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BHU PMT BHU PMT (Mains) Solved Paper-2009

  • 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 :

                     \[v_{r}^{2}=v_{m}^{2}-{{v}^{2}}\]                 \[v=\frac{320}{4}m{{\min }^{-1}}=80\,m{{\min }^{-1}}\]                 \[{{v}_{m}}=\frac{5}{3}{{v}_{r}}\]                 \[v_{r}^{2}={{\left[ \frac{5}{3}{{v}_{r}} \right]}^{2}}-{{(80)}^{2}}\]                 \[\Rightarrow \]\[\frac{16}{9}v_{r}^{2}={{(80)}^{2}}\]                 \[{{v}_{r}}=60\,m{{\min }^{-1}}\]


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