ICSE Class 11

Step 1: Analyze the equilibrium of the connected spheres.

Sphere $P$ of volume $V$ floats in liquid $L_1$, and sphere $Q$ of volume $V$ floats in liquid $L_2$. Since the string is taut, the tension $T$ acts downwards on $P$ and upwards on $Q$.

For sphere $P$ in liquid $L_1$:

$$\sigma_1 V g = \rho_1 V g + T$$

For sphere $Q$ in liquid $L_2$:

$$\sigma_2 V g + T = \rho_2 V g$$

Step 2: Relate the densities using the tension.

Solve both equations for the tension $T$.

$$T = (\sigma_1 - \rho_1) V g = (\rho_2 - \sigma_2) V g$$

Equating the terms gives:

$$\sigma_1 - \rho_1 = \rho_2 - \sigma_2$$

Rearranging the terms yields:

$$\rho_1 - \sigma_2 = -(\rho_2 - \sigma_1)$$

Taking the absolute values:

$$|\rho_1 - \sigma_2| = |\rho_2 - \sigma_1|$$

Step 3: Write the expressions for the terminal velocities.

The terminal velocity of a sphere of radius $r$ and density $\rho$ in a liquid of density $\sigma$ and viscosity $\eta$ is given by Stokes' Law.

For sphere $P$ alone in liquid $L_2$:

$$\overline{V}_P = \frac{2}{9}\frac{r^2 g(\rho_1 - \sigma_2)}{\eta_2}$$

For sphere $Q$ alone in liquid $L_1$:

$$\overline{V}_Q = \frac{2}{9}\frac{r^2 g(\rho_2 - \sigma_1)}{\eta_1}$$

Step 4: Find the ratio of the magnitudes of the terminal velocities.

Using the equality of the density differences $|\rho_1 - \sigma_2| = |\rho_2 - \sigma_1|$, we find the ratio of the speeds:

$$\frac{|\overline{V}_P|}{|\overline{V}_Q|} = \frac{\eta_1}{\eta_2}$$

Step 5: Determine the sign of the dot product of the velocities.

Since the density differences have opposite signs:

$$\rho_1 - \sigma_2 = -(\rho_2 - \sigma_1)$$

The terminal velocities $\overline{V}_P$ and $\overline{V}_Q$ point in opposite directions. Thus, their dot product is negative:

$$\overline{V}_P \cdot \overline{V}_Q < 0$$

Answer: The correct relations are $\dfrac{|\overline{V}_P|}{|\overline{V}_Q|} = \dfrac{\eta_1}{\eta_2}$ and $\overline{V}_P \cdot \overline{V}_Q < 0$.