Question

The force required to increase the length of a thin copper wire of cross-sectional area \(0.1\ \text{cm}^2\) by \(0.1\%\) is (Young’s modulus of copper is \(11 \times 10^{10}\ \text{N m}^{-2}\))

• A. $550\ \text{N}$
• B. $11 \times 10^4\ \text{N}$
• C. $10.5 \times 10^3\ \text{N}$
• D. $1100\ \text{N}$
• E. $5.5 \times 10^3\ \text{N}$

Hint:

In elasticity problems: - Always convert area to $\text{m}^2$ - Percentage strain = $\frac{\text{value}}{100}$

Answer 1

The Correct Option is D

Concept: Young’s modulus: \[ Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L} \] \[ \Rightarrow F = Y \cdot A \cdot \frac{\Delta L}{L} \]

Step 1: Convert given quantities.
\[ A = 0.1\ \text{cm}^2 = 0.1 \times 10^{-4} = 10^{-5}\ \text{m}^2 \] \[ \frac{\Delta L}{L} = 0.1\% = \frac{0.1}{100} = 10^{-3} \]

Step 2: Substitute values.
\[ F = (11 \times 10^{10})(10^{-5})(10^{-3}) \]

Step 3: Simplify.
\[ F = 11 \times 10^{2} = 1100\ \text{N} \]