The specific resistance `rho` of a circular wire is given by `rho= pi r^(2)R//l` . Given: radius `r = 0.20 pm 0.02 cm` , resistance `R= 25 +1 Omega` , and length `1 = 5.00 pm 0.01 cm`. The percentage error in `rho` is ______ %.

Explanation

To find the percentage error in the specific resistance \( \rho \) given by the formula:

\[ \rho = \frac{\pi r^2 R}{l} \]

we will follow these steps:

### Step 1: Identify the variables and their uncertainties
- Radius \( r = 0.20 \pm 0.02 \, \text{cm} \)
- Resistance \( R = 25 \pm 1 \, \Omega \)
- Length \( l = 5.00 \pm 0.01 \, \text{cm} \)

### Step 2: Write the formula for percentage error
The percentage error in \( \rho \) can be calculated using the formula for propagation of uncertainty:

\[ \frac{\Delta \rho}{\rho} \times 100 = 2 \frac{\Delta r}{r} + \frac{\Delta R}{R} + \frac{\Delta l}{l} \]

Where:
- \( \Delta r \) is the uncertainty in radius
- \( \Delta R \) is the uncertainty in resistance
- \( \Delta l \) is the uncertainty in length

### Step 3: Calculate each term
1. **Calculate \( \frac{\Delta r}{r} \)**:
\[ \frac{\Delta r}{r} = \frac{0.02}{0.20} = 0.1 \]

2. **Calculate \( \frac{\Delta R}{R} \)**:
\[ \frac{\Delta R}{R} = \frac{1}{25} = 0.04 \]

3. **Calculate \( \frac{\Delta l}{l} \)**:
\[ \frac{\Delta l}{l} = \frac{0.01}{5.00} = 0.002 \]

### Step 4: Substitute values into the percentage error formula
Now substituting these values into the percentage error formula:

\[ \frac{\Delta \rho}{\rho} \times 100 = 2 \times 0.1 + 0.04 + 0.002 \]

Calculating this gives:

\[ \frac{\Delta \rho}{\rho} \times 100 = 0.2 + 0.04 + 0.002 = 0.242 \]

### Step 5: Convert to percentage
Thus, the percentage error in \( \rho \) is:

\[ \frac{\Delta \rho}{\rho} \times 100 = 24.2 \% \]

### Final Answer
The percentage error in \( \rho \) is **24.2%**.

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