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JCECE Engineering JCECE Engineering Solved Paper-2013

  • question_answer
    A boat can go across a lake and return in time \[{{T}_{0}}\] at a speed\[v\]. On a rough day there is a uniform current at speed \[{{v}_{1}}\] to help the onward journey and impede the return journey. If the time taken to go across and return on the same day be\[T\], then \[T/{{T}_{0}}\] will be

    A) \[\frac{1}{(1-v_{1}^{2}/{{v}^{2}})}\]       

    B) \[\frac{1}{(1+v_{1}^{2}/{{v}^{2}})}\]

    C) \[(1-v_{1}^{2}/{{v}^{2}})\]                          

    D)  \[\left( 1+\frac{v_{1}^{2}}{{{v}^{2}}} \right)\]

    Correct Answer: A

    Solution :

    Let \[d\] be the width of lake to be crossed, then                 \[{{T}_{0}}=\frac{2d}{v}\] and        \[T=\frac{d}{(v+{{v}_{1}})}-\frac{d}{(v-{{v}_{1}})}=\frac{2dv}{{{v}^{2}}-v_{1}^{2}}\]                    \[=\frac{2dv}{{{v}^{2}}(1-v_{1}^{2}/{{v}^{2}})}=\frac{2d}{v(1-v_{1}^{2}/{{v}^{2}})}\] So,          \[\frac{T}{{{T}_{0}}}=\frac{1}{(1-v_{1}^{2}/{{v}^{2}})}\]


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