In a potentiometer experiment, the galvanometer shows no deflection when a cell is connected across `60 cm` of the potentiometer wire. If the cell is shunted by a resistance of `6 Omega`, the balance is obtained across `50 cm` of the wire. The internal resistance of the cell is

Options

A. `0.5Omega`
B. `0.6Omega`
C. `1.2Omega`
D. `1.5Omega`

Answer

To solve the problem step-by-step, we will use the principles of a potentiometer and the relationship between voltage, resistance, and length of the wire.

### Step 1: Understand the initial condition
When the cell is connected across `60 cm` of the potentiometer wire, the galvanometer shows no deflection. This means that the potential difference (V) across `60 cm` of the wire is equal to the electromotive force (E) of the cell.

### Step 2: Set up the equation for the initial condition
Let the potential difference across `1 cm` of the potentiometer wire be `k`. Therefore, the potential difference across `60 cm` of the wire can be expressed as:
\[ V = k \times 60 \]
Since this is equal to the EMF of the cell, we have:
\[ E = k \times 60 \quad \text{(1)} \]

### Step 3: Understand the condition when the cell is shunted
When the cell is shunted by a resistance of `6 Ω`, the balance is obtained across `50 cm` of the potentiometer wire. The effective EMF across the shunt can be expressed as:
\[ E' = \frac{E}{R + r} \times R \]
Where:
- \( R = 6 \, \Omega \) (the shunt resistance)
- \( r \) = internal resistance of the cell

The potential difference across `50 cm` of the potentiometer wire can be expressed as:
\[ E' = k \times 50 \quad \text{(2)} \]

### Step 4: Set up the equation for the shunted condition
From the above, we can equate the two expressions for EMF:
\[ \frac{E}{R + r} \times R = k \times 50 \]

### Step 5: Substitute \( E \) from equation (1) into the shunted condition
Substituting \( E = k \times 60 \) into the equation gives:
\[ \frac{k \times 60}{6 + r} \times 6 = k \times 50 \]

### Step 6: Simplify the equation
Dividing both sides by \( k \) (assuming \( k \neq 0 \)):
\[ \frac{60}{6 + r} \times 6 = 50 \]

### Step 7: Cross-multiply and solve for \( r \)
Cross-multiplying gives:
\[ 60 \times 6 = 50 \times (6 + r) \]
\[ 360 = 300 + 50r \]
\[ 360 - 300 = 50r \]
\[ 60 = 50r \]
\[ r = \frac{60}{50} = 1.2 \, \Omega \]

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