J & K CET Medical J & K - CET Medical Solved Paper-2011

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

    A)                 \[0.25kg-{{m}^{2}}\]                      

    B)  \[2.5kg-{{m}^{2}}\]

    C)                 \[4.5kg-{{m}^{2}}\]                        

    D)  \[25\text{ }kg-{{m}^{2}}\]

    Correct Answer: A

    Solution :

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


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