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Manipal Engineering Manipal Engineering Solved Paper-2013

  • question_answer
    A man crosses a \[320\,\,m\] wide river perpendicular to the current in 4 min. If in still water he can swim with a speed \[5/3\] times that of the current, then the speed of the current, in\[m{{m}^{-1}}\]in is

    A)  30                         

    B)  40     

    C)  50         

    D)         60

    Correct Answer: D

    Solution :

    \[v_{r}^{2}=v_{m}^{2}-{{v}^{2}}\]   \[v=\frac{320}{4}\text{m/min=80m/min}\]                 \[{{v}_{m}}=\frac{5}{3}v\],                 \[v_{r}^{2}=t\frac{5}{3}{{({{v}_{r}})}^{2}}-{{(80)}^{2}}\]          \[\frac{16}{9}v_{r}^{2}={{(80)}^{2}}\]                 \[{{v}_{r}}=60\text{m/min}\] \[{{u}^{2}},\,\,{{u}_{0}}\]form canours linear


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