In an experiment to measure the internal resistance of a cell by potentiometer, it is found that the balance point is at a length of 2 m when the cell is shunted by a `4Omega` resistance and at 3 m when cell is shunted by a `8Omega` resistance. The internal resistance of cell is -

Options

A. `12Omega`
B. `8Omega`
C. `16Omega`
D. `1Omega`

Answer

To find the internal resistance of the cell using the given data from the potentiometer experiment, we can follow these steps:

### Step 1: Understand the Setup
In the experiment, we have a cell with an internal resistance \( r \) and an electromotive force (EMF) \( E \). The cell is shunted by two different resistances: \( R_1 = 4 \, \Omega \) and \( R_2 = 8 \, \Omega \). The balance points on the potentiometer wire are \( L_1 = 2 \, m \) for \( R_1 \) and \( L_2 = 3 \, m \) for \( R_2 \).

### Step 2: Write the Equations for Each Case
For the first case (shunted by \( 4 \, \Omega \)):
\[ \frac{E}{r + R_1} = \frac{V_0}{L_1} \]
Substituting \( R_1 = 4 \, \Omega \) and \( L_1 = 2 \, m \):
\[ \frac{E}{r + 4} = \frac{V_0}{2} \quad \text{(1)} \]

For the second case (shunted by \( 8 \, \Omega \)):
\[ \frac{E}{r + R_2} = \frac{V_0}{L_2} \]
Substituting \( R_2 = 8 \, \Omega \) and \( L_2 = 3 \, m \):
\[ \frac{E}{r + 8} = \frac{V_0}{3} \quad \text{(2)} \]

### Step 3: Eliminate \( E \) and \( V_0 \)
From equations (1) and (2), we can eliminate \( E \) and \( V_0 \) by dividing the two equations:
\[ \frac{E}{r + 4} \div \frac{E}{r + 8} = \frac{V_0/2}{V_0/3} \]
This simplifies to:
\[ \frac{r + 8}{r + 4} = \frac{3}{2} \]

### Step 4: Cross Multiply and Solve for \( r \)
Cross multiplying gives:
\[ 2(r + 8) = 3(r + 4) \]
Expanding both sides:
\[ 2r + 16 = 3r + 12 \]
Rearranging gives:
\[ 16 - 12 = 3r - 2r \]
\[ 4 = r \]

### Step 5: Conclusion
Thus, the internal resistance of the cell is:
\[ \boxed{4 \, \Omega} \]