Verified Solution • 2025–26 Session moderate

QJEEM23Phe962: JEE Main 2023 Physics - Class 11 JEEM23Phe962 Solution: If earth has a mass nine times and radius twice to that of a…

If earth has a mass nine times and radius twice to that of a planet P. Then ve3x ms−1\frac{v_{e}}{3} \sqrt{x} \mathrm{~ms}^{-1} will be the minimum velocity required by a rocket to pull out of gravitational force of P\mathrm{P} , where vev_{e} is escape velocity on earth. The value of xx is

Multiple Choice Options

  • A 1
  • B 3
  • C 2
  • D 18

Step-by-Step Verified Solution

amp;ME=9MPamp;RE=2RP\begin{aligned}& M_E=9 M_P \\\\& R_E=2 R_P\end{aligned}

Escape velocity =2GMR=\sqrt{\frac{2GM}{R}}

For earth ve=2GMEREv_e=\sqrt{\frac{2 G M_E}{R_E}}

amp; For P,ve=2GME9RE2=2GMERE×29amp;=ve23\begin{aligned}& \text { For } P, v_e=\sqrt{\frac{\frac{2 G M_E}{9}}{\frac{R_E}{2}}}=\sqrt{\frac{2 G M_E}{R_E} \times \frac{2}{9}} \\\\& =\frac{v_e \sqrt{2}}{3}\end{aligned}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring