Why is My Young's Modulus Calculation Giving a Different Result?

[toforfiltum]

Homework Statement

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?

 

[vanoccupanther]

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.

 

[toforfiltum]

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?

 

[vanoccupanther]

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.

 

[toforfiltum]

Yes, its not meant to be solved numerically.

Ok, thanks.