A) \[19.6\times {{10}^{20}}N/{{m}^{2}}\]
B) \[19.6\times {{10}^{18}}N/{{m}^{2}}\]
C) \[19.6\times {{10}^{10}}N/{{m}^{2}}\]
D) \[19.6\times {{10}^{15}}N/{{m}^{2}}\]
Correct Answer: C
Solution :
[c] \[Y=\frac{F/A}{\Delta \ell /\ell }=\frac{\frac{250\times 10}{50\times {{10}^{-6}}}}{\frac{0.5\times {{10}^{-3}}}{2}}\] \[=\frac{250\times 9.8}{50\times {{10}^{-6}}}\times \frac{2}{0.5\times {{10}^{-3}}}\] \[\Rightarrow \]\[19.6\times {{10}^{10}}N/{{m}^{2}}\]You need to login to perform this action.
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