In a wire V = 5.0V, I = 2.00A, L=10.0cm and diameter d=5.00mm. Evaluate `(Delta(rho) xx 100)/rho`

Options

A. 3.9
B. 0.019
C. 0.029
D. 0.03

Answer

To solve the problem, we need to evaluate the maximum percentage error in resistivity, represented as \((\Delta \rho \times 100) / \rho\). We will follow these steps:

### Step 1: Understand the relationship between voltage, current, and resistance
According to Ohm's law, the resistance \(R\) can be expressed as:
\[ R = \frac{V}{I} \]
where \(V\) is the voltage and \(I\) is the current.

### Step 2: Determine the resistivity formula
The resistivity \(\rho\) can be expressed in terms of resistance \(R\) and the physical dimensions of the wire:
\[ \rho = R \cdot \frac{A}{L} \]
where \(A\) is the cross-sectional area and \(L\) is the length of the wire.

### Step 3: Calculate the cross-sectional area \(A\)
The area \(A\) of the wire can be calculated using the diameter \(d\):
\[ A = \frac{\pi d^2}{4} \]
Given that \(d = 5.00 \, \text{mm} = 0.005 \, \text{m}\), we can calculate \(A\):
\[ A = \frac{\pi (0.005)^2}{4} = \frac{\pi \times 0.000025}{4} \approx 1.9635 \times 10^{-5} \, \text{m}^2 \]

### Step 4: Substitute values into the resistivity formula
Now we can substitute the values of \(V\), \(I\), \(A\), and \(L\) into the resistivity formula:
\[ \rho = \frac{V}{I} \cdot \frac{A}{L} \]
Given \(V = 5.0 \, \text{V}\), \(I = 2.0 \, \text{A}\), and \(L = 10.0 \, \text{cm} = 0.1 \, \text{m}\):
\[ \rho = \frac{5.0}{2.0} \cdot \frac{1.9635 \times 10^{-5}}{0.1} = 2.5 \cdot 1.9635 \times 10^{-4} \approx 4.90875 \times 10^{-4} \, \Omega \cdot \text{m} \]

### Step 5: Calculate the maximum percentage error in resistivity
The maximum percentage error in resistivity can be calculated using the formula:
\[ \frac{\Delta \rho}{\rho} \times 100 = \frac{\Delta V}{V} + \frac{\Delta I}{I} + \frac{\Delta L}{L} + 2 \frac{\Delta D}{D} \]

### Step 6: Determine the least counts
- For voltage \(V = 5.0 \, \text{V}\), the least count \(\Delta V = 0.1 \, \text{V}\):
\[ \frac{\Delta V}{V} = \frac{0.1}{5.0} = 0.02 \]

- For current \(I = 2.0 \, \text{A}\), the least count \(\Delta I = 0.01 \, \text{A}\):
\[ \frac{\Delta I}{I} = \frac{0.01}{2.0} = 0.005 \]

- For length \(L = 10.0 \, \text{cm}\), the least count \(\Delta L = 0.1 \, \text{cm} = 0.001 \, \text{m}\):
\[ \frac{\Delta L}{L} = \frac{0.001}{0.1} = 0.01 \]

- For diameter \(d = 5.0 \, \text{mm}\), the least count \(\Delta D = 0.01 \, \text{mm} = 0.00001 \, \text{m}\):
\[ \frac{\Delta D}{D} = \frac{0.00001}{0.005} = 0.002 \]

### Step 7: Substitute the errors into the formula
Now substituting these values into the error formula:
\[ \frac{\Delta \rho}{\rho} \times 100 = 0.02 + 0.005 + 0.01 + 2 \times 0.002 \]
Calculating this gives:
\[ \frac{\Delta \rho}{\rho} \times 100 = 0.02 + 0.005 + 0.01 + 0.004 = 0.039 \text{ or } 3.9\% \]

### Final Answer
The maximum percentage error in resistivity is approximately \(3.9\%\).

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