Question

The graph drawn between change in length and varying load w is a straight line starting from origin. Cross-sectional area of the wire is 10^–6 m². Slope of the curve is $\frac{10^{–4}}{20}$. Young's modulus will be -(Given : length of the rod = 1 m)

• A. 2 × 10¹¹ N/m²
• B. 10¹¹ N/m²
• C. 3 × 10^–11 N/m²
• D. None

Answer 1

Answer: (A)

2 × 10¹¹ N/m²

Answer 2

Dl = $\frac{\mathcal{l}}{yA}$ W

Where, $\frac{1}{yA}$ is slope of the graph.

Dl = $\frac{\mathcal{l}}{a}$ × $\frac{1}{slope}$ = $\frac{1}{10^{–6}}$ × $\frac{20}{10^{–4}}$ = 2 × 10¹¹ N/m²