A solid copper sphere has diameter 36 cm and is drawn into a uniform cylindrical wire of diameter 4 mm (0.4 cm). Assuming no loss of material, find the length of the wire (in metres).

Introduction / Context:When a solid is reshaped without loss, its volume remains constant. Here, a copper sphere is redrawn into a thin cylindrical wire; therefore, sphere volume = wire volume. The task is to compute the wire length from these equal volumes.

Given Data / Assumptions:

• Sphere diameter = 36 cm ⇒ radius r_s = 18 cm.
• Wire diameter = 4 mm = 0.4 cm ⇒ radius r_w = 0.2 cm.
• No loss of material; volumes are equal.
• Answer required in metres.

Concept / Approach:

• Volume of sphere: V_s = (4/3)*π*r_s^3.
• Volume of cylinder (wire): V_w = π*r_w^2*L.
• Set V_s = V_w and solve for length L, then convert cm → m.

Step-by-Step Solution:

Verification / Alternative check:Quick ratio check: Length ∝ r_s^3 / r_w^2. With r_s = 18 and r_w = 0.2, length scale is huge; a result in the order of 2 km is reasonable, matching 1944 m.

Why Other Options Are Wrong:

• 2000 m: Rounds too loosely; exact computation gives 1944 m.
• 15000 m: Exaggerates by ~7.7×; inconsistent with volume equality.
• 1855 m: Underestimates the exact ratio.
• None of these: Incorrect since 1944 m is attainable.

Common Pitfalls:

• Forgetting to convert 4 mm to 0.4 cm.
• Using diameter instead of radius in formulas.
• Failing to convert final length from cm to m.

Final Answer:1944 m