A student performs an experiment to determine the Young's modulus of a wire, exactly`2m` long, by Searle's method. In a partcular reading, the student measures the extension in the length of the wire to be `0.8mm with an uncertainty of `+- 0.05mm` at a load of exactly `1.0kg`, the student also measures the diameter of the wire to be `04mm` with an uncertainty of `+-0.01mm`. Take `g=9.8m//s^(2)` (exact). the Young's modulus obtained from the reading is

Options

A. `( 2.0 +- 0.3) xx 10^(11) N m^(-2)`
B. `( 2.0 +- 0.2) xx 10^(11) N m^(-2)`
C. `( 2.0 +- 0.1) xx 10^(11) N m^(-2)`
D. `( 2.0 +- 0.5) xx 10^(11) N m^(-2)`

Explanation

To determine the Young's modulus of a wire using the given data, we will follow these steps:

### Step 1: Understand the formula for Young's Modulus
Young's modulus (Y) is defined as the ratio of stress to strain. The formula can be expressed as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
Where:
- Stress = \(\frac{F}{A}\)
- Strain = \(\frac{\Delta L}{L}\)

### Step 2: Calculate the Stress
First, we need to calculate the force (F) applied to the wire. The force can be calculated using the weight of the load:
\[ F = m \cdot g \]
Given:
- Mass \(m = 1.0 \, \text{kg}\)
- Acceleration due to gravity \(g = 9.8 \, \text{m/s}^2\)

Calculating the force:
\[ F = 1.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 9.8 \, \text{N} \]

Next, we need to calculate the cross-sectional area (A) of the wire. The diameter of the wire is given as \(0.4 \, \text{mm}\), which we convert to meters:
\[ d = 0.4 \, \text{mm} = 0.4 \times 10^{-3} \, \text{m} = 4 \times 10^{-4} \, \text{m} \]
The radius \(r\) is:
\[ r = \frac{d}{2} = \frac{4 \times 10^{-4}}{2} = 2 \times 10^{-4} \, \text{m} \]
Now, we can calculate the area:
\[ A = \pi r^2 = \pi (2 \times 10^{-4})^2 = \pi \times 4 \times 10^{-8} \approx 1.25664 \times 10^{-7} \, \text{m}^2 \]

Now we can calculate the stress:
\[ \text{Stress} = \frac{F}{A} = \frac{9.8 \, \text{N}}{1.25664 \times 10^{-7} \, \text{m}^2} \approx 7.80 \times 10^{7} \, \text{Pa} \]

### Step 3: Calculate the Strain
The original length of the wire \(L\) is given as \(2 \, \text{m}\), and the extension \(\Delta L\) is given as \(0.8 \, \text{mm}\):
\[ \Delta L = 0.8 \, \text{mm} = 0.8 \times 10^{-3} \, \text{m} = 8 \times 10^{-4} \, \text{m} \]
Now we can calculate the strain:
\[ \text{Strain} = \frac{\Delta L}{L} = \frac{8 \times 10^{-4} \, \text{m}}{2 \, \text{m}} = 4 \times 10^{-4} \]

### Step 4: Calculate Young's Modulus
Now we can substitute the values of stress and strain into the Young's modulus formula:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{7.80 \times 10^{7} \, \text{Pa}}{4 \times 10^{-4}} \approx 1.95 \times 10^{11} \, \text{Pa} \]

### Step 5: Calculate the Uncertainty in Young's Modulus
To find the uncertainty in Young's modulus, we will use the formula for fractional error:
\[ \frac{\Delta Y}{Y} = \frac{\Delta F}{F} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L} \]
Where:
- \(\Delta F = g \cdot \Delta m = 9.8 \cdot 0.05 \, \text{N}\)
- \(\Delta d = 0.01 \, \text{mm} = 0.01 \times 10^{-3} \, \text{m}\)
- \(\Delta L = 0.05 \, \text{mm} = 0.05 \times 10^{-3} \, \text{m}\)

Calculating each term:
1. \(\Delta F = 9.8 \times 0.05 = 0.49 \, \text{N}\)
2. \(\frac{\Delta F}{F} = \frac{0.49}{9.8} \approx 0.05\)
3. \(\frac{\Delta d}{d} = \frac{0.01 \times 10^{-3}}{0.4 \times 10^{-3}} = 0.025\)
4. \(\frac{\Delta L}{L} = \frac{0.05 \times 10^{-3}}{2} = 0.025\)

Now substituting into the fractional error formula:
\[ \frac{\Delta Y}{Y} = 0.05 + 2(0.025) + 0.025 = 0.125 \]

Now calculating the uncertainty in Young's modulus:
\[ \Delta Y = Y \cdot 0.125 = (1.95 \times 10^{11}) \cdot 0.125 \approx 2.44 \times 10^{10} \, \text{Pa} \]

### Final Result
Thus, the Young's modulus obtained from the readings is:
\[ Y \approx 1.95 \times 10^{11} \, \text{Pa} \pm 2.44 \times 10^{10} \, \text{Pa} \]