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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2005

  • question_answer
    A body whose moment of inertia is 3 kg-m2 is in rest. It is rotated for 20 sec by a torque of 6 Nm, angular displacement of the body will be:

    A)  400 radian         

    B)         200 radian

    C)  100 radian         

    D)         250 radian

    Correct Answer: A

    Solution :

    \[I=3\,kg\,{{m}^{2}},t=6\,Nm,t=20\,\sec \] From \[t=I\,\alpha \] Angular acceleration \[\,\alpha =\frac{t}{I}=\frac{6}{3}\]                                         \[=2\,rad/{{\sec }^{2}}\] \[\therefore \]Angular displacement                                 \[={{\omega }_{0}}t+\frac{1}{2}\alpha \,{{t}^{2}}\] \[=0\times 20+\frac{1}{2}(2){{(20)}^{2}}\] \[=400\,radian\]


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