Question

Two wires \(W_1\) and \(W_2\) of the same material and equal radius are stretched by equal forces well within their elastic limit. If the lengths of wires \(W_1\) and \(W_2\) are in the ratio \(1:3\), then what will be the ratio of strains produced in wires \(W_1\) and \(W_2\)?

• A. \(1:1\)
• B. \(1:3\)
• C. \(3:1\)
• D. \(1:6\)

Hint:

For wires of the same material and same cross-sectional area subjected to equal forces: Stress is the same and therefore \[ \text{Strain is also the same}. \] The original lengths do not affect the strain in this case.

Answer 1

The Correct Option is A

Concept: Hooke's law gives \[ Y = \frac{\text{Stress}}{\text{Strain}}. \] Hence, \[ \text{Strain} = \frac{\text{Stress}}{Y}. \]

Step 1: Compare the stresses in the two wires. The wires have \[ \text{same material} \Rightarrow Y=\text{same}, \] and \[ \text{same radius} \Rightarrow A=\text{same}. \] Also the applied forces are equal. Therefore, \[ \text{Stress} = \frac{F}{A} \] is the same for both wires.

Step 2: Compare the strains. Since \[ \text{Strain} = \frac{\text{Stress}}{Y}, \] and both stress and Young's modulus are identical, \[ \text{Strain}_1 = \text{Strain}_2. \] Hence, \[ \text{Strain ratio} = 1:1. \] \[\begin{aligned} \boxed{1:1} \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.