Please help understand scaling where 8 liquid spheres coalesce to form one large one?

[匿名]

2010-10-25 8:01 pm

Eight little spheres of mercury coalesce to form a single sphere. Compared to the combined surface areas of the eight little spheres, the surface area of the big sphere is:
A) one half
B) double
C) one quarter
D) the same
E) one eighth

The answer is A) and I'm wondering how that figure came about. The Surface Area of a sphere is 4πr², so the surface area of 8 spheres would be 8*4πr² or 32πr²

I'm having a hard time understanding how that surface area is the same as 16πr² which would be half the surface area of the of the 8 small spheres that coalesce into one larger one.

Any help to understand this principle of scaling is greatly appreciated! Thanks in advance!

回答 (1)

Randy P

2010-10-25 8:07 pm

✔ 最佳答案

You're right that the total surface area of the little spheres is 32 pi r^2.

The volumes add up. The total volume is 8 times the volume of each sphere. So the radius of the big sphere is 8^(1/3) or 2 times the radius of the little spheres. The surface area of the big sphere is therefore 4*pi*(2r)^2 = 16 pi r^2.

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