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From a circular disc of radius R and mass 9 M, a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is

  1. 4 blank M R to the power of 2 end exponent    
  2. fraction numerator 40 over denominator 9 end fraction M R to the power of 2 end exponent    
  3. 10 blank M R to the power of 2 end exponent    
  4. fraction numerator 37 over denominator 9 end fraction M R to the power of 2 end exponent    

The correct answer is: 4 blank M R to the power of 2 end exponent


    I subscript r e m a i n i n g end subscript equals I subscript w h o l e end subscript minus I subscript r e m o v e d end subscript
    or I equals fraction numerator 1 over denominator 2 end fraction open parentheses 9 M close parentheses open parentheses R to the power of 2 end exponent close parentheses minus open square brackets fraction numerator 1 over denominator 2 end fraction m open parentheses fraction numerator R over denominator 3 end fraction close parentheses to the power of 2 end exponent plus fraction numerator 1 over denominator 2 end fraction m open parentheses fraction numerator 2 R over denominator 3 end fraction close parentheses to the power of 2 end exponent close square brackets (i)
    Here, m equals fraction numerator 9 M over denominator pi R to the power of 2 end exponent end fraction cross times pi open parentheses fraction numerator R over denominator 3 end fraction close parentheses to the power of 2 end exponent equals M
    Substituting in Eq. (i), we have
    I equals 4 M R to the power of 2 end exponent

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