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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2008

  • question_answer
    A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10 s. Then the coefficient of friction is

    A)  0.02                                      

    B)  0.03

    C)  0.06                      

    D)         0.01

    Correct Answer: C

    Solution :

    Let coefficient of friction be\[\mu \], then retardation will be\[\mu g\]. From equation of motion, v = u + at \[\Rightarrow \]               \[0=6-\mu g\times 10\] \[\Rightarrow \]               \[\mu =\frac{6}{100}=0.06\]


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