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BCECE Medical BCECE Medical Solved Papers-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)  300 J          

    D)  175 J

    Correct Answer: C

    Solution :

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


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