A potentiometer wire of length 100 cm has a resistance of 30 ohms. It is connected in series with a resistance of 20 ohms and an accumulator of emf 10 V having negligible internal resistance. A source of 2.4 V is balanced against a length L of the potentiometer wire. What is the value of L?

Correct option is B

​$$\text{Given:} \\\text{Potentiometer wire length, } l = 100 \ \text{cm} \\\text{Resistance of potentiometer wire, } R_1 = 30 \ \Omega \\\text{Resistance of series resistor, } R_2 = 20 \ \Omega \\\text{Source emf, } V = 10 \ \text{V} \\\text{Source emf for balancing, } V_{\text{bal}} = 2.4 \ \text{V}$$

$$\text{The total resistance } R \text{ in the circuit is the sum of } R_1 \text{ and } R_2:$$

$$R = R_1 + R_2 = 30 \ \Omega + 20 \ \Omega = 50 \ \Omega$$

$$\text{The current passing through the potentiometer wire can be found using Ohm's law:}$$​$$I = \frac{V}{R} = \frac{10}{50} = 0.2 \ \text{A}$$

$$\text{The potential difference } V \text{ across the potentiometer wire (with resistance } R_1 = 30 \ \Omega) \text{ is:}$$$$V = I \cdot R_1 = 0.2 \times 30 = 6 \ \text{V}$$

$$\text{From the potentiometer formula:} \\\phi = \frac{V}{T} = \frac{6}{100} = 0.06$$

$$\text{Where:} \\\phi \text{ is the potential gradient,} \\T \text{ is the total length of the wire (100 cm).}\text{Now, the emf of 2.4 V is balanced against the length } L \text{ of the wire:} \\2.4 = \phi \cdot L$$$$\text{Substitute the value of } \phi: \\2.4 = 0.06 \cdot L$$

$$\text{Solving for } L: \\L = \frac{2.4}{0.06} = 40 \ \text{cm}$$​​​​​​​​​​​