Solved papers for JEE Main & Advanced JEE Main Paper (Held On 11 April 2014)

done JEE Main Paper (Held On 11 April 2014) Total Questions - 1

• question_answer1) From a sphere of mass M and radius R, a smaller sphere of radius\[\frac{R}{2}\]is carved out such that the cavity made in the original sphere is between its centre and the periphery (See figure). For the configuration in the figure where the distance between the centre of the original sphere and the removed sphere is 3R, the gravitational force between the two sphere is:   [JEE Main Online Paper ( Held On 11 Apirl  2014 )
  A)
  \[\frac{41G{{M}^{2}}}{3600{{R}^{2}}}\] 
  done clear
  B)
  \[\frac{41G{{M}^{2}}}{450{{R}^{2}}}\]
  done clear
  C)
  \[\frac{59\,G{{M}^{2}}}{450{{R}^{2}}}\]                
  done clear
  D)
  \[\frac{\,G{{M}^{2}}}{225{{R}^{2}}}\]
  done clear
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