JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Evening)

  • question_answer
              A particle of mass m is moving in a straight line with momentum p. Starting at time \[t=0,\]a force F = kt acts in the same direction on the moving particle during time interval so that its momentum changes from? to 3p. Here k is a constant. The value of T is [JEE  Main Online Paper (Held on 11-jan-2019 Evening)]

    A) \[2\sqrt{\frac{p}{k}}\]   

    B)               \[\sqrt{\frac{2k}{p}}\]

    C) \[\sqrt{\frac{2p}{k}}\]                           

    D)   \[2\sqrt{\frac{k}{p}}\]

    Correct Answer: A

    Solution :

    As\[F=\frac{dp}{dt}\]or\[kt\,dt=dp\] Integrating both sides,                            \[k\int\limits_{0}^{T}{t}dt=\int_{p}^{3p}{dp}\] \[\frac{k{{T}^{2}}}{2}=[3p-p]=2p\] \[\therefore \]\[T=2\sqrt{\frac{p}{k}}\]


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