Question

Sixty four solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S. Find the ratio of S and S.​

Answer 1

Answer:

volume of 64 iron spheres=volume of new sphere

64×4/3πr³=4/3πr'³

64r³=r'³

³√64r³=³√r'³

4r=r'

r'=4r

ratio of s and s'

s/s'=4πr²/4πr²

=(r²)/(4r')²

=r²/16r²

=1/16

Answer 2

Given:

64 solid iron spheres with radius 'r' and surface area 'S'.

They are melted to form a sphere with a  surface area of 'S'.

To Find:

the ratio of S:S

Solution:

The volume of 64 spheres = volume of the new sphere

The volume of the sphere = 4/3πr³

So,

64×4/3πr² = 4/3πR³

64r³ = R³

∛64r³ = ∛R³

4r = R

R = 4r

The ratio of 'S' and 'S' is

S/S = 4πr²/4πR²

     = (r²)/(4R²)

     = r²/16R²

     =1/16

     = 1:16

Therefore, the ratio of 'S':'S' = 1:16.