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.

- anonymous

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- anonymous

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- anonymous

@Hero @Preetha @ghazi @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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## More answers

- anonymous

can u give me the link of the book?

- AravindG

you are not from India?

- anonymous

nope

- AravindG

i suggest you google it..its easy to get NCERT CLASS 11 PHYSICS PART 2

- anonymous

do they have the solutions

- AravindG

yes i think .....

- AravindG

worth checking

- AravindG

infact the book is very nice for reference

- anonymous

http://www.vidhyarjan.com/news/ncert-solutions-for-class-11-physics

- AravindG

http://ncertbooks.prashanthellina.com/11_Physics.html

- anonymous

can u change the language?

- AravindG

ofcourse eng is below scroll down

- AravindG

gt it?

- anonymous

the question isnt there

- 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

can u show the working please:)

- anonymous

try doing it :P.. first you tell me how much will be the stress that comes on X and Y ?

- anonymous

stress = force/ area
how do i distrbte the force

- 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

hence Fx : Fy = 2:1

- anonymous

so now calculate Fx and Fy

- anonymous

Fx = 2/3 Fy
Fy = 1/3 Fx ?

- anonymous

total force is known to you.. so find the ratios now!

- anonymous

Fx = 2/3 Ftotal :P and Fy = 1/3 Ftotal

- anonymous

Fx = 20N and Fy = 10N

- 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

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

woah wait.. how did you get the strain for Y ?

- anonymous

arent they equal

- anonymous

nope :-/.. why would the strains be equal?

- anonymous

same mterial

- 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

they are NOT SAME MATERIAL.. IF THEY WERE.. THEY WOULD HAVE SAME YOUNG"S MODULUS :P

- anonymous

i dont undrstnd >< cn u shw the working

- 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

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

ok il use thnks fr ur help :)

- anonymous

No problimo amigo :D

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