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NEET Physics Rotational Motion NEET PYQ-Rotational Motion

  • question_answer
    A small object of uniform density rolls up a curved surface with an initial velocity v'. It reaches up to a maximum height of \[\frac{3{{\upsilon }^{2}}}{4g}\] with respect to the initial position. The object is        [NEET 2013]

    A)       ring

    B) solid sphere

    C) hollow sphere

    D) disc

    Correct Answer: D

    Solution :

    As         \[v=\sqrt{\frac{2gh}{1+\frac{{{k}^{2}}}{{{r}^{2}}}}}\]
    Given    \[h=\frac{3{{v}^{2}}}{4g}\]
    \[{{v}^{2}}=\frac{2gh}{1+\frac{{{k}^{2}}}{{{r}^{2}}}}=\frac{2g3{{v}^{2}}}{4g\left( 1+\frac{{{k}^{2}}}{{{r}^{2}}} \right)}=\frac{6g{{v}^{2}}}{4g\left( 1+\frac{{{u}^{2}}}{{{v}^{2}}} \right)}\]       \[1=\frac{3}{2\left( 1+\frac{{{k}^{2}}}{{{v}^{2}}} \right)}\]
    or         \[1+\frac{{{k}^{2}}}{{{r}^{2}}}=\frac{3}{2}\] or \[\frac{{{k}^{2}}}{{{r}^{2}}}=\frac{3}{2}-1=\frac{1}{2}\]
    \[{{k}^{2}}=\frac{1}{2}{{r}^{2}}\] (Equation of disc) Hence, the object is disc.


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