A potentiometer wire of length 100 cm has a resistance of `100Omega` it is connected in series with a resistance and a battery of emf 2 V and of negligible internal resistance. A source of emf 10 mV is balanced against a length of 40 cm of the potentiometer wire. what is the value of the external resistance?

Options

A. `790Omega`
B. `890Omega`
C. `990Omega`
D. `1090Omega`

Explanation

To solve the problem step by step, we will follow the information provided in the question and apply the principles of a potentiometer circuit.

### Step 1: Understand the Circuit
We have a potentiometer wire of length 100 cm and resistance 100 Ω connected in series with an external resistance \( R \) and a battery of emf 2 V. The internal resistance of the battery is negligible.

### Step 2: Determine the Current in the Circuit
The total resistance in the circuit is the sum of the resistance of the potentiometer wire and the external resistance \( R \):
\[ R_{\text{total}} = 100 \, \Omega + R \]
Using Ohm's Law, the current \( I \) in the circuit can be calculated as:
\[ I = \frac{V}{R_{\text{total}}} = \frac{2 \, \text{V}}{100 \, \Omega + R} \]

### Step 3: Calculate the Potential Difference Across the Potentiometer Wire
The potential difference across a length \( L \) of the potentiometer wire is proportional to the current and the resistance of that length. The resistance of the 40 cm length of the potentiometer wire can be calculated as:
\[ R_L = \frac{L}{100} \times 100 \, \Omega = \frac{40}{100} \times 100 \, \Omega = 40 \, \Omega \]
The potential difference \( V_L \) across this length is given by:
\[ V_L = I \times R_L = I \times 40 \, \Omega \]

### Step 4: Set the Potential Difference Equal to the Source Emf
According to the problem, this potential difference \( V_L \) is balanced against a source of emf of 10 mV (or 0.01 V):
\[ I \times 40 \, \Omega = 10 \, \text{mV} = 0.01 \, \text{V} \]

### Step 5: Substitute the Expression for Current
Substituting the expression for \( I \) from Step 2 into the equation from Step 4:
\[ \frac{2 \, \text{V}}{100 + R} \times 40 = 0.01 \]

### Step 6: Solve for \( R \)
Rearranging the equation gives:
\[ \frac{80}{100 + R} = 0.01 \]
Cross-multiplying yields:
\[ 80 = 0.01 \times (100 + R) \]
Expanding this gives:
\[ 80 = 1 + 0.01R \]
Subtracting 1 from both sides:
\[ 79 = 0.01R \]
Dividing both sides by 0.01:
\[ R = \frac{79}{0.01} = 7900 \, \Omega \]

### Final Answer
The value of the external resistance \( R \) is:
\[ \boxed{7900 \, \Omega} \]