Figure represents the extension (\(\Delta l\)) of a wire of length \(l\), suspended from the ceiling of the room at one end and with a load \(W\) connected to the other end. If the cross-sectional area of the wire is \(10^{-5}\,\text{m}^2\) then the Young’s modulus of the wire is _____ N/m\(^2\).
Step 1: Understanding the Question:
The task is to find the Young's modulus (\(Y\)) of a material based on its dimensions and a load vs. extension graph. Step 2: Key Formula or Approach:
Young's modulus is given by:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{W/A}{\Delta l/L} = \frac{W \cdot L}{A \cdot \Delta l} \]
From the graph, the slope of the line (\(W\) vs \(\Delta l\)) is \(\frac{W}{\Delta l}\). Step 3: Detailed Explanation:
Given:
Original length \(L = 1 \text{ m}\)
Area of cross-section \(A = 10^{-5} \text{ m}^2\)
From the provided graph, we pick a point on the line: at \(W = 50 \text{ N}\), the extension is \(\Delta l = 5 \times 10^{-4} \text{ m}\) (noting the y-axis scale).
Now, calculate Young's modulus:
\[ Y = \frac{50 \times 1}{10^{-5} \times 5 \times 10^{-4}} \]
\[ Y = \frac{50}{5 \times 10^{-9}} \]
\[ Y = 10 \times 10^9 = 1.0 \times 10^{10} \text{ N/m}^2 \] Step 4: Final Answer:
The value of Young's modulus is \(1.0 \times 10^{10} \text{ N/m}^2\).
