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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
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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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Other answers:

can u give me the link of the book?
you are not from India?
nope
i suggest you google it..its easy to get NCERT CLASS 11 PHYSICS PART 2
do they have the solutions
yes i think .....
worth checking
infact the book is very nice for reference
http://www.vidhyarjan.com/news/ncert-solutions-for-class-11-physics
http://ncertbooks.prashanthellina.com/11_Physics.html
can u change the language?
ofcourse eng is below scroll down
gt it?
the question isnt there
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..
can u show the working please:)
try doing it :P.. first you tell me how much will be the stress that comes on X and Y ?
stress = force/ area how do i distrbte the force
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!
hence Fx : Fy = 2:1
so now calculate Fx and Fy
Fx = 2/3 Fy Fy = 1/3 Fx ?
total force is known to you.. so find the ratios now!
Fx = 2/3 Ftotal :P and Fy = 1/3 Ftotal
Fx = 20N and Fy = 10N
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
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
woah wait.. how did you get the strain for Y ?
arent they equal
nope :-/.. why would the strains be equal?
same mterial
now you have to use the first equatoin.. that i put.. but use lx and ly notation instead of l1 and l2 notation
they are NOT SAME MATERIAL.. IF THEY WERE.. THEY WOULD HAVE SAME YOUNG"S MODULUS :P
i dont undrstnd >< cn u shw the working
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
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
ok il use thnks fr ur help :)
No problimo amigo :D

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