In a potentiometer experiment, the balancing length with a cell is 560cm. When an external resistance of 10ohms is connected in parallel to the cell the balancing length changs by 60cm. The internal resistance of the cell in ohm is

Options

A. 3.6
B. 2.4
C. 1.2
D. 0.6

Explanation

To find the internal resistance of the cell in the given potentiometer experiment, we can follow these steps:

### Step 1: Understand the Initial Setup
We have a potentiometer wire with a balancing length \( L_1 = 560 \, \text{cm} \) when connected to a cell. The potential gradient \( k \) can be expressed as:
\[ E = k \cdot L_1 \]
where \( E \) is the EMF of the cell.

### Step 2: Analyze the Change with External Resistance
When a 10 ohm external resistance is connected in parallel to the cell, the balancing length changes to \( L_2 \). Given that the change in balancing length is 60 cm, we can find \( L_2 \):
\[ L_2 = L_1 - 60 = 560 \, \text{cm} - 60 \, \text{cm} = 500 \, \text{cm} \]

### Step 3: Write the Equations for Both Conditions
From the first condition (without external resistance):
\[ E = k \cdot L_1 \]
From the second condition (with external resistance):
\[ E - I \cdot r = k \cdot L_2 \]
where \( r \) is the internal resistance of the cell and \( I \) is the current flowing through the circuit.

### Step 4: Relate Current and Resistance
The current \( I \) can be expressed in terms of the potential gradient and the external resistance \( R \):
\[ I = \frac{k \cdot L_2}{R} \]
Substituting \( R = 10 \, \Omega \) and \( L_2 = 500 \, \text{cm} \) into the equation gives:
\[ I = \frac{k \cdot 500}{10} \]

### Step 5: Substitute Current into the Second Condition
Substituting for \( I \) in the second equation:
\[ E - \left(\frac{k \cdot 500}{10}\right) \cdot r = k \cdot 500 \]
Rearranging gives:
\[ E = k \cdot 500 + \left(\frac{k \cdot 500}{10}\right) \cdot r \]

### Step 6: Equate the Two Expressions for EMF
From the first condition:
\[ E = k \cdot 560 \]
Equating both expressions for \( E \):
\[ k \cdot 560 = k \cdot 500 + \left(\frac{k \cdot 500}{10}\right) \cdot r \]

### Step 7: Simplify the Equation
Dividing through by \( k \) (assuming \( k \neq 0 \)):
\[ 560 = 500 + \frac{500}{10} \cdot r \]
This simplifies to:
\[ 560 - 500 = 50 \cdot r \]
\[ 60 = 50 \cdot r \]

### Step 8: Solve for Internal Resistance
Now, solving for \( r \):
\[ r = \frac{60}{50} = 1.2 \, \Omega \]

### Final Answer
The internal resistance of the cell is \( r = 1.2 \, \Omega \).

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