The radius of the planet is double that of the earth, but their average densities are same. \(\mathrm{V}_{\mathrm{p}}\) and \(V_E\) are the escape velocities of planet and earth respectively. If \(\frac{V_p}{V_E}=x\), the value of ' \(x\) ' is

Escape velocity is given by,
\(\begin{aligned}
\mathrm{v}_{\mathrm{c}} & =\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \\
& =\sqrt{\frac{2 \mathrm{G}}{\mathrm{R}} \times \frac{4}{3} \pi \mathrm{R}^3 \rho}=\sqrt{\frac{8 \mathrm{G}}{3} \pi \mathrm{R}^2 \rho}=\sqrt{\frac{8 \mathrm{G} \pi \rho}{3}} \times \mathrm{R}
\end{aligned}\)
As the planets have the same density;
\(\begin{array}{ll}
& \mathrm{V}_{\mathrm{e}} \propto \mathrm{R} \\
\therefore \quad & \frac{\mathrm{~V}_{\mathrm{P}}}{\mathrm{~V}_{\mathrm{E}}}=\frac{\mathrm{R}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{E}}}=\frac{2 \mathrm{R}}{\mathrm{R}}=2 \\
\therefore \quad & \frac{\mathrm{~V}_{\mathrm{P}}}{\mathrm{~V}_{\mathrm{E}}}=2 \\
\therefore \quad & \mathrm{X}=2
\end{array}\)