CMC Medical CMC-Medical VELLORE Solved Paper-2012

  • question_answer
    When a ring of mass 10 kg and diameter 0.4 m is rotated about its axis it makes 2100 rev/min. The angular momentum of the ring will be

    A)  \[0.4\,kg\,{{m}^{2}}{{s}^{-1}}\]                

    B)  \[4.4\,kg\,{{m}^{2}}{{s}^{-1}}\]

    C)  \[44\,kg\,{{m}^{2}}{{s}^{-1}}\]                 

    D)  \[88\,kg\,{{m}^{2}}{{s}^{-1}}\]

    Correct Answer: D

    Solution :

                    The moment of inertia of the ring, \[I=m{{r}^{2}}=10\times {{(0.2)}^{2}}=0.4\,kg-{{m}^{2}}\] Angular frequency ofthe ring, \[\omega =2\pi n=2\pi \times \frac{2100}{60}\,rad\text{/}s\] \[\therefore \] Angular momentum of the ring, \[L=I\omega =\frac{0.4\times 2\pi \times 2100}{60}\]\[=88\,kg\,{{m}^{2}}\text{/}s\]


You need to login to perform this action.
You will be redirected in 3 sec spinner