A) 1.16 mm
B) 0.90 mm
C) 1.36 mm
D) 1.00 mm
Correct Answer: A
Solution :
\[\frac{F}{A}=stress\] \[\frac{400\times 4}{\pi {{d}^{2}}}=379\times {{10}^{6}}\] \[{{d}^{2}}=\frac{1600}{\pi \times 379\times {{10}^{6}}}=1.34\times {{10}^{-6}}\] \[d=\sqrt{1.34}\times {{10}^{-3}}=1.15\times {{10}^{-3}}m\]
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