An ideal cell having a steady emf of 2 volt is connected across the potentiometer wire of length 10m. The potentiometer wire is of magnesium and having resistance of 11.5Ω/m. An another cell gives a null point at 6.9m. If a resistance of 5Ω is put in series with potentiometer wire, find the new position of the null point.

Answer

To solve the problem step by step, let's break it down:

### Given Data:
- EMF of the ideal cell, \( E = 2 \, \text{V} \)
- Length of the potentiometer wire, \( L = 10 \, \text{m} \)
- Resistance of the potentiometer wire per meter, \( R_{\text{wire}} = 11.5 \, \Omega/\text{m} \)
- Null point with the first cell, \( d_1 = 6.9 \, \text{m} \)
- Resistance added in series, \( R = 5 \, \Omega \)

### Step 1: Calculate the total resistance of the potentiometer wire
The total resistance of the potentiometer wire can be calculated as:
\[ R_{\text{total}} = R_{\text{wire}} \times L = 11.5 \, \Omega/\text{m} \times 10 \, \text{m} = 115 \, \Omega \]

### Step 2: Calculate the current flowing through the potentiometer wire
Using Ohm's law, the current \( I \) flowing through the circuit can be calculated as:
\[ I = \frac{E}{R_{\text{total}}} = \frac{2 \, \text{V}}{115 \, \Omega} \approx 0.01739 \, \text{A} \]

### Step 3: Calculate the potential difference across the null point
The potential difference across the length of wire up to the null point (6.9 m) can be calculated as:
\[ V_{\text{null}} = I \times R_{\text{wire}} \times d_1 = 0.01739 \, \text{A} \times (11.5 \, \Omega/\text{m} \times 6.9 \, \text{m}) \approx 1.38 \, \text{V} \]

### Step 4: Calculate the new total resistance when 5Ω is added
The new total resistance in the circuit when the 5Ω resistor is added in series is:
\[ R'_{\text{total}} = R_{\text{total}} + R = 115 \, \Omega + 5 \, \Omega = 120 \, \Omega \]

### Step 5: Calculate the new current flowing through the potentiometer wire
The new current \( I' \) can be calculated as:
\[ I' = \frac{E}{R'_{\text{total}}} = \frac{2 \, \text{V}}{120 \, \Omega} \approx 0.01667 \, \text{A} \]

### Step 6: Set up the equation for the new null point
Let \( L \) be the new position of the null point. The potential difference at the new null point can be expressed as:
\[ V_{\text{null}} = I' \times R_{\text{wire}} \times L \]
We know this potential difference must equal the EMF of the second cell, which is 1.38 V:
\[ 1.38 \, \text{V} = 0.01667 \, \text{A} \times (11.5 \, \Omega/\text{m}) \times L \]

### Step 7: Solve for \( L \)
Rearranging the equation gives:
\[ L = \frac{1.38 \, \text{V}}{0.01667 \, \text{A} \times 11.5 \, \Omega/\text{m}} \approx 7.2 \, \text{m} \]

### Final Answer:
The new position of the null point is approximately \( 7.2 \, \text{m} \).

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