question_answer

A body of mass 1 kg is rotating in a vertical circle of radius 1m. What will be (he 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 Difference in \[KE={{K}_{A}}-{{K}_{B}}\] Given, \[{{K}_{A}}=\frac{1}{2}mv_{A}^{2}=\frac{1}{2}m(5\,g\,r)\] \[{{K}_{B}}=\frac{1}{2}mv_{B}^{2}=\frac{1}{2}m(g\,r)\] \[\therefore \] \[\Delta \,KE=\frac{1}{2}m(5\,g\,r)-\frac{1}{2}m(g\,r)=2m\,g\,r\] Given,\[m=1\,kg,\,r=1g=10\,m/{{s}^{2}}\] \[\therefore \] \[\Delta \Kappa \Epsilon =2\times 1\times 10\times 1=20\,J\]