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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

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

Grade:10

1 Answers

Susmita
425 Points
4 years 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 \frac{dl1}{dl2}=\frac{1}{2}
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 F1=F2
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.

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