Volume conservation – rod hammered into wire: A solid circular rod of diameter 2 cm and length 30 cm is drawn into a uniform wire of length 3 m (i.e., 300 cm). Assuming no loss of material, find the diameter of the wire (in cm).

Introduction / Context:
When a metal rod is drawn into a wire, the volume of the material remains constant (neglecting losses). For cylinders, volume equals πr^2h. Equating the initial and final cylinder volumes allows recovery of the unknown final radius/diameter from the change in length.

Given Data / Assumptions:

• Initial: diameter 2 cm → r1 = 1 cm; length L1 = 30 cm
• Final: length L2 = 3 m = 300 cm; radius r2 unknown; diameter = 2r2
• Volume conservation: πr1^2L1 = πr2^2L2

Concept / Approach:
Cancel π and solve for r2^2 = (r1^2 * L1) / L2. Then take square root to get r2 and multiply by 2 for the diameter.

Step-by-Step Solution:
r2^2 = (1^2 * 30) / 300 = 0.1r2 = √0.1 = 1/√10 cmDiameter = 2r2 = 2/√10 cm

Verification / Alternative check:
Check volumes: initial V1 = π * 1^2 * 30 = 30π; final V2 = π * (1/√10)^2 * 300 * (2^2?) (careful: radius used above) → using r2^2 = 0.1 gives V2 = π * 0.1 * 300 = 30π, matching V1.

Why Other Options Are Wrong:
2/10 cm = 0.2 cm and 1/10 cm = 0.1 cm do not satisfy volume equality; 1/√10 cm is the radius, not the diameter; 1/5 cm is inconsistent with the computed r2.

Common Pitfalls:
Mixing up diameter and radius; forgetting to convert 3 m to 300 cm; incorrectly applying √ to the ratio; failing to conserve volume.

Final Answer:
2/√10 cm