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JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Motion on Inclined Surface

  • question_answer
    A block is lying on an inclined plane which makes 60° with the horizontal. If coefficient of friction between block and plane is 0.25 and \[g=10\,m/{{s}^{2}}\], then acceleration of the block when it moves along the plane will be        [RPET 1997]

    A)                         \[2.50\,m/{{s}^{2}}\]            

    B)                           \[5.00\,m/{{s}^{2}}\]            

    C)                         \[7.4\,m/{{s}^{2}}\]            

    D)                           \[8.66\,m/{{s}^{2}}\]

    Correct Answer: C

    Solution :

                                \[a=g(\sin \theta -\mu \cos \theta )\]\[=10(\sin 60{}^\circ -0.25\cos 60{}^\circ )\]             \[a=7.4\ m/{{s}^{2}}\]            


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