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NEET Sample Paper NEET Sample Test Paper-3

  • question_answer
    A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to \[{{F}_{1}}\] on a particle placed at A, distant 2R from the centre of the sphere. A spherical cavity of radius R/2 is now made on the sphere as shown in the figure. The sphere with cavity now applies a gravitational force \[\frac{{{F}_{2}}}{{{F}_{1}}}\] will be                        

    A)       \[\frac{1}{2}\]                                              

    B)  \[\frac{3}{4}\]

    C)                  \[\frac{7}{8}\]                                  

    D)  \[\frac{14}{9}\]

    Correct Answer: D

    Solution :

        Let the mass of solid sphere be M and when cavity is made, its mass \[M'=\frac{7M}{8}\] Then,    \[{{F}_{1}}=\frac{GMm}{{{(2R)}^{2}}}=\frac{GMm}{4{{R}^{2}}}\] and     \[{{F}_{2}}=\frac{G.\frac{7M}{8}m}{\frac{3{{R}^{2}}}{2}}=\frac{7}{8}\times \frac{4}{9}\,\,\frac{GMm}{{{R}^{2}}}\] \[\therefore \]  \[\frac{{{F}_{2}}}{{{F}_{1}}}=\frac{14}{9}\]


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