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 \[=30-10=20\,cm\]You need to login to perform this action.
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