Youngs Modulus Calculation from Extension Load Graph

To find the Young's modulus of the wire, we use the formula for Young's modulus ($Y$), which is defined as the ratio of tensile stress to tensile strain:

$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\left(\frac{W}{A}\right)}{\left(\frac{\Delta l}{L}\right)} = \frac{W \cdot L}{A \cdot \Delta l}$

where:

• $W$ is the load applied to the wire,
• $L$ is the original length of the wire,
• $A$ is the cross-sectional area of the wire,
• $\Delta l$ is the extension in the length of the wire.

From the given question:

• Length of the wire, $L = 1\text{ m}$
• Cross-sectional area, $A = 10^{-5}\text{ m}^2$

From the $\Delta l$ versus $W$ graph, taking any point on the straight line:

• At $W = 60\text{ N}$, the extension $\Delta l = 6 \times 10^{-4}\text{ m}$

Thus, the ratio $\frac{W}{\Delta l}$ is: $\frac{W}{\Delta l} = \frac{60\text{ N}}{6 \times 10^{-4}\text{ m}} = 10^5\text{ N/m}$

Substituting these values into the Young's modulus formula: $Y = \left(\frac{W}{\Delta l}\right) \cdot \frac{L}{A} = 10^5 \cdot \frac{1}{10^{-5}} = 10^{10}\text{ N/m}^2 = 1.0 \times 10^{10}\text{ N/m}^2$

Therefore, the Young's modulus of the wire is $1.0 \times 10^{10}\text{ N/m}^2$, which corresponds to Option C.