A) 1.35 mm
B) 1.35 cm
C) \[1.35\times {{10}^{-6}}m\]
D) 2mm
Correct Answer: A
Solution :
Here, stress\[=\frac{F}{A}=3.5\times {{10}^{8}}\] \[\therefore \] \[A=\frac{F}{3.5\times {{10}^{8}}}=\frac{500}{3.5\times {{10}^{8}}}\] \[=1.43\times {{10}^{-6}}{{m}^{2}}\] or \[\pi {{r}^{2}}=1.43\times {{10}^{-6}}\] \[\Rightarrow \] \[\frac{\pi {{D}^{2}}}{4}=1.43\times {{10}^{-6}}\] \[\Rightarrow \] \[{{D}^{2}}=\frac{4\times 1.43\times {{10}^{-6}}}{\pi }=\frac{5.72\times {{10}^{-6}}}{3.14}\] or \[D=1.35\times {{10}^{-3}}m=1.35\,mm\]
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