A potentiometer wire of length 1 m has a resistance of `5Omega`. It is connected to a 8 V battery in series with a resistance of `15Omega`. Determine the emf of the primary cell which gives a balance point at 60 cm.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Circuit Configuration
We have a potentiometer wire of length \( L = 1 \, \text{m} \) and resistance \( R_w = 5 \, \Omega \). It is connected in series with a resistor \( R_s = 15 \, \Omega \) and an \( 8 \, \text{V} \) battery.

### Step 2: Calculate the Total Resistance
The total resistance \( R_t \) in the circuit is the sum of the resistance of the potentiometer wire and the series resistor:
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