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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    A fly wheel of moment of inertia\[3\times {{10}^{2}}kg\text{ }{{m}^{2}}\]is rotating with uniform angular speed of 4.6 \[rad{{s}^{-1}}\]. If a torque of\[6.9\times {{10}^{2}}Nm\]retards the wheel, then the time in which the wheel comes to rest is

    A)  1.5 s                     

    B)         2 s

    C)  0.5 s                     

    D)         1 s

    E)  2.5 s

    Correct Answer: B

    Solution :

    Moment of inertia \[=3\times {{10}^{2}}kg-{{m}^{2}}\] Angular speed \[\omega =4.6\text{ }rad/s\] Torque \[\tau =6.9\,\times {{10}^{2}}\,Nm\] \[\tau =I\alpha =I\times \frac{\omega }{t}\] \[6.9\times {{10}^{2}}=3\times {{10}^{2}}\times \frac{4.6}{t}\] Or           \[t=\frac{3\times {{10}^{2}}\times 4.6}{6.9\times {{10}^{2}}}\]                 \[=2s\]


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