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BCECE Engineering BCECE Engineering Solved Paper-2012

  • question_answer
    If under the action of a force \[(4i+j+3k)N,\] a   particle   moves   from   position\[{{r}_{1}}=3i+2j-6k\]        to        position \[{{r}_{2}}=14i+13j+9k,\]then the work done will be

    A)  50 J       

    B)                         75 J                       

    C)         100 J                    

    D)         175 J

    Correct Answer: C

    Solution :

    Here, \[{{r}_{1}}=3i+2j-6k\]and \[{{r}_{2}}=14i+13j+9k\] So, displacement,\[({{r}_{2}}-{{r}_{1}})=\] \[(14i+13j+9k)-(3i+2j-6k)\] \[=11\,i+11j+15k\] Hence, work done \[=F.s\]                 \[=(4i+j+3k).(11i+11j+15k)\] \[=44+11+45=100\,J\]


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