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(B) According to the law of conservation of energy,the total energy at the surface equals the total energy at infinity.Initial energy at the surface:

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$\sqrt{3} v_{e}$

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(B) According to the law of conservation of energy,the total energy at the surface equals the total energy at infinity.Initial energy at the surface: $E_i = K_i + U_i = \frac{1}{2} m(2 v_e)^2 - \frac{G M m}{R} = 2 m v_e^2 - m v_e^2 = m v_e^2$ (since $v_e^2 = \frac{2 G M}{R}$).Final energy at infinity: $E_f = K_f + U_f = \frac{1}{2} m v^2 + 0$.Equating $E_i = E_f$:$m v_e^2 = \frac{1}{2} m v^2$$v^2 = 2 v_e^2$$v = \sqrt{2} v_e$ is incorrect based on the standard energy balance approach. Let's re-evaluate: The energy provided is $K_i = \frac{1}{2} m (2 v_e)^2 = 2 m v_e^2$. The energy required to escape is $K_{req} = \frac{1}{2} m v_e^2$. The remaining kinetic energy at infinity is $K_f = K_i - K_{req} = 2 m v_e^2 - 0.5 m v_e^2 = 1.5 m v_e^2$.Thus,$\frac{1}{2} m v^2 = \frac{3}{2} m v_e^2$,which gives $v^2 = 3 v_e^2$,so $v = \sqrt{3} v_e$.

$A$ body is projected vertically upwards from the Earth's surface with a velocity $2 v_{e}$,where $v_{e}$ is the escape velocity from the Earth's surface. The velocity of the body when it escapes the gravitational pull is

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