JCECE Medical JCECE Medical Solved Paper-2005

  • question_answer
    An object of mass m is attached to light string which passess through a hollow tube. The object is set into rotation in a horizontal circle of radius \[{{r}_{1}}.\]If the string is pulled shortening the radius to \[{{r}_{2}},\]the ratio of new kinetic energy to the original kinetic energy is:

    A) \[{{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{2}}\]

    B)  \[{{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}\]

    C)  \[\frac{{{r}_{1}}}{{{r}_{2}}}\]

    D)  \[\frac{{{r}_{2}}}{{{r}_{1}}}\]

    Correct Answer: B

    Solution :

     Kinetic energy (K) of rotation is \[K=\frac{1}{2}I{{\omega }^{2}}\] where \[\omega =\frac{v}{r},r\]is radius of circle. \[\therefore \] \[K=\frac{1}{2}I\left( \frac{{{v}^{2}}}{{{r}^{2}}} \right)\] \[\therefore \] \[\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{r_{2}^{2}}{r_{1}^{2}}\] \[\Rightarrow \] \[\frac{{{K}_{2}}}{{{K}_{1}}}\frac{r_{1}^{2}}{r_{2}^{2}}\]


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