A) \[~3.22\,c{{m}^{2}}\]
B) \[~2.38\,c{{m}^{2}}\]
C) \[~0.32\,c{{m}^{2}}\]
D) \[~8.23\,c{{m}^{2}}\]
Correct Answer: A
Solution :
Invoking Newton's second law, we get \[T-mg=ma\] Let, the area of cross seion of the cable be A \[\therefore \] \[7\times {{10}^{7}}A-mg=ma\] or, \[A=\frac{mg+ma}{7\times {{10}^{7}}}=\frac{m(g+a)}{7\times {{10}^{7}}}\] \[=\frac{2000(9.8+1.5)}{7\times {{10}^{7}}}=3.22\times {{10}^{-4}}{{m}^{2}}\] \[=3.22\,c{{m}^{2}}\] Hence, the correction option is [a].You need to login to perform this action.
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