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J & K CET Medical J & K - CET Medical Solved Paper-2011

  • question_answer
    A cubical block rests on an inclined plane of coefficient of friction \[\mu =1\sqrt{3.}\] What should be the angle of inclination so that the block just slides down the inclined plane?

    A) \[30{}^\circ \]                                   

    B) \[60{}^\circ \]

    C) \[45{}^\circ \]                                   

    D) \[90{}^\circ \]

    Correct Answer: A

    Solution :

                     \[\tan \theta =\mu \] \[\tan \theta =\frac{1}{\sqrt{3}}\] \[tan\theta =tan\text{ }{{30}^{o}}\] Angle of inclination, \[\theta ={{30}^{o}}\]


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