`R=(65 +- 1) ohm` , `l=(5+-0.1)mm` and
`d=(10+-0.5)mm`.Find error in claculation of resistivity.

Options

A. 0.21
B. 0.13
C. 0.16
D. 0.41

Answer

To find the error in the calculation of resistivity given the resistance, length, and diameter with their respective uncertainties, we can follow these steps:

### Step 1: Understand the formula for resistivity
The resistivity \( \rho \) can be expressed in terms of resistance \( R \), length \( L \), and diameter \( D \):
\[ \rho = \frac{R \cdot \pi \left(\frac{D}{2}\right)^2}{4L} \]
This can be simplified to:
\[ \rho = \frac{\pi D^2 R}{16L} \]

### Step 2: Identify the variables and their uncertainties
Given:
- \( R = 65 \pm 1 \, \Omega \)
- \( L = 5 \pm 0.1 \, \text{mm} \) (convert to meters: \( 5 \, \text{mm} = 0.005 \, \text{m} \))
- \( D = 10 \pm 0.5 \, \text{mm} \) (convert to meters: \( 10 \, \text{mm} = 0.01 \, \text{m} \))

### Step 3: Calculate the relative error in resistivity
The relative error in resistivity can be calculated using the formula:
\[ \frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \cdot \frac{\Delta D}{D} + \frac{\Delta L}{L} \]

### Step 4: Substitute the values into the formula
1. Calculate \( \Delta R/R \):
\[ \frac{\Delta R}{R} = \frac{1}{65} \approx 0.01538 \]

2. Calculate \( \Delta D/D \):
\[ \frac{\Delta D}{D} = \frac{0.5}{10} = 0.05 \]
Therefore, \( 2 \cdot \frac{\Delta D}{D} = 2 \cdot 0.05 = 0.1 \)

3. Calculate \( \Delta L/L \):
\[ \frac{\Delta L}{L} = \frac{0.1}{5} = 0.02 \]

### Step 5: Combine the relative errors
Now, sum the relative errors:
\[ \frac{\Delta \rho}{\rho} = 0.01538 + 0.1 + 0.02 = 0.13538 \approx 0.135 \]

### Step 6: Conclusion
The relative error in the resistivity calculation is approximately \( 0.135 \) or \( 13.5\% \).