A plane is drawn parallel to the base of a right circular cone dividing altitude in the ratio of 2 : 1 to chop off the conical part. The chopped off conical part is taller than the remaining portion. The ratio of the volumes of the chopped off part and the remaining part is
Correct Answer: Description for Correct answer:
\( \Large \frac{R}{r}=\frac{3}{2} \)
=> \( \Large r = \frac{2}{3}R \)
Therefore, \( \Large V_{c}=\frac{1}{3} \pi \left(\frac{2}{3}\right)^{2}.3=\frac{4}{9} \pi \)
\( \Large V_{Frustum} = \frac{ \pi h}{2}\left[ R^{2}+r^{2}+R.r \right] \)
= \( \Large \frac{ \pi }{2}\left[ 1+\frac{4}{9}+\frac{2}{3} \right]=\frac{19 \pi }{18} \)
Therefore, \( \Large \frac{V_{c}}{V_{F}} =\ \frac{8}{19} \)
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