In the experiment of Ohm's law, when potential difference of 10.0 V is applied, current measured is 1.00 A. If length of wire is found to be 10.0cm and diameter of wire 2.50 mm, then find maximum permissible percentage error in resistivity.

Explanation

To solve the problem of finding the maximum permissible percentage error in resistivity based on the given parameters, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Ohm's Law**:
Ohm's law states that \( V = I \cdot R \), where \( V \) is the potential difference, \( I \) is the current, and \( R \) is the resistance.

2. **Express Resistance in Terms of Resistivity**:
The resistance \( R \) can be expressed in terms of resistivity \( \rho \) as:
\[ R = \frac{\rho \cdot L}{A} \]
where \( L \) is the length of the wire and \( A \) is the cross-sectional area. For a wire with diameter \( D \):
\[ A = \frac{\pi D^2}{4} \]
Thus, we can rewrite the resistance as:
\[ R = \frac{\rho \cdot L}{\frac{\pi D^2}{4}} = \frac{4\rho L}{\pi D^2} \]

3. **Rearranging for Resistivity**:
Rearranging the above equation gives us:
\[ \rho = \frac{V \cdot \pi D^2}{4 \cdot I \cdot L} \]

4. **Identify the Given Values**:
- \( V = 10.0 \, \text{V} \)
- \( I = 1.00 \, \text{A} \)
- \( L = 10.0 \, \text{cm} = 0.1 \, \text{m} \)
- \( D = 2.50 \, \text{mm} = 0.0025 \, \text{m} \)

5. **Determine the Errors in Measurements**:
- The error in potential \( \Delta V = 0.1 \, \text{V} \)
- The error in current \( \Delta I = 0.01 \, \text{A} \)
- The error in length \( \Delta L = 0.1 \, \text{cm} = 0.001 \, \text{m} \)
- The error in diameter \( \Delta D = 0.01 \, \text{mm} = 0.00001 \, \text{m} \)

6. **Calculate the Relative Errors**:
The relative error in resistivity \( \rho \) can be expressed as:
\[ \frac{\Delta \rho}{\rho} = \frac{\Delta V}{V} + 2 \cdot \frac{\Delta D}{D} + \frac{\Delta I}{I} + \frac{\Delta L}{L} \]

7. **Substituting the Values**:
- Calculate \( \frac{\Delta V}{V} = \frac{0.1}{10} = 0.01 \)
- Calculate \( \frac{\Delta D}{D} = \frac{0.00001}{0.0025} = 0.004 \)
- Calculate \( \frac{\Delta I}{I} = \frac{0.01}{1} = 0.01 \)
- Calculate \( \frac{\Delta L}{L} = \frac{0.001}{0.1} = 0.01 \)

Now substituting these values into the relative error equation:
\[ \frac{\Delta \rho}{\rho} = 0.01 + 2 \cdot 0.004 + 0.01 + 0.01 \]
\[ \frac{\Delta \rho}{\rho} = 0.01 + 0.008 + 0.01 + 0.01 = 0.038 \]

8. **Convert to Percentage**:
To find the maximum permissible percentage error in resistivity:
\[ \text{Percentage Error} = \frac{\Delta \rho}{\rho} \times 100\% = 0.038 \times 100\% = 3.8\% \]

### Final Answer:
The maximum permissible percentage error in resistivity is **3.8%**.