DUMET Medical DUMET Medical Solved Paper-2009

  • question_answer
    An object moves at a constant speed along a circular path in horizontal XY plane, with the centre at the origin. When the object is a\[x=-2m,\] its velocity is \[(-4m{{s}^{-1}})\mathbf{\hat{j}}.\] What is the objects acceleration when it is \[y=2\,m\]?

    A)  \[-(8\,m{{s}^{-2}})\,\hat{j}\]     

    B)  \[(8\,m{{s}^{-2}})\,\hat{i}\]

    C)  \[(-4\,m{{s}^{-2}})\,\hat{j}\]     

    D)  \[(4\,m{{s}^{-2}})\,\hat{i}\]

    Correct Answer: B

    Solution :

     When object is at \[x=-m\,,\,\,v=-(4\,m{{s}^{-1}})\,\,\hat{j}\] As \[v=\omega \,r\] \[\therefore \] \[-4=\omega \times (-2)\]or \[\omega =2\,rad\,{{s}^{-1}}\] Now, when object is at \[y=2\,m,\,\,\omega =2\] \[\therefore \] Acceleration \[a={{\omega }^{2}}r={{(2)}^{2}}\times 2=8\,m{{s}^{-2}}\]


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