In a potentiometer arrangement a cell of emf `1.20 V` given a balance point at `30 cm` length of the wire. The cell a now replaced by another cell of unknown emf. The ratio of emfs of the two cells is `1.5 `, calculate the difference in the balancing length of the potentiometer wire in the two cases

Answer

To solve the problem step by step, we will use the principles of a potentiometer and the relationship between the lengths of the wire and the EMFs of the cells.

### Step 1: Understand the relationship in a potentiometer
In a potentiometer, the balancing length \( L \) is directly proportional to the EMF \( E \) of the cell connected. This can be expressed as:
\[ \frac{L_1}{E_1} = \frac{L_2}{E_2} \]
where \( L_1 \) and \( L_2 \) are the balancing lengths for the first and second cells, and \( E_1 \) and \( E_2 \) are their respective EMFs.

### Step 2: Assign known values
From the problem, we know:
- The EMF of the first cell \( E_1 = 1.20 \, V \)
- The balancing length for the first cell \( L_1 = 30 \, cm \)
- The ratio of the EMFs \( \frac{E_2}{E_1} = 1.5 \)

### Step 3: Calculate the EMF of the second cell
Using the ratio of the EMFs:
\[ E_2 = 1.5 \times E_1 = 1.5 \times 1.20 \, V = 1.80 \, V \]

### Step 4: Set up the equation for the second cell
Now we can use the relationship established in Step 1 for the second cell:
\[ \frac{L_1}{E_1} = \frac{L_2}{E_2} \]
Substituting the known values:
\[ \frac{30 \, cm}{1.20 \, V} = \frac{L_2}{1.80 \, V} \]

### Step 5: Solve for \( L_2 \)
Cross-multiplying gives:
\[ L_2 = \frac{30 \, cm \times 1.80 \, V}{1.20 \, V} \]
Calculating \( L_2 \):
\[ L_2 = \frac{30 \times 1.80}{1.20} = 45 \, cm \]

### Step 6: Calculate the difference in balancing lengths
Now, we find the difference between the lengths for the two cells:
\[ \Delta L = L_2 - L_1 = 45 \, cm - 30 \, cm = 15 \, cm \]

### Final Answer
The difference in the balancing length of the potentiometer wire in the two cases is \( 15 \, cm \).

---