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  3. WB JEE Medical

WB JEE Medical WB JEE Medical Solved Paper-2010

  • question_answer
    A boy of mass 40 kg is climbing a vertical pole at a constant speed. If the coefficient of  friction between his palms and the pole is 0.8 and \[g=10\,m/{{s}^{2}},\]the horizontal force that he is applying on the pole is

    A)  300 N                                   

    B)  400 N

    C)  500 N                                   

    D)  600 N

    Correct Answer: C

    Solution :

                     The coefficient of friction \[=0.8,\,g=10\,m/{{s}^{2}},mass=40\,kg\] The frictional force \[=\mu N\] Hence,    \[\mu N=mg\Rightarrow N=\frac{mg}{\mu }\]                 \[=\frac{40\times 10}{0.8}=5000\,N\]


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