2. Solved Papers
  3. CET Karnataka Medical

CET Karnataka Medical CET - Karnataka Medical Solved Paper-2000

  • question_answer
    A box is lying on an inclined plane what is the coefficient of static friction if the box starts sliding when an angle of inclination is \[60{}^\circ \]:

    A)  1.173

    B)  1732

    C)  2732

    D)  1-677

    Correct Answer: B

    Solution :

     Angle of inclination \[\theta ={{60}^{o}}\] The formula of the coefficient of static friction \[\mu \] is \[\mu =\tan \theta =\tan {{60}^{o}}=\sqrt{3}\] Thus,    \[\mu =1.732\]


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