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(A) In air,the forces acting on each sphere are tension $T$,weight $mg$,and electrostatic force $F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}$.At equi

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$B, C$

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(A) In air,the forces acting on each sphere are tension $T$,weight $mg$,and electrostatic force $F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}$.At equilibrium:$T \cos(\alpha/2) = mg$$T \sin(\alpha/2) = F$Dividing the two equations: $	an(\alpha/2) = \frac{F}{mg} = \frac{q^2}{4\pi\epsilon_0 r^2 mg}$.When immersed in a dielectric liquid of dielectric constant $K$,the electrostatic force becomes $F' = \frac{F}{K}$. The sphere also experiences an upward buoyant force $F_B = V ho_l g$,where $V$ is the volume of the sphere and $ho_l$ is the density of the liquid. The effective weight becomes $mg' = mg - V ho_l g = V ho_s g - V ho_l g = V g (ho_s - ho_l)$,where $ho_s$ is the density of the sphere.At equilibrium in the liquid:$T' \cos(\alpha/2) = V g (ho_s - ho_l)$$T' \sin(\alpha/2) = F/K$Dividing the two equations: $	an(\alpha/2) = \frac{F/K}{V g (ho_s - ho_l)}$.Since the angle $\alpha$ remains the same,$	an(\alpha/2)$ is constant:$\frac{F}{mg} = \frac{F/K}{V g (ho_s - ho_l)}$$\frac{1}{mg} = \frac{1}{K V g (ho_s - ho_l)}$Since $m = V ho_s$,we have:$\frac{1}{V ho_s g} = \frac{1}{K V g (ho_s - ho_l)}$$ho_s = K (ho_s - ho_l)$$21 (ho_s - 800) = ho_s$$21 ho_s - 16800 = ho_s$$20 ho_s = 16800 \implies ho_s = 840 \ kg \ m^{-3}$.Since $F' = F/K$,the electric force reduces (Option $B$ is correct). The density of the sphere is $840 \ kg \ m^{-3}$ (Option $C$ is correct).

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