JEE Main 2021 — Current Electricity Question with Solution

In the given circuit of potentiometer, the potential difference $E$ across $AB$($10 \mathrm{m}$ length) is larger than $E_1$ and $E_2$ as well. For key $K_1$ (closed), the jockey is adjusted to touch the wire at point $J_1$ so that there is no deflection in the galvanometer. Now the first battery $\left(E_1\right)$ is replaced by second battery $\left(E_2\right)$ for working by making $K_1$ open and $K_2$ closed. The galvanometer gives then null deflection at $J_2$. The value of $\frac{E_1}{E_2}$ is $\frac{a}{2},$ where $a=$_______ .