JEE Main & Advanced JEE Main Paper Phase-I (Held on 09-1-2020 Morning)

  • question_answer
    Consider a force\[\vec{F}=-x\hat{i}+y\hat{j}\]. The work done by this force in moving a particle from point A(1, 0) to B(0, 1) along the line segment is: (all quantities are in SI units)
    [JEE MAIN Held on 09-01-2020 Morning]

    A) 2                     

    B) 1

    C) \[\frac{1}{2}\]              

    D) \[\frac{3}{2}\]

    Correct Answer: B

    Solution :

    [b] \[W=\int{dW=\int{\left( -x\hat{i}+y\hat{j} \right).\left( dx\hat{i}+dy\hat{j} \right)}}\] \[W=-\int\limits_{1}^{0}{xdx+}\int\limits_{0}^{1}{ydy}\] \[=\frac{1}{2}+\frac{1}{2}=1J\]


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