Potentiometer wire length is 10m, having a total resistance of `10Omega`, if a battery of emf 2 volt negligible internal resistance and a rheostat is connected to it then potential gradient is 20mV/m find the resistance applied through rheostat:-

Options

A. `90Omega`
B. `990Omega`
C. `40Omega`
D. `190Omega`

Answer

To solve the problem step by step, we will follow the outlined approach:

### Step 1: Understand the given data
- Length of the potentiometer wire, \( L = 10 \, \text{m} \)
- Total resistance of the potentiometer wire, \( R_p = 10 \, \Omega \)
- EMF of the battery, \( E = 2 \, \text{V} \)
- Potential gradient, \( k = 20 \, \text{mV/m} = 20 \times 10^{-3} \, \text{V/m} \)

### Step 2: Calculate the potential difference across the potentiometer wire
The potential difference \( V_p \) across the entire length of the potentiometer wire can be calculated using the formula:
\[ V_p = k \times L \]
Substituting the values:
\[ V_p = 20 \times 10^{-3} \, \text{V/m} \times 10 \, \text{m} = 0.2 \, \text{V} = 200 \, \text{mV} \]

### Step 3: Calculate the current through the potentiometer wire
Using Ohm's law, the current \( I \) through the potentiometer wire can be calculated as:
\[ I = \frac{V_p}{R_p} \]
Substituting the values:
\[ I = \frac{0.2 \, \text{V}}{10 \, \Omega} = 0.02 \, \text{A} = 20 \, \text{mA} \]

### Step 4: Set up the equation for the total resistance in the circuit
According to Ohm's law, the total resistance in the circuit can be expressed as:
\[ E = I \times (R_p + R) \]
Where \( R \) is the resistance of the rheostat. Rearranging gives:
\[ R = \frac{E}{I} - R_p \]

### Step 5: Substitute the known values to find \( R \)
Substituting the values of \( E \), \( I \), and \( R_p \):
\[ R = \frac{2 \, \text{V}}{0.02 \, \text{A}} - 10 \, \Omega \]
Calculating the first term:
\[ R = 100 \, \Omega - 10 \, \Omega = 90 \, \Omega \]

### Final Answer
The resistance applied through the rheostat is \( R = 90 \, \Omega \).

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