A hollow metallic sphere has outer radius 10 cm and is melted to make 50 smaller solid spheres of radius 2 cm. What is the inner radius of the original sphere?

Correct — A.
Melting keeps the volume of metal the same.

Metal in the shell = (4⁄3)π(R³ − r³), with R = 10
Metal in 50 small spheres = 50 × (4⁄3)π × 2³

Cancel (4⁄3)π: 10³ − r³ = 50 × 8 = 400
r³ = 1000 − 400 = 600
r = ∛600 ≈ 8.43 cm → option (a)

Why the others are wrong

• (b)
  7.48 cm — Cube it: 7.48³ ≈ 418.5, leaving 1000 − 418.5 ≈ 581.5 for the shell (in units of (4⁄3)π). Fifty spheres of radius 2 cm need only 400.
• (c)
  5.46 cm — 5.46³ ≈ 162.8, leaving about 837 for the shell, more than double the 400 that fifty 2 cm spheres use. A cavity this small makes the wall far too thick.
• (d)
  6.44 cm — 6.44³ ≈ 267.1, leaving about 733 for the shell against the 400 needed. The metal is only 40% of a solid 10 cm ball, so the cavity must be larger.

When a solid is melted and recast, its shape changes but its volume does not. Set the volume before equal to the volume after.

A hollow sphere's metal is the full ball minus the cavity: (4⁄3)π(R³ − r³). Every sphere here carries the same (4⁄3)π, so it cancels and the question runs on cubes of radii: R³ − r³ = 50 × 2³.

Key facts

• Volume of a sphere = (4⁄3)πr³.
• Metal in a hollow sphere = (4⁄3)π(R³ − r³), with R the outer and r the inner radius.
• Melting and recasting conserves volume, not surface area.
• ∛600 ≈ 8.434, since 8.43³ ≈ 599.1 and 8.44³ ≈ 601.2.

Common traps

• Solving r³ = 400 instead of 10³ − r³ = 400: that gives r ≈ 7.37, not the inner radius.
• Multiplying 50 by the radius 2 instead of by its cube 8: volumes compare as r³, so the metal is 50 × 8 = 400.

No directly related past PYQ was found.