Understanding Wire Extension Principles

The extension (\(\delta L\)) of a wire under tensile stress is determined by its material properties (Young's modulus, Y), original length (L), cross-sectional area (A), and the applied force (F). The relationship is given by:

\(\delta L = \frac{FL}{AY}\)

In this problem:

• The material is the same for both wires (Y is constant).
• The length is the same for both wires (L is constant).
• The load (and thus the force F) is the same for both wires (F is constant).

Therefore, the extension is inversely proportional to the cross-sectional area (A):

\(\delta L \propto \frac{1}{A}\)

Calculating the Extension Ratio

The cross-sectional area A of a wire is calculated using its diameter (d) or radius (r) as \(A = \pi r^2 = \frac{\pi d^2}{4}\). Since A depends on the square of the diameter, the extension (\(\delta L\)) is inversely proportional to the square of the diameter (\(d^2\)):

\(\delta L \propto \frac{1}{d^2}\)

Let \(d_A\) and \(d_B\) be the diameters of wire A and wire B, respectively. Let \(\delta L_A\) and \(\delta L_B\) be their extensions.

We are given that the diameter of wire A is twice that of wire B:

\(d_A = 2d_B\)

The ratio of the extensions can be written as:

\(\frac{\delta L_A}{\delta L_B} = \frac{d_B^2}{d_A^2}\)

Substitute \(d_A = 2d_B\) into the equation:

\(\frac{\delta L_A}{\delta L_B} = \frac{d_B^2}{(2d_B)^2} = \frac{d_B^2}{4d_B^2}\)

Simplify the expression:

\(\frac{\delta L_A}{\delta L_B} = \frac{1}{4}\)

Thus, the ratio of the extensions produced in wires A and B is 1:4.