Recasting spheres — A 3 cm diameter sphere is recast into three spheres of diameters 1.5 cm, 2 cm, and a third unknown. Find the third diameter.

Introduction / Context:
Volume is conserved when melting and recasting. For spheres, volume depends on the cube of radius: V = (4/3)πr^3. Work exactly to avoid rounding errors.

Given Data / Assumptions:

• Initial diameter = 3 cm ⇒ r0 = 1.5 cm
• Two new diameters: 1.5 cm (r1 = 0.75 cm) and 2 cm (r2 = 1 cm)
• Third radius r3 unknown

Concept / Approach:
Set total initial volume equal to sum of volumes of the three smaller spheres and solve for r3.

Step-by-Step Solution:
V0 = (4/3)π(1.5)^3 = (4/3)π * 3.375 = 4.5πV1 = (4/3)π(0.75)^3 = (4/3)π * 0.421875 = 0.5625πV2 = (4/3)π(1)^3 = 1.333...πV3 = V0 − (V1 + V2) = 4.5π − 1.895833...π = 2.604166...π = (125/48)π(4/3)π r3^3 = (125/48)π ⇒ r3^3 = 125/64 ⇒ r3 = 5/4 = 1.25 cm ⇒ diameter = 2.50 cm

Verification / Alternative check:
Exact fraction arithmetic confirms r3^3 = 125/64.

Why Other Options Are Wrong:
2.66 cm is a rounded overestimate; 2.25 cm is too small; 3 cm and 3.5 cm exceed available volume.

Common Pitfalls:
Using diameters directly (cube of diameter/2) or rounding π prematurely.

Final Answer:
2.50 cm