A) \[4\times {{10}^{6}}\]dynes
B) \[2\times {{10}^{12}}\]dynes
C) \[2\times {{10}^{12}}\]newtons
D) \[2\times {{10}^{8}}\]dynes
Correct Answer: B
Solution :
To double the length of wire, Stress = Young's modulus \ \[\frac{F}{A}=2\times {{10}^{12}}\frac{dyne}{c{{m}^{2}}}.\] If A = 1 then F = 2 × 1012 dyneYou need to login to perform this action.
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