Potentiometer wire of length `1 m` is connected in series with `490 Omega` resistance and `2 V` battery. If `0.2 mV/cm` is the potential gradient, then resistance of the potentiameter wire is approximately

Options

A. `4.9Omega`
B. `7.9Omega`
C. `5.9Omega`
D. `6.9Omega`

Answer

To find the resistance of the potentiometer wire, we will follow these steps:

### Step 1: Understand the given data
- Length of the potentiometer wire (L) = 1 m
- Resistance in series (R₁) = 490 Ω
- Voltage of the battery (V) = 2 V
- Potential gradient (k) = 0.2 mV/cm

### Step 2: Convert the potential gradient to volts per meter
We need to convert 0.2 mV/cm to V/m:
\[ 0.2 \, \text{mV/cm} = 0.2 \times 10^{-3} \, \text{V/cm} = 0.2 \times 10^{-3} \times 100 \, \text{V/m} = 0.02 \, \text{V/m} \]

### Step 3: Calculate the potential drop across the entire length of the wire
The potential drop (V_AB) across the entire length of the potentiometer wire can be calculated using the potential gradient:
\[ V_{AB} = k \times L = 0.02 \, \text{V/m} \times 1 \, \text{m} = 0.02 \, \text{V} \]

### Step 4: Set up the voltage division equation
Using the voltage division rule, the voltage across the potentiometer wire (V_AB) can be expressed as:
\[ V_{AB} = \frac{V \cdot R}{R_1 + R} \]
Where R is the resistance of the potentiometer wire.

### Step 5: Substitute the known values into the equation
Substituting the known values into the voltage division equation:
\[ 0.02 = \frac{2 \cdot R}{490 + R} \]

### Step 6: Cross-multiply to solve for R
Cross-multiplying gives:
\[ 0.02(490 + R) = 2R \]
Expanding this:
\[ 9.8 + 0.02R = 2R \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ 9.8 = 2R - 0.02R \]
\[ 9.8 = 1.98R \]

### Step 8: Solve for R
Now, we can solve for R:
\[ R = \frac{9.8}{1.98} \approx 4.95 \, \Omega \]
This can be rounded to approximately 4.9 Ω.

### Final Answer
The resistance of the potentiometer wire is approximately **4.9 Ω**.

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