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. 3.5 m
B. 3 m
C. 2.75 m
D. 2 m

Answer

To solve the problem step by step, we will use the concept of a potentiometer, which is a device used to measure the potential difference (voltage) by balancing it against a known voltage.

### Step 1: Understand the initial conditions
We have a potentiometer wire of length \( L = 10 \, \text{m} \) and the null deflection (balancing length) for the cell in the secondary circuit is \( l = 2.5 \, \text{m} \).

### Step 2: Calculate the potential drop per unit length
The potential drop per unit length of the potentiometer wire can be expressed as:
\[ \frac{V}{L} \]
where \( V \) is the total potential difference across the potentiometer wire.

### Step 3: Set up the equation for the initial condition
From the initial condition, we can write:
\[ E = \left(\frac{V}{L}\right) \cdot l \]
Substituting the known values:
\[ E = \left(\frac{V}{10}\right) \cdot 2.5 \]

### Step 4: Consider the new conditions
Now, the length of the potentiometer wire is increased by 1 m, making the new length \( L' = 10 + 1 = 11 \, \text{m} \).

### Step 5: Set up the equation for the new condition
The potential drop across the potentiometer wire remains the same, so we can write:
\[ E = \left(\frac{V}{L'}\right) \cdot l' \]
where \( l' \) is the new balancing length we want to find.

### Step 6: Equate the two expressions for E
Since the potential drop \( E \) remains the same, we can equate the two expressions:
\[ \left(\frac{V}{10}\right) \cdot 2.5 = \left(\frac{V}{11}\right) \cdot l' \]

### Step 7: Cancel \( V \) from both sides
We can cancel \( V \) from both sides since it is common:
\[ \frac{2.5}{10} = \frac{l'}{11} \]

### Step 8: Solve for \( l' \)
Now, cross-multiply to solve for \( l' \):
\[ 2.5 \cdot 11 = 10 \cdot l' \]
\[ 27.5 = 10 \cdot l' \]
\[ l' = \frac{27.5}{10} = 2.75 \, \text{m} \]

### Conclusion
The new position of the null point will be \( 2.75 \, \text{m} \).