A potentiometer has a wire of $100 c m$ length and its resistance is $10 o h m s .$ It is connected in series with a resistance of $40 o h m s$ and a battery of emf $2 V$ and negligible internal resistance. If a source of unknown emf $E$ connected in the secondary is balanced by $40 c m$ length of potentiometer wire, the value of $‘ E ’$ is

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$0.8 V$

$0.16 V$

$0.4 V$

$0.08 V$

answer is D.

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Detailed Solution

Length $L = 100 c m$

Resistance $R = 10 Ω$

Series resistance $R_{S} = 40 Ω$

Battery of emf $E = 2 V$

Balancing length $l = 40 c m$

emf of the cell in secondary circuit,

$E_{1} = (i ρ) l$

$E_{1} = (\frac{E}{R + R_{S}}) \frac{R}{L} l$

$= (\frac{2}{10 + 40}) \frac{10}{100} \times 40$

$= \frac{2 \times 10 \times 40}{50 \times 100}$

$= \frac{8}{50}$

$∴ E_{1} = 0.16 V$

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