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NEET NEET SOLVED PAPER 2016 Phase-I

  • question_answer
    A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of \[\text{2}\text{.0 rad }{{\text{s}}^{-2}}.\] Its net  acceleration in \[\text{m}{{\text{s}}^{-2}}\]at the end of 2.0 s is approximately :                               

    A)  8.0                                        

    B)  7.0

    C)   6.0                                       

    D)  3.0

    Correct Answer: A

    Solution :

                     Particle at periphery will have both radial and tangential acceleration \[{{a}_{t}}=R\alpha =0.5\times 2=1\,m/{{s}^{2}}\] \[\omega ={{\omega }_{0}}+\alpha t\] \[\omega =0+2\times 2=4\,\,rad/\sec \] \[{{a}_{c}}={{\omega }^{2}}R={{(4)}^{2}}\times 0.5=16\times 0.5=8\,m/{{s}^{2}}\] \[{{a}_{total}}=\sqrt{a_{p}^{2}+a_{c}^{2}}=\sqrt{{{1}^{2}}+{{8}^{2}}}\approx \,\,8m/{{s}^{2}}\] *In this question we have assumed the point to be located at periphery of the disc.


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