Finding Young's Modulus

  • #1
341
4

Homework Statement


upload_2015-7-8_22-54-55.png


Homework Equations


E= (F/A) x (L/ΔL)

The Attempt at a Solution


I know that since the material is the same, the Young modulus should be the same. However, when I try to find the ratio of the second wire to the first, I get the answer C. For the first wire, E= 4FL / d2Δl, since A = d2/4.
For the second wire, the value of E I obtain is F x ½L / (d2/16) x Δl , which is twice the first value. I can't see what's wrong with my working. Can someone point it out?
 

Answers and Replies

  • #2
This is a conceptual question, so we know for a particular material at a certain temperature Young's modulus will be a constant. Using the above equation we see that area does not reduce linearly so you're probably wondering how do I compare a thicker wire to a thinner one. Remember that to get the same ratio of change in length from original length in the thinner wire will require less force. So while your area is a quarter of the size of the original wire, the force needed is also reduced.
 
  • #3
341
4
This is a conceptual question, so we know for a particular material at a certain temperature Young's modulus will be a constant. Using the above equation we see that area does not reduce linearly so you're probably wondering how do I compare a thicker wire to a thinner one. Remember that to get the same ratio of change in length from original length in the thinner wire will require less force. So while your area is a quarter of the size of the original wire, the force needed is also reduced.
Oh I see, so from the information given in the question above, there's no way of obtaining the same value of E as the first wire without knowing the change in the value of F is it?
 
  • #4
Oh I see, so from the information given in the question above, there's no way of obtaining the same value of E as the first wire without knowing the change in the value of F is it?

Yes, its not meant to be solved numerically.
 
  • #5
341
4

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