UPSC NDA I 2022 General Ability Solved Paper

Concept:When a wire is cut into equal parts, the resistance of each piece is proportional to its length.So each part has $\frac{1}{n}$ of the original resistance if cut into $n$ equal pieces.Connecting identical resistors in parallel gives an equivalent resistance that is $\frac{1}{n}$ of each individual resistance.Explanation:Original resistance $R = 50\,\Omega$.The wire is cut into 5 equal pieces.Resistance of each piece $= \frac{50}{5} = 10\,\Omega$.All 5 pieces are connected in parallel.For $n$ identical resistors each of resistance $r$ in parallel, equivalent resistance $R_{eq} = \frac{r}{n}$.Here $r = 10\,\Omega$ and $n = 5$.So $R_{eq} = \frac{10}{5} = 2\,\Omega$.Verification using parallel formula:$\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{5}{10} = \frac{1}{2}$.Hence $R_{eq} = 2\,\Omega$.Answer:$2\,\Omega$ (Option A).