Two bodies of masses \(m_{1}\) and \(m_{2}\) are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual a gravitational attraction Their relative velocity of approach at separation distance \(\mathrm{r}\) between them is (A) $\left[\left\\{2 \mathrm{G}\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right)\right\\} / \mathrm{r}\right]^{-1 / 2}$ (B) $\left[\left\\{2 \mathrm{G}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)\right\\} / \mathrm{r}\right]^{1 / 2}$ (C) $\left[\mathrm{r} /\left\\{2 \mathrm{G}\left(\mathrm{m}_{1} \mathrm{~m}_{2}\right)\right\\} / \mathrm{r}\right]^{1 / 2}$ (D) $\left[\left(2 \mathrm{Gm}_{1} \mathrm{~m}_{2}\right) / \mathrm{r}\right]^{1 / 2}$

First, we need to recall the expression for gravitational force between the two masses \(m_1\) and \(m_2\) at distance \(r\) apart. It is given by: \[F = G \frac{m_{1}m_{2}}{r^2}\] where \(F\) represents the gravitational force, \(G\) is the gravitational constant, and \(r\) is the distance between the two masses.

Next, let's find the expression for the gravitational potential energy (GPE) of the system as a function of distance \(r\). The GPE, represented as \(U(r)\), can be obtained by integrating the gravitational force with respect to distance: \[U(r) = -\int F dr\] \[U(r) = -\int G \frac{m_{1}m_{2}}{r^2} dr\] Upon integrating, we get: \[U(r) = -G \frac{m_{1}m_{2}}{r} + C\] where \(C\) is the constant of integration. Since the potential energy is zero at infinite separation, we get \(C = 0\). Thus, the expression for GPE becomes: \[U(r) = -G \frac{m_{1}m_{2}}{r}\]

Now, let's apply the conservation of mechanical energy to the system. The total mechanical energy can be written as the sum of the kinetic energy (KE) and the potential energy (PE): \[E_{total} = KE + U(r)\] Initially, the masses are at rest and have zero kinetic energy. As the masses move, their kinetic energy increases while their gravitational potential energy decreases. Since mechanical energy is conserved, the final energy remains the same as the initial energy: \[0 = KE + (-G \frac{m_{1}m_{2}}{r})\]

We can represent the kinetic energy of the system in terms of the relative velocity between the two masses. The relative velocity, denoted as \(v_r\), can be written as: \[KE = \frac{1}{2} m_{1} \left(\frac{m_{2}}{m_{1}+m_{2}}\right) v_r^2+ \frac{1}{2} m_{2} \left(\frac{m_{1}}{m_{1}+m_{2}}\right) v_r^2\] Substituting the expression for KE into the conservation of mechanical energy equation, we get: \[0 = \frac{1}{2} m_{1} \left(\frac{m_{2}}{m_{1}+m_{2}}\right) v_r^2+ \frac{1}{2} m_{2} \left(\frac{m_{1}}{m_{1}+m_{2}}\right) v_r^2 - G \frac{m_{1}m_{2}}{r}\] Now, let's solve this equation for the relative velocity \(v_r\): \[v_r^2 = \frac{2G(m_{1}+m_{2})}{r}\] \[v_r = \sqrt{\frac{2G(m_{1}+m_{2})}{r}}\] The expression for the relative velocity of approach between the two masses at separation distance \(r\) is: \[v_r = \left[\frac{2G(m_{1}+m_{2})}{r}\right]^{\frac{1}{2}}\] Comparing the final expression with the given options, the correct answer is: (B) \(\left[\frac{2G(m_{1}+m_{2})}{r}\right]^{\frac{1}{2}}\)