A potentiometer wire AB is 100 cm long. The EMF of the test cell is 1.5 V. At a balance length of 60 cm, calculate the potential difference across the entire wire AB.

The problem provides the following information from Figure 5:

• The EMF of the test cell, $E_2 = 1.5 \, \text{V}$.
• The total length of the potentiometer wire AB, $L_{AB} = 100 \, \text{cm}$.
• The balance length AC, $L_{AC} = 60 \, \text{cm}$.

According to the principle of a potentiometer, at the balance point, the EMF of the test cell ($E_2$) is equal to the potential difference across the balance length of the wire ($V_{AC}$).
So, $E_2 = V_{AC}$.

The potential gradient ($k$) along the potentiometer wire is uniform. It is defined as the potential difference per unit length:

$$k = \frac{V_{AB}}{L_{AB}}$$

The potential difference across the balance length $AC$ can also be expressed using the potential gradient:

$$V_{AC} = k \times L_{AC}$$

Substituting the expression for $k$:

$$V_{AC} = \frac{V_{AB}}{L_{AB}} \times L_{AC}$$

Since $E_2 = V_{AC}$, we can write:

$$E_2 = \frac{V_{AB}}{L_{AB}} \times L_{AC}$$

Now, plug in the given values:

$$1.5 \, \text{V} = \frac{V_{AB}}{100 \, \text{cm}} \times 60 \, \text{cm}$$

To find $V_{AB}$, rearrange the equation:

$$V_{AB} = \frac{1.5 \, \text{V} \times 100 \, \text{cm}}{60 \, \text{cm}}$$ $$V_{AB} = \frac{150}{60} \, \text{V}$$ $$V_{AB} = \frac{15}{6} \, \text{V}$$ $$V_{AB} = 2.5 \, \text{V}$$

The potential difference between the ends A and B of the potentiometer is $2.5 \, \text{V}$.

Comparing this result with the given options:
A 1.5 V
B 3.75 V
C 2.5 V
D 4 V

The calculated value matches option C.