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A material ‘B’ is having twice the specific resistance of another material ‘A’. A circular wire made of ‘B’ is having twice the diameter of a wire made of ‘A’. Then for ‘A’ and ‘B’ to have the same resistance, which of the following ratio is $\frac{1 B}{1 A}$ of their respective lengths?

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\frac{1}{2}

\frac{1}{4}

answer is A.

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Detailed Solution

A material ‘B’ is having twice the specific resistance of another material ‘A’. A circular wire made of ‘B’ is having twice the diameter of a wire made of ‘A’. Then for ‘A’ and ‘B’ to have the same resistance, the ratio

\frac{1 B}{1 A}

of their respective lengths must be 2.

R = ρ \frac{L}{A}

⟹

R = ρ \frac{4 \times L}{D^{2}}

where D is diameter
Given that

\frac{ρB}{ρA}

= 2
and,

\frac{DA}{DB}

\frac{1}{2}

For same resistance-
ρA

\frac{L_{A}}{D^{2} \times A}

= ρB

\frac{L_{B}}{D^{2} \times B}

\frac{L_{B}}{L_{A}}

\frac{ρB}{ρA}

\times \frac{D^{2} B}{D^{2} A}

\frac{L_{B}}{L_{A}}

\frac{1}{2}

\times 2^{2}

\frac{L_{B}}{L_{A}}

= 2

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