NEET Sample Paper NEET Sample Test Paper-14

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

    A)  2L                  

    B)  4L

    C)  \[\frac{L}{2}\]                                  

    D) \[\frac{L}{4}\]

    Correct Answer: D

    Solution :

    Angular momentum \[=L=mvr=m{{r}^{2}}\omega \] Also kinetic energy is \[K=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{r}^{2}}{{\omega }^{2}}\] or         \[K=\frac{1}{2}\times \frac{L}{\omega }.\,{{\omega }^{2}}=\frac{L\omega }{2}\] \[\therefore \]    \[L=\frac{2k}{\omega }\] Given, \[\omega '=2\omega ,\,k'=\frac{1}{2}k\] \[\therefore \]    \[L'=\frac{2k'}{\omega '}=\frac{2({\scriptstyle{}^{k}/{}_{2}})}{2\omega }=\frac{L}{4}\] Hence, the correction option is [d].

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