The specific resistance `rho` of a circular wire of radius `r`, resistance `R`, and length `l` is given by `rho = pi r^(2) R//l`. Given : ` r = 0.24 +- 0.02 cm , R = 30 +- 1 Omega` , and `l = 4.80 +- 0.01 cm`. The percentage error in `rho` is nearly

Options

A. `7 %`
B. `9%`
C. `13 %`
D. `20%`

Explanation

To find the percentage error in the specific resistance \( \rho \) of a circular wire, we start with the formula given:

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

Where:
- \( r \) is the radius of the wire,
- \( R \) is the resistance,
- \( l \) is the length of the wire.

### Step 1: Identify the variables and their uncertainties
We have:
- \( r = 0.24 \pm 0.02 \, \text{cm} \)
- \( R = 30 \pm 1 \, \Omega \)
- \( l = 4.80 \pm 0.01 \, \text{cm} \)

### Step 2: Calculate the percentage error in \( r^2 \)
The formula for the percentage error in \( r^2 \) is given by:

\[ \text{Percentage error in } r^2 = 2 \times \frac{\Delta r}{r} \times 100 \]

Substituting the values:

\[ \Delta r = 0.02 \, \text{cm}, \quad r = 0.24 \, \text{cm} \]

\[ \text{Percentage error in } r^2 = 2 \times \frac{0.02}{0.24} \times 100 = 16.67\% \]

### Step 3: Calculate the percentage error in \( R \)
The percentage error in \( R \) is given by:

\[ \text{Percentage error in } R = \frac{\Delta R}{R} \times 100 \]

Substituting the values:

\[ \Delta R = 1 \, \Omega, \quad R = 30 \, \Omega \]

\[ \text{Percentage error in } R = \frac{1}{30} \times 100 = 3.33\% \]

### Step 4: Calculate the percentage error in \( l \)
The percentage error in \( l \) is given by:

\[ \text{Percentage error in } l = \frac{\Delta l}{l} \times 100 \]

Substituting the values:

\[ \Delta l = 0.01 \, \text{cm}, \quad l = 4.80 \, \text{cm} \]

\[ \text{Percentage error in } l = \frac{0.01}{4.80} \times 100 = 0.2083\% \]

### Step 5: Combine the percentage errors
The total percentage error in \( \rho \) can be calculated by adding the individual percentage errors:

\[ \text{Total percentage error in } \rho = \text{Percentage error in } r^2 + \text{Percentage error in } R + \text{Percentage error in } l \]

Substituting the values:

\[ \text{Total percentage error in } \rho = 16.67\% + 3.33\% + 0.2083\% \approx 20\% \]

### Conclusion
Thus, the percentage error in \( \rho \) is nearly \( 20\% \).