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Two uniform wires A and B are of the same metal and have equal masses. The radius of wire A is twice that of wire B. The total resistance of A and B when connected in parallel is

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4 Ω when the resistance of wire A is 4.25 Ω

5 Ω when the resistance of wire A is 4.25 Ω

4 Ω when the resistance of wire B is 4.25 Ω

5 Ω when the resistance of wire B is 4.25 Ω

answer is A.

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Detailed Solution

$\frac{R_{A}}{R_{B}} = {(\frac{r_{B}}{r_{A}})}^{4} \Rightarrow \frac{R_{A}}{R_{B}} = {(\frac{1}{2})}^{4} = \frac{1}{16} \Rightarrow R_{B} = 16 R_{A}$

When R_A and R_B are connected in parallel then equivalent resistance

$R_{e q} = \frac{R_{A} R_{B}}{(R_{A} + R_{B})} = \frac{16}{17} R_{A}$

$If R_{A} = 4.25 Ω then R_{e q} = 4 Ω i.e.option (a) is correct.$

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