anonymous
  • anonymous
A uniform steel wire X and Y, suspended at the same horizontal level, are connected by a light and cross-sectional area A, A weight of 30 N is placed at O, on the road OX : OY = 1 : 2. Both wires are stretched and the rod XY then remains horizontal. If he wire X has a young modulus E1 of 1.0 * 10^11 pa, calculate the yound modulus of E2 of the wire Y assuming the elastic limit is not exceeded for both wires.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1358779489492:dw|
anonymous
  • anonymous
@Hero @Preetha @ghazi @AravindG
AravindG
  • AravindG
I think I saw a similiar question in NCERT text chapter 11 under topic of Young's modulus and all..take a look

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anonymous
  • anonymous
can u give me the link of the book?
AravindG
  • AravindG
you are not from India?
anonymous
  • anonymous
nope
AravindG
  • AravindG
i suggest you google it..its easy to get NCERT CLASS 11 PHYSICS PART 2
anonymous
  • anonymous
do they have the solutions
AravindG
  • AravindG
yes i think .....
AravindG
  • AravindG
worth checking
AravindG
  • AravindG
infact the book is very nice for reference
anonymous
  • anonymous
http://www.vidhyarjan.com/news/ncert-solutions-for-class-11-physics
AravindG
  • AravindG
http://ncertbooks.prashanthellina.com/11_Physics.html
anonymous
  • anonymous
can u change the language?
AravindG
  • AravindG
ofcourse eng is below scroll down
AravindG
  • AravindG
gt it?
anonymous
  • anonymous
the question isnt there
anonymous
  • anonymous
Ok.. so since the xy is horizontal.. it means after stretching, both the bars have exactly the same length.. so i can say \[l1+\Delta l1 = l2 + \Delta l2 \] next step we find the stress that is coming on each of them.. that should be easy.. since the ration of ox: oy is given to us..
anonymous
  • anonymous
can u show the working please:)
anonymous
  • anonymous
try doing it :P.. first you tell me how much will be the stress that comes on X and Y ?
anonymous
  • anonymous
stress = force/ area how do i distrbte the force
anonymous
  • anonymous
first find the force.. force equals the weight that each one would experience.. the weight is at a ratio of 2:1 .. hence the distribution will be in inverse ratio right? 1:2.. cause if weight is far away, its effect is smaller!
anonymous
  • anonymous
hence Fx : Fy = 2:1
anonymous
  • anonymous
so now calculate Fx and Fy
anonymous
  • anonymous
Fx = 2/3 Fy Fy = 1/3 Fx ?
anonymous
  • anonymous
total force is known to you.. so find the ratios now!
anonymous
  • anonymous
Fx = 2/3 Ftotal :P and Fy = 1/3 Ftotal
anonymous
  • anonymous
Fx = 20N and Fy = 10N
anonymous
  • anonymous
ok.. :D.. next.. find the stress on X and since young's modulus of X is given.. find the strain.. Similarly do the same thing for Y.. but young's modulus is not given.. divide the two equations to get rid of A
anonymous
  • anonymous
thanks alot Strain = F/A*E1 strain in X = 20/A* 1.0 * 10^11 strain is equal so E2 = stress/ strain E2 = 10/A *A*1.0*10^11/20 E2 = 5 * 10^11 pa
anonymous
  • anonymous
woah wait.. how did you get the strain for Y ?
anonymous
  • anonymous
arent they equal
anonymous
  • anonymous
nope :-/.. why would the strains be equal?
anonymous
  • anonymous
same mterial
anonymous
  • anonymous
now you have to use the first equatoin.. that i put.. but use lx and ly notation instead of l1 and l2 notation
anonymous
  • anonymous
they are NOT SAME MATERIAL.. IF THEY WERE.. THEY WOULD HAVE SAME YOUNG"S MODULUS :P
anonymous
  • anonymous
i dont undrstnd >< cn u shw the working
anonymous
  • anonymous
use the first equation .. try doing it.. see you need to get rid of \[\Delta ly/ ly\] in that equation i asked you to build by dividing the two equations
anonymous
  • anonymous
so here are the two equations \[Fx / Fy = Ex \Delta x ly / Ey \Delta ylx \] and \[lx + \Delta x = ly + \Delta y\] somehow use them
anonymous
  • anonymous
ok il use thnks fr ur help :)
anonymous
  • anonymous
No problimo amigo :D

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