Two spheres P and Q of equal radii have densities ρ1 and ρ2, respectively. The spheres are connected

Q: Why does the string being taut mean that the tension is greater than zero?

A taut string is under tension. If the tension were zero, the string would go slack. For the string to remain taut, the upper sphere must have a tendency to float upwards (buoyancy > weight) and the lower sphere must have a tendency to sink downwards (weight > buoyancy), pulling the string tight from both ends.

Q: How do we know that sphere P will rise in liquid L2?

From the equilibrium and taut string conditions, we established the density hierarchy: $\rho_1 < \sigma_1 < \sigma_2 < \rho_2$. Since the density of sphere P ($\rho_1$) is strictly less than the density of liquid L2 ($\sigma_2$), sphere P is lighter than liquid L2 and will experience a net upward buoyant force, causing it to rise.

Q: Why do the density difference terms cancel out when taking the ratio of terminal velocities?

The combined equilibrium equation $(\rho_1 + \rho_2) = (\sigma_1 + \sigma_2)$ can be rearranged to $\sigma_2 - \rho_1 = \rho_2 - \sigma_1$. Since these two density differences are mathematically identical, they cancel out completely when we divide the terminal velocity of P by that of Q.

Q: What is the physical meaning of the dot product of the two terminal velocities being negative?

The dot product of two vectors is negative when they point in opposite directions (angle between them is $180^\circ$). Here, sphere P rises upwards ($+\hat{j}$) and sphere Q sinks downwards ($-\hat{j}$), so their velocity vectors are anti-parallel, making their dot product negative.

Q: Does the viscosity of the liquids affect the static equilibrium of the floating spheres?

No, viscosity is a dynamic property that only opposes relative motion (viscous drag). Since the spheres are in static equilibrium in the first part of the problem, viscosity plays no role in the force balance until the spheres are separated and allowed to move individually.

Question

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

Options:

(A) $\left|\frac{\vec{v}_P}{\vec{v}_Q}\right| = \frac{\eta_1}{\eta_2}$

(B) $\left|\frac{\vec{v}_P}{\vec{v}_Q}\right| = \frac{\eta_2}{\eta_1}$

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

(D) $\vec{v}_P \cdot \vec{v}_Q < 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine a beautifully layered cocktail of two immiscible liquids, $L_1$ and $L_2$, resting in a container. Liquid $L_1$ is the lighter upper layer with density $\sigma_1$ and viscosity $\eta_1$, while $L_2$ is the denser lower layer with density $\sigma_2$ and viscosity $\eta_2$. Suspended in this fluid system are two identical spheres, $P$ and $Q$, connected by a massless string. Sphere $P$ floats in the upper liquid $L_1$, and sphere $Q$ floats in the lower liquid $L_2$. The string connecting them is perfectly vertical and taut, indicating a non-zero tension $T$ pulling them together. Let us dive deep into the physics of this equilibrium state and uncover the hidden mathematical relationships. ### The Master Equation of Equilibrium Since the combined system of both spheres is in static equilibrium, the total upward buoyant force must perfectly balance the total downward gravitational force. Let $V$ be the volume of each sphere. The total weight of the spheres is: $$W_{ ext{total}} = ( ho_1 V + ho_2 V)g$$ The total upthrust exerted by the liquids is: $$U_{ ext{total}} = (\sigma_1 V + \sigma_2 V)g$$ Equating these two forces gives us our master density relation: $$( ho_1 + ho_2)Vg = (\sigma_1 + \sigma_2)Vg \implies ho_1 + ho_2 = \sigma_1 + \sigma_2$$ We can rearrange this equation to group the density differences: $$ ho_2 - \sigma_1 = \sigma_2 - ho_1$$ Let this common density difference be $\Delta ho$. This symmetry is the key to solving the problem. ### Analyzing the Taut String Condition For the string to remain taut, sphere $P$ must have a tendency to rise in liquid $L_1$, meaning its density must be less than the liquid's density: $$ ho_1 < \sigma_1$$ Similarly, sphere $Q$ must have a tendency to sink in liquid $L_2$, meaning its density must be greater than the liquid's density: $$ ho_2 > \sigma_2$$ Since $L_2$ is the lower liquid, we also know that $\sigma_2 > \sigma_1$. Combining these inequalities gives us a clear picture of the density hierarchy: $$ ho_1 < \sigma_1 < \sigma_2 < ho_2$$ This hierarchy ensures that if sphere $P$ is placed alone in liquid $L_2$, it will rise, and if sphere $Q$ is placed alone in liquid $L_1$, it will sink. ### Calculating Terminal Velocities When sphere $P$ is placed alone in liquid $L_2$, it experiences an upward buoyant force that exceeds its weight because $ ho_1 < \sigma_2$. As it rises, it reaches a constant terminal velocity $\vec{v}_P$ when the upward buoyancy is balanced by the downward gravity and viscous drag: $$6\pi \eta_2 r v_P = (\sigma_2 - ho_1) V g \implies v_P = \frac{2 r^2 g}{9 \eta_2} (\sigma_2 - ho_1)$$ Since sphere $P$ is rising, its velocity vector points upwards: $$\vec{v}_P = v_P \hat{j}$$ Similarly, when sphere $Q$ is placed alone in liquid $L_1$, it sinks because $ ho_2 > \sigma_1$. It reaches a constant terminal velocity $\vec{v}_Q$ when the downward gravity is balanced by the upward buoyancy and viscous drag: $$6\pi \eta_1 r v_Q = ( ho_2 - \sigma_1) V g \implies v_Q = \frac{2 r^2 g}{9 \eta_1} ( ho_2 - \sigma_1)$$ Since sphere $Q$ is sinking, its velocity vector points downwards: $$\vec{v}_Q = -v_Q \hat{j}$$ ### Final Calculation and Vector Directions Let us find the ratio of the magnitudes of these terminal velocities: $$\left|\frac{\vec{v}_P}{\vec{v}_Q} ight| = \frac{\frac{2 r^2 g}{9 \eta_2} (\sigma_2 - ho_1)}{\frac{2 r^2 g}{9 \eta_1} ( ho_2 - \sigma_1)}$$ Using our master density difference relation $\sigma_2 - ho_1 = ho_2 - \sigma_1 = \Delta ho$, we can cancel out the density terms completely: $$\left|\frac{\vec{v}_P}{\vec{v}_Q} ight| = \frac{\eta_1}{\eta_2}$$ This confirms that **option (a)** is correct. Furthermore, because $\vec{v}_P$ points upwards ($+\hat{j}$) and $\vec{v}_Q$ points downwards ($-\hat{j}$), they are in opposite directions. Their dot product must therefore be negative: $$\vec{v}_P \cdot \vec{v}_Q < 0$$ This confirms that **option (d)** is also correct.

Select Answer:

Visualized Solution

Analyzing the Combined System in Equilibrium

Combined Equilibrium Equation

Taut String Condition & Density Relations

Equating the Density Differences

Terminal Velocity of Sphere $P$ alone in $L_{2}$

Terminal Velocity of Sphere $Q$ alone in $L_{1}$

Finding the Ratio of Terminal Velocities

Direction of Terminal Velocities and Dot Product

The Sigma Insight: Viscosity and Stokes' Law

Analyzing the Setup

The Master Equation of Equilibrium

Analyzing the Taut String Condition

Calculating Terminal Velocities

Final Calculation and Vector Directions

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