Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2011

  • question_answer
    A solid sphere is rotating about a diameter at an angular velocity \[\omega \] If it cools so that its radius reduces two\[\frac{1}{n}\] of its original value, its angular velocity becomes

    A) \[\frac{\omega }{n}\]                                    

    B) \[\frac{\omega }{{{n}^{2}}}\]

    C) \[n\omega \]                                    

    D) \[{{n}^{2}}\omega \]

    Correct Answer: D

    Solution :

    On applying law of conservation of angular momentum. \[{{I}_{1}}{{\omega }_{1}}={{I}_{2}}{{\omega }_{2}}\] For solid sphere,                 \[I=\frac{2}{5}m{{r}^{2}}\] \[\therefore \]  \[\frac{2}{5}mr_{1}^{2}{{\omega }_{1}}=\frac{2}{5}mr_{2}^{2}{{\omega }_{2}}\]                 \[{{r}^{2}}\omega ={{\left( \frac{r}{n} \right)}^{2}}\omega _{2}^{2}\]                 \[{{\omega }_{2}}={{n}^{2}}\omega \]


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