Consider two solid spheres P and Q each of density $8\,\text{gm cm}^{-3}$ and diameters $1\,\text{cm}$ and $0.5\,\text{cm}$, respectively. Sphere P is dropped into a liquid of density $0.8\,\text{gm cm}^{-3}$ and viscosity $\eta = 3$ poiseuilles. Sphere Q is dropped into a liquid of density $1.6\,\text{gm cm}^{-3}$ and viscosity $\eta = 2$ poiseuilles. The ratio of the terminal velocities of P and Q is

Step 1: Identify the given parameters.

The properties of the spheres and the liquids are given as follows.

$$\rho_s = 8\text{ g cm}^{-3}$$ $$d_P = 1\text{ cm} \implies r_P = 0.5\text{ cm}$$ $$d_Q = 0.5\text{ cm} \implies r_Q = 0.25\text{ cm}$$ $$\rho_{L_P} = 0.8\text{ g cm}^{-3}$$ $$\eta_P = 3\text{ poiseuilles}$$ $$\rho_{L_Q} = 1.6\text{ g cm}^{-3}$$ $$\eta_Q = 2\text{ poiseuilles}$$

Step 2: State the formula for terminal velocity.

The terminal velocity $v_t$ of a sphere of radius $r$ and density $\rho_s$ falling through a liquid of density $\rho_L$ and viscosity $\eta$ is given by Stokes' Law.

$$v_t = \frac{2}{9}\frac{r^2(\rho_s - \rho_L)g}{\eta}$$

Step 3: Express the ratio of the terminal velocities.

Taking the ratio of the terminal velocities for sphere P and sphere Q yields the following relation.

$$\frac{v_P}{v_Q} = \left(\frac{r_P}{r_Q}\right)^2 \left(\frac{\rho_s - \rho_{L_P}}{\rho_s - \rho_{L_Q}}\right) \left(\frac{\eta_Q}{\eta_P}\right)$$

Step 4: Calculate the ratio.

Substitute the given values into the ratio equation.

$$\frac{v_P}{v_Q} = \left(\frac{0.5}{0.25}\right)^2 \left(\frac{8 - 0.8}{8 - 1.6}\right) \left(\frac{2}{3}\right)$$ $$\frac{v_P}{v_Q} = (2)^2 \left(\frac{7.2}{6.4}\right) \left(\frac{2}{3}\right)$$ $$\frac{v_P}{v_Q} = 4 \times \frac{9}{8} \times \frac{2}{3}$$ $$\frac{v_P}{v_Q} = 3$$

Answer: The ratio of the terminal velocities of P and Q is $3$ (or $3:1$).