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(D) The potential gradient $x$ is defined as the potential drop per unit length,$x = \frac{V}{L} = \frac{IR}{L}$.Since the wires are in series,the cur

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$1 : 9$

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(D) The potential gradient $x$ is defined as the potential drop per unit length,$x = \frac{V}{L} = \frac{IR}{L}$.Since the wires are in series,the current $I$ flowing through both wires is the same.Given that the lengths $L$ are equal and the material is the same (resistivity $\rho$ is constant),the resistance $R$ is given by $R = \rho \frac{L}{A}$,where $A$ is the cross-sectional area.Thus,the potential gradient $x = \frac{I \rho L}{A L} = \frac{I \rho}{A}$.Since $I$ and $\rho$ are constant,$x \propto \frac{1}{A}$.Since the area $A = \pi r^2 = \pi (d/2)^2$,we have $A \propto d^2$,where $d$ is the diameter.Therefore,$x \propto \frac{1}{d^2}$.Given the ratio of diameters $d_{AC} : d_{CB} = 3 : 1$,the ratio of potential gradients is $\frac{x_{AC}}{x_{CB}} = \frac{d_{CB}^2}{d_{AC}^2} = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$.Thus,the ratio is $1 : 9$.

$A$ potentiometer consists of two wires $AC$ and $CB$ of the same material and of equal lengths but diameters in the ratio $3 : 1$. Then the potential gradients on the two wires will be in the ratio :-

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