In a potentiometer experiment , the balancing length for a cell was found to the 1.20 m . Now , a resistance of `10 Omega` is connected across the terminals of this cell and the balancing length becomes 80 cm . The internal resistance of this cell is .

Options

A. `2 Omega`
B. `5 Omega`
C. `1.5 Omega`
D. `15Omega`

Answer

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

### Step 1: Understand the Setup
In a potentiometer experiment, we have two scenarios:
1. When the switch is open, the balancing length \( L_1 \) is measured.
2. When a resistance \( R \) is connected across the cell and the switch is closed, the new balancing length \( L_2 \) is measured.

### Step 2: Write Down the Given Values
From the problem statement, we have:
- \( L_1 = 1.20 \, \text{m} \) (or 120 cm)
- \( L_2 = 0.80 \, \text{m} \) (or 80 cm)
- \( R = 10 \, \Omega \) (the resistance connected across the terminals of the cell)

### Step 3: Apply the Potentiometer Principle
According to the potentiometer principle, the voltage drop across the lengths of the wire is proportional to the lengths measured. Therefore, we can write the following equations based on the two scenarios:

1. When the switch is open:
\[ E = k \cdot L_1 \]

2. When the switch is closed:
\[ E = (R + r) \cdot I \quad \text{where } I = \frac{E}{R + r} \]
The voltage drop across the potentiometer wire is:
\[ V' = k \cdot L_2 \]

### Step 4: Set Up the Ratio of the Two Scenarios
Since both scenarios measure the same electromotive force \( E \), we can equate them:
\[ k \cdot L_1 = (R + r) \cdot I \]
And for the second scenario:
\[ k \cdot L_2 = R \cdot I \]

### Step 5: Form the Equations
From the first scenario:
\[ E = k \cdot L_1 \]
From the second scenario:
\[ E = k \cdot L_2 + R \cdot I \]

### Step 6: Take the Ratio of the Two Equations
Taking the ratio of the two equations gives:
\[ \frac{L_1}{L_2} = \frac{R + r}{R} \]

### Step 7: Substitute Known Values
Substituting the known values into the ratio:
\[ \frac{120 \, \text{cm}}{80 \, \text{cm}} = \frac{10 + r}{10} \]
This simplifies to:
\[ \frac{3}{2} = \frac{10 + r}{10} \]

### Step 8: Cross Multiply and Solve for \( r \)
Cross multiplying gives:
\[ 3 \cdot 10 = 2 \cdot (10 + r) \]
\[ 30 = 20 + 2r \]
\[ 30 - 20 = 2r \]
\[ 10 = 2r \]
\[ r = 5 \, \Omega \]

### Final Answer
The internal resistance of the cell is \( r = 5 \, \Omega \).

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