In the determination if Young's modulus `((Y = (4MLg)/(pild^(2))` by using searle's method, a wire of length `L=2m` and diameter `d=0.5mm` is used. For a load `M=2.5kg`, an extension `l=0.25mm` in the length of the wire is observed. Quantites `D and l` are measured using a screw gauge and a micrometer, respectively. they have the same pitch of `0.5mm`. the number of divisions on their circular scale is `100`. the contrubution to the maximum probable error of the `Y` measurement

Options

A. due to the error in the measurements of `d and l` are the same.
B. due to the error in the measurements of `d` is twice that due to the error in the measurement of `l`.
C. due to the error in the measurements of `l` is twice that due to the error in the measurement of `d`.
D. due to the error in the measurements of `d` is four times that due to the error in the measurement of `l`.

Explanation

To determine the contribution to the maximum probable error of Young's modulus \( Y \) in the given scenario, we will follow these steps:

### Step 1: Understand the formula for Young's modulus
The formula for Young's modulus is given by:
\[ Y = \frac{4MLg}{\pi d^2} \]
where:
- \( M \) = mass (in kg)
- \( L \) = original length of the wire (in m)
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( d \) = diameter of the wire (in m)

### Step 2: Identify the parameters and their values
From the problem:
- \( L = 2 \, \text{m} \)
- \( d = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} = 5 \times 10^{-4} \, \text{m} \)
- \( M = 2.5 \, \text{kg} \)
- Extension \( l = 0.25 \, \text{mm} = 0.25 \times 10^{-3} \, \text{m} = 2.5 \times 10^{-4} \, \text{m} \)

### Step 3: Calculate the possible errors in measurements
The error in measurement for both \( L \) and \( d \) can be calculated using the pitch and the number of divisions on the circular scale.

Given:
- Pitch = \( 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} \)
- Number of divisions = 100

The least count (LC) for both the screw gauge and micrometer is:
\[ \text{LC} = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{0.5 \times 10^{-3}}{100} = 5 \times 10^{-6} \, \text{m} \]

Thus, the possible errors in measurements \( \Delta L \) and \( \Delta d \) are:
\[ \Delta L = \Delta d = 5 \times 10^{-6} \, \text{m} \]

### Step 4: Calculate the percentage error in \( L \) and \( d \)
The percentage error in \( L \) is calculated as:
\[ \text{Percentage error in } L = \left( \frac{\Delta L}{L} \right) \times 100 = \left( \frac{5 \times 10^{-6}}{2} \right) \times 100 = 0.00025 \times 100 = 0.025\% \]

The percentage error in \( d \) is calculated as:
\[ \text{Percentage error in } d = \left( \frac{\Delta d}{d} \right) \times 100 = \left( \frac{5 \times 10^{-6}}{5 \times 10^{-4}} \right) \times 100 = 0.01\% \]

### Step 5: Calculate the contribution to the error in \( Y \)
The contributions to the error in \( Y \) are based on the powers of \( L \) and \( d \) in the formula for Young's modulus:
- The power of \( L \) in the formula is 1.
- The power of \( d \) in the formula is 2.

Thus, the contributions to the percentage error in \( Y \) are:
\[ \text{Contribution from } L = 1 \times \text{Percentage error in } L = 1 \times 0.025\% = 0.025\% \]
\[ \text{Contribution from } d = 2 \times \text{Percentage error in } d = 2 \times 0.01\% = 0.02\% \]

### Step 6: Total contribution to the maximum probable error in \( Y \)
Now, we sum the contributions:
\[ \text{Total contribution to error in } Y = 0.025\% + 0.02\% = 0.045\% \]

### Final Answer
The contribution to the maximum probable error of the Young's modulus \( Y \) measurement is approximately \( 0.045\% \).