A cone and a sphere have equal volumes. If the cone’s height is twice its radius, what is the ratio of the cone’s radius to the sphere’s radius?

Correct — A.
Cone, with h = 2r: V = (1⁄3)πr² × 2r = (2⁄3)πr³
Sphere of radius R: V = (4⁄3)πR³
Set equal and cancel π: (2⁄3)r³ = (4⁄3)R³

Multiply both sides by 3⁄2: r³ = 2R³
Take cube roots: r ⁄ R = ∛2, so r : R = ∛2 : 1 → option (a)

Why the others are wrong

• (b)
  ∛4 : 1 — ∛4 : 1 is the answer if the cone's height equalled its radius: (1⁄3)πr³ = (4⁄3)πR³ gives r³ = 4R³. Here h = 2r doubles the cone's volume, so r³ halves to 2R³.
• (c)
  ∛6 : 1 — ∛6 : 1 needs r³ = 6R³, which is what you get by writing the sphere's volume as 4πR³ and dropping its 1⁄3. With (4⁄3)πR³ the result is r³ = 2R³.
• (d)
  ∛9 : 1 — ∛9 : 1 needs r³ = 9R³. Equal volumes give (2⁄3)r³ = (4⁄3)R³, so r³ is only 2R³.

When two solids have equal volume, write each volume in its own radius, set them equal and cancel what is common (π here).

The stem ties the cone's height to its radius (h = 2r), so the cone's volume becomes a pure r³ term. Both sides are then cubes, and a cube root turns the volume ratio into a radius ratio.

Key facts

• Volume of a cone = (1⁄3)πr²h.
• Volume of a sphere = (4⁄3)πr³.
• If r³ = k × R³, then r : R = ∛k : 1.
• A cone with h = 2r has volume (2⁄3)πr³, half that of a sphere of the same radius.

Common traps

• Using h = r instead of h = 2r, which gives ∛4 : 1.
• Writing the sphere's volume as 4πR³ (dropping the 1⁄3), which gives ∛6 : 1.
• Inverting the ratio: the question asks cone radius to sphere radius, and the cone's is the larger.

No directly related past PYQ was found.