The length of a potentiometer wire is 10 m and carries steady current on connecting the sliding jockey to standard cell of 1.018 volt, the null point is obtained at the distance of 850 cm.
The potential gradient along wire is:

Options

A. `1.2xx10^(-2)Vm^(-1)`
B. `1.2xx10^(-2)V//cm`
C. `1.2xx10^(-3)Vcm^(-1)`
D. `1.2xx10^(-3)Vcm`

Explanation

To find the potential gradient along the potentiometer wire, we will follow these steps:

### Step 1: Understand the given data
- Length of the potentiometer wire (L) = 10 m
- EMF of the standard cell (E) = 1.018 V
- Distance at which the null point is obtained (d) = 850 cm = 8.5 m (since 1 m = 100 cm)

### Step 2: Use the formula for potential gradient
The potential gradient (k) is defined as the potential difference (E) divided by the length (d) at which the null point is obtained. The formula is:
\[ k = \frac{E}{d} \]

### Step 3: Substitute the values into the formula
Now, substituting the values we have:
\[ k = \frac{1.018 \, \text{V}}{8.5 \, \text{m}} \]

### Step 4: Calculate the potential gradient
Perform the calculation:
\[ k = \frac{1.018}{8.5} \approx 0.1197 \, \text{V/m} \]

### Step 5: Convert the result to the appropriate units
Since the options are in volts per centimeter, we need to convert volts per meter to volts per centimeter:
\[ 1 \, \text{V/m} = 0.01 \, \text{V/cm} \]
Thus,
\[ k \approx 0.1197 \, \text{V/m} = 0.1197 \times 0.01 \, \text{V/cm} = 0.001197 \, \text{V/cm} \]

### Step 6: Express in scientific notation
To express this in scientific notation:
\[ 0.001197 \, \text{V/cm} = 1.197 \times 10^{-3} \, \text{V/cm} \]

### Step 7: Round to significant figures
Rounding this to two significant figures gives:
\[ k \approx 1.2 \times 10^{-3} \, \text{V/cm} \]

### Conclusion
The potential gradient along the wire is:
\[ \text{Potential Gradient} = 1.2 \times 10^{-3} \, \text{V/cm} \]