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A solid metallic sphere of radius 4 cm is melted and recast into spheres of 2 cm each. What is the ratio of the surface area of the original sphere to the sum of surface area of the spheres, so formed?

A solid metallic sphere of radius 4 cm is melted and recast into spheres of 2 cm each. What is the ratio of the surface area of the original sphere to the sum of surface area of the spheres, so formed? Correct Answer 1 ∶ 2

Given:

Radius of solid metallic sphere (R) = 4 cm 

Radius of small spheres (r) = 2 cm 

Formula Used:

The surface area of the sphere = 4πr2

The volume of the sphere = 4πr3/3

Calculation:

According to the questio 

Number of spheres = Volume of (Bigger sphere ÷ Small spheres)

⇒ Number of spheres = 4πR3/3 ÷ 4πr3/3

⇒ Number of spheres = (R/r)3

⇒ Number of spheres = (4/2)3 = 8 

Now, Surface area of original sphere = 4π (4)2 

⇒ 64π 

And sum of sufrace area of small spheres = 8 × 4π (2)2 

⇒ 128π 

∴ The required ratio is 64π : 128π = 1 : 2.

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