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WB JEE Medical WB JEE Medical Solved Paper-2014

  • question_answer
    A uniform solid spherical ball is rolling down a smooth inclined plane from a height \[h.\]The velocity attained by the ball when it reaches the bottom of the inclined plane is \[\upsilon .\]If the ball is now thrown vertically upwards with the same velocity \[\upsilon ,\]the maximum height to which the ball will rise is

    A)  \[\frac{5h}{8}\]

    B)  \[\frac{3h}{5}\]

    C)  \[\frac{5h}{7}\]

    D)  \[\frac{7h}{9}\]

    Correct Answer: C

    Solution :

     We know that total kinetic energy of a body rolling without slipping \[{{K}_{total}}={{K}_{rot}}+{{K}_{trans}}\] For solid spherical ball, \[I=\frac{2}{5}m{{R}^{2}}\](along to diameter) and\[v=R\omega ,\]where R is radius of spherical ball So,     \[{{K}_{total}}=\frac{1}{2}\left( \frac{2}{5}m{{R}^{2}} \right){{\omega }^{2}}+\frac{1}{2}m{{R}^{2}}{{\omega }^{2}}\] \[=\frac{7}{10}m{{R}^{2}}{{\omega }^{2}}\] \[K=\frac{7}{10}m{{v}^{2}}\] Potential energy = Kinetic energy \[mgh=\frac{7}{10}m{{v}^{2}}\] \[{{v}^{2}}=\frac{10}{7}gh\] ?(i) For vertical projection, \[{{v}^{2}}={{u}^{2}}+2gh\] \[\frac{10}{7}gh=0+2gh\Rightarrow h=\frac{5}{7}h\]


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