Uttarakhand PMT Uttarakhand PMT Solved Paper-2004

  • question_answer
    The torque of force \[\text{\vec{F}=(2\hat{i}-3\hat{J}+4\hat{k})}\] newton    acting    at    a    point \[\text{\vec{r}=(3\hat{i}+2\hat{J}+3\hat{k})}\] metre about origin is:

    A)  \[6\vec{i}-6\vec{J}+12\vec{k}N-m\]

    B)  \[-6\vec{i}+6\vec{J}-12\vec{k}N-m\]

    C)  \[17\vec{i}-6\vec{J}-13\vec{k}N-m\]

    D) \[-17\vec{i}-6\vec{J}-13\vec{k}N-m\]

    Correct Answer: C

    Solution :

     Here: \[\overrightarrow{F}=2\hat{i}-3\hat{j}+4\hat{k}N\] Position vector of a point \[\overrightarrow{r}=3\hat{i}+2\hat{j}+3\hat{k}\,m\] The torque acting at a point about the origin is given by \[\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=(3\hat{i}+2\hat{j}+3\hat{k})\times (2\hat{i}-3\hat{j}+4\hat{k})\] \[=\left| \begin{matrix}    {\hat{i}} & {\hat{j}} & {\hat{k}}  \\    3 & 2 & 3  \\    2 & -3 & 4  \\ \end{matrix} \right|\] \[=\hat{i}[8-(-9)]-\hat{j}(12-6)+\hat{k}(-9-4)\] \[=17\hat{i}-6\hat{j}-13\hat{k}N-m\]


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