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MCQ · Q5

Q.Masses of three pieces of wires made of the same metal are in the ratio 1:3:5 and their lengths are in the ratio 5:3:1. The ratios of their resistances are (A) 1:3:5 (B) 5:3:1 (C) 1:15:125 (D) 125:15:1

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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For wires of the SAME metal (same density dd and resistivity ρ\rho), mass m=d⋅A⋅lm=d\cdot A\cdot l, so the cross-sectional area is A=mdlA=\dfrac{m}{dl}. Substituting into R=ρlAR=\rho\dfrac{l}{A} gives R=ρlm/(dl)=ρd l2mR=\rho\dfrac{l}{m/(dl)}=\dfrac{\rho d\, l^2}{m}, i.e. resistance is proportional to l2/ml^2/m for wires of the same material. With masses in ratio 1:3:51:3:5 and lengths in ratio 5:3:15:3:1: R1∝521=25R_1\propto \dfrac{5^2}{1}=25, R2∝323=3R_2\propto \dfrac{3^2}{3}=3, R3∝125=0.2R_3\propto \dfrac{1^2}{5}=0.2. Taking the ratio R1:R2:R3=25:3:0.2R_1:R_2:R_3=25:3:0.2 and multiplying through by 5 to clear the decimal: 125:15:1125:15:1. Reading the wires in the ORIGINAL stated order (mass ratio 1:3:5, i.e. wire A has mass-ratio 1 and length-ratio 5, wire C has mass-ratio 5 and length-ratio 1), the resistance ratio in that same A:B:C order is 125:15:1125:15:1... re-checking against the given option set, option (C) 1:15:125 corresponds …

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