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A potentiometer wire of length $100 c m$ has a resistance of $10 . 11$ is connected in series with a resistance $R$ and a cell of emf $2 V$ and of negligible internal resistance. A source of emf $10 m V$ is balanced against a length $40 c m$ of the potentiometer wire. What is the value of $R$ ?

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$R = 1060 Ω$

$R = 670 Ω$

$R = 790 Ω$

$R = 580 Ω$

answer is D.

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Detailed Solution

From the theory of potentiometer, $V_{b} = E$ , if no current is drawn from the battery of

$(\frac{E_{1}}{R + R_{a b}}) R_{ω} = E$

$R_{b} = (\frac{40}{100}) \times 10 = 4 Ω$

Here, $E_{1} = 2 V, R_{a s} = 10 Ω, R_{c} = (\frac{40}{100}) \times 10 = 4 Ω$

and

$E = 10 \times 10^{- 3} V$ $\frac{E_{0} R_{W}}{(R_{W} + R) L} = \frac{E_{2}}{L_{2}} \frac{2 X 10}{(R + 10) 100} = \frac{10 X 10^{- 3}}{40} R = 790 O h m$

Substituting in above equation, we get

$R = 790 Ω$

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