2. Solved Papers
  3. CMC Medical

CMC Medical CMC-Medical Ludhiana Solved Paper-2013

  • question_answer
    A particle of man m is projected at an angle \[45{}^\circ \] with velocity v. When the particle is a highest point, the angular momentum with respect to point of projection

    A)  0                                            

    B)  \[\frac{m{{v}^{3}}}{4\sqrt{2g}}\]

    C)  \[\frac{m{{v}^{2}}}{\sqrt{2g}}\]                               

    D)  \[\frac{\sqrt{2}\,m{{v}^{3}}}{4g}\]

    Correct Answer: B

    Solution :

                    Angular momentum J = r x P \[=mv\cos \theta p\] \[=mv\cos \theta \frac{{{v}^{2}}{{\sin }^{2}}\theta }{2g}\] at            \[\theta =45{}^\circ \] \[J=\frac{m{{v}^{3}}}{4\sqrt{2g}}\]


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