JEE Challenger
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PhyHardAdv 2015 • Paper 2

Terminal Velocity and Equilibrium of Connected Spheres Floating in Viscous Liquids

Two spheres PP and QQ having identical radii possess densities ρ1\rho_1 and ρ2\rho_2, respectively. These spheres are connected via a light massless string and placed inside two fluids L1L_1 and L2L_2 with densities σ1\sigma_1 and σ2\sigma_2 along with viscosities η1\eta_1 and η2\eta_2, respectively. In static equilibrium, sphere PP floats in L1L_1 and sphere QQ floats in L2L_2, with the connecting string held taut (refer to figure). When sphere PP alone is placed in L2L_2, its terminal velocity is V⃗P\vec{V}_P, while sphere QQ alone placed in L1L_1 achieves terminal velocity V⃗Q\vec{V}_Q. Which of the following relationship(s) is/are correct?

Question Diagram 1

Options

A

∣V⃗P∣∣V⃗Q∣=η1η2\frac{|\vec{V}_P|}{|\vec{V}_Q|} = \frac{\eta_1}{\eta_2}

Correct
B

∣V⃗P∣∣V⃗Q∣=η2η1\frac{|\vec{V}_P|}{|\vec{V}_Q|} = \frac{\eta_2}{\eta_1}

C

V⃗P⋅V⃗Q>0\vec{V}_P \cdot \vec{V}_Q > 0

D

V⃗P⋅V⃗Q<0\vec{V}_P \cdot \vec{V}_Q < 0

Correct

Topics & Concepts

Mechanical Properties of FluidsFluid DynamicsNewton's Laws

Step-by-Step Solution

To determine the correct relationships, let us analyze the equilibrium of the system and the motion of each sphere independently.

1. Equilibrium of the Connected Spheres

Let VV be the volume and RR be the radius of each sphere.

For the two-liquid system in static equilibrium:

  • Liquid L1L_1 floats on top of liquid L2L_2, which implies σ1<σ2\sigma_1 < \sigma_2.
  • Sphere PP is in liquid L1L_1 and sphere QQ is in liquid L2L_2.
  • The connecting string is taut, so the tension T>0T > 0.

For sphere PP (in liquid L1L_1): The forces acting on it are:

  • Buoyant force upward: FB1=Vσ1gF_{B1} = V \sigma_1 g
  • Weight downward: WP=Vρ1gW_P = V \rho_1 g
  • Tension downward: TT

Equating the vertical forces: FB1=WP+T  ⟹  T=V(σ1−ρ1)gF_{B1} = W_P + T \implies T = V(\sigma_1 - \rho_1)g

Since T>0T > 0, we have: σ1>ρ1\sigma_1 > \rho_1

For sphere QQ (in liquid L2L_2): The forces acting on it are:

  • Buoyant force upward: FB2=Vσ2gF_{B2} = V \sigma_2 g
  • Weight downward: WQ=Vρ2gW_Q = V \rho_2 g
  • Tension upward: TT

Equating the vertical forces: FB2+T=WQ  ⟹  T=V(ρ2−σ2)gF_{B2} + T = W_Q \implies T = V(\rho_2 - \sigma_2)g

Since T>0T > 0, we have: ρ2>σ2\rho_2 > \sigma_2

Equating the expressions for tension TT: V(σ1−ρ1)g=V(ρ2−σ2)gV(\sigma_1 - \rho_1)g = V(\rho_2 - \sigma_2)g σ1−ρ1=ρ2−σ2  ⟹  σ2−ρ1=ρ2−σ1— (1)\sigma_1 - \rho_1 = \rho_2 - \sigma_2 \implies \sigma_2 - \rho_1 = \rho_2 - \sigma_1 \quad \text{--- (1)}

2. Terminal Velocities of the Spheres

Case 1: Sphere PP alone in liquid L2L_2 Since σ2>σ1>ρ1\sigma_2 > \sigma_1 > \rho_1, we have σ2>ρ1\sigma_2 > \rho_1. Thus, the buoyant force on PP in L2L_2 exceeds its weight, causing sphere PP to move vertically upwards. At terminal velocity ∣V⃗P∣|\vec{V}_P|, the viscous drag force balances the net upward buoyant force: 6πη2R∣V⃗P∣=V(σ2−ρ1)g6\pi \eta_2 R |\vec{V}_P| = V(\sigma_2 - \rho_1)g ∣V⃗P∣=V(σ2−ρ1)g6πη2R(directed upward)|\vec{V}_P| = \frac{V(\sigma_2 - \rho_1)g}{6\pi \eta_2 R} \quad (\text{directed upward})

Case 2: Sphere QQ alone in liquid L1L_1 Since ρ2>σ2>σ1\rho_2 > \sigma_2 > \sigma_1, we have ρ2>σ1\rho_2 > \sigma_1. Thus, the weight of sphere QQ exceeds the buoyant force in L1L_1, causing sphere QQ to move vertically downwards. At terminal velocity ∣V⃗Q∣|\vec{V}_Q|, the viscous drag force balances the net downward force: 6πη1R∣V⃗Q∣=V(ρ2−σ1)g6\pi \eta_1 R |\vec{V}_Q| = V(\rho_2 - \sigma_1)g ∣V⃗Q∣=V(ρ2−σ1)g6πη1R(directed downward)|\vec{V}_Q| = \frac{V(\rho_2 - \sigma_1)g}{6\pi \eta_1 R} \quad (\text{directed downward})

3. Comparing the Terminal Velocities

Using equation (1), σ2−ρ1=ρ2−σ1\sigma_2 - \rho_1 = \rho_2 - \sigma_1, we can take the ratio of the magnitudes of the terminal velocities: ∣V⃗P∣∣V⃗Q∣=V(σ2−ρ1)g6πη2RV(ρ2−σ1)g6πη1R=η1η2\frac{|\vec{V}_P|}{|\vec{V}_Q|} = \frac{\frac{V(\sigma_2 - \rho_1)g}{6\pi \eta_2 R}}{\frac{V(\rho_2 - \sigma_1)g}{6\pi \eta_1 R}} = \frac{\eta_1}{\eta_2}

Since V⃗P\vec{V}_P points vertically upward and V⃗Q\vec{V}_Q points vertically downward, the two velocity vectors are in opposite directions. Therefore: V⃗P⋅V⃗Q<0\vec{V}_P \cdot \vec{V}_Q < 0

Thus, the correct options are A and D.

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