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JEE Main & Advanced AIEEE Solved Paper-2003

  • 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 10s. Then, the coefficient of friction is     AIEEE  Solved  Paper-2003

    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, we get     \[0=6-\mu g\times 10\]                 (\[\because a=\mu g\]) \[\Rightarrow \]               \[\mu =\frac{6}{100}=0.06\]


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