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Potentiometer wire of length 1 m is connected in series with 490 $Ω$ resistance and 2 V battery. If 0.2 mV/cm is the potential gradient then resistance of the potentiometer wire is:

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4.9 $Ω$

7.9 $Ω$

5.9 $Ω$

6.9 $Ω$

answer is A.

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Detailed Solution

$Current in the circuit is$

$i = \frac{2}{490 + R_{P}}$

$potential gradient$

$\frac{i R_{P}}{L} = \frac{0.2 \times 10^{- 3}}{10^{- 2}} = 0.02 V / m$

$\frac{2}{490 + R_{P}} \frac{R_{P}}{1} = 0.02$

$solving for R_{p} we get R_{p} = 4.9 Ω$

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