A cell in the secondary circuit gives null deflection for 2.5 m length of a potentiometer having 10m length of wire. If the length of the potentiometer wire is increased by 1m without changing the cell in the primary, the position of the null point wil be

Options

A. 3.5m
B. 3m
C. 2.75m
D. 2.0m

Explanation

To solve the problem step by step, we will analyze the situation using the principles of a potentiometer and the relationship between voltage, length, and resistance.

### Step 1: Understand the initial conditions
We have a potentiometer wire of length 10 m, and the null deflection occurs at a length of 2.5 m. This means that the potential difference across the 2.5 m length of the wire is equal to the EMF (E) of the secondary cell.

### Step 2: Determine the potential gradient (k)
The potential gradient (k) is defined as the voltage per unit length of the wire. Initially, we can express k as:
\[ k = \frac{V}{L} \]
where \( V \) is the voltage of the primary cell and \( L \) is the total length of the wire (10 m).

### Step 3: Calculate the EMF of the secondary cell
Using the null deflection condition:
\[ E = k \times 2.5 \]
Substituting for k:
\[ E = \left( \frac{V}{10} \right) \times 2.5 \]
Thus, we can express E as:
\[ E = \frac{2.5V}{10} = \frac{V}{4} \]

### Step 4: Increase the length of the potentiometer wire
Now, the length of the potentiometer wire is increased by 1 m, making the new length \( L' = 11 \) m. We need to find the new potential gradient \( k' \):
\[ k' = \frac{V}{11} \]

### Step 5: Establish the new null point condition
The new null point (let's call it \( l' \)) will satisfy:
\[ E = k' \times l' \]
Substituting for \( k' \):
\[ E = \left( \frac{V}{11} \right) \times l' \]

### Step 6: Set the two expressions for E equal
From the previous steps, we have:
1. \( E = \frac{V}{4} \)
2. \( E = \left( \frac{V}{11} \right) \times l' \)

Setting these equal gives:
\[ \frac{V}{4} = \left( \frac{V}{11} \right) \times l' \]

### Step 7: Cancel V and solve for \( l' \)
Assuming \( V \neq 0 \), we can cancel V from both sides:
\[ \frac{1}{4} = \frac{l'}{11} \]
Now, cross-multiplying gives:
\[ 11 = 4l' \]
Thus,
\[ l' = \frac{11}{4} = 2.75 \text{ m} \]

### Conclusion
The position of the null point after increasing the length of the potentiometer wire by 1 m is **2.75 m**.

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