In the given figure, there is a circuit of potentiometer of length AB = 10 m. The resistance per unit length is 0.1 $$\Omega$$ per cm. Across AB, a battery of emf E and internal resistance r is connected. The maximum value of emf measured by this potentiometer is:

We need to find the maximum value of EMF ($$E_{\text{max}}$$) that can be measured by the given potentiometer setup.

1. Analyze the Components of the Potentiometer

Let's break down the information provided in the circuit diagram and the text:

• Total Length of Potentiometer Wire ($$AB$$): $$L = 10\text{ m} = 1000\text{ cm}$$
• Resistance per Unit Length ($$\rho$$): $$0.1\ \Omega/\text{cm}$$
• Driver Cell Battery (Primary Circuit): $$\text{EMF } (V_{\text{driver}}) = 6\text{ V}$$
• Series Resistance in Primary Circuit ($$R_h$$): $$20\ \Omega$$

2. Calculate the Total Resistance of Wire AB

Using the resistance per unit length and the total length of the wire in centimeters:

$$R_{AB} = \rho \times L$$

$$R_{AB} = 0.1\ \Omega/\text{cm} \times 1000\text{ cm} = 100\ \Omega$$

3. Determine the Current in the Primary Circuit

The total current ($$I$$) flowing through the main driving circuit loop (comprising the $$6\text{ V}$$ battery, the $$20\ \Omega$$ resistor, and the potentiometer wire $$AB$$) is given by Ohm's Law:

$$I = \frac{V_{\text{driver}}}{R_{AB} + R_h}$$

$$I = \frac{6\text{ V}}{100\ \Omega + 20\ \Omega} = \frac{6}{120} = 0.05\text{ A}$$

4. Find the Maximum Measurable EMF

A potentiometer can only balance an external cell if its EMF is less than or equal to the total potential drop across the entire length of the potentiometer wire ($$V_{AB}$$). Therefore, the maximum value of EMF that can be measured occurs when the balancing length extends to the maximum limit of the wire ($$10\text{ m}$$):

$$E_{\text{max}} = V_{AB}$$

Using Ohm's Law across the wire $$AB$$:

$$V_{AB} = I \times R_{AB}$$

$$V_{AB} = 0.05\text{ A} \times 100\ \Omega = 5\text{ V}$$

Final Answer: Option A ($$5\text{ V}$$)