Two solid spheres A and B of equal volumes but of different densities $\mathrm{d}_{\mathrm{A}}$ and $\mathrm{d}_{\mathrm{B}}$ are connected by a string. They are fully immersed in a fluid of density $\mathrm{d}_{\mathrm{F}}$. They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if

(A) $\mathrm{d}_{\mathrm{A}} \mathrm{d}_{\mathrm{F}}$; (D) $\mathrm{d}_{\mathrm{A}}+\mathrm{d}_{\mathrm{B}}=2 \mathrm{~d}_{\mathrm{F}}$

(C) $\mathrm{d}_{\mathrm{A}}>\mathrm{d}_{\mathrm{F}}$

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Fluid Mechanics

Two solid spheres A and B of equal volumes but of different densities

d_{A}

and

d_{B}

are connected by a string. They are fully immersed in a fluid of density

d_{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

Answer: (A), (B), (D)

For sphere A to float upward (tension pulling it down), buoyant force must exceed its weight:

d_{A} < d_{F}

. For sphere B to sink (tension pulling it up), its weight must exceed buoyant force:

d_{B} > d_{F}

. For equilibrium of the system, considering both spheres together:

V d_{A} g + V d_{B} g = 2 V d_{F} g

, giving

d_{A} + d_{B} = 2 d_{F}

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · One or more correct options