A material B has twice the specific resistance (resistivity) of A . A circular wire made of B has twice the diameter of a wire made of A . For the two - Physics MCQ AIEEE 2006 2026 | ExamTest.live

Solution & Step-by-step Explanation

Resistance

R = ρ \frac{L}{A} = ρ \frac{L}{π ( d /2 ) ^{2}} = ρ \frac{4 L}{π d ^{2}}

.Given:

ρ_{B} = 2 ρ_{A}

d_{B} = 2 d_{A}

, and

R_{A} = R_{B}

ρ_{A} \frac{L _{A}}{d _{A}^{2}} = ρ_{B} \frac{L _{B}}{d _{B}^{2}}

ρ_{A} \frac{L _{A}}{d _{A}^{2}} = (2 ρ_{A}) \frac{L _{B}}{( 2 d _{A} ) ^{2}}

\frac{L _{A}}{d _{A}^{2}} = \frac{2 L _{B}}{4 d _{A}^{2}} = \frac{L _{B}}{2}

\frac{L _{A}}{L _{B}} = \frac{1}{2}

Self-Correction: Checking the FIITJEE key provided in text (

2

). Let's re-read: "twice the specific resistance... twice the diameter".

R_{A} = ρ \frac{L _{A}}{A _{A}}

R_{B} = 2 ρ \frac{L _{B}}{4 A _{A}} = \frac{1}{2} ρ \frac{L _{B}}{A _{A}}

.If

R_{A} = R_{B} ⟹ L_{A} = \frac{1}{2} L_{B} ⟹ L_{A} / L_{B} = 1/2

.Wait, the FIITJEE text says (1) which is 2. Let's re-verify:

\frac{ρ _{A} L _{A}}{r _{A}^{2}} = \frac{2 ρ _{A} L _{B}}{( 2 r _{A} ) ^{2}} ⟹ L_{A} = \frac{2}{4} L_{B} ⟹ L_{A} = 0.5 L_{B}

.Likely the source key has

L_{B} / L_{A}

or a typo. I will stick to the calculation.

A material $B$ has twice the specific resistance (resistivity) of $A$ . A circular wire made of $B$ has twice the diameter of a wire made of $A$ . For the two wires to have the same resistance, the ratio $L_{A} / L_{B}$ of their respective lengths must be: