General
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Chemistry-

From threactions: C(s)+2H2(g)→CH4(g)DH=–Xcal C(g)+4H(g)→CH4(g),DH=–X1Kcal CH4(g)→CH3(g)+H(g)DH=+Y(Kcal) Bond energof C–H bond is-

Chemistry-General

  1. fraction numerator X over denominator 4 end fraction Kcal.mol    
  2. YKcal.mol    
  3. fraction numerator X 1 over denominator 4 end fraction Kcal.mol    
  4. X1Kcal.mol    

    Answer:The correct answer is: fraction numerator X 1 over denominator 4 end fraction Kcal.mol

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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

    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
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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

    physics-General
    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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    Four bodies of equal mass start moving with same speed are shown in the figure. In which of the following combination the centre of mass will remain at origin?

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    equals fraction numerator m cross times 0 plus m cross times 1 plus m cross times 2 over denominator m plus m plus m end fraction equals 1
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    stack x with minus on top equals fraction numerator not stretchy sum m subscript i end subscript x subscript i end subscript over denominator not stretchy sum m subscript i end subscript end fraction
    equals fraction numerator m cross times 0 plus m cross times 1 plus m cross times 2 over denominator m plus m plus m end fraction equals 1
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