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MATCH THE COLUMN
A particle is taken to a distance \( r(R) \) from centre of the earth. \( \mathrm{R} \) is radius of the earth. It is given velocity \( \mathrm{V} \) which is perpendicular to radius. With the given values of \( \mathrm{V} \) in column-I you have to match the resultant path of particle in column-II. Here \( \mathrm{G} \) is the universal gravitational constant and \( M \) is the mass of the earth.
\begin{tabular}{|l|l|r|l|}
\hline \multicolumn{2}{|c|}{\( \begin{array}{c}\text { Column-I } \\
\text { (Velocity) }\end{array} \)} & \multicolumn{2}{|c|}{\( \begin{array}{c}\text { Column-II } \\
\text { (Path) }\end{array} \)} \\
\hline A. & \( V=\sqrt{\mathrm{GM} / \mathrm{r}} \) & p. & Elliptical \\
\hline B. & \( V=\sqrt{2 \mathrm{GM} / \mathrm{r}} \) & q. & Parabolic \\
\hline C. & \( V\sqrt{2 \mathrm{GM} / \mathrm{r}} \) & r. & Hyperbolic \\
\hline D. & \( \sqrt{\mathrm{GM} / \mathrm{r}}V\sqrt{2 \mathrm{GM} / \mathrm{r}} \) & s. & Circular \\
\hline
\end{tabular}
- (1) A-(s); B-(q);C-(r); D-(p)
(2) A-(p); B-(s); C-(q); D-(r)
(3) A-(q); B-(r); C-(p); D-(s)
(4) A-(s); B-(r); C-(q); D-(p)
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