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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 4 times the radius of its base, then find the radius of the ice-cream cone.
Solution
We have
the base radius of the cylindrical containers, R=6 cm,
the height of the container, H = 15 cm,
Let the base radius and the height of the ice-cream cone be r an
Also, h = 4r
Now, the volume of the cylindrical container = πR2H
`= 22/7xx6xx6xx15`
`= 11880/7` cm3
`rArr "the volume of the ice -cream distributed to 10 children" = 11880/7 cm`
`rArr 10xx "Volume of a ice - creame cone = 11880/7" `
`rArr 10xx ("Volume of the cone + Volume of the hemisphere") = 11880/7`
`rArr 10xx(1/3 pir^2h + 2/3 pir^3) = 11880/7`
`rArr 10xx(1/3pir^2hxx4r + 2/3pir^3)= 11880/7`
`rArr 10xx(1/3pir^2xx4r+2/3pir^3)= 11880/7`
`rArr 10xx(6/3pir^3)=11880/7`
`rArr 10xx2xx22/7xxr^3 = 11880/7`
`rArr r^3 = (11880xx7)/(7xx10xx2xx22)`
`rArr r^3 = 27`
`rArr r = root(3)(27)`
∴ r = 3 cm
So, the radius of the ice-cream cone is 3 cm.
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