In a potentiometer arrangement , a cell of EMF 2 V gives a balance point at 40 cm length of the wire . If this cell is replaced by another cell and the balance point shifts to 60 cm , then the EMF of the second cell is

Options

A. 1 V
B. 2 V
C. 3 V
D. 4 V

Answer

To solve the problem step by step, we will use the principles of a potentiometer, which states that the potential difference (EMF) across a wire is directly proportional to its length when a constant current flows through it.

### Step 1: Understand the relationship between EMF and length
In a potentiometer, the EMF (E) of a cell is directly proportional to the length of the wire (L) over which it is balanced. This can be expressed as:
\[ E_1 \propto L_1 \]
\[ E_2 \propto L_2 \]

### Step 2: Write the equations for both cells
From the above relationship, we can write:
\[ E_1 = k \cdot L_1 \]
\[ E_2 = k \cdot L_2 \]
where \( k \) is the constant of proportionality (potential gradient).

### Step 3: Set up the ratio of the EMFs
Dividing the two equations gives us:
\[ \frac{E_1}{E_2} = \frac{L_1}{L_2} \]

### Step 4: Substitute the known values
From the problem:
- \( E_1 = 2 \, \text{V} \) (the EMF of the first cell)
- \( L_1 = 40 \, \text{cm} \) (the length for the first cell)
- \( L_2 = 60 \, \text{cm} \) (the length for the second cell)

Substituting these values into the ratio:
\[ \frac{2}{E_2} = \frac{40}{60} \]

### Step 5: Simplify the ratio
We can simplify \( \frac{40}{60} \) to \( \frac{2}{3} \):
\[ \frac{2}{E_2} = \frac{2}{3} \]

### Step 6: Cross-multiply to solve for \( E_2 \)
Cross-multiplying gives:
\[ 2 \cdot 3 = 2 \cdot E_2 \]
\[ 6 = 2E_2 \]

### Step 7: Solve for \( E_2 \)
Dividing both sides by 2:
\[ E_2 = \frac{6}{2} = 3 \, \text{V} \]

### Conclusion
The EMF of the second cell is \( 3 \, \text{V} \).

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