Question #85d89

1 Answer
Michael
Mar 18, 2015

Young's Modulus :

#E="stress"/"strain"="force per unit area"/"extension per unit length"#

So #"stress"# # =F/A#

The X sectional area of the wire #A=pir^2#

Since #d=2"mm", r=1"mm" = 10^(-3)"m"#

#A=pixx10^(-3)xx10^(-3)=pi10^(-6)m^(2)#

So #"stress"=(20)/(pi10^(-6))=6.36xx10^(6)kg.m^(-1).s^(-2)#

strain #=x/l=(0.24xx10^(-3))/4=6xx10^(-5)#

So #E="stress"/"strain"=(6.36xx10^(6))/(6xx10^(-5))=1.06xx10^(11)kg.m^(-1).s^(-2)#

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