Lead balls from a larger sphere — count by volume ratio: How many solid lead balls of diameter 2 cm can be made by melting a sphere of diameter 16 cm (no loss of material)?

Introduction / Context:
When a large sphere is recast into smaller spheres, the number produced equals the ratio of the large volume to the small volume. Because sphere volume is proportional to the cube of radius (or diameter), the count reduces to a simple cube ratio of diameters.

Given Data / Assumptions:

• Large sphere diameter D_L = 16 cm ⇒ radius R_L = 8 cm.
• Small ball diameter D_s = 2 cm ⇒ radius r_s = 1 cm.
• No wastage or gaps; exact volume conservation.

Concept / Approach:
Number N = V_L / V_s = ( (4/3)πR_L^3 ) / ( (4/3)πr_s^3 ) = (R_L / r_s)^3 = (8/1)^3.

Step-by-Step Solution:
N = (8)^3 = 512

Verification / Alternative check:
Compute actual volumes: V_L = (4/3)π*512 = (2048/3)π; V_s = (4/3)π*1 = (4/3)π; their ratio is 512.

Why Other Options Are Wrong:
2048, 2055, 2058 are far larger than the correct cube ratio; they appear to confuse volume with surface area or linear scaling.

Common Pitfalls:
Using diameter squared instead of cubed; forgetting to convert diameters to radii consistently (though it cancels here).

Final Answer:
512 (not listed) ⇒ None of the above