A battery is connected to a potentiometer and a balance point is obtained at `84 cm` along the wire. When its terminals are connected by a `5 Omega` resistor, the balance point changes to `70 cm`.
Find the new position of the balance point when `5 Omega` resistor is changed by `4 Omega` resistor.

Options

A. `26.5 cm`
B. `52 cm`
C. `67.2 cm`
D. `83.3 cm`

Explanation

To solve the problem step by step, we will use the principles of potentiometers and the relationship between resistance, balance points, and lengths along the wire.

### Step 1: Understand the initial conditions
- The initial balance point (L1) when the battery is connected without any load is at 84 cm.
- When a 5 Ω resistor is connected across the battery terminals, the new balance point (L2) is at 70 cm.

### Step 2: Use the formula for internal resistance
The internal resistance (R) of the battery can be calculated using the formula:
\[ R = \frac{L1 - L2}{L2} \times R1 \]
Where:
- \(L1 = 84 \, \text{cm}\)
- \(L2 = 70 \, \text{cm}\)
- \(R1 = 5 \, \Omega\)

### Step 3: Calculate the internal resistance
Substituting the values into the formula:
\[ R = \frac{84 - 70}{70} \times 5 \]
\[ R = \frac{14}{70} \times 5 = \frac{1}{5} \times 5 = 1 \, \Omega \]

### Step 4: Find the new balance point with a 4 Ω resistor
Now, we need to find the new balance point (L3) when the 5 Ω resistor is replaced by a 4 Ω resistor. We will use the same formula:
\[ R = \frac{L1 - L3}{L3} \times R3 \]
Where:
- \(R3 = 4 \, \Omega\)

### Step 5: Substitute known values into the formula
Using the internal resistance we found (1 Ω):
\[ 1 = \frac{84 - L3}{L3} \times 4 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ L3 = \frac{84 \times 4}{4 + 1} = \frac{336}{5} \]

### Step 7: Calculate L3
Now, calculating L3:
\[ L3 = \frac{336}{5} = 67.2 \, \text{cm} \]

### Final Answer
The new position of the balance point when the 5 Ω resistor is replaced by a 4 Ω resistor is **67.2 cm**.

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