Find the percentage error in specific resistance given by `rho = (pir^2R)/(l)` where r is the radius having value `(0.2+-0.02)`cm, R is the resistance of `(60+-2) ohm` and l is the length of `(150+-0.1)`cm.

Explanation

To find the percentage error in specific resistance (or resistivity) given by the formula:

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

where:
- \( r \) is the radius with a value of \( 0.2 \pm 0.02 \) cm,
- \( R \) is the resistance with a value of \( 60 \pm 2 \) ohm,
- \( l \) is the length with a value of \( 150 \pm 0.1 \) cm.

We will follow these steps:

### Step 1: Identify the formula for percentage error
The percentage error in a quantity can be calculated using the formula:

\[ \text{Percentage Error} = \frac{\Delta \rho}{\rho} \times 100 \]

where \( \Delta \rho \) is the absolute error in \( \rho \).

### Step 2: Determine the contributions to the error
Using the formula for \( \rho \), we can find the contributions to the error from each variable:

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

### Step 3: Calculate the individual errors
1. **For radius \( r \)**:
- \( r = 0.2 \) cm, \( \Delta r = 0.02 \) cm
- Contribution to error:
\[ 2 \frac{\Delta r}{r} = 2 \times \frac{0.02}{0.2} = 2 \times 0.1 = 0.2 \text{ or } 20\% \]

2. **For resistance \( R \)**:
- \( R = 60 \) ohm, \( \Delta R = 2 \) ohm
- Contribution to error:
\[ \frac{\Delta R}{R} = \frac{2}{60} \approx 0.0333 \text{ or } 3.33\% \]

3. **For length \( l \)**:
- \( l = 150 \) cm, \( \Delta l = 0.1 \) cm
- Contribution to error:
\[ \frac{\Delta l}{l} = \frac{0.1}{150} \approx 0.00067 \text{ or } 0.067\% \]

### Step 4: Sum the contributions to find total percentage error
Now, we can sum the contributions to find the total percentage error in \( \rho \):

\[ \text{Total Percentage Error} = 20\% + 3.33\% + 0.067\% \approx 23.397\% \]

### Step 5: Round the result
Finally, rounding to two decimal places, we get:

\[ \text{Total Percentage Error} \approx 23.40\% \]

Thus, the percentage error in specific resistance is approximately **23.40%**.

---