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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 16

  • question_answer
    A particle is projected from a point A, which is at a distance 4R from the centre of the earth and with speed \[{{V}_{1}}\] in a direction making \[30{}^\circ \] with the \[{{V}_{1}}\] line joining the centre of the earth and point A, as shown. Find the speed \[{{V}_{1}}\] if particle passes grazing the surface of the earth. Consider gravitational interaction only between these two. (Use \[\frac{GM}{R}=6.4\times {{10}^{7}}{{m}^{2}}/{{s}^{2}}\])

    A) \[8000\text{ }m/s\]                    

    B) \[\frac{8000}{\sqrt{2}}\text{ }m/s\]     

    C) \[\text{8000}\sqrt{2}\text{ }m/s\]          

    D) None of these

    Correct Answer: B

    Solution :

           [b] Conserving angular momentum \[m({{V}_{1}}\cos 60{}^\circ ).4R=m{{V}_{2}}R\] \[\frac{{{V}_{2}}}{{{V}_{1}}}=2\] Conserving energy of the system, we get \[-\frac{GMm}{4R}+\frac{1}{2}mV_{1}^{2}=-\frac{GMm}{R}+\frac{1}{2}mV_{2}^{2}\] \[\frac{1}{2}V_{2}^{2}-\frac{1}{2}V_{1}^{2}=\frac{3}{4}\frac{GM}{R}\] or \[V_{1}^{2}=\frac{1}{2}\frac{GM}{R}\] \[\Rightarrow \,\,\,{{V}_{1}}=\frac{1}{\sqrt{2}}\sqrt{64\times {{10}^{6}}}=\frac{8000}{\sqrt{2}}m/s\]


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