A potentiometer wire of length 100 cm has a resistance 5 ohm. It is connected in series with a resistance and a cell of emf 2v and of negligible internal resistance. A source of emf 5mv balanced by 10cm length of potentiometer wire. The value of external resistance is_____

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195 $Ω$

540 $Ω$

990 $Ω$

190 $Ω$

answer is B.

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Detailed Solution

$\begin{array}{l} E = \frac{E_{0} R_{w} ℓ}{(R + r + R_{w}) L} \\ \Rightarrow 5 \times 10^{- 3} = \frac{2 (5) (10)}{(R + 5) 100} \\ \Rightarrow R + 5 = 200; R = 195 Ω \end{array}$

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