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The length of the potentiometer wire is 4m and its resistance is 8 ohm. A battery of emf. 2 volt and internal resistance 2 $Ω$ is connected in series with the wire. A cell in the secondary circuit gets balanced at 2.5 m length of the wire. On drawing a current of 0.2 A from the cell, the null point is obtained at 2 m. The internal resistance of the cell will be

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0.5 $Ω$

1 $Ω$

1.5 $Ω$

2 $Ω$

answer is B.

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Detailed Solution

E $\propto$ 2.5

(E-Ir) $\propto$ 2

$E = (\frac{2}{8 + 2}) \frac{8}{4} \times 2.5 = 1 v o l t$

$\frac{E}{E - i r} = \frac{2.5}{2}$

$\frac{1}{1 - 0.2 r} = \frac{2.5}{2}$

$r = 1 Ω$

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