A potentiometer wire of 5m length and having 20`Omega `resistance is connected in series with a battery and a resistance of `3980 Omega `. A cell of emf 1.1V is balanced across the potential difference of the external resistance. If the emf of the thermocouple is 2.2mV, the corresponding balancing length is

Options

A. 398cm
B. 200cm
C. 450cm
D. 306cm

Answer

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

### Step 1: Identify Given Values
- Length of the potentiometer wire (L) = 5 m
- Resistance of the potentiometer wire (R_wire) = 20 Ω
- External resistance (R_ext) = 3980 Ω
- EMF of the cell (E_cell) = 1.1 V
- EMF of the thermocouple (E_thermocouple) = 2.2 mV = 2.2 × 10^(-3) V

### Step 2: Calculate Total Resistance
The total resistance in the circuit (R_total) is the sum of the resistance of the potentiometer wire and the external resistance:
\[ R_{total} = R_{wire} + R_{ext} = 20 \, \Omega + 3980 \, \Omega = 4000 \, \Omega \]

### Step 3: Calculate the Current (I)
Using Ohm's Law, the current (I) flowing through the circuit can be calculated as:
\[ I = \frac{E_{cell}}{R_{total}} \]
Substituting the values:
\[ I = \frac{1.1 \, V}{4000 \, \Omega} = 0.000275 \, A = 2.75 \, mA \]

### Step 4: Calculate the Potential Gradient (K)
The potential gradient (K) is given by the formula:
\[ K = \frac{I \cdot R_{wire}}{L} \]
Substituting the values:
\[ K = \frac{(0.000275 \, A) \cdot (20 \, \Omega)}{5 \, m} \]
\[ K = \frac{0.0055 \, V}{5 \, m} = 0.0011 \, V/m = 1.1 \, mV/m \]

### Step 5: Calculate the Balancing Length (L_bal)
The balancing length (L_bal) can be calculated using the formula:
\[ L_{bal} = \frac{E_{thermocouple}}{K} \]
Substituting the values:
\[ L_{bal} = \frac{2.2 \times 10^{-3} \, V}{1.1 \times 10^{-3} \, V/m} = 2 \, m \]

### Step 6: Convert to Centimeters
Since the problem asks for the balancing length in centimeters:
\[ L_{bal} = 2 \, m \times 100 = 200 \, cm \]

### Final Answer
The corresponding balancing length is **200 cm**.

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