JEE Challenger
More from Mechanical Properties of Fluids

Equilibrium of Connected Spheres Immersed in Fluid

Two solid spheres A and B of equal volumes but of different densities dAd_A and dBd_B are connected by a string. They are fully immersed in a fluid of density dFd_F. They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if

Diagram 0

Options

A

dA<dFd_A < d_F

Correct
B

dB>dFd_B > d_F

Correct
C

dA>dFd_A > d_F

D

dA+dB=2dFd_A + d_B = 2 d_F

Correct

Step-by-Step Solution

To determine the conditions for the system of two solid spheres AA and BB to remain in static equilibrium fully immersed in a fluid, we analyze the forces acting on each sphere individually and on the combined system.

Let:

  • VV be the volume of each sphere (given that both have equal volumes).
  • dAd_A and dBd_B be the densities of spheres AA and BB, respectively.
  • dFd_F be the density of the fluid.
  • TT be the tension in the string connecting the two spheres.
  • gg be the acceleration due to gravity.

1. Equilibrium of Sphere A (Upper Sphere)

The forces acting on sphere AA are:

  • Upward buoyant force: FBA=VdFgF_{BA} = V d_F g
  • Downward gravitational force (weight): WA=VdAgW_A = V d_A g
  • Downward tension force exerted by the string: TT

For sphere AA to be in equilibrium:
FBA=WA+TF_{BA} = W_A + T
VdFg=VdAg+TV d_F g = V d_A g + T
T=Vg(dF−dA)T = V g (d_F - d_A)

For the string to be taut and under tension, we must have T>0T > 0:
Vg(dF−dA)>0  ⟹  dA<dFV g (d_F - d_A) > 0 \implies d_A < d_F

Hence, option (A) is correct.


2. Equilibrium of Sphere B (Lower Sphere)

The forces acting on sphere BB are:

  • Upward buoyant force: FBB=VdFgF_{BB} = V d_F g
  • Upward tension force exerted by the string: TT
  • Downward gravitational force (weight): WB=VdBgW_B = V d_B g

For sphere BB to be in equilibrium:
FBB+T=WBF_{BB} + T = W_B
VdFg+T=VdBgV d_F g + T = V d_B g
T=Vg(dB−dF)T = V g (d_B - d_F)

For the tension to be strictly positive (T>0T > 0):
Vg(dB−dF)>0  ⟹  dB>dFV g (d_B - d_F) > 0 \implies d_B > d_F

Hence, option (B) is correct.


3. Equilibrium of the Combined System (Spheres A + B)

For the entire system suspended in the fluid without accelerating vertically, the total buoyant force must equal the total weight of both spheres:
∑Fup=∑Fdown\sum F_{\text{up}} = \sum F_{\text{down}}
FBA+FBB=WA+WBF_{BA} + F_{BB} = W_A + W_B
VdFg+VdFg=VdAg+VdBgV d_F g + V d_F g = V d_A g + V d_B g
2VdFg=Vg(dA+dB)2 V d_F g = V g (d_A + d_B)
dA+dB=2dFd_A + d_B = 2 d_F

Hence, option (D) is correct.


Conclusion

The arrangement shown is possible with a non-zero tension in the string if and only if:

  • dA<dFd_A < d_F
  • dB>dFd_B > d_F
  • dA+dB=2dFd_A + d_B = 2 d_F

Correct Options: A, B, D