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EAMCET Medical EAMCET Medical Solved Paper-2008

  • question_answer
    Three forces \[\overrightarrow{A}=(\widehat{i}+\widehat{j}+\widehat{k}),\] \[\overrightarrow{B}=(2\widehat{i}-\widehat{j}+3\widehat{k})\,and\,\overrightarrow{C}\]acting on a body to keep it in equilibrium. Then \[\overrightarrow{C}\] is

    A)  \[-(3\widehat{i}+4\widehat{k})\]                           

    B)  \[-(4\widehat{i}+3\widehat{k})\]

    C)   \[3\widehat{i}+4\widehat{j}\]                                

    D)  \[2\widehat{i}-3\widehat{k}\]

    Correct Answer: A

    Solution :

                     Three forces \[\vec{A},\vec{B},\vec{C}\]are acting on a body which is in equilibrium position, then \[\vec{A}+\vec{B}+\vec{C}=0\] \[\Rightarrow \]  \[(\hat{i}+\hat{j}+\hat{k})+(2\hat{i}-\hat{j}+3\hat{k})+\vec{C}=0\] or            \[(3\hat{i}+4\hat{k})+\vec{C}=0\] \[\therefore \]  \[\vec{C}=-(3\hat{i}+4\hat{k})\]


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