Q108 / 150 · MHT CET 2026 (13 April Shift 2) · Physics
If earth has a mass nine times and radius twice to the planet P. Then $\dfrac{v_e}{3}\sqrt{x}$ ms$^{-1}$ 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
- Correct answerA$2$
- B$3$
- C$18$
- D$1$
Solution
The escape velocity on Earth is given by $v_e = \sqrt{\dfrac{2GM_e}{R_e}}$
The escape velocity on planet P is given by $v_p = \sqrt{\dfrac{2GM_p}{R_p}}$
Given that $M_e = 9M_p \Rightarrow M_p = \dfrac{M_e}{9}$ and $R_e = 2R_p \Rightarrow R_p = \dfrac{R_e}{2}$
Substituting these values into the expression for $v_p$:
$v_p = \sqrt{\dfrac{2G \left( \dfrac{M_e}{9} \right)}{\dfrac{R_e}{2}}}$
$v_p = \sqrt{\dfrac{2GM_e}{R_e} \times \dfrac{2}{9}}$
$v_p = \sqrt{\dfrac{2GM_e}{R_e}} \times \dfrac{\sqrt{2}}{3}$
$v_p = \dfrac{v_e}{3}\sqrt{2}$
Comparing this with the given expression $\dfrac{v_e}{3}\sqrt{x}$, we get $x = 2$.
Answer: $2$
The escape velocity on planet P is given by $v_p = \sqrt{\dfrac{2GM_p}{R_p}}$
Given that $M_e = 9M_p \Rightarrow M_p = \dfrac{M_e}{9}$ and $R_e = 2R_p \Rightarrow R_p = \dfrac{R_e}{2}$
Substituting these values into the expression for $v_p$:
$v_p = \sqrt{\dfrac{2G \left( \dfrac{M_e}{9} \right)}{\dfrac{R_e}{2}}}$
$v_p = \sqrt{\dfrac{2GM_e}{R_e} \times \dfrac{2}{9}}$
$v_p = \sqrt{\dfrac{2GM_e}{R_e}} \times \dfrac{\sqrt{2}}{3}$
$v_p = \dfrac{v_e}{3}\sqrt{2}$
Comparing this with the given expression $\dfrac{v_e}{3}\sqrt{x}$, we get $x = 2$.
Answer: $2$
Hence, the correct answer is option: 'A'.