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JEE Main & Advanced JEE Main Paper (Held On 11 April 2015)

  • question_answer
    A large number (n) of identical beads, each of mass m and radius r are strung on a thin smooth rigid horizontal rod of length \[L(L>>r)\] and are at rest at random positions. The rod is mounted between two rigid supports (see figure). If one of the beads is now given a speed v, the average force experienced by each support after a long time is (assume all collisions are elastic) : [JEE Main Online Paper (Held On 11 April 2015)]  

    A) \[\frac{m{{\upsilon }^{2}}}{L-nr}\]

    B) \[\frac{m{{\upsilon }^{2}}}{L-2nr}\]

    C) \[\frac{m{{\upsilon }^{2}}}{2(L-nr)}\]

    D)  zero

    Correct Answer: B

    Solution :

    As collisions are elastic and masses are equal, velocities of colliding particles get exchanged. \[\Delta \vec{P}\]in each collision with the supports \[=2mv\] Time interval between consecutive collisions with one support \[=\frac{\left( L-2nr \right)\times 2}{v}\] \[{{F}_{avg}}=\frac{\Delta P}{T}=\frac{2mv}{\left( L-2nr \right)2/v}=\frac{m{{v}^{2}}}{L-2nr}\]


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