A student finds the balancing length to be 'l' with a cell of constant emf in the secondary circuit. Another student connects the same cell in the secondary circuit of potentiometer of double the length but with a cell of half the emf in the primary circuit. The balancing length will be [cell in primary is ideal and no series resistance is present in primary circuit.]

Options

A. `4l`
B. `l//4`
C. `2l`
D. `l//2`

Answer

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understanding the First Case
In the first case, a student finds the balancing length to be \( L \) with a cell of constant EMF \( E \) in the secondary circuit. The potential difference across the potentiometer wire is given by:
\[ V = I \times L \]
where \( I \) is the current flowing through the potentiometer.

### Step 2: Setting Up the Second Case
In the second case, the same cell is connected in the secondary circuit of a potentiometer that has double the length (i.e., \( 2L \)). However, the EMF in the primary circuit is now half the original value, \( \frac{E}{2} \).

### Step 3: Analyzing the Potential Difference in the Second Case
For the second case, the potential difference across the new potentiometer wire of length \( 2L \) is:
\[ V' = I' \times (2L) \]
where \( I' \) is the new current flowing through the circuit.

### Step 4: Relating the Currents and EMFs
Since the EMF in the secondary circuit remains the same as the first case (which is \( E \)), we can set up the following relationship:
\[ I \times L = E \quad \text{(from the first case)} \]
\[ I' \times (2L) = E \quad \text{(from the second case)} \]

### Step 5: Finding the Current in the Second Case
From the first case, we can express the current \( I \) as:
\[ I = \frac{E}{L} \]
For the second case, substituting \( I' \) gives:
\[ I' = \frac{E}{2L} \]

### Step 6: Setting Up the Balancing Length Equation
Now, we can relate the two cases using the balancing condition:
\[ I \times L = I' \times L' \]
Substituting the expressions for \( I \) and \( I' \):
\[ \frac{E}{L} \times L = \frac{E}{2L} \times L' \]
This simplifies to:
\[ E = \frac{E}{2L} \times L' \]

### Step 7: Solving for the Balancing Length \( L' \)
Cancelling \( E \) from both sides (since \( E \neq 0 \)):
\[ 1 = \frac{L'}{2L} \]
Thus, we find:
\[ L' = 2L \]

### Step 8: Final Calculation for Balancing Length
Now, since the EMF in the primary circuit is half, we need to adjust our previous equation:
\[ I' = \frac{E/2}{2L} = \frac{E}{4L} \]
Substituting this back gives us:
\[ L' = 4L \]

### Conclusion
The balancing length in the second case will be \( 4L \).

### Final Answer
The balancing length will be \( 4L \).

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