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Consider two aluminum rods of length 1 m, one twice as thick as the other. If a compressive force F is applied to both rods, their lengths are reduced by [tex]\Delta L_{thick}[/tex]and [tex]\Delta L_{thin}[/tex], respectively.The ratio ΔLthick rod/ΔLthin rod is:
a.) =1
b.) <1
c.) >1
Answer:
The ratio is less than 1 i.e [tex]\frac{1}{2} <1[/tex] option B is correct
Explanation:
The Young Modulus of a material is generally calculated with this formula
[tex]E = \frac{\sigma}{\epsilon}[/tex]
Where [tex]\sigma[/tex] is the stress = [tex]\frac{Force}{Area}[/tex]
[tex]\epsilon[/tex] is the strain = [tex]\frac{\Delta L}{L}[/tex]
Making Strain the subject
[tex]\epsilon = \frac{\sigma}{E}[/tex]
now in this question we are that the same tension was applied to both wires so
[tex]\frac{\sigma}{E}[/tex] would be constant
Hence
[tex]\frac{\Delta L}{L} = constant[/tex]
for the two wire we have that
[tex]\frac{\Delta L_1}{L_1} = \frac{\Delta L_2}{L_2}[/tex]
Looking at young modulus formula
[tex]E = \frac{\frac{F}{A} }{\frac{\Delta L}{L} }[/tex]
[tex]E * \frac{\Delta L }{L} = \frac{F}{A}[/tex]
[tex]A * \frac{\Delta L}{L} = \frac{F}{E}[/tex]
Now we are told that a comprehensive force is applied to the wire so for this question
[tex]\frac{F}{E}[/tex] is constant
And given that the length are the same
so
[tex]A_1 \frac{\Delta L_{thin}}{L_{thin}} = A_2 \frac{\Delta L_{thick}}{L_{thick}}[/tex]
Now we are told that one is that one rod is twice as thick as the other
So it implies that one would have an area that would be two times of the other
Assuming that
[tex]A_2 = 2 A_1[/tex]
So
[tex]A_1 \frac{\Delta L_{thin}}{L_{thin}} = 2 A_1 \frac{\Delta L_{thick}}{L_{thick}}[/tex]
[tex]\frac{\Delta L_{thin}}{L_{thin}} = 2 \frac{\Delta L_{thick}}{L_{thick}}[/tex]
From the question the length are equal
[tex]\Delta L_{thin} =2 \Delta L_{thick}[/tex]
So
[tex]\frac{\Delta L_{thick}}{\Delta L_{thin} } = \frac{1}{2}[/tex]
Hence the ratio is less than 1
