Explanation: Electrical resistance ($R$) is directly proportional to the length ($L$) of a conductor and inversely proportional to its cross-sectional area ($A$). This relationship is given by the formula $R = \rho L/A$, where $\rho$ is resistivity. For a circular conductor, the cross-sectional area ($A$) is calculated as $A = \pi r^2$ or $A = \pi (d/2)^2 = \pi d^2/4$, where $r$ is the radius and $d$ is the diameter. If the diameter ($d$) is doubled, the new diameter becomes $2d$. The new cross-sectional area will be $A' = \pi (2d)^2/4 = \pi (4d^2)/4 = 4(\pi d^2/4) = 4A$. So, doubling the diameter quadruples the cross-sectional area. Since resistance is inversely proportional to the cross-sectional area, if the area becomes four times larger, the resistance will become one-fourth ($1/4$) its original value. This confirms option A. Option C states that resistance varies inversely with the cross-sectional area of the conductor, which is a fundamental principle of resistance as shown in the formula $R = \rho L/A$. Since both statements A and C are correct descriptions of the effect, D is the correct answer.