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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2009

  • question_answer
    A door 1.6 m wide requires a force of 1 N to be applied at the free end to open or close it. The force that is required at a point 0.4 m distance from the hinges for opening or closing the door is

    A)  1.2 N             

    B)  3.6 N

    C)  2.4 N             

    D)  4 N

    Correct Answer: D

    Solution :

    Here, torque \[\tau =1.6\times 1=1.6N-m\] So, when \[d=0.4\text{ }m,\]\[F=\frac{\tau }{d}=\frac{1.6}{0.4}=4N\]


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