A current of `(2.5+-0.05)` A flows through a wire and develops a potential difference of `(10+-0.1)` volt. Resistance of th wire in ohm, is

Options

A. `4+-0.12`
B. `4+-0.04`
C. `4+-0.08`
D. `4+-0.02`

Explanation

To find the resistance of the wire using the given current and potential difference, we can follow these steps:

### Step 1: Use Ohm's Law
Ohm's Law states that \( R = \frac{V}{I} \), where \( R \) is the resistance, \( V \) is the potential difference, and \( I \) is the current.

### Step 2: Substitute the values
Given:
- Current \( I = 2.5 \, \text{A} \) with uncertainty \( \Delta I = 0.05 \, \text{A} \)
- Potential difference \( V = 10 \, \text{V} \) with uncertainty \( \Delta V = 0.1 \, \text{V} \)

Now, substituting the values into the formula:
\[ R = \frac{10 \, \text{V}}{2.5 \, \text{A}} = 4 \, \Omega \]

### Step 3: Calculate the percentage uncertainties
To find the total uncertainty in resistance, we first need to calculate the percentage uncertainties in \( V \) and \( I \).

1. **Percentage uncertainty in \( V \)**:
\[ \frac{\Delta V}{V} \times 100 = \frac{0.1}{10} \times 100 = 1\% \]

2. **Percentage uncertainty in \( I \)**:
\[ \frac{\Delta I}{I} \times 100 = \frac{0.05}{2.5} \times 100 = 2\% \]

### Step 4: Combine the uncertainties
Since the uncertainties are independent, we can add the percentage uncertainties to find the total percentage uncertainty in \( R \):
\[ \text{Total percentage uncertainty} = 1\% + 2\% = 3\% \]

### Step 5: Calculate the absolute uncertainty in \( R \)
Now, we can find the absolute uncertainty in \( R \):
\[ \Delta R = \frac{3}{100} \times 4 = 0.12 \, \Omega \]

### Step 6: Write the final result
Thus, the resistance of the wire is:
\[ R = 4 \pm 0.12 \, \Omega \]

---