UPSC NDA I 2019 General Ability Solved Paper

Concept:The resistance of a wire is given by $R = \frac{\rho l}{A}$, where $\rho$ is the resistivity (material constant), $l$ is the length, and $A$ is the cross-sectional area. For a cylindrical wire, $A = \pi r^2$.Explanation:Since both wires are copper, $\rho$ is the same.Let the first wire have radius $r$ and length $l$. So $R_1 = R = \frac{\rho l}{\pi r^2}$.The second wire has radius $2r$ and length $l/2$. Its area becomes $A_2 = \pi (2r)^2 = 4\pi r^2$.Thus, $R_2 = \frac{\rho (l/2)}{4\pi r^2} = \frac{\rho l}{8\pi r^2}$.Comparing, $R_2 = \frac{1}{8} \cdot \frac{\rho l}{\pi r^2} = \frac{R}{8}$.Alternatively, using ratio method: $\frac{R_1}{R_2} = \frac{l_1}{l_2} \cdot \frac{A_2}{A_1} = \frac{l}{(l/2)} \cdot \frac{4\pi r^2}{\pi r^2} = 2 \times 4 = 8$, so $R_2 = R/8$.Answer:$R_2 = \frac{R}{8}$, which corresponds to option D.