A wire of length 6 m is stretched such that its radius is reduced by 20%. Which one of the following is the value of increase in its length ?

When a wire is stretched, its volume remains constant. The volume of a cylindrical wire is given by V = A × L, where A is the cross-sectional area (πr²) and L is the length. If the radius (r) is reduced by 20%, the new radius (r') becomes 0.8r. Consequently, the new cross-sectional area (A') becomes π(0.8r)² = 0.64A. Since volume is conserved (V = A × L = A' × L'), the new length (L') is calculated as L' = (A / A') × L = (A / 0.64A) × L = 1.5625L. The increase in length is L' - L = 1.5625L - L = 0.5625L. Expressed as a percentage, the increase is 0.5625 × 100 = 56.25%. This principle of volume conservation is fundamental in material science and electrical resistance calculations.