In the following diagram,the lengths of wires | vedclass

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(B) Let the length of both wires be $l$. Let the radius of wire $BC$ be $r$,so the radius of wire $AB$ is $2r$.The resistance of a wire is given by $R

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$1: 4$

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(B) Let the length of both wires be $l$. Let the radius of wire $BC$ be $r$,so the radius of wire $AB$ is $2r$.The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$.For wire $BC$,$R_{BC} = \rho \frac{l}{\pi r^2} = R$ (let).For wire $AB$,$R_{AB} = \rho \frac{l}{\pi (2r)^2} = \rho \frac{l}{4\pi r^2} = \frac{R}{4}$.Since the wires are in series,the same current $I$ flows through both.The potential gradient $x$ is given by $x = \frac{V}{l} = \frac{IR}{l}$.Since $I$ and $l$ are the same for both wires,the potential gradient is proportional to the resistance: $x \propto R$.Therefore,the ratio of potential gradients is $\frac{x_{AB}}{x_{BC}} = \frac{R_{AB}}{R_{BC}} = \frac{R/4}{R} = \frac{1}{4}$.Thus,the ratio is $1: 4$.

In the following diagram,the lengths of wires $AB$ and $BC$ are equal,but the radius of wire $AB$ is double that of $BC$. The ratio of the potential gradient on wires $AB$ and $BC$ will be (wires are made of the same material):

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