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Question
Two solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed.
Solution 1
Radius of small sphere = r = 2 cm
Radius of big sphere = R = 4 cm
Volume of small sphere = `4/3 pir^3`
= `(4pi)/3 xx (2)^3`
= `(32pi)/3 cm^3`
Volume of big sphere = `4/3 piR^3`
= `(4pi)/3 xx (4)^3`
= `(256pi)/3 cm^3`
Volume of both the spheres = `(32pi)/3 + (256pi)/3`
= `(288pi) /3cm^3`
We need to find R1.h = 8 cm ...(Given)
Volume of the cone = `1/3 piR_1^2 xx (8)`
Volume of the cone = Volume of both the sphere
`=> 1/3 piR_1^2 xx (8) = (288pi)/3`
`=> R_1^2 xx (8) = 288`
`=> R_1^2 = 288/8`
`=> R_1^2 = 36`
`=>` R1 = 6 cm
Solution 2
We have,
Volume of the cone = Sum of volumes of the two melted spheres
`=> 1/3 pi(r)^2 xx 8 = 4/3 pi xx (2)^3 + 4/3 pi xx (4)^3`
`=>` 8r2 = 4 × 8 + 4 × 64
`=>` 8r2 = 32 + 256
`=>` 8r2 = 288
`=>` r2 = 36
`=>` r = 6
Thus, the radius of the cone so formed is 6 cm.
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