question_answer

A body of mass \[1\,\,kg\] is rotating in a vertical circle of radius \[1\,\,m\]. What will be the difference in kinetic energy at the top and at the bottom of the circle? (Take\[g=10\,\,m/{{s}^{2}})\]

A) \[50\,\,J\]                                           

B) \[30\,\,J\]

C) \[20\,\,J\]                           

D) \[10\,\,J\]

Correct Answer: C

Solution :

Key Idea: Velocity at bottom is \[\sqrt{5}\] times the velocity at top. The energy possessed by a body due to velocity \[v\] is given by                 \[K=\frac{1}{2}m{{v}^{2}}\] Difference in\[KE={{K}_{A}}-{{K}_{B}}\] Given,\[{{K}_{A}}=\frac{1}{2}mv_{A}^{2}=\frac{1}{2}m(5g\,\,r)\]                 \[{{K}_{B}}=\frac{1}{2}mv_{B}^{2}=\frac{1}{2}m(g\,\,r)\] \[\therefore \]\[\Delta KE=\frac{1}{2}m(5g\,\,r)-\frac{1}{2}m(g\,\,r)=2m\,\,g\,\,r\] Given,\[m=1\,\,kg,\,\,r=1\,\,m,\,\,g=10\,\,m/{{s}^{2}}\] \[\therefore \]  \[\Delta KE=2\times 1\times 10\times 1=20\,\,J\]