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Question

A body of mass 1 kg is rotating in a vertical circle of radius 1m. What will be the difference in its kinetic energy at the top and bottom of the circle? (Take g equals 10 blank m s to the power of negative 2 end exponent)

  1. 10 J  
  2. 20 J  
  3. 30 J  
  4. 50 J  

The correct answer is: 20 J


    Difference in K E equals fraction numerator 1 over denominator 2 end fraction m open square brackets open parentheses square root of 5 g r end root close parentheses to the power of 2 end exponent minus square root of g r end root close square brackets to the power of 2 end exponent
    equals 2 m g r equals 2 cross times 1 cross times 10 cross times 1 equals 20 blankJ

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