The radius of a long cylindrical wire is reduced to one-third of its original value, while the volume of the metal remains the same. By what factor does the length of the wire increase?

Introduction / Context:
This question deals with the relationship between radius, length, and volume of a cylindrical wire. When the radius changes but the volume of metal remains constant, the length must change accordingly. The problem asks for the factor by which the length increases when the radius is reduced to one third of its original value.

Given Data / Assumptions:
• Original radius of the wire is r.
• New radius after drawing the wire thinner is r / 3.
• Original length is L.
• Volume of the wire before and after remains the same.
• The wire is approximated as a perfect cylinder.

Concept / Approach:
Volume of a cylinder is V = πr2L. With volume constant, any change in radius must be compensated by a change in length to keep πr2L unchanged. Let the new length be Lnew and new radius be r / 3. Equating original volume and new volume gives an equation relating L and Lnew, which can be solved to find the factor by which length changes.

Step-by-Step Solution:
Step 1: Original volume V1 = πr2L. Step 2: After drawing the wire, new radius rnew = r / 3 and new length Lnew is unknown. Step 3: New volume V2 = π(rnew)2Lnew = π(r / 3)2Lnew = π(r2 / 9)Lnew. Step 4: Since volume is conserved, V1 = V2. Step 5: So πr2L = π(r2 / 9)Lnew. Step 6: Cancel π and r2 from both sides to obtain L = (1 / 9)Lnew. Step 7: Thus Lnew = 9L, so the length becomes nine times the original length.

Verification / Alternative check:
Consider a numeric example. Suppose r = 3 units and L = 1 unit. Original volume V = π × 9 × 1 = 9π. New radius r / 3 = 1, and to preserve volume, π × 1 × Lnew must equal 9π, giving Lnew = 9 units. This numerical check confirms the ninefold increase in length.

Why Other Options Are Wrong:
• 1.5 times or 3 times are far too small to compensate for reducing radius to one third, because area depends on radius squared.
• 6 times would correspond to wrong algebra or confusing radius with diameter.

Common Pitfalls:
A common error is to assume length is inversely proportional to radius, whereas volume depends on radius squared, so length is inversely proportional to the square of the radius. Remember that halving the radius multiplies length by four for constant volume, and reducing radius to one third multiplies length by nine.

Final Answer:
The length of the wire increases by a factor of 9 times.