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Q.

An aluminium wire and a steel wire of the same length and cross section are joined end to end. The composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in length of the composite wire is 1.35 mm, find the ratio of the i) stress in the two wires and ii) strains in the two wires. 

YAI=7×1010Nm-2, Ysteel=2×1011Nm-2

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a

1 : 1, 20 : 7

b

 1 : 1, 10 : 7

c

 20 : 7, 1 : 1

d

10 : 7, 1 : 1

answer is A.

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Detailed Solution

i) From the question as the length is same 

l1=l2

also, the area is same A1=A2

The same load will act on both wires in a composite wire.

Thus the ratio of the stress = 1:1.

ii)

Given,

total elongation e = 1.35mm = eAl+eS

YAI=7×1010Nm-2 and  Ysteel=2×1011Nm-2

Elongation 

e=FlAY

As F, l, A are same

e1Y

eAleS=YSUAl=20×10107×1010=207

Hence the correct answer is 1:1, 20:7.

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