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, we start with the formula for specific resistance:

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

Where:
- \( r \) = radius of the wire = \( 0.26 \pm 0.02 \) cm
- \( l \) = length of the wire = \( 156.0 \pm 0.1 \) cm
- \( R \) = resistance of the wire = \( 64 \pm 2 \, \Omega \)

### Step 1: Identify the powers of each variable in the formula

In the formula for specific resistance, we can see:
- \( r \) has a power of 2
- \( R \) has a power of 1
- \( l \) has a power of -1

### Step 2: Calculate the fractional errors for each variable

The fractional error in a quantity \( x \) is given by:

\[ \text{Fractional error} = \frac{\Delta x}{x} \]

Where \( \Delta x \) is the absolute error in \( x \).

1. **For \( r \)**:
\[ \Delta r = 0.02 \, \text{cm}, \quad r = 0.26 \, \text{cm} \]
\[ \text{Fractional error in } r = \frac{0.02}{0.26} \approx 0.0769 \]

2. **For \( R \)**:
\[ \Delta R = 2 \, \Omega, \quad R = 64 \, \Omega \]
\[ \text{Fractional error in } R = \frac{2}{64} = 0.03125 \]

3. **For \( l \)**:
\[ \Delta l = 0.1 \, \text{cm}, \quad l = 156.0 \, \text{cm} \]
\[ \text{Fractional error in } l = \frac{0.1}{156} \approx 0.000641 \]

### Step 3: Calculate the total percentage error in \( \rho \)

Using the formula for the percentage error in \( \rho \):

\[ \text{Percentage error in } \rho = \left(2 \times \text{Fractional error in } r + \text{Fractional error in } R - \text{Fractional error in } l\right) \times 100 \]

Substituting the values we calculated:

\[ \text{Percentage error in } \rho = \left(2 \times 0.0769 + 0.03125 + 0.000641\right) \times 100 \]

Calculating each term:

\[ 2 \times 0.0769 \approx 0.1538 \]
\[ 0.1538 + 0.03125 \approx 0.18505 \]
\[ 0.18505 + 0.000641 \approx 0.185691 \]

Now, multiplying by 100:

\[ \text{Percentage error in } \rho \approx 18.57\% \]

### Final Answer

The percentage error in specific resistance \( \rho \) is approximately **18.57%**.

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