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An ideal battery of emf 2V and a series resistance R are connected in the primary circuit of a potentiometer of length 1m and resistance $5 Ω$ . The value of R to give a potential difference of 5mV across the 10cm of potentiometer wire is

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$180 Ω$

$200 Ω$

$190 Ω$

$195 Ω$

answer is C.

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Detailed Solution

$\begin{array}{l} 5 \times 10^{- 3} = \frac{V}{L} l = \frac{iR}{L} l \\ 5 \times 10^{- 3} = (\frac{2}{R + 5}) \frac{5}{1} \times 10 \times 10^{- 2} \end{array}$

on solving, we get

$R = 195 Ω$

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