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 section made?

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 of smaller cone. From triangles AOB and AMN. \[\angle A\] is common and MN ||OB \[\therefore \] Triangles AOB and AMN are similar, \[\therefore \]      \[\frac{AO}{AM}=\frac{BO}{MN}\] \[\Rightarrow \]   \[\frac{30}{h}=\frac{R}{r}\] 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,             \[\frac{1}{3}\pi {{r}^{2}}h=\left( \frac{1}{3}\pi {{R}^{2}}H \right)\times \frac{1}{27}\] \[\Rightarrow \]   \[{{r}^{2}}h=\frac{{{R}^{2}}H}{27}\] \[\Rightarrow \]   \[27{{r}^{2}}h={{R}^{2}}H\] \[\Rightarrow \]   \[\frac{27h}{H}=\frac{{{R}^{2}}}{{{r}^{2}}}\] \[\Rightarrow \]   \[\frac{27h}{H}={{\left( \frac{30}{h} \right)}^{2}}\]                  [From Eq. (i)] \[\Rightarrow \]   \[\frac{27h}{H}=\frac{900}{{{h}^{2}}}\] \[\Rightarrow \]   \[27{{h}^{3}}=900H=900\times 30\] \[\Rightarrow \]   \[{{h}^{3}}=\frac{900\times 30}{27}=1000\] \[\Rightarrow \]   \[h=\sqrt[3]{1000}=10\,cm\] \[\therefore \]  Required height \[=30-10=20\,cm\]