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 \[\frac{1}{2}m\left( 5\operatorname{v}_{e}^{2} \right)-mgR=\frac{1}{2}mv_{\infty }^{2}\]\[\Rightarrow \]\[\frac{5v_{e}^{2}}{2}-gR=\frac{1}{2}v_{\infty }^{2}\]\[\Rightarrow \]\[\frac{5v_{e}^{2}}{2}-\frac{2gR}{2}=\frac{1}{2}v_{\infty }^{2}\] |
| \[\frac{5v_{e}^{2}}{2}-\frac{v_{e}^{2}}{2}=\frac{1}{2}v_{\infty }^{2}\left( \because v_{e}^{2}=2gR \right)\]\[\Rightarrow \]\[\frac{1}{2}\left( 4v_{e}^{2} \right)=\frac{1}{2}v_{\infty }^{2}\]\[\Rightarrow \]\[{{\left( \frac{{{v}_{\infty }}}{{{v}_{e}}} \right)}^{2}}=4\Rightarrow \frac{{{v}_{\infty }}}{{{v}_{e}}}=2\] |
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