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two metal wires have the resistivity of 1.62 × 10^-18 ωm and 4.86 × 10^-18 ωm. compare the resistance of 2 wires if a is twice as long as b and half of its diameter |
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Answer» riend,◆ Answer -R1/R2 = 1/6◆ Explaination-# Given-ρ1 = 1.62×10^-18 ohm-mρ2 = 4.86×10^-18 ohm-ml1/l2 = 2d1/d2 = 1/2# Solution-Resistance of wire A,R1 = ρ1l1/A1Resistance of wire B,R2 = ρ2l2/A2Ratio of RESISTANCES is -R1/R2 = (ρ1l1/A1) / (ρ2l2/A2)R1/R2 = (ρ1/ρ2) × (l1/l2) × (d2/d1)^2R1/R2 = (1.62×10^-18 / 4.86×10^-18) × (2) × (1/2)^2R1/R2 = (1/3) × (2) × (1/4)R1/R2 = 1/6 THEREFORE, resistance of wire B is SIX times that of A.Hope it helps... |
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