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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 $ε_{1} = 1 V$ and $ε_{2} = 2 V$ are connected as shown in the figure.

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The null point is at A

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

The null point is at distance of 8/15 m from A

When Jockey is connected to point A no current is flowing through 1V battery

answer is A, B.

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Detailed Solution

For null point current flow in the loop CD only.

$i = \frac{3 V}{2 Ω + 1 Ω} = 1 A, V_{C D} = 1 V - 1 (1) = 0$

$∴$ option (A) is correct is $V_{A} > V_{B} .$

When Jockey touches to B current flow from A to B to increase the P.D. across the secondary circuit.

$∴$ option (B) is correct.

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