UPSC NDA I 2025 General Ability Paper

Concept:For a constant voltage, heat produced is inversely proportional to resistance. So the ratio of heat depends on the equivalent resistance in each configuration.Explanation:Let the resistance of each wire be $R$.When connected in parallel, equivalent resistance $R_p = \frac{R}{2}$.Heat produced $H_p \propto \frac{1}{R_p} \propto \frac{2}{R}$.When connected in series, equivalent resistance $R_s = 2R$.Heat produced $H_s \propto \frac{1}{R_s} \propto \frac{1}{2R}$.Thus, the ratio $H_p : H_s = \frac{2}{R} : \frac{1}{2R} = 4:1$? Wait, recalc: $\frac{2/R}{1/(2R)} = 4$, so ratio is 4:1. But the original solution gave 2:1? Let's verify: Original says "H_{parallel} ∝ 1 / R_{parallel} ⇒ ∝ 1 / (R/2) = 2/R" and "H_{series} ∝ 1/(2R)". Then ratio (2/R) : (1/(2R)) = (2/R) * (2R/1) = 4:1. The original solution mistakenly concluded 2:1. Actually, (2/R) / (1/(2R)) = (2/R)*(2R) = 4. So correct ratio is 4:1. The options include 4:1 as B. So the existing solution had an error. We must correct it. The question asks to rewrite the given solution, but we are to verify internally. The problem statement says "Two conducting wires of same material... equal lengths and equal diameters" so each has same resistance R. Parallel: R/2, series: 2R. Heat = V^2 t / R, so H_parallel / H_series = (R_series)/(R_parallel) = 2R / (R/2) = 4. So ratio is 4:1. The given solution incorrectly says 2:1. As an expert tutor, we should provide the correct solution. The user says "Verify the result internally before presenting". So we must correct it. The answer should be option B: 4:1.Thus, rewrite with correct ratio.Keep formatting: Concept, Explanation, Answer. Use HTML. No extra text.Concept:Heat produced in a resistor is given by $H = \frac{V^2}{R}t$. For the same voltage and time, heat is inversely proportional to resistance.Explanation:Each wire has the same resistance $R$ (same material, length, and diameter).In parallel: equivalent resistance $R_p = \frac{R}{2}$. So heat $H_p \propto \frac{1}{R_p} = \frac{2}{R}$.In series: equivalent resistance $R_s = 2R$. So heat $H_s \propto \frac{1}{R_s} = \frac{1}{2R}$.Ratio $H_p : H_s = \frac{2}{R} : \frac{1}{2R} = \frac{2/R}{1/(2R)} = 4:1$.Answer:4:1 (Option B)