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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    A force of magnitude \[\sqrt{6}\] acting along line joining the points\[A(2,-1,1)\]and B (3, 1, 2) displaces a particle from A to B. The work done by ±e force is:

    A)  6                            

    B)         \[6\sqrt{6}\]

    C) \[\sqrt{6}\]                        

    D)         12

    E)  \[2\sqrt{6}\]

    Correct Answer: A

    Solution :

    Displacement \[=(3-2)\hat{i}+(1+1)\hat{j}+(2-1)\hat{k}\] \[=\hat{i}+2\hat{j}+\hat{k}\] Force    \[=\sqrt{6}(\hat{i}+2\hat{j}+\hat{k})\] \[=(\hat{i}+2\hat{j}+\hat{k})\] \[\therefore \]Work done\[=(\hat{i}+2\hat{j}+\hat{k}).(\hat{i}+2\hat{j}+\hat{k})\] \[=6\]


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