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JEE Main & Advanced Physics Wave Mechanics Question Bank Self Evaluation Test - Mechanical Properties of Solids

  • 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\[1.5m/{{s}^{2}}\]. 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) \[3.28c{{m}^{2}}\]

    B) \[2.28c{{m}^{2}}\]

    C) \[0.328c{{m}^{2}}\]

    D) \[0.823c{{m}^{2}}\]

    Correct Answer: A

    Solution :

    [a] Given, the breaking strength of cable \[{{f}_{u}}=7\times {{10}^{7}}N/{{m}^{2}}\] The force carried by the cable, \[F=m(g+a)=2000(9.8+1.5)22600N\] The area of cross-section,  \[A=\frac{F}{{{f}_{u}}}=\frac{22600}{7\times {{10}^{7}}}\] \[=3.28\times {{10}^{-4}}{{m}^{2}}\]


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