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Question

Masses of three wires of the same metal are in the ratio 1:2:3 and their lengths in the ratio 3:2:1. Electrical resistances of these wires will be in the ratio

A
1:2:3
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B
3:2:1
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C
1:6:27
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D
27:6:1
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Solution

The correct option is C 27:6:1
Here the ratio of masses of the wires is 1:2:3. So,
m1=k,m2=2k and m3=3k, where k is the proportionality constant.
And the ratio of there lengths is 3:2:1.
So, l1=3s,l2=2s,l3=s, where s is also a proportionality constant.

Now the resistance of a wire is given by,

R=ρlA=ρl2V=ρl2dM.

Using, density(d)=Mass(m)Volume(V). Or, Rl2m.
R1(3s)2k
R2(2s)22k
R3s23k

Using it the ratio of their resistance is 27:6:1.

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