question_answer

The height of the cone is 40 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume is\[\frac{1}{64}\]of the volume of the cone. At what height, above the base, is the section made?

A)  6 cm                           

B)  8 cm

C)  10 cm                         

D)  20 cm

Correct Answer: D

Solution :

(d): Let H and R be the height and radius of bigger cone respectively and h and r that of smaller cone. From triangles AOB and AMN. \[\angle A\] is common and \[MN\parallel OB\]. \[\therefore \] Triangles AOB and AMN are similar, \[\therefore \frac{AO}{MA}=\frac{BO}{MN}\] \[\Rightarrow \frac{40}{h}=\frac{R}{r}\]              ???.(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 \[\frac{1}{3}\pi {{r}^{2}}h=\left( \frac{1}{3}\pi {{R}^{2}}h \right)\times \frac{1}{64}\] \[\Rightarrow {{r}^{2}}h=\frac{{{R}^{2}}H}{64}\Rightarrow 64{{r}^{2}}h={{R}^{2}}H\] \[\Rightarrow \frac{64h}{H}={{\left( \frac{40}{h} \right)}^{2}}\] [form (i)] \[\Rightarrow \frac{64h}{H}=\frac{1600}{{{h}^{2}}}\] \[\Rightarrow 64{{h}^{3}}=1600H=1600\times 40\] \[\Rightarrow {{h}^{3}}=\frac{1600\times 40}{64}=1000\] \[\Rightarrow {{h}^{3}}=\sqrt[3]{1000}=10cm\] \[\therefore \] Required height \[=30-10=20cm\]