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JEE Main & Advanced Sample Paper JEE Main Sample Paper-13

  • question_answer
    If g is the acceleration due to gravity at the surface of the earth, then the energy required to launch a satellite of mass m from the surface of the earth into a circular orbit at an altitude of 2 R where R being the radius of the earth, is

    A)  \[\frac{mgR}{6}\]                           

    B)  \[\frac{mgR}{3}\]

    C)  \[\frac{2mgR}{3}\]                         

    D)  \[\frac{5}{6}\,mgR\]

    Correct Answer: D

    Solution :

     Total initial energy \[\therefore \] Total final energy \[{{K}_{a}}=\frac{[HCO{{O}^{-}}]\,[{{H}^{+}}]}{[HCOOH]}\] So.    \[[{{K}_{a}}=\]        \[(-3.7)=1.995\times {{10}^{-4}}]\]           (\[\because \] height = 2R) Energy required \[x=\frac{1.995\times {{10}^{-4}}\times 0.15}{0.1}\] \[[HCO{{O}^{-}}]=2.99\times {{10}^{-4}}\]            \[\therefore \] \[=\frac{2.99\times {{10}^{-4}}}{0.15}\times 100\]            \[I=\frac{pV}{RT}=\frac{5\times 9}{RT}\]


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