Calculate the percentage error in specific resistance , `rho = pi r^(2) R // l `, where r = radius of wire `= 0.26 +- 0.02 cm` , l = length of wire `= 156.0 +- 0.1 cm`, and R = resistance of wire `= 64 +- 2 Omega`.

Explanation

To calculate the percentage error in specific resistance (ρ), we will follow these steps:

### Step 1: Understand the formula for specific resistance
The specific resistance (ρ) is given by the formula:
\[ \rho = \frac{\pi r^2 R}{l} \]
where:
- \( r \) = radius of the wire
- \( R \) = resistance of the wire
- \( l \) = length of the wire

### Step 2: Identify the values and their uncertainties
We are given:
- \( r = 0.26 \pm 0.02 \) cm
- \( l = 156.0 \pm 0.1 \) cm
- \( R = 64 \pm 2 \) Ω

### Step 3: Calculate the percentage error formula
The percentage error in ρ can be calculated using the formula for propagation of uncertainty:
\[ \text{Percentage Error} = \left( \frac{\Delta \rho}{\rho} \right) \times 100 \]
Where:
- \(\Delta \rho\) is the total uncertainty in ρ.

Since ρ depends on \( r^2 \), \( R \), and \( l \), we can express the percentage error as:
\[ \frac{\Delta \rho}{\rho} = 2 \frac{\Delta r}{r} + \frac{\Delta R}{R} + \frac{\Delta l}{l} \]

### Step 4: Substitute the values into the error formula
Now we substitute the values into the formula:
- \(\Delta r = 0.02\) cm, \( r = 0.26\) cm
- \(\Delta R = 2\) Ω, \( R = 64\) Ω
- \(\Delta l = 0.1\) cm, \( l = 156.0\) cm

Calculating each term:
1. For \( r \):
\[ 2 \frac{\Delta r}{r} = 2 \times \frac{0.02}{0.26} \approx 0.1538 \]

2. For \( R \):
\[ \frac{\Delta R}{R} = \frac{2}{64} = 0.03125 \]

3. For \( l \):
\[ \frac{\Delta l}{l} = \frac{0.1}{156.0} \approx 0.000641 \]

### Step 5: Combine the errors
Now, we add these percentage errors together:
\[ \frac{\Delta \rho}{\rho} = 0.1538 + 0.03125 + 0.000641 \approx 0.185691 \]

### Step 6: Convert to percentage
Now, multiply by 100 to get the percentage error:
\[ \text{Percentage Error} = 0.185691 \times 100 \approx 18.57\% \]

### Step 7: Round off the result
Rounding off to three significant figures, we get:
\[ \text{Percentage Error} \approx 18.6\% \]

### Final Answer
The percentage error in specific resistance is approximately **18.6%**.

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