A) \[4.8\times {{10}^{6}}\text{ }N{{m}^{2}}\]
B) \[5.2\times {{10}^{6}}\text{ }N{{m}^{2}}\]
C) \[6.2\times {{10}^{6}}\text{ }N{{m}^{2}}\]
D) \[3.1\times {{10}^{6}}\text{ }N{{m}^{2}}\]
Correct Answer: D
Solution :
Tensile stresss in wire will be \[=\frac{\text{Tensile}\,\,\text{force}}{\text{Cross}\,\,\text{section}\,\,\text{Area}}\] \[=\frac{mg}{\pi {{R}^{2}}}=\frac{4\times 3.1\pi }{\pi \times 4\times {{10}^{-6}}}N{{m}^{-2}}=3.1\times {{10}^{6}}N{{m}^{-2}}\]
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