Question

# A cylindrical container of radius $$6 cm$$ and height $$15 cm$$ is filled with ice-cream. The whole ice-cream has to be distributed to $$10$$ children in equal cones with hemispherical tops. If the height of the conical portion is four times the radius of its base, find the radius of the ice-cream cone.

A
4 cm
B
6 cm
C
3 cm
D
None of the above

Solution

## The correct option is A $$3\ cm$$Let $$R$$ and $$H$$ be the radius and height of the cylindrical container respectively.Given, $$R = 6 cm$$ and $$H = 15 cm$$Now, volume of ice-cream in the cylindrical container $$=\pi R^{ 2 }H=\pi \times 6^{ 2 }\times 15=540\pi cm^{ 3 }$$Suppose the radius of the cone be $$r$$ cm.Height of the cone $$= h = 2(2r) = 4r$$     ....(given)Radius of the hemispherical portion $$= r$$ cm Now, volume of ice-cream in the cylinder $$=$$ volume of cone $$+$$ volume of hemisphere$$= \cfrac { 1 }{ 3 } \pi r^{ 2 }h+\cfrac { 2 }{ 3 } \pi r^{ 3 }\\ = \cfrac { 1 }{ 3 } \pi r^{ 2 }(h+2r)=\frac { 1 }{ 3 } \pi r^{ 2 }(4r+2r)=2\pi r^{ 3 }$$ Given, number of ice-cream cones distributed to the children $$= 10$$Therefore, $$10 \times$$ Volume of ice-cream in the cone $$=$$ Volume of ice-cream in the cylindrical container$$\Rightarrow 10\times 2\pi r^{ 3 }=540\pi \\ \Rightarrow r^{ 3 }=27\\ \Rightarrow r=3cm$$Hence, the radius of the cone is 3 cm.Mathematics

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