A sphere is melted and recast into 8 identical cones, each with radius 3 cm and height 4 cm. What was the radius of the original sphere?

Correct — A. Melting keeps the volume, so the sphere holds exactly what the eight cones hold.

One cone = (1⁄3)πr²h = (1⁄3)π × 9 × 4 = 12π cm³
Eight cones = 8 × 12π = 96π cm³

Sphere: (4⁄3)πR³ = 96π
R³ = 96 × 3⁄4 = 72

R = ∛72 ≈ 4.16 cm, since 4.16³ ≈ 71.99 → option (a)

Why the others are wrong

• (b)
  5.28 cm — 5.28 cm cubes to about 147, roughly double the 72 the cones' volume allows. A sphere that size would hold about 196π cm³, not 96π.
• (c)
  6.54 cm — 6.54 cm cubes to about 280, nearly four-fold the 72 needed. Test any radius by cubing it: R³ must come to 72.
• (d)
  7 .56cm — 7 .56cm, read as 7.56 cm, cubes to about 432, which is 6 × 72. That sphere would hold the volume of 48 such cones, not 8.

Volume is conserved when a solid is melted and recast: total volume before = total volume after, whatever the shapes.

Here the cones' total is 8 × (1⁄3)π(3²)(4) = 96π. Setting (4⁄3)πR³ equal to it, π cancels, leaving R³ = 72.

The cube root of 72 is not a whole number, so the options are decimals: 4³ = 64 and 4.2³ ≈ 74.09 bracket it.

Key facts

• Volume of a cone = (1⁄3)πr²h.
• Volume of a sphere = (4⁄3)πR³.
• ∛72 ≈ 4.16, because 4.16³ ≈ 71.99.

Common traps

• Dropping the 1⁄3 from the cone's volume, which makes R³ = 216 and R = 6 cm.
• Using one cone's 12π instead of all eight, which gives R³ = 9 and R ≈ 2.08 cm.

No directly related past PYQ was found.