KVPY Sample Paper KVPY Stream-SX Model Paper-5

  • question_answer
    A particle moves on a rough horizontal ground with some initial velocity say \[{{v}_{0}}.\] If 3/4th of its kinetic energy is lost in friction in time \[{{t}_{0}}.\] Then coefficient of friction between the particle and the ground is -

    A) \[\frac{{{v}_{0}}}{2g{{t}_{0}}}\]        

    B) \[\frac{{{v}_{0}}}{4g{{t}_{0}}}\]

    C) \[\frac{3{{v}_{0}}}{4g{{t}_{0}}}\]                  

    D) \[\frac{{{v}_{0}}}{g{{t}_{0}}}\]

    Correct Answer: A

    Solution :

    [A] \[\frac{3}{4}th\] energy is lost i.e.,\[\frac{1}{4}th\] kinetic energy is left. Hence, its velocity becomes \[\frac{{{v}_{0}}}{2}\] under a retardation of \[\mu g\] in time to. \[\therefore \,\,\frac{{{v}_{0}}}{2}={{v}_{0}}-\mu g\,\,{{t}_{0}}\] or \[\mu g\,\,{{t}_{0}}=\frac{{{v}_{0}}}{2}\] or \[\mu =\frac{{{v}_{0}}}{2g{{t}_{0}}}\]           

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