J & K CET Engineering J and K - CET Engineering Solved Paper-2011

  • question_answer
    A constant torque of \[3.14\text{ }Nm\] is exerted on a pivoted wheel. If the angular acceleration of the wheel is \[4\pi \text{ }rad/{{s}^{2}},\] then the moment of Inertia of the wheel is

    A)  \[0.25\text{ }kg-{{m}^{2}}\]

    B)  \[2.5\text{ }kg-{{m}^{2}}\]

    C)  \[4.5\text{ }kg-{{m}^{2}}\]

    D)  \[25\text{ }kg-{{m}^{2}}\]

    Correct Answer: A

    Solution :

    Torque, \[\tau =I\omega \] Moment of inertia,   \[I=\frac{\tau }{\omega }\] \[=\frac{3.14}{4\times \pi }\] \[=\frac{3.14}{4\times 3.14}=\frac{1}{4}\] \[=0.25kg-{{m}^{2}}\]


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