CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

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

    E)  1

    Correct Answer: B

    Solution :

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


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