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.

Step 1: Identify the given values and the principle of a potentiometer. The problem provides the following information from Figure 5: • The EMF of the test cell, E_2 = 1.5 \, V. • The total length of the potentiometer wire AB, L_AB = 100 \, cm. • The balance length AC, L_AC = 60 \, 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. Step 2: Relate the potential difference across the balance length to the potential difference across the entire wire. The potential gradient (k) along the potentiometer wire is uniform. It is defined as the potential difference per unit length: k = V_ABL_AB The potential difference across the balance length AC can also be expressed using the potential gradient: V_AC = k × L_AC Substituting the expression for k: V_AC = V_ABL_AB × L_AC Step 3: Substitute the known values and solve for V_AB. Since E_2 = V_AC, we can write: E_2 = V_ABL_AB × L_AC Now, plug in the given values: 1.5 \, V = V_AB100 \, cm × 60 \, cm To find V_AB, rearrange the equation: V_AB = 1.5 \, V × 100 \, cm60 \, cm V_AB = (150)/(60) \, V V_AB = (15)/(6) \, V V_AB = 2.5 \, V The potential difference between the ends A and B of the potentiometer is 2.5 \, 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. The final answer is 2.5 V