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ETwo vertical wires X and Y, suspended at the same horizontal level, are connected by a light rod XY at their lower ends. The wire have the same length l and cross- sectional area A. A weight of 30N is placed at O on the rod, where XO:OY =1;2.  Both wires are stretched and the rod XY then remains horizontal.If the wire X has a young modulus E1 of 1.0×10^11N/m^2, Calculate the young modulus kf E2
one year ago

I cannot give you any quantitative answer.But may be I can give you some qualitative idea so that you can think further.
Look that X and Y are made of different material.So the amount of stretch in X,Y for the same amount of force are different.
So when we attach the rod ,X and Y will stretch to different amount and the rod will not be horizontal.
To make the rod horizontal we are applying a force of 30 N.But see that we are applying more of it towards X than Y.This means if we want a X and Y to stretch to l+dl then we have to apply more force on X than on Y.This means that for the same amount of force X stretches less than Y.
Or we can say that
In the end the rod is horizontal.That means the centre of mass of the rod is at the centre of it.That means the amount of force on X and Y is the same.
Or
Given A1=A2 and l1=l2.
From E=(Fl)/(Adl) we get
E2/E1=dl1/dl2=1/2
Or E2=E1/2=0.5×10^11 N/m^2
I hope this helps you.
one year ago
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