We need to analyze the relationship between load, elongation, Young's modulus, length, and cross-sectional area to determine the correct statements.
Step 1: Recall the Formula for Young's Modulus
Young's modulus \(Y\) is defined as: \[Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L_0}\] where:
We can rearrange this formula to express the elongation \(x\) as: \[x = \frac{F L_0}{A Y}\] or, equivalently, \[F = \frac{AY}{L_0} x\]
Step 2: Analyze the Graph
The graph plots \(F\) (Load) vs. \(x\) (elongation). This represents a linear relationship, and the slope of the graph is given by: \[\text{Slope} = \frac{F}{x} = \frac{AY}{L_0}\]
From the graph, we can observe that the slope of line \(A\) is greater than the slope of line \(B\). Therefore: \[\frac{A_A Y_A}{L_A} > \frac{A_B Y_B}{L_B}\] where \(A_A\) and \(A_B\) are the cross-sectional areas of wires \(A\) and \(B\), and \(Y_A\) and \(Y_B\) are their respective Young's moduli.
Step 3: Apply Given Conditions
We are given that the wires are made of the same material, which means \(Y_A = Y_B\). We are also given that the wires have the same length, so \(L_A = L_B\). Therefore, the inequality simplifies to: \[A_A > A_B\]
Step 4: Determine the Correct Statements
Conclusion
The correct statements are:
If the given graph shows the load (W) attached to and the elongation ($\Delta l$) produced in a wire of length 1 meter and cross-sectional area 1 mm$^2$, then the Young's modulus of the material of the wire is