UPSC NDA I 2023 General Ability Solved Paper

Concept:The resistance of a wire depends on its length and cross-sectional area.Stretching the wire changes both, which affects the current.Explanation:Given original resistance $R_1 = 10\ \Omega$, length $l_1$, and area $A_1$.The wire is stretched to double its length, so new length $l_2 = 2l_1$.Volume remains constant: $A_1 l_1 = A_2 l_2$.Thus $A_2 = \frac{A_1 l_1}{l_2} = \frac{A_1}{2}$.Resistance is given by $R = \frac{\rho l}{A}$, where $\rho$ is resistivity.Original resistance: $R_1 = \frac{\rho l_1}{A_1}$.New resistance: $R_2 = \frac{\rho (2l_1)}{A_1/2} = \frac{4\rho l_1}{A_1} = 4R_1 = 40\ \Omega$.The resistance becomes four times the original value.For a constant applied voltage $V$, Ohm's law gives $I = V/R$.Current is inversely proportional to resistance.New current $I_2 = \frac{V}{4R_1} = \frac{1}{4} I_1$.Therefore, the current becomes one‑fourth of its original value.Answer:The correct option is C: one‑fourth of its original value.