AIIMS AIIMS Solved Paper-2001

  • question_answer
    A constant torque of 31.4 N-m is exerted on a pivoted wheel. If the angular acceleration of the wheel is \[4\pi \,\,rad/{{s}^{2}},\] then the moment of inertia, will be:

    A)  \[5.8\,kg-{{m}^{2}}\] 

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

    C)             \[5.6\,kg-{{m}^{2}}\]

    D)  \[2.5\,kg-{{m}^{2}}\]

    Correct Answer: D

    Solution :

                    The moment of inertia (I) of a body about an axis is equal to the torque \[{{k}_{1}}\] required to produce unit angular acceleration \[{{F}_{1}}=-{{k}_{1}}{{y}_{1}}\] in the body about that axis is \[{{F}_{2}}=-{{k}_{2}}{{y}_{2}}\]    \[{{k}_{2}}\] Given, \[{{k}_{1}}\] \[{{k}_{2}}{{F}_{1}}+{{k}_{1}}{{F}_{2}}=-{{k}_{1}}{{k}_{2}}({{y}_{1}}+{{y}_{2}})=-{{k}_{1}}{{k}_{2}}y\]   \[{{F}_{1}}={{F}_{2}}=F(say)\]


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