Two spheres P and Q of equal radii have densities $\rho_1$ and $\rho_2$, respectively. The spheres are connected by a massless string and placed in liquids $L_1$ and $L_2$ of densities $\sigma_1$ and $\sigma_2$ and viscosities $\eta_1$ and $\eta_2$, respectively. They float in equilibrium with the sphere P in $L_1$ and sphere Q in $L_2$ and the string being taut (see figure). If sphere P alone in $L_2$ has terminal velocity $\vec{V}_P$ and Q alone in $L_1$ has terminal velocity $\vec{V}_Q$, then

(A) $\frac{|\vec{V}_P|}{|\vec{V}_Q|}=\frac{\eta_1}{\eta_2}$; (D) $\vec{V}_P \cdot \vec{V}_Q < 0$

(B) $\frac{|\vec{V}_P|}{|\vec{V}_Q|}=\frac{\eta_2}{\eta_1}$

(C) $\vec{V}_P \cdot \vec{V}_Q > 0$

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Fluid Mechanics

Two spheres P and Q of equal radii have densities

ρ_{1}

and

ρ_{2}

, respectively. The spheres are connected by a massless string and placed in liquids

L_{1}

and

L_{2}

of densities

σ_{1}

and

σ_{2}

and viscosities

η_{1}

and

η_{2}

, respectively. They float in equilibrium with the sphere P in

L_{1}

and sphere Q in

L_{2}

and the string being taut (see figure). If sphere P alone in

L_{2}

has terminal velocity

V_{P}

and Q alone in

L_{1}

has terminal velocity

V_{Q}

, then

Answer: (A), (D)

For equilibrium with taut string:

(ρ_{1} + ρ_{2}) V = (σ_{1} + σ_{2}) V

This gives

ρ_{1} + ρ_{2} = σ_{1} + σ_{2}

.

Since the string is taut, P experiences upward buoyancy (so

ρ_{1} < σ_{1}

) and Q experiences downward net force (so

ρ_{2} > σ_{2}

).

From the constraint:

σ_{1} - ρ_{1} = ρ_{2} - σ_{2}

.

Terminal velocity of P in

L_{2}

V_{P} = \frac{2 r ^{2} ( σ _{2} - ρ _{1} ) g}{9 η _{2}}

(upward, since

ρ_{1} < σ_{1}

and

σ_{2} > ρ_{2} > σ_{2} - (σ_{1} - ρ_{1})

)

Actually, since

ρ_{1} < σ_{1}

, P would rise in

L_{2}

ρ_{1} < σ_{2}

.

Terminal velocity of Q in

L_{1}

V_{Q} = \frac{2 r ^{2} ( ρ _{2} - σ _{1} ) g}{9 η _{1}}

(downward)

Using

σ_{1} - ρ_{1} = ρ_{2} - σ_{2}

∣ \frac{V _{P}}{V _{Q}} ∣ = \frac{η _{1}}{η _{2}}

✓

Since P moves upward and Q moves downward, their velocities are in opposite directions:

V_{P} \cdot V_{Q} < 0

✓

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · One or more correct options