MGIMS WARDHA MGIMS WARDHA Solved Paper-2004

  • question_answer
    A sphere rolls down an inclined plane of inclination \[\theta .\] What is the acceleration a the sphere reaches bottom?

    A) \[\frac{5}{7}g\,\sin \,\theta \]                   

    B) \[\frac{3}{7}g\,\sin \,\theta .\]

    C) \[\frac{2}{7}g\,\sin \,\theta .\]                  

    D) \[\frac{2}{5}g\,\sin \,\theta .\]

    Correct Answer: A

    Solution :

    Acceleration of the sphere \[a=\frac{g\sin \theta }{1+\frac{{{k}^{2}}}{{{r}^{2}}}}=\frac{g\sin \theta }{1+\frac{2}{5}}\] \[\left[ \because For\text{ }sphere\frac{{{k}^{2}}}{{{r}^{2}}}=\frac{2}{5} \right]\] \[=\frac{5}{7}g\sin \theta \]


You need to login to perform this action.
You will be redirected in 3 sec spinner