JEE Main & Advanced JEE Main Solved Paper-2016

  • question_answer
    A particle of mass m is moving along the side of a square of side 'a', with a uniform speed u in the x-y plane as shown in the figure : Which of the following statement is false for the angular momentum \[\vec{L}\] about the origin ? [JEE Main Solved Paper-2016 ]

    A) \[\vec{L}=\frac{m\upsilon }{\sqrt{2}}R\hat{k}\]when the particle is moving from D to A

    B) \[\vec{L}=-\frac{m\upsilon }{\sqrt{2}}R\hat{k}\] when the particle is moving from  to B

    C)  \[\vec{L}=m\upsilon \left[ \frac{R}{\sqrt{2}}-a \right]\hat{k}\]when the particle is moving from C to D

    D) \[\vec{L}=m\upsilon \left[ \frac{R}{\sqrt{2}}+a \right]\hat{k}\]when the particle is moving from B to C

    Correct Answer: A , C

    Solution :

                    \[\vec{L}=\vec{r}\times \vec{P}\]or\[\vec{L}=rp\sin \theta \hat{n}\]or\[\vec{L}={{r}_{\bot }}(P)\hat{n}\]                For D to A\[\vec{L}=\frac{R}{\sqrt{2}}mV(-\hat{k})\] For A to B\[\vec{L}=\frac{R}{\sqrt{2}}mV(-\hat{k})\] For C to D\[\vec{L}=\left( \frac{R}{\sqrt{2}}+a \right)mV(\hat{k})\] For B to C\[\vec{L}=\left( \frac{R}{\sqrt{2}}+a \right)mV(\hat{k})\]

    Solution :

                    \[\vec{L}=\vec{r}\times \vec{P}\]or\[\vec{L}=rp\sin \theta \hat{n}\]or\[\vec{L}={{r}_{\bot }}(P)\hat{n}\]                For D to A\[\vec{L}=\frac{R}{\sqrt{2}}mV(-\hat{k})\] For A to B\[\vec{L}=\frac{R}{\sqrt{2}}mV(-\hat{k})\] For C to D\[\vec{L}=\left( \frac{R}{\sqrt{2}}+a \right)mV(\hat{k})\] For B to C\[\vec{L}=\left( \frac{R}{\sqrt{2}}+a \right)mV(\hat{k})\]


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