A potentiometer wire of 10m length and 20 ohm resistance is connected in series with resistance R ohms and a battery of emf 2V, negligible internal resistance. The potential gradient on the wire is 0.16 millivolt/centimetre then R is---ohms

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230

46.6

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{dV}{dx} = \frac{iρ}{A} = \frac{E}{R + R^{1}} \times \frac{R}{L} \\ 0.16 \times \frac{10^{- 3}}{10^{- 2}} = \frac{2 \times 20}{(20 + R) 10} \\ R = 230 Ω \end{array}$

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