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JEE Main & Advanced JEE Main Paper Phase-I Held on 07-1-2020 Morning

  • question_answer
    A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to: \[(1\text{ }HP=746\text{ }W,\text{ }g=10\text{ }m{{s}^{2}})\] [JEE MAIN Held on 07-01-2020 Morning]

    A) \[1.5\text{ }m{{s}^{1}}\]        

    B) \[1.9\text{ }m{{s}^{1}}\]

    C) \[1.7\text{ }m{{s}^{1}}\]        

    D) \[2.0\text{ }m{{s}^{1}}\]

    Correct Answer: B

    Solution :

    [b] \[{{F}_{total}}=Mg+\text{ }friction\] \[=2000\times 10+4000\] \[=20,000+4000=24000\text{ }N\] \[P=F\times v\] \[60\times 746=24000\times v\] \[\Rightarrow \text{ }v\text{ =}1.86\text{ }m/s\approx 1.9\text{ }m/s\]


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