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A potentiometer wire of 10 m length and having a resistance of 1 ohm $m^{- 1}$ is connected to an accumulator of emf 2.2 volt and a high resistance box. To obtain a potential gradient of 2.2 mili volt/ meter, the value of resistance used from the resistance box will be

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1000 ohm

990 ohm

790 ohm

810 ohm

answer is C.

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Detailed Solution

$R_{P m} = 10 Ω$

Potential gradient

$= \frac{V_{p I}}{L} = \frac{I R_{P M}}{L} = \frac{E R_{P M}}{(R + R_{P M_{}}) L}$

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