In the below circuit, the battery E1 has an emf of 12 V and zero internal resistance. While the battery E2 has an emf of 2 V. If the galvanometer G reads zero, then the value of the resistance X in ohms is:

250 100 50 200

100

We have, I1 = 0 ∴  VX = E2 = 2V $\frac{V_{500 \Omega}}{V_{X}}=\frac{(12-2)}{2}=\frac{500}{X}$ ∴  X = 100 Ω