JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer A particle of mass m is moving along a trajectory given by
    \[x={{x}_{0}}+a\cos {{\omega }_{1}}t\]
    \[y={{y}_{0}}+b\sin {{\omega }_{2}}t\]
    The torque, acting on the particle about the origin, at t = 0 is :
    [JEE Main 10-4-2019 Morning]

    A) \[m(-{{x}_{0}}b+{{y}_{0}}a)\omega _{1}^{2}\hat{k}\]

    B) \[+m{{y}_{0}}a\omega _{1}^{2}\hat{k}\]

    C) \[-m({{x}_{0}}b\omega _{2}^{2}-{{y}_{0}}a\omega _{1}^{2})\hat{k}\]

    D) Zero

    Correct Answer: B

    Solution :

    \[F=-m(a\omega _{1}^{2}\cos {{\omega }_{1}}t\hat{i}+b\omega _{2}^{2}sin{{\omega }_{2}}t\hat{j})\]           \[\vec{r}=({{x}_{0}}+a\cos {{\omega }_{1}}t)\hat{i}+({{y}_{0}}+b\sin {{\omega }_{2}}t)\hat{j}\]           \[\vec{T}=\vec{r}\times \vec{F}=-m({{x}_{0}}+a\cos {{\omega }_{1}}t)b\omega _{2}^{2}sin{{\omega }_{2}}t\hat{k}\]           \[+m({{y}_{0}}+b\sin {{\omega }_{2}}t)a\omega _{1}^{2}\cos {{\omega }_{1}}t\hat{k}\] \[=ma\omega _{1}^{2}{{y}_{0}}\hat{k}\]


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