A 6 volt battery of internal resistance $1 Ω$ is connected across a uniform wire AB of length 100 cm.The positive terminal of another battery of emf 4V and internal resistance $1 Ω$ is joined to the point A as shown. Find the distance of point D from A if potential of C is equal to potential of D.(resistance of $AB = 5 Ω$ )

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60 cm

80 cm

40 cm

75 cm

answer is A.

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Detailed Solution

Potential difference in the wire at length $l$ is

$\Rightarrow V = \frac{E}{R + r} \cdot \frac{R}{L} l$

$4 = \frac{6}{6} \times \frac{5}{100} \times 1 \Rightarrow l = 80 cm$

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