VMMC VMMC Medical 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) \[\frac{L}{2}\]                   

    D)        \[\frac{L}{4}\]

    Correct Answer: D

    Solution :

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


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