Uttarakhand PMT Uttarakhand PMT Solved Paper-2010

  • question_answer
    A solid sphere of mass 2 kg rolls up a \[30{}^\circ \] incline with an initial speed 10 m/s. The maximum height reached by the sphere is\[(g=10m/{{s}^{2}})\]

    A)  3.5m           

    B)  7.0m

    C)  10.Sm         

    D)  14.0m

    Correct Answer: B

    Solution :

     By conservation of mechanical energy \[mgh=\frac{1}{2}m{{v}^{2}}\left( 1+\frac{{{K}^{2}}}{{{R}^{2}}} \right)\] \[\Rightarrow \] \[\frac{7}{10}m{{v}^{2}}=mgh\] \[\Rightarrow \] \[h=\frac{7}{10}\left( \frac{{{v}^{2}}}{g} \right)\] Given, \[v=10\text{ }m/s,g=10\text{ }m/{{s}^{2}}\] \[\therefore \]Maximum height, \[h=\frac{7}{10}\times \frac{10\times 10}{10}\Rightarrow h=7.0m\]


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