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JEE MAIN 2021
25-02-2021 S2
Question
The initial velocity $v_i$ required to project a body vertically upward from the surface of the earth to reach a height of $10 R$, where $R$ is the radius of the earth, may be described in terms of escape velocity $v_e$ such that $v_i=\sqrt{\frac{x}{y}} \times v_e$. The value of $x$ will be $\_\_\_\_$ .
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Solution
$\begin{aligned} & \frac{-G M m}{11 R}=\frac{-G M m}{R}+\frac{1}{2} m v^2 \\ & v=\sqrt{\frac{20 G M}{11 R}}\end{aligned}$
Question Tags
JEE Main
Physics
Easy
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