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The resistances of two metallic wires are R₁ = (5± 0.2)Ω and R₂ = (10 ± 0.1) Ω when these two wires are connected in parallel, the equivalent resistance is found to be 3.3Ω with a maximum percentage error is

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answer is B.

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Detailed Solution

$\frac{{ΔR}_{P}}{R_{P}} \times 100 = R_{P} [\frac{{ΔR}_{1}}{R_{1}^{2}} + \frac{{ΔR}_{2}}{R_{2}^{2}}] \times 100 = 3 . Where R_{P} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} .$

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