A) 0.16%
B) 0.32%
C) 0.08%
D) 0.24%43
Correct Answer: A
Solution :
[a] Given: \[F=100kN={{10}^{5}}N\] \[Y=2\times {{10}^{11}}N{{m}^{-2}}\] \[{{\ell }_{0}}=1.0m\] radius \[r=10mm={{10}^{-2}}m\] \[From\,formula,\,Y=\frac{Stress}{strain}\] \[\Rightarrow Strain=\frac{Stress}{Y}=\frac{F}{AY}\] \[=\frac{{{10}^{5}}}{\pi {{r}^{2}}Y}=\frac{{{10}^{5}}}{3.14\times {{10}^{-4}}\times 2\times {{10}^{11}}}=\frac{1}{628}\] Therefore % strain \[=\frac{1}{628}\times 100=0.16%\]You need to login to perform this action.
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