Question

The force required to stretch a steel wire of area of cross-section \(1 \ \text{mm}^2\) to double its length is
(Young’s modulus of steel \(= 2 \times 10^{11} \ \text{N/m}^2\))

• A. \(2 \times 10^3 \ \text{N}\)
• B. \(2 \times 10^5 \ \text{N}\)
• C. \(2 \times 10^2 \ \text{N}\)
• D. \(2 \times 10^4 \ \text{N}\)

Hint:

When the length is doubled, \( \Delta L = L \). Use \( F = YA \) directly since the \( L \) cancels out in the expression.

Answer 1

The Correct Option is B

Step 1: Use Young’s modulus formula
Young’s modulus \( Y = \dfrac{F L}{A \Delta L} \Rightarrow F = \dfrac{Y A \Delta L}{L} \) Given: The length is doubled, so \( \Delta L = L \), hence:
\[ F = \dfrac{Y A L}{L} = Y A \] Step 2: Substitute the values
\[ Y = 2 \times 10^{11} \ \text{N/m}^2, \quad A = 1 \ \text{mm}^2 = 1 \times 10^{-6} \ \text{m}^2 \]
\[ F = Y A = 2 \times 10^{11} \times 10^{-6} = 2 \times 10^5 \ \text{N} \]