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BHU PMT BHU PMT (Mains) Solved Paper-2006

  • question_answer
    A person walked up a stalled escalator in 90 s. When standing on the same escalator, now moving, he is carried up in 60 s. How much time would it take him to walk up the moving escalator?

    A)  36s                                       

    B)  30s

    C)  60s                                       

    D)  26s

    Correct Answer: A

    Solution :

                     Let the length of escalator be L. If v is the velocity of man (relative to escalator) and v' of escalator. Then according to given problem \[\frac{L}{v}=90\,s\]                         ...(1) and         \[\frac{L}{v'}=60\,s\]                         ...(2) Now if the person walks up on the moving escalator his velocity relative to the ground will be\[v+v'\]. So, time taken by him to move a distance L relative to the ground will be: \[t=\frac{L}{v+v'}\] \[\Rightarrow \]               \[\frac{1}{t}=\frac{v'}{L}+\frac{v}{L}\] which in the light of Eqs. (1) and (2), gives                 \[\frac{1}{t}=\frac{1}{60}+\frac{1}{90}\]       i.e.,  \[t=36s\]


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