\]are similar.

Then, by basic proportionality theorem,

\[\therefore \]      \[\frac{AO}{AM}=\frac{BO}{MN}\]

\[\Rightarrow \]   \[\frac{30}{h}=\frac{R}{r}\]                               (i)

Volume of smaller cone

\[=\frac{1}{3}\pi {{r}^{2}}h\]

Volume of bigger cone \[=\frac{1}{3}\pi {{R}^{2}}H\]

According to the question,

\[=\frac{1}{3}\pi {{r}^{2}}h=\left( \frac{1}{3}\pi {{R}^{2}}H \right)\times \frac{1}{27}\]

\[\Rightarrow \]   \[{{r}^{2}}h=\frac{{{R}^{2}}H}{27}\]\[\Rightarrow \]\[27{{r}^{2}}h=30{{R}^{2}}\]

\[\Rightarrow \]   \[\frac{27h}{30}=\frac{{{R}^{2}}}{{{r}^{2}}}\]

\[\Rightarrow \]   \[\frac{27h}{30}={{\left( \frac{30}{h} \right)}^{2}}\]     [from Eq. (i)]

\[\Rightarrow \]   \[\frac{27h}{30}=\frac{900}{{{h}^{2}}}\]\[\Rightarrow \]\[27{{h}^{3}}=900\times 30\]

\[\Rightarrow \]   \[{{h}^{3}}=\frac{900\times 30}{27}=1000\]

\[\Rightarrow \]   \[h=\sqrt[3]{1000}=10\,\,cm\]

\[\therefore \]Required height above which cut is made

\[=30-10=20\,\,cm\]