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Manipal Medical Manipal Medical Solved Paper-2014

  • question_answer
    A box is lying on an inclined plane. If the box starts sliding when the angle of inclination is 45°, then its coefficient of friction will be

    A) 1                                                

    B) 2     

    C) \[\sqrt{3}\]                                           

    D) \[\frac{\sqrt{3}}{2}\]

    Correct Answer: A

    Solution :

    The angle of inclination \[\theta ={{45}^{\circ }}\] The coefficient of static friction is given by \[\mu =\tan \,\theta =\tan \,{{45}^{\circ }}=1\]


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