Question

A wire of length 2 m and cross-sectional area \(1 \, \text{mm}^2\) is stretched by a force of 10 N. If Young's modulus of the material is \(2 \times 10^{11} \, \text{N/m}^2\), what is the extension in the wire?

• A. 0.01 mm
• B. 1 mm
• C. 0.1 mm
• D. 2 mm

Hint:

Remember to convert units carefully in elasticity problems: - \(1 \, \text{mm}^2 = 10^{-6} \, \text{m}^2\)
- Extension is often very small, so convert to millimeters after calculation.

Answer 1

The Correct Option is C

Step 1: Write down the known values 
Length, \( L = 2\, \text{m} \) 
Cross-sectional area, \( A = 1\, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \) 
Force, \( F = 10\, \text{N} \) 
Young's modulus, \( Y = 2 \times 10^{11} \, \text{N/m}^2 \) 
Step 2: Recall the formula for extension \(\Delta L\) 
\[ Y = \frac{F L}{A \Delta L} \quad \Rightarrow \quad \Delta L = \frac{F L}{A Y} \] Step 3: Substitute the values 
\[ \Delta L = \frac{10 \times 2}{1 \times 10^{-6} \times 2 \times 10^{11}} = \frac{20}{2 \times 10^{5}} = 1 \times 10^{-4} \, \text{m} \] Step 4: Convert meters to millimeters 
\[ 1 \times 10^{-4} \, \text{m} = 0.1 \, \text{mm} \] Wait, this contradicts the option selected as correct — let's verify carefully: Calculate: \[ \Delta L = \frac{F L}{A Y} = \frac{10 \times 2}{1 \times 10^{-6} \times 2 \times 10^{11}} = \frac{20}{2 \times 10^{5}} = 1 \times 10^{-4} \text{m} \] \(1 \times 10^{-4}\, \text{m} = 0.1\, \text{mm}\), which corresponds to option (1).
 Correct Answer: (1) 0.1 mm