The relative error in resistivity of a material where
resistance = 1.05 + 0.01 W
diameter = 0.60 + 0.01 mm
length = 75.3 + 0.1 cm is

Options

A. `0.04`
B. `0.40`
C. `0.08`
D. `0.01`

Answer

To find the relative error in the resistivity of a material given the resistance, diameter, and length, we can follow these steps:

### Step 1: Understand the formula for resistivity
The resistivity (ρ) of a material is given by the formula:
\[ \rho = R \cdot \frac{\pi D^2}{4L} \]
where:
- \(R\) is the resistance,
- \(D\) is the diameter,
- \(L\) is the length.

### Step 2: Identify the given values and their uncertainties
From the problem, we have:
- Resistance \(R = 1.05 \, \Omega\) with an uncertainty of \(\Delta R = 0.01 \, \Omega\)
- Diameter \(D = 0.60 \, \text{mm} = 0.0006 \, \text{m}\) with an uncertainty of \(\Delta D = 0.01 \, \text{mm} = 0.00001 \, \text{m}\)
- Length \(L = 75.3 \, \text{cm} = 0.753 \, \text{m}\) with an uncertainty of \(\Delta L = 0.1 \, \text{cm} = 0.001 \, \text{m}\)

### 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 \(\frac{\Delta R}{R}\):
\[ \frac{\Delta R}{R} = \frac{0.01}{1.05} \approx 0.00952 \]

2. Calculate \(\frac{\Delta D}{D}\):
\[ \frac{\Delta D}{D} = \frac{0.00001}{0.0006} \approx 0.01667 \]

3. Calculate \(\frac{\Delta L}{L}\):
\[ \frac{\Delta L}{L} = \frac{0.001}{0.753} \approx 0.00133 \]

4. Now substitute these values into the relative error formula:
\[ \frac{\Delta \rho}{\rho} = 0.00952 + 2 \cdot 0.01667 + 0.00133 \]
\[ = 0.00952 + 0.03334 + 0.00133 \approx 0.04419 \]

### Step 5: Finalize the result
The relative error in resistivity is approximately:
\[ \frac{\Delta \rho}{\rho} \approx 0.04419 \text{ or } 0.044 \]

### Conclusion
Thus, the relative error in the resistivity of the material is approximately \(0.044\).

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