A metal sphere having a radius of 10 centimeters is melted down and molded into 8 identical smaller solid spheres. What is the ratio of the surface area of the original sphere to the total surface area of all 8 smaller spheres?

Correct — A. Melting keeps the volume, so find the small radius first.

Volume: 8 × (4⁄3)πr³ = (4⁄3)π × 10³
r³ = 1000 ÷ 8 = 125, so r = 5 cm

Original area: 4π × 10² = 400π
Eight small: 8 × 4π × 5² = 800π

Ratio: 400π : 800π = 1 : 2 → option (a)

Why the others are wrong

• (b)
  2:1 — 2 : 1 is the eight small spheres (800π) against the original (400π). The question names the original sphere first, so the ratio is 400π : 800π.
• (c)
  1:4 — 1 : 4 is one small sphere (100π) against the original (400π). The question compares the original with all eight small spheres together, 800π in total.
• (d)
  1:1 — 1 : 1 assumes melting keeps the surface area. Only the volume is kept, and eight small spheres expose more surface: 800π against 400π.

Melting and recasting conserves volume, not surface area. Equal volumes fix the small radius: eight spheres of radius r hold 8r³ in place of 10³, so r = 10 ÷ ∛8 = 5.

Area goes with r². Each small sphere has a quarter of the original area, and eight of them together have twice it. In general, recasting into n equal spheres multiplies the total area by ∛n.

Key facts

• Surface area of a sphere = 4πr², volume = (4⁄3)πr³.
• n equal spheres recast from one sphere of radius R each have radius R ÷ ∛n.
• Their total surface area is ∛n times the original: ∛8 = 2 here.

Common traps

• Assuming the total surface area stays the same because the amount of metal does.
• Dividing the radius by 8 instead of by ∛8 = 2.

No directly related past PYQ was found.