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.
Answers
Given : An iron spherical ball has been melted and recast into smaller balls of equal size and the radius of each of the smaller balls is 1/4 of the radius of the original ball.
Solution :
Let the radius of the smaller ball be r.
Given : Radius of each smaller ball = 1/4 Radius of original ball.
r = ¼ × Radius of original ball.
Radius of original ball, r1 = 4r
Volume of original ball ,V1 = 4/3 πr1³
Volume of original ball ,V1 = 4/3 π(4r)³
Volume of Smaller ball , V2= 4/3 π(r)³
Number of balls = Volume of original ball / Volume of Smaller ball
Number of balls = 4/3 π(4r)³/ 4/3 π(r)³
= 64r³/r³ = 64 / 1
Number of balls = 64
Surface area of sphere = 4πr²
Surface area of original ball (S1) = 4π(4r1)²
Surface area of each Smaller ball ,(S2) = 4πr²
Total surface area of 64 small balls , S2 = 64 × 4πr²
Total surface area of 64 small balls / Surface area of original ball = 64 × 4πr² / 4π(4r)²
S2/ S1 = (64 × 4πr²) / (4π × 16r²)
S2/ S1 = 64/16 = 4
S2/ S1 = 4
S2 : S1 = 4 : 1
S2 = 4S
Hence, Number of balls made is 64 and the total surface area of small balls is equal to 4 times the surface area of the original ball.
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