Question

Work done in stretching a wire of length L, area A, Young's modulus Y by x is:

• A. YAxL
• B. YAx^2L
• C. YAx^22L
• D. 2YAx^2L

Hint:

Elastic work done is always: \[ W = \frac{1}{2} \times \text{Force} \times \text{extension} \]

Answer 1

The Correct Option is C

Step 1: Understand the situation.
A wire of length $L$ and area of cross section $A$ is stretched by an amount $x$. The material has Young's modulus $Y$. We want the work done in stretching it.

Step 2: Find the stretching force.
Young's modulus links stress and strain. From it, the force needed to stretch the wire by a small amount $x'$ is: \[ F = \frac{Y A x'}{L} \] So the force is not constant. It grows as the wire stretches more.

Step 3: Why we must integrate.
Because the force keeps increasing while we stretch, we cannot just multiply force by distance. We add up the work over each tiny stretch using integration.

Step 4: Set up the work integral.
Work is force times small displacement, added up from $0$ to $x$: \[ W = \int_0^x \frac{Y A x'}{L}\,dx' \]

Step 5: Carry out the integration.
Pull the constants out and integrate $x'$: \[ W = \frac{Y A}{L}\int_0^x x'\,dx' = \frac{Y A}{L}\cdot \frac{x^2}{2} \]

Step 6: Write the final result.
Combining the terms gives: \[ W = \frac{Y A x^2}{2L} \] which matches the third option. \[ \boxed{\dfrac{Y A x^{2}}{2L}} \]