JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Afternoon)

  • question_answer
    A thin smooth rod of length L and mass M is rotating freely with angular speed\[{{\omega }_{0}}\]about an axis perpendicular to the rod and passing through its center. Two beads of mass m and negligible size are at the center of the rod initially. The beads are free to slide along the rod. The angular speed of the system , when the beads reach the opposite ends of the rod, will be :- [JEE Main 9-4-2019 Afternoon]

    A) \[\frac{M{{\omega }_{0}}}{M+3m}\]                     

    B) \[\frac{M{{\omega }_{0}}}{M+m}\]

    C) \[\frac{M{{\omega }_{0}}}{M+2m}\]                     

    D) \[\frac{M{{\omega }_{0}}}{M+6m}\]

    Correct Answer: D

    Solution :

    Applying angular momentum conservation, about axis of rotation \[{{L}_{i}}={{L}_{f}}\] \[\frac{M{{L}^{2}}}{12}{{\omega }_{0}}=\left( \frac{M{{L}^{2}}}{12}+m{{\left( \frac{L}{2} \right)}^{2}}\times 2 \right)\omega \] \[\Rightarrow \]\[\omega =\frac{M{{\omega }_{0}}}{M+6m}.\]


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