Question

An iron spherical ball has been melted and recast into smaller balls of equal size. If the radius of each of the smaller balls is 1/4 of the radius of the original ball, how many such balls are made? Compare the surface area, of all the smaller balls combined together with that of the original ball.

Answer 1

Let the radius of bigger spherical be R 

Volume of bigger spherical ball =\(\frac{4}{3}πR^3\)

Radius of smaller spherical ball = \(\frac{1}{4}R\)

Volume of smaller ball = \(\frac{4}{3}π(\frac{1}{4}R)^3\)

Let number of equal size spherical balls be n 

Volume of n equal spherical ball =Volume of bigger spherical ball

⇒ n x \(\frac{4}{3}π(\frac{1}{4}R)^3\) = \(\frac{4}{3}πR^3\)

⇒ n × (\(\frac{R}4\))³ R³ 

⇒ n = 4³ 

⇒ n = 64 balls 

Surface area of bigger spherical ball = 4π R² 

Surface area of smaller spherical ball = \(4π(\frac{1}{4}R)^2\)

Ratio between the surface area of bigger and 64 smaller spherical ball