The resistivity of a potentiometer wire is 40 | vedclass

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(B) The resistance $R$ of a wire is given by $R = \frac{\rho l}{A}$, where $\rho$ is resistivity, $l$ is length, and $A$ is the area of cross-section.

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$10^{-2} 	ext{ V m}^{-1}$

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(B) The resistance $R$ of a wire is given by $R = \frac{ho l}{A}$, where $ho$ is resistivity, $l$ is length, and $A$ is the area of cross-section.Potential gradient is defined as the potential drop per unit length, given by $x = \frac{V}{l}$.Using Ohm's law, $V = IR$, so $x = \frac{IR}{l} = I \left( \frac{R}{l} ight)$.From the resistance formula, $\frac{R}{l} = \frac{ho}{A}$.Substituting the given values: $\frac{R}{l} = \frac{40 	imes 10^{-8} \Omega 	ext{ m}}{8 	imes 10^{-6} 	ext{ m}^2} = 5 	imes 10^{-2} \Omega 	ext{ m}^{-1}$.Now, calculating the potential gradient: $x = I 	imes \left( \frac{R}{l} ight) = 0.2 	ext{ A} 	imes 5 	imes 10^{-2} \Omega 	ext{ m}^{-1} = 10^{-2} 	ext{ V m}^{-1}$.

The resistivity of a potentiometer wire is $40 \times 10^{-8} \Omega \text{ m}$ and its area of cross-section is $8 \times 10^{-6} \text{ m}^2$. If $0.2 \text{ A}$ current is flowing through the wire, the potential gradient of the wire is:

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