#### 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.

#### Solution

Given that radius of each of smaller ball

=1/4 Radius of original ball.

Let radius of smaller ball be . r

Radius of bigger ball be 4r

Volume of big spherical ball =`4/3pir^3` (∵ r = 4r)

`V_1 = 4/3pi(4r)^3` ............(1)

Volume of each small ball = `4/3pir^3`

`v_2=4/3PIR^3` .............(2)

Let no of balls be 'n'

`n = V_1/V_2`

⇒ `n=(4/3pi(4r)^3)/(4/3pi(r)^3)`

⇒ n = 4^{3} = 64

∴ No of small balls = 64

Curved surface area of sphere = 4πr^{2}

Surface area of big ball (S_{1}) = 4π(4r)^{2} ............(3)

Surface area of each small ball(S_{1}) = 4πr^{2}

Total surface area of 64 small balls

`(S_2)=64xx4pir^2` ..............(4)

By combining (3) and (4)

⇒ `S_2/3=4`

⇒ S_{2} = 4s

∴Total surface area of small balls is equal to 4 times surface area of big ball.