If specific resistance of a potentiometer wire is $10^{- 7} Ω m$ and current flowing through it is $0.1 a m p$ ., cross-sectional area of wire is $10^{- 6} m^{2}$ then potential gradient will be

10^{- 2} v o l t / m

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10^{- 4} v o l t / m

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10^{- 6} v o l t / m

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10^{- 8} v o l t / m

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Solution

The correct option is A $10^{- 2} v o l t / m$
Resistance of the potentiometer wire $R = ρ \frac{l}{A}$
Potential difference across the wire $V = I l$
So. potential gradient $\frac{V}{l} = \frac{I R}{l} = \frac{I ρ l}{A l}$
Or $\frac{V}{l} = \frac{I ρ}{A} = \frac{0.1 \times 10^{- 7}}{10^{- 6}} = 0.01 = 10^{- 2} V / m$ .

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