2. Solved Papers
  3. BCECE Engineering

BCECE Engineering BCECE Engineering Solved Paper-2003

  • question_answer
    Forces acting on a particle have a magnitude 5, 3 and 1 unit and act in the direction of the vectors   \[6\hat{i}+2\hat{j}+3\hat{k},3\hat{i}-4\hat{j}+6\hat{k}\] and \[2\hat{i}-3\hat{j}-6\hat{k}\]respectively. Then remain constant while the particle is displaced from the point \[A(2,-1,-3)\]to \[B(5,-1,-1),\]The work done is:

    A)  11 unit                                

    B)                   33 unit                

    C)                   10 unit                                

    D)                   30 unit

    Correct Answer: B

    Solution :

    Key Idea: If the force \[\vec{F}\]and displacement \[\vec{d},\]then work done \[=\vec{F}.\vec{d}.\] Since,    \[{{\vec{F}}_{1}}=\frac{5(6\hat{i}+2\hat{j}+3\hat{k})}{7}\]                 \[{{\vec{F}}_{2}}=\frac{3(3\hat{i}-2\hat{j}+6\hat{k})}{7}\] and        \[{{\vec{F}}_{3}}=\frac{1(2\hat{i}-3\hat{j}-6\hat{k})}{7}\] \[\therefore \]Resultant force, \[\vec{F}={{\vec{F}}_{1}}+{{\vec{F}}_{2}}+{{\vec{F}}_{3}}\] \[=\frac{1}{7}(30\hat{i}+10\hat{j}+15\hat{k}+9\hat{i}-6\hat{j}+18\hat{k}+2\hat{i}-3\hat{j}-6\hat{k})\]                \[=\frac{1}{7}(41\hat{i}+\hat{j}+27\hat{k})\]                 Now,     \[\overrightarrow{AB}=(5i-\hat{j}-\hat{k})-(2\hat{i}-\hat{j}-3\hat{k})\]                                 \[=3\hat{i}+0\hat{j}+4\hat{k}\]                 \[\therefore \]Work done \[\mathbf{=\vec{F}}\mathbf{.}\overrightarrow{\mathbf{AB}}\]                                 \[=\frac{1}{7}(41\hat{i}+\hat{j}+27\hat{k}).(3\hat{i}+4\hat{k})\]                                 \[=\frac{1}{7}(123+108)=33unit\]


You need to login to perform this action.
You will be redirected in 3 sec spinner