Question

A wire of length \(L\) and cross-sectional area \(A\) has a Young's modulus \(Y\). If the wire is stretched by a force \(F\), the elongation produced in the wire is:

• A. \( \frac{F}{A \cdot Y} \)
• B. \( \frac{F \cdot L}{A \cdot Y} \)
• C. \( \frac{A \cdot Y}{F} \)
• D. \( \frac{L}{A \cdot Y} \)

Hint:

To calculate the elongation of a wire under a force, use the formula: \( \Delta L = \frac{F \cdot L}{A \cdot Y} \). Remember, the elongation increases with more force or a longer wire but decreases with a larger cross-sectional area or a higher Young's modulus.

Answer 1

The Correct Option is B

The elongation (\( \Delta L \)) of a wire under a stretching force is governed by the Young's modulus (\( Y \)) of the material. Young's modulus is defined as the ratio of stress to strain. Stress is the force applied per unit area, and strain is the relative change in length. Mathematically, Young's modulus is given by: \[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\frac{F}{A}}{\frac{\Delta L}{L}} \] Where: - \( F \) is the force applied to the wire, - \( A \) is the cross-sectional area of the wire, - \( \Delta L \) is the change in length (elongation), - \( L \) is the original length of the wire, - \( Y \) is the Young's modulus of the material. Rearranging this formula to solve for the elongation \( \Delta L \), we get: \[ \frac{\Delta L}{L} = \frac{F}{A \cdot Y} \] Multiplying both sides by \( L \), we obtain the expression for elongation \( \Delta L \): \[ \Delta L = \frac{F \cdot L}{A \cdot Y} \] This formula shows that the elongation is: - Directly proportional to the force (\( F \)) applied and the original length (\( L \)), - Inversely proportional to the cross-sectional area (\( A \)) and the Young's modulus (\( Y \)). Thus, the correct answer is \( \frac{F \cdot L}{A \cdot Y} \).