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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    A wheel of moment of inertia\[2.5\text{ }kg-{{m}^{2}}\]has an initial angular velocity of\[40\text{ }rad\text{ }{{s}^{-1}}\]. A constant torque of 10 Nm acts on the wheel. The time during which the wheel is accelerated to\[60\text{ }rad\text{ }{{s}^{-1}}\] is

    A)  4s                          

    B)         6s

    C)  5s                          

    D)         2.5s

    E)  4.5s

    Correct Answer: C

    Solution :

    Given:       \[MI=2.5\text{ }kg\text{ }{{m}^{-2}}\] \[\omega =40\text{ }rad\text{ }{{s}^{-1}}\] \[\tau =10\text{ }Nm\] As          \[\tau =I\alpha \] \[\therefore \]  \[10=2.5\alpha \] \[\alpha =4\text{ }rad\text{ }{{s}^{-2}}\] Now,     \[\omega ={{\omega }_{0}}+\alpha t\] \[\therefore \]  \[60=40+4\times t\] or          \[20=4t\] \[t=5s\]


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