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JEE Main & Advanced AIEEE Solved Paper-2002

  • question_answer
    Initial angular velocity of a circular disc of mass M is \[{{\omega }_{\,1}}\]. Then, two small spheres of mass m are attached gently to two diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc?   AIEEE  Solved  Paper-2002

    A) \[\left( \frac{M+m}{M} \right){{\omega }_{1}}\]  

    B)           \[\left( \frac{M+m}{m} \right){{\omega }_{1}}\]

    C) \[\left( \frac{M}{M+4m} \right){{\omega }_{1}}\]               

    D)           \[\left( \frac{M}{M+2m} \right){{\omega }_{1}}\]

    Correct Answer: C

    Solution :

    Conservation of angular momentum gives              \[\frac{1}{2}M{{R}^{2}}{{\omega }_{1}}=\left( \frac{1}{2}M{{R}^{2}}+2M{{R}^{2}} \right){{\omega }_{2}}\]              \[(\because {{l}_{1}}{{\omega }_{1}}={{l}_{2}}{{\omega }_{2}}\] and \[{{l}_{1}}=\frac{1}{2}m{{R}^{2}}\],                                 \[{{l}_{2}}=\frac{1}{2}m{{R}^{2}}+2m{{R}^{2}})\]              \[\Rightarrow \,\,\frac{1}{2}M{{R}^{2}}{{\omega }_{1}}=\frac{1}{2}{{R}^{2}}(M+4m){{\omega }_{2}}\] \[\therefore \]     \[{{\omega }_{2}}=\left( \frac{M}{M+4m} \right){{\omega }_{1}}\]


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