A potentiometer wire PQ of 1m length is connected to a standard cell $$E_1$$. Another cell $$E_2$$ of emf 1.02 V is connected with a resistance 'r' and switch S (as shown in figure). With switch S open, the null position is obtained at a distance of 49 cm from Q. The potential gradient in the potentiometer wire is:

Total length of potentiometer wire $$PQ$$: $$L = 1\text{ m} = 100\text{ cm}$$

Given null point distance measured from end Q: $$l_Q = 49\text{ cm}$$

Balancing length measured from the reference zero-potential end P: $$l_P = L - l_Q = 100 - 49 = 51\text{ cm}$$

With switch S open, no current is drawn from cell $$E_2$$, making the balancing voltage equal to its open-circuit emf:

$$V_{\text{balance}} = E_2 = 1.02\text{ V}$$

Potential gradient $$k$$ along the wire: $$k = \frac{E_2}{l_P} = \frac{1.02}{51} = 0.02\text{ V/cm}$$