A current of `(2.5 pm 0.05)` A flows through a wire and develops a potential difference of `(10 pm 0.07)` volt. Resistance of the wire in ohm, is

Options

A. `4 pm 0.108`
B. `4 pm 0.04`
C. `4 pm 0.1104`
D. `4 pm 0.02`

Answer

To find the resistance of the wire, we can use Ohm's Law, which states that:

\[ V = I \times R \]

Where:
- \( V \) is the potential difference (voltage),
- \( I \) is the current, and
- \( R \) is the resistance.

### Step 1: Rearranging Ohm's Law
From Ohm's Law, we can rearrange the equation to solve for resistance \( R \):

\[ R = \frac{V}{I} \]

### Step 2: Substituting the Values
Now we substitute the given values into the equation. The current \( I \) is \( 2.5 \pm 0.05 \) A and the potential difference \( V \) is \( 10 \pm 0.07 \) V.

\[ R = \frac{10 \, \text{V}}{2.5 \, \text{A}} \]

### Step 3: Calculating the Resistance
Now we perform the division:

\[ R = \frac{10}{2.5} = 4 \, \Omega \]

### Step 4: Calculating the Uncertainty in Resistance
To find the uncertainty in resistance \( \Delta R \), we can use the formula for the propagation of uncertainties. The formula for the relative uncertainty in \( R \) is:

\[ \frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I} \]

Where:
- \( \Delta V = 0.07 \) V (uncertainty in voltage),
- \( \Delta I = 0.05 \) A (uncertainty in current).

### Step 5: Substituting the Uncertainties
Now we substitute the uncertainties into the equation:

\[ \frac{\Delta R}{4} = \frac{0.07}{10} + \frac{0.05}{2.5} \]

### Step 6: Calculating Each Term
Calculating each term:

1. \( \frac{0.07}{10} = 0.007 \)
2. \( \frac{0.05}{2.5} = 0.02 \)

Now, adding these two values:

\[ \frac{\Delta R}{4} = 0.007 + 0.02 = 0.027 \]

### Step 7: Solving for \( \Delta R \)
Now we can find \( \Delta R \):

\[ \Delta R = 4 \times 0.027 = 0.108 \]

### Step 8: Final Result
Thus, the resistance of the wire with its uncertainty is:

\[ R \pm \Delta R = 4 \pm 0.108 \, \Omega \]

### Final Answer
The resistance of the wire is:

\[ 4 \pm 0.11 \, \Omega \]

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