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J & K CET Engineering J and K - CET Engineering Solved Paper-2006

  • question_answer
    A disc of moment of inertia \[5\text{ }kg-{{m}^{2}}\]is acted upon by a constant torque of\[40\text{ }Nm\]. Starting from rest the time taken by it to acquire an angular velocity of 24 rad/s is

    A)  \[3\text{ }s\]           

    B)  \[\text{4 }s\]

    C)  \[2.5\,\,s\]         

    D)  \[120\,\,s\]

    Correct Answer: A

    Solution :

    Torque is denned as rate of change of angular momentum \[\therefore \] \[\tau =\frac{dJ}{dt}=\frac{d\,(I\omega )}{dt}\] Given, \[\tau =40N-m,\,\,\,I=5kg-{{m}^{2}},\,\,\,\omega =24rad/s\] \[dt=\frac{d(I\omega )}{\tau }=\frac{5\times 24}{40}=3s\]


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