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Question

Masses of three wires of copper are in the ratio 1 : 3 : 5 and their lengths are in the ratio of 5 : 3 : 1. The ratio of their electrical resistances are

A
1:3:5
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B
5:3:1
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C
1:15:125
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D
125:15:1
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Solution

The correct option is D 125:15:1
Mass, M =Volume × Density
M=Al×d , where A is area of cross-section and l is the length.
or A=Mld
Resistance of a wire, R=ρlA=ρl(Mld)=ρl2dM
As all the three wire are made up of same material (i.e. copper) therefore ρ and d are same for all the three wires.
Rl2MR1:R2:R3=l21M1:l22M2:l23M3=521:323:125=25:3:15
=125:15:1

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