JEE Main & Advanced Sample Paper JEE Main Sample Paper-34

  • question_answer
    A particle of mass 1 kg is projected from ground with velocity \[20\text{ }m{{s}^{-1}}\]at \[{{45}^{o}}\] from horizontal. Magnitude of its angular momentum just before it hits the ground about the point of projection will be \[\left( g=10m{{s}^{-2}} \right)\]

    A)  \[40\sqrt{2}\frac{kg{{m}^{2}}}{\sec }\] 

    B)  \[400\sqrt{2}\frac{kg{{m}^{2}}}{\sec }\]

    C)  \[800\frac{kg{{m}^{2}}}{\sec }\]                              

    D)  None

    Correct Answer: B

    Solution :

    \[R=\frac{{{u}^{2}}\sin 2\theta }{g}=40\] \[{{L}_{0}}=\] line of linear moment. shortest distance \[=(mv\sin \theta )\times R\]

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