Question

A wire of length \(100 \text{ cm}\) and area of cross-section \(2 \text{ mm}^2\) is stretched by two forces of each \(440 \text{ N}\) applied at the ends of the wire in opposite directions along the length of the wire. If the elongation of the wire is \(2 \text{ mm}\), the Young’s modulus of the material of the wire is:

• A. \( 4.4 \times 10^{11} \text{ Nm}^{-2} \)
• B. \( 1.1 \times 10^{11} \text{ Nm}^{-2} \)
• C. \( 2.2 \times 10^{11} \text{ Nm}^{-2} \)
• D. \( 3.3 \times 10^{11} \text{ Nm}^{-2} \)

Hint:

Young’s modulus is a measure of the stiffness of a material. It is given by: \[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F L}{A \Delta L} \] where stress is \( \frac{F}{A} \) and strain is \( \frac{\Delta L}{L} \).

Answer 1

The Correct Option is B

Step 1: Understanding Young’s modulus formula Young’s modulus \( Y \) is given by: \[ Y = \frac{F L}{A \Delta L} \] where: - \( F = 440 \text{ N} \) (applied force),
- \( L = 100 \text{ cm} = 1 \text{ m} \) (original length),
- \( A = 2 \text{ mm}^2 = 2 \times 10^{-6} \text{ m}^2 \) (cross-sectional area),
- \( \Delta L = 2 \text{ mm} = 2 \times 10^{-3} \text{ m} \) (elongation).
Step 2: Substituting the values
\[ Y = \frac{440 \times 1}{(2 \times 10^{-6}) \times (2 \times 10^{-3})} \] Step 3: Simplifying the expression \[ Y = \frac{440}{4 \times 10^{-9}} \] \[ Y = 1.1 \times 10^{11} \text{ Nm}^{-2} \] Thus, the correct answer is option (B) \(1.1 \times 10^{11} \text{ Nm}^{-2}\).