2. Solved Papers
  3. MGIMS WARDHA

MGIMS WARDHA MGIMS WARDHA Solved Paper-2006

  • question_answer
    A body of mass 1 kg is moving in a vertical r circular path of radius 1 m. The difference between the kinetic energies at its highest                 and lowest position is :

    A) 20 J                                        

    B)  10 J

    C) \[4\sqrt{5}\] J                                   

    D) 10\[(\sqrt{5}\]-1)J

    Correct Answer: A

    Solution :

     At highest point velocity \[{{v}_{1}}=\sqrt{rg}\] At lowest point velocity\[{{v}_{2}}=\sqrt{5rg}\] \[\therefore \]Difference in kinetic energies \[=\frac{1}{2}m[v_{2}^{2}-v_{1}^{2}]\] \[=\frac{1}{2}m[5rg-rg]\] \[=2m(rg)\] \[=2\times 1\times 1\times 10\] \[=20J\]


You need to login to perform this action.
You will be redirected in 3 sec spinner