The resistivity of a potentiometer wire is, if the area of cross section of the wire is `4cm^(2)`. The current flowing in the circuit is 1A, the potential gradient is 7.5 v/m _______

Options

A. `3xx10^(-3)Omega-m`
B. `2xx10^(-6)Omega-m`
C. `4xx10^(-2)Omega-m`
D. `5xx10^(-4)Omega-m`

Answer

To find the resistivity of the potentiometer wire, we can follow these steps:

### Step 1: Write down the given data
- Area of cross-section (A) = 4 cm²
- Current (I) = 1 A
- Potential gradient (k) = 7.5 V/m

### Step 2: Convert the area from cm² to m²
Since the area is given in cm², we need to convert it to m² for consistency in SI units.
\[ A = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \]

### Step 3: Use the formula for potential gradient
The potential gradient (k) is defined as:
\[ k = \frac{V}{L} \]
Where V is the potential difference and L is the length of the wire.

### Step 4: Relate resistance to resistivity
The resistance (R) of the wire can be expressed as:
\[ R = \frac{\rho L}{A} \]
Where:
- \(\rho\) is the resistivity,
- L is the length of the wire,
- A is the area of cross-section.

### Step 5: Substitute R in the potential gradient formula
From Ohm's law, we know:
\[ V = I \times R \]
Substituting R into the potential gradient formula gives:
\[ k = \frac{I \times R}{L} = \frac{I \times \left(\frac{\rho L}{A}\right)}{L} \]
This simplifies to:
\[ k = \frac{I \rho}{A} \]

### Step 6: Rearranging to find resistivity
Rearranging the equation to solve for resistivity (\(\rho\)):
\[ \rho = \frac{k \times A}{I} \]

### Step 7: Substitute the known values
Now we substitute the known values into the equation:
\[ \rho = \frac{7.5 \, \text{V/m} \times (4 \times 10^{-4} \, \text{m}^2)}{1 \, \text{A}} \]

### Step 8: Calculate the resistivity
Calculating the above expression:
\[ \rho = \frac{7.5 \times 4 \times 10^{-4}}{1} = 3 \times 10^{-3} \, \Omega \cdot \text{m} \]

### Final Answer
The resistivity of the potentiometer wire is:
\[ \rho = 3 \times 10^{-3} \, \Omega \cdot \text{m} \]

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