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Question

The height of a cone is $$40 cm$$. A small cone is cut off at the top by a plane parallel to the base. If the volume of the small cone be $$\dfrac{1}{64} $$ of the volume of the given cone, at what height above the base is the section made ?


A
35cm
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B
30cm
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C
25cm
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D
20cm
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Solution

The correct option is B $$30 cm$$
Let $${ r }_{ 1 }$$ and $$h$$ be the radius and height of the upper cone.
Let $${ r }_{ 2 }$$ be the radius of the bigger cone.
Let $${ v }_{ 1 }$$ & $${ v }_{ 2 }$$ be the volumes of upper cone and bigger cone respectively.
Now, $${ v }_{ 1 }=\cfrac { 1 }{ 64 } { v }_{ 2 }$$
$$\Rightarrow \cfrac { 1 }{ 3 } \pi { r }_{ 1 }^{ 2 }h=\cfrac { 1 }{ 64 } \times \cfrac { 1 }{ 3 } \pi { r }_{ 2 }^{ 2 }\times 40$$
$$\Rightarrow { \left( \cfrac { { r }_{ 1 } }{ { r }_{ 2 } }  \right)  }^{ 2 }=\cfrac { 40 }{ 64 } \cfrac { 1 }{ h } \rightarrow (1)$$.

$$\triangle ADE\sim \triangle ABC$$ thus
$$\cfrac { AD }{ AB } =\cfrac { DE }{ BC } \Rightarrow \cfrac { h }{ 40 } =\cfrac { { r }_{ 1 } }{ { r }_{ 2 } } \rightarrow (2)$$.

From $$(1)$$ and $$(2)$$, we get,
$${ \left( \cfrac { h }{ 40 }  \right)  }^{ 2 }=\cfrac { 40 }{ 64 } \times \cfrac { 1 }{ h }$$ 
$$\Rightarrow { h }^{ 3 }=\cfrac { 40\times 40\times 40 }{ 4\times 4\times 4 }$$ 
$$\Rightarrow h=\sqrt [ 3 ]{ \cfrac { 40\times 40\times 40 }{ 4\times 4\times 4 }  } $$
$$\Rightarrow h=\cfrac { 40 }{ 4 } =10cm$$.
Height of upper cone$$=10cm$$.
At $$\left( 40-10=30cm \right) $$ above the base the section is made.

928615_247639_ans_2302a46502cb4c73b220d71ebbf35447.png

Mathematics

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