JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    A body is released from a point at distance r from the centre of earth. If R is the radius of the earth and \[r>R\], then the velocity of the body at the time of striking the earth will be:

    A)  \[\sqrt{gR}\]                             

    B)  \[\sqrt{2gR}\]

    C)  \[\sqrt{\frac{2gR}{r-R}}\]                      

    D)  \[\sqrt{\frac{2gR(r-R)}{r}}\]

    Correct Answer: D

    Solution :

    Using law of conservation of energy, \[-\frac{GMm}{r}=\frac{1}{2}m{{v}^{2}}-\frac{GMM}{R}\] \[=\frac{{{v}^{2}}}{2}\,=\frac{GM}{R}\,-\frac{GM}{r}\] \[=GM\,\left( \frac{r-R}{rR} \right)=gR\left( \frac{r-R}{r} \right)\] \[v=\,\sqrt{\frac{2gR(r-R)}{r}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner