In a potentiometer the balancing with a cell is at length of 220cm. On shunting the cell with a resistance of `3Omega` balance length becomes 130cm. What is the internal resistance of this cell.

Options

A. `4.5Omega`
B. `7.8Omega`
C. `6.3Omega`
D. `2.08Omega`

Explanation

To solve the problem of finding the internal resistance of the cell in the potentiometer setup, we can follow these steps:

### Step 1: Understand the given information
- The initial balancing length (without shunt) is \( L_1 = 220 \, \text{cm} \).
- The balancing length after shunting with a resistance of \( R = 3 \, \Omega \) is \( L_2 = 130 \, \text{cm} \).

### Step 2: Use the formula for internal resistance
The formula for calculating the internal resistance \( r \) of the cell when a shunt resistor is connected is given by:

\[ r = R \left( \frac{L_1}{L_2} - 1 \right) \]

Where:
- \( R \) is the shunt resistance (3 Ω).
- \( L_1 \) is the balancing length without the shunt (220 cm).
- \( L_2 \) is the balancing length with the shunt (130 cm).

### Step 3: Substitute the values into the formula
Substituting the values into the formula:

\[ r = 3 \left( \frac{220}{130} - 1 \right) \]

### Step 4: Calculate the ratio \( \frac{L_1}{L_2} \)
Calculating the ratio:

\[ \frac{220}{130} = 1.6923 \quad (\text{approximately } 1.69) \]

### Step 5: Subtract 1 from the ratio
Now subtract 1 from the ratio:

\[ 1.69 - 1 = 0.69 \]

### Step 6: Multiply by the shunt resistance
Now, multiply this result by the shunt resistance:

\[ r = 3 \times 0.69 = 2.07 \, \Omega \]

### Step 7: Conclusion
Thus, the internal resistance of the cell is:

\[ \boxed{2.07 \, \Omega} \]

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