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