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.

Answer 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

Answer 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.