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