The length of a potentiometer wire is 10 cm. A cell of emf E is balanced at a length 10/3 cm from the positive end of the wire. If the length of the wire is increased by 5 cm, at what distance (in cm) from positive end will the same cell give a balance point?

Answer

To solve the problem step by step, we will use the concept of potential gradient in a potentiometer.

### Step 1: Understand the initial setup
We are given:
- Length of the potentiometer wire, \( L_1 = 10 \, \text{cm} \)
- The cell of emf \( E \) is balanced at a length of \( 10/3 \, \text{cm} \) from the positive end.

### Step 2: Calculate the potential gradient for the initial wire
The potential gradient (\( k \)) is defined as the emf per unit length of the wire. For the initial setup, we can denote the standard emf as \( E_0 \).

\[ k_1 = \frac{E_0}{L_1} = \frac{E_0}{10 \, \text{cm}} \]

### Step 3: Relate the emf \( E \) to the potential gradient
Since the cell of emf \( E \) is balanced at \( 10/3 \, \text{cm} \):

\[ E = k_1 \times \frac{10}{3} = \frac{E_0}{10} \times \frac{10}{3} = \frac{E_0}{3} \]

### Step 4: Consider the new length of the potentiometer wire
Now, the length of the wire is increased by 5 cm:

\[ L_2 = L_1 + 5 \, \text{cm} = 10 \, \text{cm} + 5 \, \text{cm} = 15 \, \text{cm} \]

### Step 5: Calculate the new potential gradient
The new potential gradient (\( k_2 \)) will be:

\[ k_2 = \frac{E_0}{L_2} = \frac{E_0}{15 \, \text{cm}} \]

### Step 6: Set up the equation for the new balance point
Let the new balance point be \( x \, \text{cm} \) from the positive end. The emf \( E \) will now balance at this new length:

\[ E = k_2 \times x = \frac{E_0}{15} \times x \]

### Step 7: Equate the two expressions for \( E \)
From the previous steps, we have two expressions for \( E \):

1. \( E = \frac{E_0}{3} \)
2. \( E = \frac{E_0}{15} \times x \)

Setting them equal gives:

\[ \frac{E_0}{3} = \frac{E_0}{15} \times x \]

### Step 8: Solve for \( x \)
We can cancel \( E_0 \) from both sides (assuming \( E_0 \neq 0 \)):

\[ \frac{1}{3} = \frac{x}{15} \]

Cross-multiplying gives:

\[ x = 15 \times \frac{1}{3} = 5 \, \text{cm} \]

### Conclusion
The new balance point when the length of the wire is increased by 5 cm is \( 5 \, \text{cm} \) from the positive end.

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