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The height of a cone is 32 cm. A small cone is cut off at the top by a plane

parallel to its base . If its volume is ⅟64 of the volume of the given cone, at
what height above the base , is the cone cut? (use π = 22/7 )



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Let the height, radius and voleume of larger cone be H, R and V
and the height, radius and volume of smaller cone is h, r and v.

Given that H = 32 cm, v/V =1/64, 
We need to find (H-h).

Note that  \frac{h}{H} = \frac{r}{R}

 \frac{v}{V} = \frac{(1/3) \pi r^2h}{(1/3) \pi R^2H} \\ \frac{1}{64} = \frac{r^2h}{R^2H} \\ \frac{1}{64} =  ( \frac{r}{R} )^{2} \frac{h}{H} \\ \frac{1}{64} =  ( \frac{h}{H} )^{2} \frac{h}{H} = (\frac{h}{H})^3 \\  \sqrt[3]{\frac{1}{64}}  = \frac {h}{H}\\\frac {h}{H}= \frac{1}{4} \\ \\h= \frac{H}{4}= \frac{32}{4}= 8\ cm\\ \\ thus\ (H-h) =32-8 = \boxed {24\ cm}

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