2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Evening)

  • question_answer
    The position co-ordinates of a particle moving in a 3-D coordinate system is given by             \[~x=a\text{ }cos\omega t\]             \[y=a\text{ }sin\omega t\]       and \[z=a\omega t\] The speed of the particle is: [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

    A) \[a\omega \]    

    B)                           \[\sqrt{3}\,a\omega \]

    C) \[\sqrt{2}\,a\omega \]

    D)               \[2\,a\omega \]

    Correct Answer: C

    Solution :

    \[{{V}_{x}}=a\omega sin\omega t\] \[{{V}_{y}}=a\omega cos\omega t\] \[{{V}_{z}}=a\omega \] \[\left| V \right|=\sqrt{V_{x}^{2}+V_{y}^{2}+V_{z}^{2}}\] \[=\,\sqrt{{{a}^{2}}{{\omega }^{2}}\,{{\sin }^{2}}\,\omega t+{{a}^{2}}{{\omega }^{2}}\,{{\cos }^{2}}\,\omega t\,+{{a}^{2}}\,{{\omega }^{2}}}\] \[\left| V \right|\,\,=\,\,\sqrt{2}\,a\omega \]          


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