In a potentiometer, the area of cross-section | vedclass

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(A) Given: Area of cross-section $A = 4 \, cm^2 = 4 	imes 10^{-4} \, m^2$, Current $I = 1 \, A$, Potential gradient $k = 7.5 \, V/m$.The potential gr

Vedclass · Accepted Answer

$3 	imes 10^{-3} \, \Omega \cdot m$

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(A) Given: Area of cross-section $A = 4 \, cm^2 = 4 	imes 10^{-4} \, m^2$, Current $I = 1 \, A$, Potential gradient $k = 7.5 \, V/m$.The potential gradient $k$ is defined as $k = \frac{V}{l}$, where $V$ is the potential difference across length $l$.Using Ohm's law, $V = I \cdot R$, where $R = ho \frac{l}{A}$.Substituting $V$ in the expression for $k$: $k = \frac{I \cdot ho \cdot l}{A \cdot l} = \frac{I \cdot ho}{A}$.Rearranging for resistivity $ho$: $ho = \frac{k \cdot A}{I}$.Substituting the values: $ho = \frac{7.5 	imes 4 	imes 10^{-4}}{1} = 30 	imes 10^{-4} \, \Omega \cdot m = 3 	imes 10^{-3} \, \Omega \cdot m$.

In a potentiometer, the area of cross-section of the wire is $4 \, cm^2$, the current flowing in the circuit is $1 \, A$ and the potential gradient is $7.5 \, V/m$, then the resistivity of the potentiometer wire is

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