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Question:

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

dA < dF

dA > dF

dB > dF

dA + dB = 2dF

Solution:

For sphere A to be at the top in liquid, the density of sphere A should be less than the density of liquid. And for sphere B to be at the bottom in liquid, the density of sphere B should be greater than the density of the liquid. The FBD of the top sphere gives us VdFg = T + VdAg and of the bottom sphere gives us T + VdFg = VdBg Substituting T from the first equation in the second we get 2VdFg = VdAg + VdBg or 2dF = dA + dB