Q 11-07-114JEE MainJEE Main 2023 (1 Feb, Shift 1)Medium
If earth has a mass nine times and radius twice that of a planet $P$, then $\dfrac{v_e}{3}\sqrt x\ \text{m s}^{-1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of $P$, where $v_e$ is escape velocity on earth. The value of $x$ is
Answer: (A) 2
$v\propto\sqrt{\dfrac MR}$: $\dfrac{v_P}{v_e}=\sqrt{\dfrac{1/9}{1/2}}=\dfrac{\sqrt2}{3}$, so $v_P=\dfrac{v_e}{3}\sqrt2$ and $x=2$.
Solution by Sreeraj P, M.Sc Physics