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JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    Block A of mass 2 kg is placed over block B of mass, 8 kg. The combination is placed over a rough horizontal surface. Coefficient of friction between B and the floor is \[0.5\]. Coefficient of friction between A and B is \[0.4\]. A horizontal force of 10 N is applied on block B. The force of friction between A and B is:

    A)  zero                 

    B)  50 N

    C)  40 N                           

    D)  100 N

    Correct Answer: A

    Solution :

                Net friction force between block and surface is: \[F=\mu \,R=0.5\,\times 10\times 10=50\,N\] Applied force is 10 N and it is less than 50 N System is at rest no friction between A and B


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