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AMU Medical AMU Solved Paper-2006

  • question_answer
    A particle performs uniform circular motion with an angular momentum L. If the frequency of particle motion is doubled and its KE is halved the angular momentum becomes

    A)  2 L                                         

    B)  4 L

    C)  L / 2                                      

    D)  L / 4

    Correct Answer: D

    Solution :

                     \[L=mvr=m{{r}^{2}}\omega \]                   ... (i) Also, kinetic energy \[KE=\frac{1}{2}m{{v}^{2}}\] or            \[KE=\frac{1}{2}m\,{{(r\omega )}^{2}}=\frac{1}{2}m{{r}^{2}}{{\omega }^{2}}\] \[\Rightarrow \]               \[L=\frac{2KE}{\omega }\] Here,     \[\omega =2\omega \]                 \[KE=\frac{1}{2}KE\] \[\therefore \]  \[L=\frac{2KE}{\omega }=\frac{2\left( \frac{1}{2}KE \right)}{2\omega }=\frac{L}{4}\]


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