Step-by-step Solution

1. Analyze the setup and identify forces: We have two solid spheres, A and B, of equal volume V and different densities $d_A$ and $d_B$. They are connected by a string and fully immersed in a fluid of density $d_F$. The system is in equilibrium, with sphere A above sphere B, and there is tension T in the string. This implies T > 0.

2. Free Body Diagram for Sphere A: Three forces act on sphere A:

   • Weight $W_A$ acting downwards: $W_A = m_A g = d_A V g$.
   • Buoyant force $F_{B,A}$ acting upwards: $F_{B,A} = d_F V g$ (since it's fully immersed).
   • Tension T from the string acting downwards.

   For equilibrium of sphere A, the net force is zero: $F_{B,A} - W_A - T = 0$ $d_F V g - d_A V g - T = 0$ $T = (d_F - d_A) V g \quad ... (1)$

3. Free Body Diagram for Sphere B: Three forces act on sphere B:

   • Weight $W_B$ acting downwards: $W_B = m_B g = d_B V g$.
   • Buoyant force $F_{B,B}$ acting upwards: $F_{B,B} = d_F V g$.
   • Tension T from the string acting upwards.

   For equilibrium of sphere B, the net force is zero: $F_{B,B} + T - W_B = 0$ $d_F V g + T - d_B V g = 0$ $T = (d_B - d_F) V g \quad ... (2)$

4. Apply the condition of tension T > 0: The problem states there is tension in the string, which means the string is taut, so T > 0.

   • From equation (1): $T = (d_F - d_A) V g > 0$. Since V and g are positive, this implies $d_F - d_A > 0$, which means $d_A < d_F$. This supports Option A. This makes physical sense: for sphere A to be pulled down by the string, it must have a tendency to float up.

   • From equation (2): $T = (d_B - d_F) V g > 0$. This implies $d_B - d_F > 0$, which means $d_B > d_F$. This supports Option B. This also makes sense: for sphere B to be pulled up by the string, it must have a tendency to sink.

   • Option C ($d_A > d_F$) contradicts our finding from equation (1), so it is incorrect.

5. Combine the equilibrium equations: Since the tension T is the same in both equations, we can equate equations (1) and (2): $(d_F - d_A) V g = (d_B - d_F) V g$ $d_F - d_A = d_B - d_F$ $2d_F = d_A + d_B$ This supports Option D.

   Alternatively, we can consider the system of both spheres together. For the whole system to be in equilibrium (not accelerating up or down), the total buoyant force must balance the total weight.

   • Total Weight $W_{total} = W_A + W_B = (d_A + d_B) V g$.
   • Total Buoyant Force $F_{B, total} = F_{B,A} + F_{B,B} = d_F V g + d_F V g = 2d_F V g$.
   • For equilibrium, $W_{total} = F_{B, total}$.
   • $(d_A + d_B) V g = 2d_F V g$.
   • $d_A + d_B = 2d_F$.

Conclusion

Based on the analysis, for the given equilibrium arrangement to be possible, the following conditions must be met:

1. $d_A < d_F$
2. $d_B > d_F$
3. $d_A + d_B = 2d_F$

Therefore, options A, B, and D are all correct.