Area & Volume

Thevolume of the original sphere
= (4/3) π r³

The volume of each small sphere
= (4/3) π r³ / n
= (4/3) π (r / ³√n)³

The radius of each small sphere
= r / ³√n

The original surface area
= 4 π r²

The new total surface area
= n x [4 π (r / ³√n)²]
= 4 π r² ³√n

The percentage increase in surface area
= [(4 π r² ³√n - 4 π r²) /4 π r²] x 100%
= ³√n- 1
= 100(³√n - 1)%

2014-08-05 08:14:01 補充：
The percentage increase in surface area is independent of r.

2014-08-05 08:45:49 補充：
兄台是英雄，小弟不是。

2014-08-06 00:27:27 補充：
你在說我嗎？怎麼我不知道。

小心！「聊天猜謎或其他非知識交流的行為」是檢舉的一個理由。哈哈！

重點：volume 不變。

A metal sphere of radius r has a volume of (4/3)πr³.

That is re-casted into n smaller spheres, suppose the radius is s.
Then, the total volume is n (4/3)πs³.

That is, (4/3)πr³ = n (4/3)πs³, or r³ = ns³

2014-08-05 08:08:56 補充：
The question wants to eliminate s, so we make s as the subject.
s³ = r³/n
s = ∛(r³/n) = r/∛n

Old surface area = 4πr²
New surface area = n 4πs² = n 4π (r/∛n)² = n 4πr²/n^(2/3) = 4πr² n^(1/3)

Thus, the percentage increase in surface area in terms of r and n is
　(New - Old)/(Old) × 100%

2014-08-05 08:11:32 補充：
= [4πr² n^(1/3) - 4πr²]/(4πr²) × 100%
= [n^(1/3) - 1] × 100% (Ans)

(Actually no need to be in terms of r. Only n is enough.)
(Think of a special case. When n = 1, nothing is change, it makes sense.)

2014-08-05 08:13:51 補充：
pingshek 兄，早晨～

英雄所見略同！

﹝｡◕‿◕｡◕‿◠｡﹞

2014-08-05 18:18:11 補充：
pingshek 兄，當日你幫助、遷就在下，再加入知識友，可見閣下的胸襟和風度。

閣下用心解答，回覆精準，對大家十分有幫助！

我是很感謝和欣賞你的！

☆ヾ(◕‿◕)ノ

2014-08-06 16:35:11 補充：
pingshek：對呀，是在說您。

謝謝提點，惡意檢舉的人的確很麻煩。

但我們也是在交流知識，這是禮儀文化的知識交流，正正是現今青少年最需要的。

唉我真係好想答架~但我竟然連f2的Maths都唔識做。好自卑!

2014-08-05 06:24:33 補充：
可否贊助你2點?我覺得答到嗰個好勁呀!