KVPY Sample Paper KVPY Stream-SX Model Paper-22

  • question_answer
    The escape velocity of a body on the Earth's surface is \[{{v}_{e}}\]. A body is thrown up with a speed \[\sqrt{5{{v}_{e}}}.\] Assuming that the Sun and planets do not influence the motion of the body, the velocity of the body at infinite distance is \[{{v}_{\infty }}\]. Then, the value of \[\frac{{{v}_{\infty }}}{{{v}_{e}}}\] is

    A) Zero

    B) 1

    C) 2

    D) 3

    Correct Answer: C

    Solution :

    Applying conservation of mechanical energy.
    Energy at initial position = Energy at Infinity
    \[\Rightarrow \]\[\frac{1}{2}mv_{i}^{2}-\frac{GmM}{R}=\frac{1}{2}mv_{\infty }^{2}+GPE\] at infinity Or \[\frac{1}{2}m{{\left( \sqrt{5}{{V}_{e}} \right)}^{1}}-\left( \frac{GM}{{{R}^{2}}} \right)mR=\frac{1}{2}mv_{\infty }^{2}+0\]
    Or \[\frac{1}{2}m\left( 5\operatorname{v}_{e}^{2} \right)-mgR=\frac{1}{2}mv_{\infty }^{2}\]\[\Rightarrow \]\[\frac{5v_{e}^{2}}{2}-gR=\frac{1}{2}v_{\infty }^{2}\]\[\Rightarrow \]\[\frac{5v_{e}^{2}}{2}-\frac{2gR}{2}=\frac{1}{2}v_{\infty }^{2}\]
    \[\frac{5v_{e}^{2}}{2}-\frac{v_{e}^{2}}{2}=\frac{1}{2}v_{\infty }^{2}\left( \because v_{e}^{2}=2gR \right)\]\[\Rightarrow \]\[\frac{1}{2}\left( 4v_{e}^{2} \right)=\frac{1}{2}v_{\infty }^{2}\]\[\Rightarrow \]\[{{\left( \frac{{{v}_{\infty }}}{{{v}_{e}}} \right)}^{2}}=4\Rightarrow \frac{{{v}_{\infty }}}{{{v}_{e}}}=2\]

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