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(D) Let the potential at point $A$ be $V_A$ and at point $B$ be $V_B$. Let the potential at the top junction be $V_1$ and at the bottom junction be $V

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$(1)$,$(2)$ and $(3)$

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(D) Let the potential at point $A$ be $V_A$ and at point $B$ be $V_B$. Let the potential at the top junction be $V_1$ and at the bottom junction be $V_2$.By applying Kirchhoff's Current Law $(KCL)$ at the top junction:$\frac{V_1 - V_A}{R} + \frac{V_1 - V_B}{R} + \frac{V_1 - V_2}{G} = 0$By applying $KCL$ at the bottom junction:$\frac{V_2 - V_A}{2R} + \frac{V_2 - V_B}{2R} + \frac{V_2 - V_1}{G} = 0$Adding these two equations,the terms involving $G$ cancel out:$\frac{V_1 - V_A}{R} + \frac{V_1 - V_B}{R} + \frac{V_2 - V_A}{2R} + \frac{V_2 - V_B}{2R} = 0$$\frac{2(V_1 - V_A) + 2(V_1 - V_B) + (V_2 - V_A) + (V_2 - V_B)}{2R} = 0$$2V_1 + 2V_1 + V_2 + V_2 = 3V_A + 3V_B$$4V_1 + 2V_2 = 3(V_A + V_B)$Since the ratio of resistances in the two branches is $\frac{R}{2R} = \frac{1}{2}$,the bridge is balanced if the potentials at the nodes connected by $G$ are equal. However,here the ratio is $\frac{R}{2R} = \frac{1}{2}$ on both sides,so the bridge is balanced. Thus,$V_1 = V_2$,and the current through $G$ is zero. Statement $(3)$ is true.Since the current through $G$ is zero,we can remove $G$. The circuit becomes two parallel branches: one with $R+R = 2R$ and one with $2R+2R = 4R$.$R_{	ext{eff}} = \frac{(2R)(4R)}{2R + 4R} = \frac{8R^2}{6R} = \frac{4}{3} R$. Statement $(2)$ is true.Since the equivalent resistance is independent of $G$ when the bridge is balanced,statement $(1)$ is also true.

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