BHU PMT BHU PMT (Mains) Solved Paper-2007

  • question_answer
    A ring of mass 0.8 kg and radius 0.1 m makes \[\frac{5}{\pi }\]rotations per second about axis perpendicular to its plane through centre. Calculate angular momentum and kinetic energy of ring.

    A)  \[0.08kg-{{m}^{2}}/s,0.2J\]

    B)  \[0.85kg-{{m}^{2}}/s,0.2J\]

    C)  \[0.85kg-{{m}^{2}}/s,0.4J\]

    D)  \[0.08kg-{{m}^{2}}/s,0.4J\]

    Correct Answer: D

    Solution :

                     Angular momentum \[L=I\omega =m{{r}^{2}}\omega \]                    (as\[I=m{{r}^{2}}\])                 \[=0.8\times {{(0.1)}^{2}}\times (2\pi \times 5/\pi )\] \[=0.08kg-{{m}^{2}}/s\] Kinetic energy of ring                 \[K=\frac{1}{2}I{{\omega }^{2}}\]                 \[=\frac{1}{2}\times 0.8\times {{(0.1)}^{2}}\times {{(2\pi \times 5/\pi )}^{2}}\]                 \[=0.4J\]


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