Question

The adjacent graph shows the extension (Dl) of a wire of length l m suspended from the top of a roof at one end and with a load W connected to the other end. If the cross-sectional area of the wire is 10^–6 m², calculate the Young’s modulus of the material of the wire –

• A. 2 × 10¹¹ N/m²
• B. 2 × 10^–11 N/m²
• C. 3 × 10¹² N/m²
• D. 2 × 10¹³ N/m²

Answer 1

Answer: (A)

2 × 10¹¹ N/m²

Answer 2

Dl =$\left( \frac{\mathcal{l}}{YA} \right)$.W

i.e., graph is a straight line passing through origin

(as shown in question also), the slope of which is $\frac{\mathcal{l}}{YA}$.

\ Slope = $\left( \frac{\mathcal{l}}{YA} \right)$

\ Y = $\left( \frac{\mathcal{l}}{A} \right)$ $\left( \frac{1}{slope} \right)$

= $\left( \frac{1.0}{10^{–6}} \right)$ $\frac{(80–20)}{(4–1) \times 10^{–4}}$ = 2.0 × 10¹¹ N/m².