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A disc of radius R and mass M is pivoted at the rim and is set for small oscillations. If simple pendulum has to have the same period as that of the disc, the length of the simple pendulum should be

  1. fraction numerator 5 over denominator 4 end fraction R    
  2. fraction numerator 2 over denominator 3 end fraction R    
  3. fraction numerator 3 over denominator 4 end fraction R    
  4. fraction numerator 3 over denominator 2 end fraction R    

The correct answer is: fraction numerator 3 over denominator 2 end fraction R



    Time period of a physical pendulum
    T equals 2 pi square root of fraction numerator I subscript 0 end subscript over denominator m g d end fraction end root equals 2 pi square root of fraction numerator open parentheses fraction numerator 1 over denominator 2 end fraction m R to the power of 2 end exponent plus m R to the power of 2 end exponent close parentheses over denominator m g R end fraction end root
    equals 2 pi square root of fraction numerator 3 R over denominator 2 g end fraction end root (i)
    T subscript s i m p l e   p e n d u l u m end subscript equals 2 pi square root of fraction numerator l over denominator g end fraction end root (ii)
    Equating (i) and (ii), l equals fraction numerator 3 over denominator 2 end fraction R.

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