JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    A particle is acted upon by constant forces \[4\hat{i}+\hat{j}-3\hat{k}\]and\[3\hat{i}+\hat{j}-\hat{k}\]which displace it from a point\[\hat{i}+2\hat{j}-3\hat{k}\]to the point \[5\hat{i}+4\hat{j}+\hat{k}\]. The work done in standard units by the forces is given by

    A) 40     

    B)                        30                          

    C)        25                          

    D)        15

    Correct Answer: A

    Solution :

    Total force, \[F=(4\hat{i}+\hat{j}-3\hat{k})+(3\hat{i}+\hat{j}-\hat{k})\] \[F=7\hat{i}+2\hat{j}-4\hat{k}\] The particle is displaced from \[A(\hat{i}+2\hat{j}+3\hat{k})\] to\[B(5\hat{i}+4\hat{j}+\hat{k})\]. \[\therefore \]Displacement \[AB=(5\hat{i}+4\hat{j}+\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})\] \[=4\hat{i}+2\hat{j}-2\hat{k}\] Work done\[=F.AB\] \[=(7\hat{i}+2\hat{j}-4\hat{k})-(4\hat{i}+2\hat{j}-2\hat{k})\] \[=28+4+8=40\]units


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