A) 6 cm
B) 8 cm
C) 10 cm
D) 20 cm
Correct Answer: D
$ \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=( \frac{1}{3}\pi R^{2}H )\times \frac{1}{27} $
$ \Rightarrow $ $ r^{2}h=\frac{R^{2}H}{27} $
$ \Rightarrow $ $ 27r^{2}h=30R^{2} $
$ \Rightarrow $ $ \frac{27h}{30}=\frac{R^{2}}{r^{2}} $
$ \Rightarrow $ $ \frac{27h}{30}={{( \frac{30}{h} )}^{2}} $ [from Eq. (i)]
$ \Rightarrow $ $ \frac{27h}{30}=\frac{900}{h^{2}} $
$ \Rightarrow $ $ 27h^{3}=900\times 30 $
$ \Rightarrow $ $ h^{3}=\frac{900\times 30}{27}=1000 $
$ \Rightarrow $ $ h=\sqrt[3]{1000}=10cm $
$ \therefore $ Required height above which cut is made $ =30-10=20cm $
BETA
