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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    A block of mass 10 kg is placed on an inclined plane. When the angle of inclination is\[30{}^\circ ,\] the block just begins to slide down the plane. The force of static friction is:

    A)  10 kg-wt                             

    B)  9.8 kg-wt           

    C)         49 kg-wt             

    D)         5 kg-wt

    E)  15 kg-wt

    Correct Answer: D

    Solution :

    \[{{f}_{k}}=mg\,\sin \theta =10\times g\times \sin 30{}^\circ \] \[=5g=5\,kg-wt\]


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