Verified Solution • 2025–26 Session moderate

QJEEM24Ph64a4: JEE Main 2024 Physics - Class 11 JEEM24Ph64a4 Solution: Escape velocity of a body from earth is . If the radius of a…

Escape velocity of a body from earth is 11.2 km/s11.2 \mathrm{~km} / \mathrm{s} . If the radius of a planet be onethird the radius of earth and mass be one-sixth that of earth, the escape velocity from the planet is :

Multiple Choice Options

  • A 7.9 km/s
  • B 8.4 km/s
  • C 4.2 km/s
  • D 11.2 km/s

Step-by-Step Verified Solution

amp;RP=RE3,MP=ME6amp; Vc=2GMeRe.... (i)amp;VP=2GMPRP.... (ii)amp;VeVp=2amp; VP=Ve2=11.22=7.9 km/sec\begin{aligned}& \mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}_{\mathrm{E}}}{3}, \mathrm{M}_{\mathrm{P}}=\frac{\mathrm{M}_{\mathrm{E}}}{6} \\& \mathrm{~V}_{\mathrm{c}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}} \quad \text{.... (i)}\\& \mathrm{V}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}}} \quad \text{.... (ii)}\\& \frac{\mathrm{V}_{\mathrm{e}}}{\mathrm{V}_{\mathrm{p}}}=\sqrt{2} \\& \mathrm{~V}_{\mathrm{P}}=\frac{\mathrm{V}_{\mathrm{e}}}{\sqrt{2}}=\frac{11.2}{\sqrt{2}}=7.9 \mathrm{~km} / \mathrm{sec}\end{aligned}

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