JEE Main 2021 — Current Electricity Question with Solution

In the given circuit of potentiometer, the potential difference E across AB (10 m length) is larger than E₁ and E₂ as well. For key K₁ (closed), the jockey is adjusted to touch the wire at point J₁ so that there is no deflection in the galvanometer. Now the first battery (E₁) is replaced by second battery (E₂) for working by making K₁ open and K₂ closed. The galvanometer gives then null deflection at J₂. The value of $$\frac{E_1}{E_2}$$ is $$\frac{a}{b}$$, where a = _________.