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

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 = 4³ = 64

∴ No of small balls = 64

Curved surface area of sphere = 4πr²

Surface area of big ball (S₁) = 4π(4r)²    ............(3)

Surface area of each small ball(S₁) = 4πr²

Total surface area of 64 small balls

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

By combining (3) and (4)

⇒ `S_2/3=4`

⇒ S₂ = 4s

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