1.

A battery of emf epsilon_0 = 5V and internal resistance 5 Omega is connected across a long uniform wire AB of length 1m and resistance per unit length5 Omegam^(-1). Two cells of epsilon_1 = 1 V and epsilon_2 = 2V are connected as shown in the figure.

Answer»

the null point is at A.
If the jockey is touched to point B, the current in the galvanometer will be going towards B.
When jockey is connected to point A, no current flows through 1 V battery.
The null point is at distance of `8//15` m from `A_2`.

Solution :a., b.
For null point, current flows in the
loop CD only.

`i = (3V)/(2OMEGA + 1Omega) = 1A`
`V_(CD) = 1V - 1(1) = 0`
Therefore, OPTION (a) is correct. That
is, `V_(A) gt V_(B)`. When jockey touches B, current from A to B increases the
PD across the secondary CIRCUIT.
Therefore, option(b) is correct.


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