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A uniform solid sphere of radius R has a mass M.A spherical position of the sphere centred at distance R/2 from the centre C of the sphere and of radius R/2 is cut away and removed. The force of gravity due to the remaining mass of the sphere on unit mass at a point P at a distance 2R from C along CO will be (in magnitude)

  1. fraction numerator 7 G M over denominator 32 R to the power of 2 end exponent end fraction    
  2. fraction numerator 7 G M over denominator 36 R to the power of 2 end exponent end fraction    
  3. fraction numerator 7 G M over denominator 8 R to the power of 2 end exponent end fraction    
  4. fraction numerator 7 G M over denominator 16 R to the power of 2 end exponent end fraction    

The correct answer is: fraction numerator 7 G M over denominator 36 R to the power of 2 end exponent end fraction


    mass of cut-out area : M divided by 8
    F subscript 1 end subscript = force due to complete disc = fraction numerator G M over denominator open parentheses 2 R close parentheses to the power of 2 end exponent end fraction equals fraction numerator G M over denominator 4 R to the power of 2 end exponent end fraction
    F subscript 2 end subscript equals Force due to cavity equals fraction numerator G open parentheses M divided by 8 close parentheses over denominator open parentheses 2 R minus R divided by 2 close parentheses to the power of 2 end exponent end fraction equals fraction numerator G M over denominator 18 R to the power of 2 end exponent end fraction
    F subscript 2 end subscript = Required force = F subscript 1 end subscript minus F subscript 2 end subscript equals fraction numerator 7 G M over denominator 36 R to the power of 2 end exponent end fraction

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