2. Sample Papers
  3. NEET

NEET Sample Paper NEET Sample Test Paper-14

  • question_answer
    An elevator cable is to have a maximum stress of \[7\times {{10}^{7}}\,N/{{m}^{2}}\]to allow for appropriate safety factors. Its maximum upward acceleration is \[1.5\text{ }m/{{s}^{2}}.\] If the cable has to support the total weight of 2000 kg of a loaded elevator, the area of cross-section of the cable should be:

    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.
You will be redirected in 3 sec spinner