A wire of radius 0 2 mm is extended by 0 5 of i... - JAMB Physics 2023 Question

Given values:
- Radius of the wire (\(r\)) = 0.2 mm = \(0.0002 \, m\) (convert to meters)
- Change in length (\(\Delta L\)) = 0.5% of the original length (\(L_0\))
- Load (\(F\)) = Weight (\(mg\)), where \(m\) is the mass and \(g\) is the acceleration due to gravity. Given \(g = 10 \, \text{m/s}^2\).

1. Convert radius to meters:
   \[ r = 0.0002 \, m \]

2. Calculate cross-sectional area (\(A\)) using the formula for the area of a circle:
   \[ A = \pi r^2 \]

3. Convert the percentage change in length to a decimal:
   \[ \text{Strain} = \frac{\Delta L}{L_0} = \frac{0.5\%}{100} \]

4. Calculate Young's Modulus (\(Y\)):
   \[ F = mg \]
   \[ Y = \frac{F/A}{\text{Strain}} \]

Let's plug in the values and calculate:

\[ r = 0.0002 \, \text{m} \]

\[ A = \pi \times (0.0002)^2 \]

\[ \text{Strain} = \frac{0.5\%}{100} \]

\[ F = 1.5 \, \text{kg} \times 10 \, \text{m/s}^2 \]

\[ Y = \frac{F/A}{\text{Strain}} \]

Now, calculate \(Y\) and compare it to the provided options.

The results of the calculations are as follows:

\[ A \approx 1.2566 \times 10^{-7} \, \text{m}^2 \]

\[ \text{Strain} = 0.005 \]

\[ F = 1.5 \, \text{kg} \times 10 \, \text{m/s}^2 = 15 \, \text{N} \]

\[ Y \approx \frac{15 \, \text{N} / 1.2566 \times 10^{-7} \, \text{m}^2}{0.005} \]

\[ Y \approx 2.3838 \times 10^{10} \, \text{N/m}^2 \]

Therefore, 

The Young's modulus for the material of the wire is approximately \(2.4 \times 10^{10} \, \text{N/m}^2\).