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  • The escape velocity of a body on a planet is <img alt="\mathrm{12 \mathrm{kms}^{-1}}" src="/latex-ima

The escape velocity of a body on a planet \mathrm{' A^{\prime}} is \mathrm{12 \mathrm{kms}^{-1}}. The escape velocity of the body on another planet \mathrm{' \mathrm{B} ', } whose density is four times and radius is half of the planet \mathrm{' \mathrm{A} ', } is :
 

Option: 1

12 \mathrm{kms}^{-1}


Option: 2

24 \mathrm{kms}^{-1}


Option: 3

36 \mathrm{kms}^{-1}


Option: 4

6 \mathrm{kms}^{-1}


Answers (1)

best_answer

\begin{aligned} &\mathrm{\text { Vescape }=\sqrt{\frac{2 G M}{R}}} .\\ \\&\mathrm{=\sqrt{\frac{2 G}{R} \times \rho \times \frac{4}{3} \pi R^{3}}} \end{aligned}

\mathrm{\text { Vescape } \alpha \; R \; \sqrt{\rho}}

\mathrm{\frac{(\text { Vescape })_{B}}{(\text { Vescape })_{A}}=\frac{R_{B} \sqrt{\rho_{B}}}{R_{A} \sqrt{\rho_{A}}}=\frac{1}{2} \times \sqrt{4}}

                                                    =1

Hence, the correct answer is Option (1)

Posted by

manish painkra

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