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(B) Let the two masses $M$ be at positions $(-L, 0)$ and $(L, 0)$. The particle $m$ is at the origin $(0, 0)$.The initial potential energy $U_i$ of th

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$b, d$

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(B) Let the two masses $M$ be at positions $(-L, 0)$ and $(L, 0)$. The particle $m$ is at the origin $(0, 0)$.The initial potential energy $U_i$ of the particle $m$ is the sum of the gravitational potential energy due to both masses:$U_i = -\frac{GMm}{L} - \frac{GMm}{L} = -\frac{2GMm}{L}$The initial kinetic energy is $K_i = \frac{1}{2}mv^2$.To escape the gravitational field,the final total energy must be at least $0$ at infinity.By the Law of Conservation of Mechanical Energy:$K_i + U_i = K_f + U_f$$\frac{1}{2}mv^2 - \frac{2GMm}{L} = 0 + 0$$\frac{1}{2}mv^2 = \frac{2GMm}{L}$$v^2 = \frac{4GM}{L}$$v = 2\sqrt{\frac{GM}{L}}$Thus,statement $(b)$ is correct. Since the particle is moving in a gravitational field,its potential energy changes as it moves,so its total energy remains constant,but its kinetic energy changes. Therefore,statement $(d)$ is also correct as the total mechanical energy is conserved.

$\text{LIST-I}$	$\text{LIST-II}$
$A$. Gravitational constant	$I$. $[LT^{-2}]$
$B$. Gravitational potential energy	$II$. $[L^2 T^{-2}]$
$C$. Gravitational potential	$III$. $[ML^2 T^{-2}]$
$D$. Acceleration due to gravity	$IV$. $[M^{-1} L^3 T^{-2}]$

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