Question

A weight \( W \) is connected to one end of a wire and the other end is fixed, length of wire is 1m. Area of cross-section is \( 10^{-5} \, \text{m}^2 \). Graph between change in length and weight is shown. Then calculate Young's modulus.

• A. \( 10^{11} \, \text{N/m}^2 \)
• B. \( 10^9 \, \text{N/m}^2 \)
• C. \( 10^8 \, \text{N/m}^2 \)
• D. \( 10^{10} \, \text{N/m}^2 \)

Hint:

To calculate Young's modulus from a graph, find the slope of the graph (\( \Delta L \) vs. \( W \)), then apply the formula \( Y = \frac{F/A}{\Delta L/L} \).

Answer 1

The Correct Option is D

Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \] Where:
- \( F \) is the force (weight \( W \))
- \( A \) is the area of cross-section
- \( \Delta L \) is the change in length
- \( L \) is the original length of the wire
From the graph, we can find the slope (change in length \( \Delta L \) versus weight \( W \)). Step 1: Calculate the slope.
From the graph, the slope \( m \) is given by:
\[ m = \frac{\Delta L}{W} = \frac{2 \times 10^{-4}}{20} = 10^{-5} \, \text{m/N} \]
Step 2: Calculate Young's modulus.
Using the formula for Young's modulus:
\[ Y = \frac{F/A}{\Delta L/L} = \frac{W/A}{\Delta L/L} \] We know:
- \( W = 20 \, \text{N} \)
- \( A = 10^{-5} \, \text{m}^2 \)
- \( L = 1 \, \text{m} \)
- \( \Delta L = 2 \times 10^{-4} \, \text{m} \)
So: \[ Y = \frac{20/10^{-5}}{2 \times 10^{-4}} = \frac{2 \times 10^6}{2 \times 10^{-4}} = 10^{10} \, \text{N/m}^2 \] Final Answer: \( 10^{10} \, \text{N/m}^2 \)