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A battery of emf $ε_{0} = 5 V$ and internal resistance $5 Ω$ is connected across a long uniform wire AB of length 1 m and resistance per unit length $5 Ω m^{- 1} .$ Two cells of $ε_{1} = 1 V a n d ε_{2} = 2 V$ are connected as shown in the figure.

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The null point is at A

The null point is at distance of 8/15 m from $A_{2}$

When jockey is connected to point A, no current flows through 1 V battery

If the jockey is touched to point B, the current in the galvanometer will be going towards B

answer is A, B.

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Detailed Solution

For null point, current flows in the loop CD only.

$i = \frac{3 V}{2 Ω + 1 Ω} = 1 A$

Therefore, option 1 is correct. That is, $V_{A} > V_{B} .$ When jockey touches B, current from A to B increases the PD across the secondary circuit. Therefore, option 2 is correct

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