A battery is connected to a potentiometer and a balance point is obtained at `84 cm` along the wire. When its terminals are connected by a `5 Omega` resistor, the balance point changes to `70 cm`.
Calculate the internal resistance of the cell.

Options

A. `4 Omega`
B. `2 Omega`
C. `5 Omega`
D. `1 Omega`

Explanation

To solve the problem of calculating the internal resistance of the cell, we can follow these steps:

### Step 1: Identify the given values
- Initial balance point (L1) = 84 cm
- Final balance point (L2) = 70 cm
- Resistance connected (R) = 5 Ω

### Step 2: Use the formula for internal resistance
The internal resistance (r) of the cell can be calculated using the formula:
\[ r = \frac{L1 - L2}{L2} \times R \]

### Step 3: Substitute the values into the formula
Substituting the known values into the formula:
\[ r = \frac{84 \, \text{cm} - 70 \, \text{cm}}{70 \, \text{cm}} \times 5 \, \Omega \]

### Step 4: Calculate the difference in balance points
Calculate \( L1 - L2 \):
\[ L1 - L2 = 84 \, \text{cm} - 70 \, \text{cm} = 14 \, \text{cm} \]

### Step 5: Substitute the difference back into the formula
Now substituting back into the formula:
\[ r = \frac{14 \, \text{cm}}{70 \, \text{cm}} \times 5 \, \Omega \]

### Step 6: Simplify the fraction
Simplifying \( \frac{14}{70} \):
\[ \frac{14}{70} = \frac{1}{5} \]

### Step 7: Calculate the internal resistance
Now substituting this back into the equation:
\[ r = \frac{1}{5} \times 5 \, \Omega = 1 \, \Omega \]

### Conclusion
The internal resistance of the cell is:
\[ \boxed{1 \, \Omega} \]

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