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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 2

  • question_answer
    A lift is tied with thick iron wire and its mass is 1000 kg. The minimum diameter of the wire if the maximum acceleration of the lift is \[1.2m{{s}^{-2}}\]and the maximum safe stress is \[1.4\times {{10}^{8}}N\text{ }{{m}^{-}}^{2}\], is (Take \[g=9.8\text{ }m\text{ }{{s}^{-}}^{2}\])       

    A) 0.00141 m

    B) 0.00282 m       

    C) 0.005m           

    D) 0.01m           

    Correct Answer: D

    Solution :

    [d]: When the lift is accelerated upwards with acceleration a, then tension in the wire is \[T=m\left( g+a \right)=1000\left( 9.8+1.2 \right)=11000\text{ }N\] Now, stress\[=\frac{F}{A}=\frac{T}{\pi {{r}^{2}}}\]or\[r=\frac{1}{200}\] \[\therefore \]Minimum diameter of the wire \[D=2r=\frac{1}{100}=0.01m\]


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