In a potentiometer experiment the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of `2Omega`, the balancing length becomes 120 cm.The internal resistance of the cell is

Options

A. `1 Omega`
B. `0.5 Omega`
C. `4 Omega`
D. `2 Omega`

Answer

To solve the problem step by step, we will use the principles of a potentiometer and Ohm's law.

### Step 1: Understand the initial conditions
In the potentiometer experiment, the balancing length with the cell is given as 240 cm. This means the EMF (E) of the cell is proportional to this length.

### Step 2: Set up the relationship for the first balancing length
Using the proportionality of EMF to length, we can write:
\[ E \propto 240 \text{ cm} \]

### Step 3: Understand the conditions after shunting
When a resistance of \(2 \Omega\) is shunted across the cell, the new balancing length becomes 120 cm. This means the new potential (V) across the potentiometer wire is proportional to this new length.

### Step 4: Set up the relationship for the second balancing length
Similarly, we can write:
\[ V \propto 120 \text{ cm} \]

### Step 5: Establish the relationship between E and V
From the above relationships, we can express:
\[ \frac{E}{V} = \frac{240}{120} = 2 \]
This implies:
\[ E = 2V \]

### Step 6: Express V in terms of E and resistances
When the cell is shunted, the voltage V can be expressed using Ohm's law:
\[ V = E - I \cdot R \]
Where:
- \(I\) is the current flowing through the circuit
- \(R\) is the shunt resistance (2 Ω)

### Step 7: Find the expression for current I
The current \(I\) can be expressed as:
\[ I = \frac{E}{R + r} \]
Where:
- \(R\) is the shunt resistance (2 Ω)
- \(r\) is the internal resistance of the cell

### Step 8: Substitute I into the expression for V
Substituting for \(I\) in the equation for \(V\):
\[ V = E - \left(\frac{E}{R + r}\right) \cdot R \]
This simplifies to:
\[ V = E - \frac{E \cdot R}{R + r} \]
\[ V = E \left(1 - \frac{R}{R + r}\right) \]
\[ V = E \left(\frac{r}{R + r}\right) \]

### Step 9: Substitute V into the equation E = 2V
Now we can substitute \(V\) back into the equation \(E = 2V\):
\[ E = 2 \cdot E \left(\frac{r}{R + r}\right) \]

### Step 10: Simplify the equation
Cancelling \(E\) from both sides (assuming \(E \neq 0\)):
\[ 1 = 2 \cdot \frac{r}{R + r} \]
\[ R + r = 2r \]
\[ R = 2r - r \]
\[ R = r \]

### Step 11: Substitute the value of R
Given that \(R = 2 \Omega\) (the shunt resistance), we can substitute:
\[ r = 2 \Omega \]

### Conclusion
Thus, the internal resistance of the cell is:
\[ r = 2 \Omega \]